Student Solutions Manual for use with
Complex Variables and Applications Seventh Edition
Selected Solutions to Exercises in Chapters 1-7
by
James Ward Brown Professor of Mathematics
The University of Michigan- Dearborn
Ruel
V. Churchill Late Professor of Mathematics
The University of Michigan
Mc Graw Higher Education Hill
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Table of Contents
Chapter
1
1
Chapter 2
22
Chapter 3
35
Chapter 4
53
Chapter 5
75
Chapter 6
94
Chapter 7
118
;
COMPLEX VARIABLES AND APPLICATIONS" (7/e) by Brown and Churchill
Chapter 1 SECTION 2
1.
2.
3.
(a)
(V2-/)-/(l-V2i) = V2-/-/-V2=-2i;
(b)
(2 -3)(-2,l)
(0
(3 >
(a)
Re(iz)
= Re[i(x + iy)] = Re(-y + ix) = -y = - Im z
Im(i'z)
= Im[i*(;t + iy)] = Im(-y + a) = x = Re z.
(l
= (-4 + 3,6 + 2) = (-1,8);
D(3,-l)(^) = (10,0)(i, L) = (2,l). 1
+ z) 2 =(l + z)(l + z) = (l + zH + (l + z)z = l-(l + z) + z(l + z) = l+z+z + z 2 = l + 2z + z 2
z=4±i,
4.
If
5.
To prove
then z
2
.
2 -2z + 2 = (l±i) -2(l±t) + 2 = ±2i-2T2i' + 2 = 0.
that multiplication is commutative, write
Z1Z2
= (*i.yi)(*2.y2 ) = ~ Wa» y^2 + *iy2 ) = (*2*i -yzVp^ H-x^) = (x2 ,y2 )(x ,y ) = z2 z l
6.
(a)
To
verify the associative
(Zj
+ z2 ) + z3 = l(x
l
»
,y l ) + (x2 , y2 )]
yi )
(y,
+ (*2 + ^3
= z +(z2 +z3 ). 1
.
law for addition, write
= ((*i + x2 ) + jcj, = (^1
l
l
.
+ (x 3 ,y 3 ) = (x, + x2
+ y2 ) + y3 ) = y%
+ v3 ) = (*i
,
vt
+ y2 ) + (*
+ (x2 + x3 ), .
yi )
+ [( x2
>
y* )
y,
3
,y3 )
+ (y2 + y3 ))
+ (*3
.
^
)]
2
(b)
To
verify the distributive law, write
z(zl
+ z2 ) = (*,30K*i.yi) + (x2 ,y2 )] = (jc,30(*i + x2 = (ja, + xx2 - yy, - yy2 = (xx ~yy + xx2 -yy2 l
l
=
,
,
+ y2 )
,
yjc,
+ yx2 + xy + xy2 )
yx,
+xy +yx2 +xy2 )
x
l
~ yyv yx + xy,) + (xx2 - yy2 yx2 + xy2 ) = (x, y)(x ,y ) + (x, y)(x2 ,y2 ) = zz +zz2 l
l
10.
The problem here
is to
,
l
.
x
solve the equation z
2
+z+1=
(x,y)(x,y) + (x,y)
for z
= {x,y) by writing
+ (1,0) = (0,0).
Since (x
2
- y 2 + x + 1, 2xy + y) = (0,0),
follows that
it
x
2
-y 2 +;e + l =
and
2xy + y = 0.
By writing the second of these equations as (2x + l)y = 0, we see that either 2x + 1 = or 2 If y = 0, the first equation becomes x +x + l = 0, which has no real roots y = 0. (according to the quadratic formula). Hence 2x + 1 = 0, or x = -1/2. In that case, the first 2 equation reveals that y = 3/4, or y = ±V3/2. Thus
SECTION 3 + 2/
1
!
(a )
2-/
|
3-4/
5/
(1
- 0(2 - 0(3 4
(c)
(l-/)
(a)
(-l)z
(b)
77- =
1
1/z
~
(1
=[(l-0d-0]
2
=(-2/)
— £=7= 1
z
z
z
1
-
- Q(-5Q _ -5 + lOj
2
_ 5/ _1 ~ " ~ ^10/ 2
5
=-4.
= z[l + (-1)] = z
z fe*0).
25
(5/)(-5/)
~ 3/)(3 -
since z + (-l)z
= -z
z
(2 |
5/
5/
>
2.
+ 2Q(3 + 4Q (3-4/)(3 + 4/) (1
=
•
= 0;
+
-5-10/ 25
=
2. 5'
3
3.
(z^Xzj^) = ^[ZaCz^)] = z^iztzjzj = zJC^K)] = z^z^zj] = (z&XZiZj.
1
6.
'
Z\Z2
= z.
—
£2
Z3 Z4
ZyZ 7.
V
_
Z2 Z
\ Z 2j
SECTION 4
w
^=(-73,1),
z2 =(V3,o)
Z\
^
|=
1
Z K l)
c2
/
=
(z 3
(z2
*0,z4 * 0).
*0,z*0).
4
(c)
^=(-3,1),
z2 =(l,4)
Z.
(d)
2.
z,
=
+ i>i
,
z2
= *i -
%
Inequalities (3), Sec. 4, are
Rez
if
we write them
In order to verify the inequality
Imz< llmzl< Izl.
and
y
2
2
+y 2
>l*l
2 .
2
-2l;tllyl
ways:
+ lyl,
+ y 2 )>l;cl 2 + 2|jtllyl + lyl 2 lxl
,
+ lyl 2 >0,
(l*l-lyl)
2
This last form of the inequality to be verified perfect square.
2
V2 Izl > IRezI + llmzl, we rewrite it in the following V2~V*
2(*
and
as
x<\x\<^x 2 + y 2
3.
y
+ Z,
>0. is
obviously true since the left-hand side
is
a
+ 1
1
;
5
4.
(a)
Rewrite It is
\z
- 1 + il= 1
as
- (1 - 1)| = 1.
|z
Write z I
- 4il +
z+
4/1
points z such that the
curve (b)
the circle centered at 1
-i
with radius
1.
IX\ \-i
I
(a)
is
shown below.
(
5.
This
Write
is
\z
= 10
as
I
J
- 4il
z
sum of the
- (- 4i)l = 10
\z
to see that this is the locus of all
distances from z to 4i and
-4i
is
a constant. Such a
an ellipse with foci ±4i.
- \\=\z + i\
as \z
- ll=lz - (-i)l
to see that this is the locus
of
all
points z such
always the same as the distance to -i. The curve then, the perpendicular bisector of the line segment from 1 to -i.
that the distance
from z
to 1 is
is,
SECTION 5 1.
2.
+ 3i' = z + 3i=z-3i;
(a)
z
(b)
iz
= iz=-iz;
(c)
(2
+ if = (2 + if = (2 - if = 4 - 4i +
M
I(2z
(a)
Rewrite Re(z-/)
2
= 4 - 4/ - 1 = 3 - 4/
+ 5)(V2-0I=I2z + 5IIV2-jI=I2TT5IV2+T = V3I2z + 5I.
through the point
= 2 as Re[* + i(-y - 1)] = 2, z = 2, shown below.
or
3>
2
x
x=
2.
This
is
the vertical line
y
6
(b)
Rewrite I2z - i\= 4 as 2 z
--
= 4,
= 2.
or
This
is
the circle centered at
2
2
shown below.
radius 2,
i/2, j
3.
Write
= x + iy and
z,
l
l
Zr
z2
= x2 + ry2
Then
.
+%) = (*i -^ +
~z2 = (*, + (y )-(x2
2)
1
= (*i - *2 ) -
- ft ) =
I
'(>'i
-y2 )
- Vi )~(x2 - iy2 ) = z -z2
(*i
l
and ZyZ 2
+ iy2 ) = {x x2 - yj2 ) + i(y^2 + xj2 ) = (X& - y2 ) - i{y x2 + x y2 ) = (x - fy,)^ - iy2 ) = z,z2 = (x + iy
l
l
)(x2
x
x
4.
6.
(a)
zfaZs
W
z
4
1
2
z
2
= z 2 z 2 = zzzz = (z
z)(z z)
.
;
= zzzz = z 4
.
fa) ZyZ2^3
J
Z2 Z3
__lz l_ i
(b)
z2 z3
8.
t
x
= (ziZ2 )z3 = zfo z3 = (*i £2)^3 = *i 22 z3
=z
lz2Z3
In this problem,
l
Z2 Z3
= Jzi l_ lz2 llz3
l
we shall use the inequalities IRezl
Specifically,
when
|Re(2
lzl<
and
\zx
(see Sec. 4)
+z2 + z3 |^|£i| + |z2 + M|
1,
+ z + z 3 < 12 + z + z 3 < 2+lzl )|
—
l
+lz
3 l
= 2+lzl+lzl 3 < 2 + 1 + 1 = 4.
with
7
10.
First write z
4
- 4z 2 + 3 = (z 2 - l)(z 2 - 3). Then
2
-ll>|lz
Iz
observe that
when
Izl
=2,
= |lzl 2 -l| = l4-ll=3
2
l-lll|
and
Thus,
when
= |lzl2 -3| = l4-3l=l.
2
2
-3l>|lz M3l|
lz
lzl=2, 4
lz
Consequently,
when
z
2 2 2 -4z +3l=lz -lHz -3l>3-l = 3.
on the
lies
circle lzl= 2,
1 4
-4z +3
z
11.
(a)
Prove that z (<=)
is real <=>
Suppose
that z
= x,
Thus z = * + iO
(=>) Suppose that z
Prove that z
that z
only
x=
either
2
-
2 -4z +3l
3
- iy = x + iy. This means
so that x
that i'2y
= 0,
or y
= 0.
is real.
is real,
is either real
lz
= z.
= z,
or z
(^) Suppose if
z
1 4
2
so that z
= x + iO. Then
or pure imaginary <=» z
=z
2
or y
.
Then
= 0,
2
(jr-i'y)
or possibly
2
z
= z2
= x - iO = jc + i'O = z. .
2
=(x + iy) or i4xy = 0. But this can be jc = y = 0. Thus z is either real or pure ,
imaginary.
(^) Suppose that z is either real or pure imaginary. If z is real, so that z-x, 2 2 2 2 2 2 2 If z is pure imaginary, so that z = iy, then z = (-iy) = (iy) = z z =x =z .
.
12. fa)
then
We shall use mathematical induction to show that Zi+z2 +-"+z„ =z +z2 +---+zn
(n
1
This
is
known when n = 2
(Sec. 5).
Assuming now
that
it is
write
Zt
+Z2 +*"+Zm +Zm+1 - (Z +Z2 + -,- +£m ) + Zm+l t
= (z +z2 +-+zm ) + zm+1 = Z + Z2 H l"Z ) + Z m m+1 1
t
= Zj + Z2 H
l"Z
m
+ Zm+1
.
true
= 2,3,...).
when n = m, we may
1
(b)
In the
same way, we can show
that
^z1 ---zn =z This
is
true
when n = 2 ^l Z 2
'
' '
Assuming
(Sec. 5).
(n
z1 --zn
l
that
it is
true
~ (^2 " = (^1^2 " 2m )^m+l = ^2 ~
£m Zm+l
"'Zm ) Zm+l
( Z1 Z 2
'
*
'
'
14.
The identities
2
= -^-^
and Rez
(Sec. 5) zz =\z\
'
= 2,3,...).
when n = m, we write
Zm ) ^nAr
enable us to write
lz-z
l=
R
as
2
(z-Z )(z-Z zz-(z5, 2
15.
Since x
- -^-^ and y =
= i? 2
,
+ z!Q+Zo5>
=/?2
2
= /? 2
-2Re(zz
lzl
)
)
+ lz
x
the hyperbola
l
2
'
.
-y 2 = 1
can be written in the following
ways: .
_n2
_\2
/ fz-z
z+z\ 2 2
+ 2zz + z 2
z
= 1,
1 2i
;
2 .
z
2
2z + 2z
2
z
-2zz+z
2
2
= 1,
_
2 +z =2.
SECTION 7 1.
(a)
Since arg l
*'
one value of
argj
—j
_2 -2i
is
= 3X8 '
~ afg(
y ~f~~£~J'
or
~2
~ 2/)
'
Consequently, the principal value
is
9
(b)
Since 6
arg(V3-i) =6arg(V3-0,
one value of arg(V3
4.
The
evident
we recall
if
the points e
5.
d = n of
solution
ie
and
1.
6
-i*)
is
6f™\ or e
the equation
that e'
e
lies
\e'
on the
-it.
-11= 2
So the
principal value
and
-it + 2n, or
it.
< < In is geometrically - II is the distance between
in the interval
circle \z\= 1
is
that le'*
See the figure below.
We know from de Moivre's formula that + isin 0) 3 = cos30 + isin30,
(cos or
cos
+ 3cos 2 0(/'sin 0) + 3cos 0(isin 0) 2 + (/sin 0) 3 = cos30 + i sin 30.
3
That is, (cos
By equating
- 3cos 0sin z
3
real parts
z
0) + i'(3cos 0sin
- sin 3 0) = cos30 + isin30.
and then imaginary parts here,
we
arrive at the desired trigonometric
identities:
(a)
8.
3 cos30 = cos 0-3cos0sin
Here z = Also,
9
re'
is
2
0;
(b)
2 3 sin30 = 3cos 0sin0-sin
0.
any nonzero complex number and n a negative integer (n =
m = -n = l,2
-l,-2,...).
By writing
and m
l
(z-
we
see that m
z"=(z
T
l
-
m (z
)
=(z
)
6)
=[l^~ m )
.
Thus
J
=[-Je
K - mB)
the definition
=\e z"
i(-~
= {z
me
\
m )
can also be written as
10
9.
two nonzero complex numbers zi and z 2 suppose that there are complex numbers c and c 2 such that Zj = qc2 and z^ = c c2 Since First of all, given
,
x
.
t
and
IzJHqllcJ it
follows that IzJ^ZjI.
Suppose, on the other hand, that
Zi
If
IzjIHcJIqIMcJIql,
we
=r
{
we know only that 1^1=^1.
exp(i 6l )
and
We may write
= r, exp(/02 ).
z2
introduce the numbers
q =r
x
expl
2
and
i
= expl
c2
we find
*
i
J
2
J'
that
c,c2
= r exp^i-^- Jexpl
= r exp(i^) = z,
t
t
J
and
qc2 =
rt exp|
i
i+il v i ,.0i-0 Jexp^-i
-^y*
|
= n exp
2
= z2
.
That is,
Zi
,
10. If 5
= l + z+z 2 +"-+z",
= qc2
l-z" —
z2
= c,c2
.
then
5 - zS = (1 + z + z 2 +•
Hence S =
and
•
.+z" )
- ( Z + z 2 + z 3 +.
. .
)
= i _ £"+1
+1
.provided z*l. That is,
1-z
H-z + z 2 +-+z"=
— 7" +1
— 1
(z*l).
1-z
Putting z
= e' e
(0
<
< 2tt)
l
in this identity,
we have
+ e m +e i20 +-+e ine =
1
l
~ eJ(n+l)6 \-e ie
——
—
11
Now the real part of the left-hand side here is evidendy l
+ cos0 + cos20+-+cosn0;
we write that side in the form
and, to find the real part of the right-hand side,
l-exp|7(n + l)0]
—
\(2n + l)0"
-exp
exp^-i
exp
i-
f)
l-exp(/0)
exp
exp
(-!)
2
H) -exp ®
which becomes
— — 9
cos
(2n + l)0
.
.
i
sin
2
.
2
.
ism
cos
(2n + l)0 •
1
2
2 -2i'sin—
2 or
r I
(2n + l)0l
sm — + sin .
,
^
J
+1
1
cos
— - cos -(2n + l)0l
j
^
j
2 sin—
The
real part of this is clearly
.
sin
(2/i
+ l)0
i+
-ir-
2
2sin^ 2
and
we arrive at
Lagrange's trigonometric identity:
1
1
+ cos + cos20+- -+cosn0 = — + •
2
sm
(2n + l)0
^ 2sin—
(0< 0<2;r).
12
SECTION 9 1.
(a)
Since 2/
= 2exp i\^ + 2kn
1/2
(2/)
(k
= 0,±1,±2,...),
the desired roots are
=V2exp
(*
=
That is,
= i+i and
c being the principal root These are sketched below.
(b)
Observe that 1-V3i
= 2exp
(l-V3/)
The
1/2
i\
— j + 2Jbr
=V2exp
i'I
(£
= 0,±1,±2,...). Hence
-^ + kn
(A:
principal root is
c
= V2V''*' 6 = V2[ cos| - /sin | ] = V2
and the other root
V3-i
,2
is
V2 These roots are shown below.
'
2J
V2
= 0,1).
13
2.
(a)
Since
-16 = 16exp[i(;r + 2*7r)]
(-16)
1/4
(k
= 0,±1,±2,...),
= 2exp
kit —+— it
,
i
4 The
the needed roots are
(*
2
= 0,1,2,3).
principal root is
c
=2,- = 2(cosf +
The other three
/S
i„l] =
2(-^ + -^j = V2(l +i ).
roots are
cx
= {ij^y*11 = cQ i = V2 (i + /)/ = - V2 (i - o, c2
=(2O*'* = -c =-V2(l +
i),
and c3
The four roots
(b)
First write
= (2e'*/4 )e»*12 = c are
= V2 (1 + OH) = V2(l - /).
shown below.
-8 -8V3/ = 16exp
(-8-8V3i)
The
(-i)
^-~ + 2kn^
l/4
=2exp
,
it
(*
= 0,+l,±2,...). Then
kit
(*
principal root is
c
=2e-
,Vr/6
=2 cos^-/sin- =2
= V3-i. I 2
2
= 0,1,2,3).
The
others are
c2
c3
These roots are
(a)
all
=(2,-/6 )e = -c =-(V3-0, i'r
wy
= (2e-'
3 *' 2
=c
shown below.
By writing -1 = lexp[i(?r + 2kn)]
(-1)
1/3
= exp
(k
.,
i
= 0,±1,±2,.
.
.),
we see that
—% + 2kit 3
The
= -(1 + V3/).
(-/)
3
principal root is
c
The other two
=e
Jt jc 1 + V3i = cos— + jsin— = .
.
roots are
c,
= e'* = -l
and „ c 2
K _ = ejsxii _ = cos„ =e nx-ixn e
All three roots are
shown below.
7t 1-V3i isin— = .
.
,
15
(b)
Since 8
= 8exp[i(0 + 2for)]
8
the principal
(k
1/6
= 0,±1,±2,...),
=V2exp(/^
(*
= 0,1,2,3,4,5),
one being c
The
the desired roots of 8 are
=V2.
others are
= V2V- = V2(cos| + /sin|] = Vlfi + v2
:
^/V^. V2 2
3 (V^"*' )** = V^cosy - isin-0-1) = - V2
c3
1
V3
2"T
N
1-V3/ *
V2
=j2e'*=-j2,
=(j2e M3 )e b, = -c
c4
,
=-
1
+ V3/ '
i
V2
and c5
All six roots are
The
three
=(V2V 2 *V* = -c2
=i^. V2
shown below.
cube roots of the number z
(zo
= -A^2 + A-Jli = 8exp^/-^j
r=2exp[
I
+ 2|E)]
In particular,
= 2exp^-|j = V2(l + i).
are evidently
(*
= 0,U).
16
1 H"
o =
With the aid of the number
"^3/
2
v
c2
5.
(a)
=c ©3 =(c
Q)3 )fi) 3
•(V3
=
we obtain the
,
3
two
other
roots:
V2
J
+ 1) + (V3-1)/
-1
V2
+ VIA _(V3-1)- (V3 + l)i 2
V2
J
Let a denote any fixed real number. In order to find the two square roots of a + i in exponential form, we write
A = \a +
i\
=Var+T
a = Arg(a + i).
and
Since
a+i
= Aexp[i(a + 2kn)]
(*
=
0,±1,±2,...),
we see that V2
(a + i)
That
is,
=VAexp
Since a +
tells
(*
= 0,1).
the desired square roots are
VaV " 2
(b)
y + Jbr
if
i
lies
above the
us that cosf
^
]
>
and
real axis,
and
VaV^V* = -4Ae ia, \
sinf
)
>
0.
\2
\2
a
/1
< a<
we know that
+ cosor
Since cos a
1
[.
n.
=—
A
,
Thus
it
<
and
.
a
1
- cos a
1
|^__a
sin
2
V
2
-7211
A
_
VA-a 42-fA'
Consequently,
VA ±VAVa/2 =±VA^cos| + isin|j = ±VA^
+a
^Va
= ±-^(VA + a+iVA-a).
.
+i
cc
it
2
2
follows that
^fA + a
~a
—<—
VA-fl
V2VA
^|
J
,
and
this
c
17
6.
The four
roots of the equation z
4
+4=
are the four fourth roots of the
we write -4 = 4expD'(;r + 2ibr)]
find those roots,
(-4)
1/4
=V2exp
(f
number -4. To
(& = 0,±1,±2,...). Then
+ Y)l = V2^/4 e^ 2
(*
= 0.1,2,3).
To be specific,
c
= V2V™ = V2f cos^ + /sin^l = V2 f-L + i * = l + i, 4 4j V V2j VV2 c/2
q =c
e"
c2
=c
e"
c3
= Co e i3 */2 = (1 + /)(-/) = 1-/.
r
=(l + i)i = -l + i,
= (l + 0(-l) = -l-i,
This enables us to write 4
z
+4 = (z-
)(z
- q)(z - c2 )(z - c3 )
= [(z - q)(z - c2 )]
•
[(z
-c
= Kz + 1) - i][(z + 1) + 1]
•
)(z
[(z
- c3 )]
- 1) - i][(z - 1) + 1]
= [(z + l) 2 + l][(z-l) 2 + l] = (z 2 + 2z + 2)(z 2 -2z + 2).
7.
Let c be any nth root of unity other than unity Exercise 10, Sec. 7),
l+z + z.2z +-+z",
,
I
itself.
With the aid of the
identity (see
l-Z" =-^-
(z*l),
l-z we find that
l
9.
Observe
+ c + c 2 + ... +c «-i = l_£l =: J__L = 1-c 1-c
first that
(z-r^^exp^^)
1
W
i(-0 -2tor) i(-2kn) = -7=exp i(-0)-expeXP_ m 1
"ilr
m
m
18
and
—
i(-e + 2kn) 1 i(-8) ,.-wm_Ji i(2kn) = - =rexp- -exp-T-exp -, )
(z
where
Jfe
= 0, 1, 2,
7
. . . ,
m-L
Since the set
i{-2kn) L exp-i
(*
m is
the
same
as the set
i{2kn)
exp ^
but in reverse order,
SECTION 1.
(a)
= 0,l,2,...,ro-l)
we find that
1/m
(z
1
)
L
(*
m = (z_1 ) 1/m
= 0,l,2,...,m-l),
.
10
Write z - 2 I
+ £1 il
as
I
z
- (2 - i)l £ 1 to see that this 2-i with radius 1. It is
circle centered at the point
is
the set of points inside and
on the
not a domain.
O
(b)
Write I2z + 3I>4 as
>2
exterior to the circle with center at
to see that the set in question consists of all points
-3/2 and radius
2. It is
a domain.
19
(c)
Write
Imz >
1
as y
>1
to see that this is the half plane consisting of all points lying
above the horizontal line y
= 1.
It is
a domain.
y
y =i
X
(d)
The set Imz =
1 is
simply the horizontal line y
= 1.
It is
not a domain.
y
y=
\
X
(e)
The set
(f)
The
< argz < —
(z
* 0)
is
indicated below.
It is
not a domain.
4
set Iz
- 4l£l zl can be
written in the form (x
- 4) 2 + y 2 1 x 2 + y\ which
reduces to
indicated below, is not a domain. The set is also geometrically evident since it consists of all points z such that the distance between z and 4 is greater than or equal to the distance between z and the origin.
*^2.
This
set,
which
is
— 20 4.
(a)
The closure of the
set -it
< argz < n
(z
# 0)
is
the entire plane.
1
(b)
We
first
write the set IRezklzl
inequality is the
same
as
2
y >
as
2
\x\<^Jx
0, or \y\> 0.
+y
Hence
2 ,
or
x
2
.
But
this last
the closure of the set IRezklzl
is
the
entire plane.
(c)
Since
-= z
2
— = 7-7 = -5x 2
zz
2
\z\
-rr,
+y
2
the set
Re - < - can be VzJ 2
2
written as —.2
x
+y
= < -, or 2 2
-2x) + y >0. Finally, by completing the square, we arrive at the inequality 2 2 2 which describes the circle, together with its exterior, that is centered (x - 1) + y > l (x
,
at
z=1
with radius
1.
The closure of this
set
is itself.
21
(d)
Since z
2
2 2 = (x + iyf =x -y + ilxy,
The closure of shown below.
\y\<\x\.
region
5.
The
set
this
2
2
S consists of all points z such
that lzl< 1 or lz
)
>
,
- 2I< 1,
as
shown below.
Since every polygonal line joining zt and z2 must contain at least one point that is clear that S is not connected.
8.
2
can be written as y < x or set consists of the lines y = ±x together with the shaded the set Re(z
not in S,
is
it
We are given that a set S contains each of its accumulation points. The problem here is to show that S must be closed. We do this by contradiction. We let Zq be a boundary point of S and suppose
that
it is
not a point in S.
The
fact that z
is
a boundary point
every neighborhood of zQ contains at least one point in S; and, since z that every deleted
neighborhood of S must contain
accumulation point of S, and that Zq is not in S. is,
S is
closed.
it
follows that z
is
at least
is
one point in
means
not in
S.
we
see
Zq is
an
S,
Thus
that
a point in S. But this contradicts the fact
We may conclude, then, that each boundary point
Zq
must be
in S.
That
22
Chapter 2 SECTION 1.
(a)
11 *
The function f(z) =
+1
z points
(fc)
z
(d)
z
is
z
The function /(z) =
is
z+z
imaginary axis. This
is
The function /(z) =
-
circle
]
defined throughout the entire finite plane except for the
= 0.
Izl
= 1, where
and y =
Using x =
f(z)
defined everywhere in the finite plane except for the
because the equation z + z
—
1
3.
+ 1 = 0.
The function /(z) = Arg( point z
(c)
2
= ±i, where
defined everywhere in the finite plane except at the
is
2
is
the
same
as
x = 0.
defined everywhere in the finite plane except on the
2
1 -Izl
^—-,
= 0.
write
= x*-y 2 -2y+ i(2x - 2xy)
_2
-2
_
2
2
+
+
_ fet Sfe - o
-2
2
= ir + 4r + 2iz-^- + ^- = z 2
5.
is
— Izl
. fa±tf + ftdg, +
SECTION
=
2
+2/z.
2
17
Consider the function '
/w-if where z = x + iy. Observe
that if z
-
x + iy Y
l
= (jc,0),
then
x + iO N2 and
if
z
= (0,y),
(z*0),
—
23 But
if
z
=
(*,*),
/fe)
= l,I^
This shows that /(z) has value value -1 at
cannot
10. (a)
all
nonzero points on the line y
4z
= x. Thus
the limit of /(z) as z tends to
To show that lim
2
= 4, we use statement
-r
- 1)
-P
4 lim
To establish the limit lim
kZ ,
——
V-»i
(z-l)
lira
-7—^
2
To
verify that lim
z
+l
= lim
\-,
j J
(2), Sec. 16,
and write
2
——
-?
= <», we refer to
—
*-»il/(z-l)
(c)
nonzero points on the real and imaginary axes but
1 at all
exist.
(z
(b)
=137.1
r 3
= lim «-»
(z
= 4.
statement (1), Sec. 16, and write
- 1) 3 = 0.
= °°, we apply statement (3),
Sec. 16, and write
z-1
i-i = lim
lim —^-j
«^fly
11. In this problem,
j
=-
^» l+z 2
= 0.
we consider the function 7-( z )
= f£±A CZ +
(ad- be *0).
fi?
fa)
Suppose
that c
= 0.
Statement
(3), Sec. 16, tells
= hm
lim «-»°
T(l/z)
»-»o
us that lim 7(z)
= — = 0. a + bz
a
= °°
since
24 (b)
Suppose
that c
* 0.
Statement
(2), Sec. 16, reveals that
lim T(z)
=—
*-»-
limT|'il = lim
\zJ
Also,
we know from statement
(1),
a + fe
a
+ dz
c
c
lim T(z)
Sec. 16, that
since
c
= 00
since
l-*-d/c
Um -i-=
SECTION 1.
(a^
If
£+4,0. az + &
19
/(z)
= 3z 2 -2z + 4,then
/'(z)
= ^(3z 2 -2z + 4) = 3^-z 2 -2-^-2 + 4-4 = 3(2z) - 2(1) + = 6z - 2. az
fife
If
lim z -»-rf/c
7(2)
/(z)
az
az
= (l-4z 2 ) 3 ,then
/'(z)
= 3(1 - 4zT T-d - 4 *
>
= 3 (! " 4z 2 f(-8z) = -24z(l - 4z 2 )
dz
(2z +
l)f (z-D-(z-l)f (2z + l)
rL) JK)
+ l)
(2z
(d)
If
/(z)
=
(2,
2
+ 1)(1) . (z _ 1)2 _ 2 (2z + l)
3 (2z
+ l) 2
)4
(1+
(z^0),then
f z
^An^^_ ~ a + Zz 2yd + dz nz) = ^ '
Z
<1
(l
)
2
(z
_
2z(l + z
2 3 )
2
2
)
g
2 2
=
2 3 )
(2z)-(lH-z 2
(z
)
[4z -(l
z 4(lH-z
+ z 2 )] _
2(l
+ z 2 ) 3 (3z 2 - 1)
2
)
2 4 )
2z
25 3.
If
/(z)
= l/z (z*0),
then 1
Aw = f(z + Az)-f(z) = z
Hence /'(z)
=
lim A*->o
We
are given that
According
f(z
)
= g(z
)
—
=
=
+ Az
that
z
(z
+ Az)z
—=—
lim *z-*0( z
and
-Az
1
+ Az)z
f\z
)
\r.
z
and g'(z
)
exist,
where g'(z )*0.
to the definition of derivative,
lira rfe) = «-*«o
Similarly,
g-(z
)
=
lim
/Mzfe) = lim /w. Z - Zq
Z-Zq
^ -^ )
)
= lim-^-.
Thus
=
lim *-*zog(z)
Urn
fWb-**) s(z)/(z-z
-_
\^
Z)I{Z
-^
lim^(z)/(z-z
)
)
_ f(Zo) ^'(z,,)'
SECTION 22 1.
(a)
= z= x — iy. So u = x, v = -y. = Vj, => 1 = -1, the Cauchy-Riemann Inasmuch as /(z)
equations are not satisfied
anywhere. (b)
= z-z=(x + iy)-(x-iy) = + i2y. So u = 0, v = 2y. Since mx = vy => = 2, the Cauchy-Riemann equations are not satisfied anywhere. f(z)
2
( c)
Here u = 2x, v f( z ) = 2x + ixy ux = vy => 2 = 2xy => xy = 1.
w
= -vT =>
Substituting
= xy 2
.
= -y 2 => y = 0. y = into Ay = 1, we have
= 1. Thus
the
Cauchy-Riemann equations do
not hold anywhere.
(d)
= e x e~'y - e*(cosy - isiny) = e x cosy - ie x siny. So u = e x cosy, = vy => cosy = -e x cosy => 2e* cosy = => cosy = 0. Thus
/(z)
«x
y= uy
^ + nn
= -v* =* -e'siny = e'siny => 2e*siny =
v
= -e x siny.
(n
=> siny =
0.
= 0,±l,±2,...).
Hence
y-nit (« = 0,±1,±2,...). Since these are two different sets of values of y, the Cauchy-Riemann equations cannot be satisfied anywhere.
26 3.
(aj
/(z ) =
- = -.- =
—
z
Izl
z
z
=
—
_+
+y
x
u
= —2 :
2
x
x
x +y
2
+y
a and
t 2
So
2
v
-y
=
, 2
x +y
, 2
.
Since
/'(z) exists
when
z
»
Moreover, when z
0.
y y
.
,
2
2
(*
(*-»)
_
(x
(fc)
/(z)
2
+y
-* 2 2xy 2 2+ +v ) V+/) 2 " ,
2
.
= vy
v
So /'(z) exists only when y =
x,
2
+y
2 2 )
2-
2
z
(z) (z)
= y 2 Now .
= 2y => y = jc
=> 2x
(x
2
\_
2
(zz)
)
2
x -i'2jcy-y
ay
2
= jc 2 + /y 2 Hence u = x 2 and ux
.
(gr _
=
2
* 0,
and
and
M
= 0.
= -v, =>
y
we find that
f\x + ix) = ux (x,x) + ivx (x,x) = 2x + 1O = 2x.
W
f(z)
= z Im z = (x + iy)y = xy + iy 2 Here .
wx
Hence /'(z)
= vy => y = 2y => y =
exists only
when
/'(0)
4.
(flj
/(z)
z
= 0.
u
= xy and v = y 2
and In
«y
= -vx
=
—
4
= Mje (0, 0) + iv, (0, 0) =
cos 40
= vg
and
We observe that
* = 0.
fact,
+ iO = 0.
= -^ = ^cos4^ + ^~sin40j (z*0).
rur
.
ue
=
Since
—
4
-sin 40
= -rvr
/is analytic in
its
domain of definition. Furthermore,
f'(z)
= e- w (ur + ivr ) = e- w ^~cos40 + i^-sin4aj = ~e- ie (cos46-isin4d) = --^"'V' 48
r
r
-4 3
r e
(b)
f(z)
'2
= -Jre'
,e
(re
rur
its
e-
w
(ur
— 2^
«" w
2^e
ien
M
'
V
U
rur
r
= ~
rT--cos| + irVsiiif
= A=e- W e wn 2V^
1
2f(z)'
y
(r>0,0< 0<2tt).
Since
1
=-e~ e sin(ln r) = ve its
and
=-e~ e cos(ln r) = -rvr
ug
domain of definition. Also,
ie w f'(z) = e- (ur +ivr ) = e-
a
e~ sin(lnr)
g
cos(lnr)
^. |
r
r
= -^f e~ e cos(ln r) + ie~ e sin(lnr)l = re
/(z)
ux
,
V
/is analytic in
When
Since
.
ue
fcos- + /sin-) 2 2J I
= e~ e cos(lnr) + ie- e sin(lnr) „
a + In).
=-—sm- = -rv
,
and
e
+ ivr ) =
1
V
<
domain of definition. Moreover,
-
f(z)
> 0, a <
(r
2*
^ cos- = v =—
f\z) =
(c)
f
= Vr cos — + n/r sin— 2*
/is analytic in
4
4
,5B
L
= * 3 + i(l - yf, we have k = x 3 and
J
v
z
= (1 - y)\ Observe that
2 = vy => 3x = -3(1- yf=*x 2 + (1- yf=0 and wy
=- =>0 Vje
Evidently, then, the
That
is,
Cauchy-Riemann equations are satisfied only when x = when z = i. Hence the expression
and y =
1.
they hold only
= ux + ivx = 3x 2 + iO = 3x 2 in which case we see that /'(/) = 0. f'(z)
is
valid only
when z =
i,
Here u and v denote the
and imaginary components of the function / defined by means
real
of the equations
t —
Az) =
when z*0,
z
when
z
= 0.
Now x
when z * 0, and
l
+y 2
the following calculations
",(0,0)
2
x +y
show
= v, (0,0) and
2
that
«y (0,0)
= -v,(0,0):
w(0 + Ajc,0)--«(0,0) AjC Ax
Ax-»0
w(0,0
+ Ay)-
A*->0 AjC
«(0,0)
Ay
Ay-»0
v(0 + Ajc,0)- v(0,0)
Ay-»0
Ay
A"->0
Ax
Ay->0
£y
_
Ax v(0,0 + Ay)- v(0,0)
Ay
Ay-»0
Equations
(2),
Sec. 22, are
ux cos
+ uy sin
-w,rsin0 + iycos0
= "r>
v
Solving these simultaneous linear equations for ux and u
ux
a = ur cosO-u g
6
sin
y
,
we find that
uy
=ur smdn + u e
v
= vr sm0 + vg
Cauchy-Riemann equations
in polar form,
,
.
and
cos0 r
Likewise,
sin0
= vr cos a -
Assume now
that the
and
e
=vg
rur
y
ue
,
=-rv
r
cos0 .
,
are satisfied at z„. It follows that
wx
=
i/
r
a cos0-w e
sin
6
= ve
cos 6 r
r
a+u = «r sin d e .
My
(a)
,
Write f(z)
cos0
= ve
_ + vsin0 = vr smd + vg cos0 = vv y .
.
,
,
r
sin0
_
vr cos0
.
= -lf vr cos0-ve
sin0\ \
= ~ vx-
= u(r, 0) + iv(r, 0) Then recall the polar form .
r«r
=ve
M„=-rvr
,
of the Cauchy-Riemann equations, which enables us to rewrite the expression (Sec. 22)
f'(z
for the derivative of /at a point z
f'(Zo )
)
= e- ie (ur +ivr )
= (r
= e~ i0 (-vg --u (A = \r
r
J
O)
,
—
lg
re
in the following
(u e
way:
+ ivg ) = —(u e + ivg ). zn
30 (b)
Consider
now the function 1 _ f {^\ _ = j\Z)--
z
_ —ig=-e 1
1
re
-is
1 _ /„ „ 6 Q -ism Q \ = = -(cos 6) .
r
•
cos
sin
6
1
r
r
.
r
With ,.(*. Q\ u{r, d)
At
C0S &
=
and
as
v(r, 0)
S"VS\B
=
,
r
the final expression for /'(z
£K(z)\ =
~ ( —
sin
z \
r
l
/
)
\
r
in part (a) tells us that
#
.cos0
1
1
z\
r
zKre'l
r J
cos
f
z
- /sin 6 r
2
when z*0.
10.
(a)
We consider a function
F(x,y), where
z
X=
+z
= y '
and
2
z-z 2i
Formal application of the chain rule for multivariable functions yields
dxdz
dz
(b)
dy dz
+l dx \2 J
dy \
2\dx
dy j
2i)~ 2{dx
dy
Now define the operator dz suggested by part
(a),
and formally apply
2{dx
dz
=
dy)
2dx
fa + *J + fa +
it
to
a function f(z)
= u(x,y) + iv(x,y):
2dy
iv
y )=
\[{ux - v
y)
If the
Cauchy-Riemann equations uz = vv u = -v*r y y
df/dz
= 0.
,
+ i(vx + uy )].
are satisfied, this tells us that
31
SECTION 24 1.
(a)
= 3x + y + i(3y - x)
f(z)
U
is
V
=3 = vy
ux
(b)
f(z)
= sin x cosh y + icos x sinh y v
'
„
>
uy
=l = -vx
.
entire since
V
= cosx cosh y = vy
ux
f(z)
is
and
,
v
U
(c)
entire since
and
uy
= sinxsinhy = -vx
= e~ y sinx - ie~ y cos x = e~ y s'mx + i(-e~ y cosx) '
v
'
>
is
.
entire since
,
,
V
ux
(d)
f(z)
= e~ y cosx - vy
= (z 2 - 2)e' x e"y = z 2 -2
g(z)
and
is
and
entire since
h(z)
uy
it is
= -e~ y sinx = -vx
.
the product of the entire functions
= e~ x e~ = e~ x (cosy - isiny) = e~ x cosy + i(-e~ x siny). ty
'
>
'
v
v
U
The function g
is entire
ux
2.
(a)
/(z)
= xy + iy U
is
= -e~
since x
it is
cos y
nowhere
a polynomial, and h
= vy
and
uy
= -e~
x
which means
sin y
= - vx
.
analytic since
that the
= vy =>y = l
and
uy
=-vx =>x = 0,
Cauchy-Riemann equations hold only
lx
f(z)
entire since
V
ux
(c)
is
'
V
= e y e = e y (cosx + /sin x) = g y cosj: + Vsinjc i
U
Mx
= vy =* -e y sinx = e y sinx
is
at the point
z
= (0,1) =
i.
nowhere analytic since
V
2e y sinx
=
=> sinx
=
and
= -v, => e y cosx = -e y cosx => 2^ cosx = More
precisely, the roots of the equation sinx
cos /iff
= (-1)" * 0. Consequently,
the
=
=> cosx are nit
= 0. (n
= 0,±l,±2,...), and
Cauchy-Riemann equations are not
satisfied
anywhere.
7.
(a)
Suppose
= u(x,y) + iv(x,y)
is analytic and real- valued in a domain D. Since f(z) is real- valued, it has the form f(z) = u(x,y) + i0. The Cauchy-Riemann equations ux =vy ,uy = -vx thus become ux =0,u = 0; and this means that u{x,y) = a, y where a is a (real) constant. (See the proof of the theorem in Sec. 23.) Evidently, then, f(z) = a. That is, / is constant in D.
that
a function f(z)
32 (b)
Suppose that a function Write
constant there.
f(z)
-
/
|/(z)|
throughout D.
If,
analytic in a
is
= c, where
Example 3
(a)
=
C
•
24 then
tells
its
a (real) constant.
* 0,
modulus
is
= 0, we
see that
If c
write f(z)7(z)
=c
2 ,
or
"
m
analytic and never zero in
in Sec.
and that
D,
the conjugate f(z)
us that f(z) must be constant in
must be D.
analytic in
D.
(Yl
SECTION 25 1.
is
is
on the other hand, c /(*)
Since f(z)
c
D
domain
straightforward to
It is
show
+uyy =0 when
we start with
harmonic conjugate v(x,y),
ux
that u xx
u(x,y)
= 2x(l-y). To
find a
= 2-2y. Now
ux (x,y)
=vy =>vy =2-2y=> v(x,y) = 2y-y 2 + (x).
Then uy
= -vx
=>
= * 2 + c.
-2x- -
0'(jc) = 2*
Consequently,
= 2y-y 2 + (jc 2 + c) = x 2 - y 2 + 2y + c.
v(;c,y)
(b)
It is
straightforward to
show
harmonic conjugate v(x,y),
ux
that
uxx
+ uyy =0 when
we start with
= Vy ^>vy =2-3x 2 + 3y 2 =»
w^Oc.y)
v(*,y)
u(x,y)
= 2x - x 3 + 3xy 2 To find a .
= 2-3* 2 + 3y 2 Now .
= 2y- 3x 2 y + y 3 +
Then uy
= -vx => 6xy = 6xy -
= 0=> 0O) = c.
Consequendy, v(jc,y)
fcj
It is
straightforward to
show
harmonic conjugate v(x,y),
= vy
vy
= 2y - 3x 2 y + y 3 + c.
that u xx
+uyy =0 when
«(jc,y)
= sinhjcsiny. To
we start with ux (x,y) = cosh x sin y. Now
= coshxsiny => v(x,y) = -coshxcosy +
Then «j,
= -vx => sinhxcosy = sinhxcosy -
Consequently, v(*,y)
= - cosh x cosy + c.
0'(jc)
=
=> 0(*) =
c.
find a
(d)
It is
show
straightforward to
m„+w=0
that
when
u{x,y)
we start with ux (x,y) =
harmonic conjugate v(x,y),
(x
= v,
2
=~
=> vy
^
+
(jc2
2)2
=
-
,
.
To
find a
Now
r~^~TT+ y2 ) 2 2
=> v(*,y) = -^£-5.+
,
^(jc).
Then 2
2
"y
= "Vx
^ (x
2
+y
2
2
2
=
2
(x
)
2
^ *'
+ y2 ) 2 ~
(jc)
= 0=
*
= °-
Consequendy, v(x,y)
=
— x
Suppose
that
v and
w = v - V,
= vy
uy
,
= -vx
and
23).
w(jc,y)
= c, where
that
uy
,
= -Vx
.
and
wy =vy -Vy =ux -ux = 0.
= c.
u and v are harmonic conjugates of each other in a domain D. Then ux
It
= Vy
c is a (real) constant (compare the proof of the theorem in Sec.
That is, v(x,y)-V(x,y)
Suppose
ux
This means that
then,
wx =vx -Vx =-uy +Uy=0 Hence
T + c.
V are harmonic conjugates of u in a domain D. ux
If
+y
= vy
Uy
,
= -vx
and
vx
= uy
vx
= 0,
v
,
y
= -ux
.
follows readily from these equations that
ux
— 0,
uy
=
and
vy
= 0.
Consequently, w(*,;y) and v(x,y) must be constant throughout/) (compare the proof of the theorem in Sec. 23).
The Cauchy-Riemann equations
in polar coordinates are
rur
=ve
and
ue - -rvT
Now ru r
= ve =>
run
+ ur = ver
.
K
and
Thus
ru n + ru + u gg =rv6r - rvr6 r
and, since v^.
= vr6 we have ,
r
which
is
;
the polar
2
un
+ rur + Ugg = 0,
form of Laplace's equation. To show
that v satisfies the
same
equation,
we observe that "e
= -rvr =>
vr
=
— 1
1
u g => v„
= —u g
r
r
— 1
fr
r
and rur Since u Br
= urg
,
w(r,0)
= lnr,
.
then,
r
If
=vg =*v gg = rur0
2
vn
+ rvr + vee = u e - ru^ -ue + rur8 = 0.
then
rV + m
r
+ u gg = r 2
[
~
+ rfi + 1
= 0.
J
This
tells
us that the function w
= In r
is
harmonic
follows from the Cauchy-Riemann equation rur thus v(r,6)
=
+ (r), where
0(r)
is at
in the
domain r>O,O<0<27T.
= vg and
the derivative u
r~~
Now it
mat ve ~ ^
present an arbitrary differentiable function of
r.
The other Cauchy-Riemann equation u g =-rvr then becomes O = -r0'(r). That is, Hence \r) = 0; and we see that (r) = c, where c is an arbitrary (real) constant. v(r, 6)
= 8+c
is
a harmonic conjugate of u(r, 6)
= In r.
35
Chapter 3 SECTION 28 2
1.
exj>(2±3m) = e ex$(±3m) = -e
(b)
( r ( cos- + isinexp—^— = exp- exp— = V«l
(c,)
exp(z
2+
3.
2
(a)
IV
m
m\
II
1
since exp(±3;ri)
,
n
.
.
n\
1
+ 7ri) = (expz)(exp7ri) = -expz,
= -l.
I
since exp7ri
= -l.
First write
exp(z )
where z
= x + iy.
This
= exp(* - iy) = e x e~ = e x cos y - ie x sin y, !y
tells
us that exp(z)
u{x,y)
- e x cosy
= u(x,y) + iv(x,y), where
and
v(x,y)
= -e x smy.
Suppose that the Cauchy-Riemann equations ux = vy and uy = -vx are satisfied point z
= x + iy.
cos y =
conclude
« and v here, these equations no value of y satisfying this pair of equations. We may since the Cauchy-Riemann equations fail to be satisfied anywhere, the It is
and sin y that,
some become
at
easy
to see that, for the functions
= 0. But there
is
function exp(z) is not analytic anywhere.
4.
The function exp(z
2 ) is
entire since
and the chain rule for derivatives
-|exp(z
2
it is
tells
2
a composition of the entire functions z and expz;
us that
2 2 2 =exp(z )-|z = 2zexp(z ). )
Alternatively,
one can show
exp(z
that exp(z
2 ) is
entire
by writing
2
2 )
= exp[(* + iy) j = exp(* 2 - y 2 )exp(i2*y) = exp(x 2 -y 2 )cos(2ry) + iexp(jc 2 - y 2 )sin(2;ty) >
w
<
»
U
and using the Cauchy-Riemann equations. ux
„
/
V
To be
specific,
=2xexp[x 2 -y 2 )cos(2;ty)-2yexp(;t 2 -y 2 )sin(2xy) = vy
and uy
= - 2y exp(x 2 - y 2 ) cos(2 xy ) - 2x exp(x 2 - y 2 )sin(2.xy ) = -vx
.
y
y
,
36
Furthermore,
-y-exp(z
2 )
= ux + ivx = 2(x + iy)[exp(* 2 -
2
)cos(2xy) + i exp(x
2
-
2
)sin(2xy)]
= 2zexp(z 2 ).
5.
We first write |exp(2z + i)\
= \exp[2x + i(2y + 1)]| = e 2x
and 2
|exp(iz
)|
= |exp[-2xy + i(x 2 - y 2 )]| = e
2xy .
Then, since |exp(2z
it
+ 1) + exp(iz 2 < |exp(2z + 1)| + |exp(iz 2 )|
follows that
|exp(2z
6.
)|
+ 1) + exp(/z 2 < e 2x + e~2xy )|
.
First write
|exp(z
2 )|
= |exp[(x + ry) 2 = |exp(j: 2 -/) + i2xy| = expU 2 -y 2 ) ]|
and exp(lzl
Since the
2 2 2 2 x -y
,
it is
2 )
= exp(jc 2 + y 2 ).
clear that exp(*
2
- y 2 ) < exp(* 2 +y 2 ). Hence
2
|exp(z )|
7.
To prove that
|exp(-2z)|
< 1 <=> Re z > 0,
|exp(-2z)|
It is
it
follows from
above that 2 ).
write
= |exp(-2;c - /2y)| = exp(-2*).
then clear that the statement to be proved
is
the
same
as
exp(-2*) <
obvious from the graph of the exponential function in calculus.
1 <=>
x>
0,
which
is
37
(a)
Write e
That
l
= -2
eV = 2*'*.
as
This
tells
us that
n + 2nn
e*=2
and
y=
* = ln2
and
y = (2n + l);r
(n
= 0,±l,±2,...).
(n
= 0,±1,±2,...).
(n
= 0,±l,±2,...).
(n
= 0,±1,±2,...).
(n
= 0,±1,±2,...).
(«
= 0,±1,±2,...).
(n
= 0,±l,+2,...).
(n
= 0,±1,±2,...);
(n
= 0,±l,±2,...).
is,
Hence z
(b)
Write *
That
l
= 1 + V3
1
as
= ln2 + (2n + l)«f
,(
* /3) ,
from which
'
we see that
e*=2
and
y=
* = ln2
and
y = ^2» + |j«
j + 2nn
is,
Consequently,
z
(cj
Write exp(2z
- 1) = 1
as e
2x_l
= \xi2 + {ln + ^jti
2x ~ l i2y
e
=l
= le i0 and note how
and
2y =
+ 2nx
it
follows that
Evidendy, then,
x=\
and
y
= n^
2
and
this
means
that
z
This problem
is
= -^ + nm
actually to find all roots of the equation
exp(iz)
= exp(/z).
38
To do
this, set
z
= jc +
iy
and rewrite the equation as y
e
Now, according
a
e' =e y e".
to the statement in italics at the beginning of Sec.8 in the text, y
e~
= ey
-x = x + Inn,
and
= 0,±l,±2,.„.
where n may have any one of the values n
y=
The
Thus
and x = nn
Suppose
10. (a)
that e
z
that
Since e
is real.
Moreover, since e
=
x
never zero,
is
0,±l,±2,...).
= e x cosy + ie x sin y, this means that e x siny = 0. siny = 0. Consequently, y = nn(n = 0,±1,±2,...);
Imz = n^(« = 0,±l,±2,...).
is,
On the
other hand, suppose that e
z
is
pure imaginary.
follows that cosy
It
= 0,
or that
— + /i7r(n = 0,±l,±2,...). 2
21
We start by writing 1
_
z
Because Re(e z )
Re(
)
= exp
Since
is
ensures that Re(e
If
f(z)
) is
e
Sec.
25
is
that
fit)
2
+y
2
x 2
-y
.
x +y
2
+ i-
'
x
2
+y
2
follows that
2
cos
domain
^
2
+ y2
that
= exp
.2 jc
+y..2
COS
,
J
,
does not contain the origin, Theorem
is
analytic in
= e u(x,y) CQS
some domain D,
1
in Sec.
25
^
y)
then
+ ie u(*,y) sinv (xy)
a composition of functions that are analytic in D, component functions
it
follows from Theorem
its
U(x,y)
2
\x +y..2
harmonic in such a domain.
= u(x,y) + i'v(jc,y)
Since e fU)
.
analytic in every 1/z
x
Izl
it
2
x-iy
z 2
zz
= e* cosy,
1/Z
_
z
\x +y
13.
(n
l
y = ^- + nn(n = Q,±\,±2,...). Thatis, Inu =
12.
= 0,±1,±2,...).
roots of the original equation are, therefore,
z=nn
(b)
(n
= e u(x y) cosvU.y), -
V(x,y) = e
u(x ' y)
sinv(jc,y)
1
in
39
are harmonic in
D. Moreover, by Theorem 2
in Sec. 25, V(x,y) is a
harmonic conjugate of
U(x,y).
14.
The problem here
is to
establish the identity
(expz)"
(a)
To show
that
= exp(nz)
(n
when n = 0,1,2,..., we = when n 0. Suppose that it is
it is
true
obviously true nonnegative integer. Then (expz)
(b)
m+l
= (expz)"
1
Suppose now that n view of part (a),
(expz)"
is
(expz)
when n - m, where
true
(expz)
m
= -1, -2,
. .
.
),
and write
1
1
exp(znz)
exp(-nz)
1
^expz;
use mathematical induction.
m = -n =
= exp(nz).
SECTION 30 1.
2.
3.
= \n\-ei\+iArg(-ei) = lne-^i = l-^i.
(a)
Log(-ei)
(b)
Log(l-
fa;
loge
= lne + i(0 + 2n7f) = l + 2/i7Fi
(b)
logi
=
(c)
log(-l + V3i)
(a)
Observe that
= lnll - i1+i'Arg(l -
lnl +
—
= In V2
(n
= |ln2 - -ji.
= 0,±l,±2,...).
i^ + 2nnj = ^2n + ^m = ln2 +
(n
= 0,±l,±2,...).
i^ + 2n^ = ln2 + 2^n + |jm
Log(l + 1)
2
m
It is is
any
= exp(mz)expz = exp(mz + z) = exp[(m + l)z],
a negative integer (n
=
= 0,±1,±2,...).
= Log(2i) = In 2 + ~/
and
2Log(l + 1) = 2^1n V2
+ ijj = In 2 +
Thus 2
Log(l + i) =2Log(l + i).
(n
= 0,±1,±2,...).
1,
2, .
. .
.
Li
(b)
On the other hand, 2
Log(-l + 1) = Log(-2/) =
In 2
- -/
and
2Log(-l + i) =
2(
lnV2
3n
+i^J = ln2 + y/.
Hence 2
Log(-l + /) *2Log(-l + /).
(a)
Consider the branch
logz
= lnr + /0
r>0,
— <6< 4
— 4
Since
2
log(i
)
= log(-l) = lnl + i;r=7B
and
21og/
= 2flnl +
= m, /-|J
we
(b)
find that log(/
2 )
= 2 log/ when this
branch of logz
is
taken.
Now consider the branch logz
= lnr + /0
r
> 0,
— 4
< 9< 4
Here
log(i
2 )
= log(-l) = In 1 + in =
m
and
2 log/
=2
In
1
+ /,5^ = 5m. 2
Hence, for
(a)
2
this particular branch, log(/
The two values of
i
m are
log(€'*
W4
e'
and e
i5tt/
)
,
* 2 log/.
\ Observe that
K
/4
)
l\ ( = lnl + /|-^ + 2«7r| = ( 2n + - \jti
= 0,±1,±2,...)
(«
and log(*
,9W4 )
5n = lnl + i|^ + 2/«rh
(2n
+ l) + -
7n
(n
= 0,±l,±2,...).
(n
= 0,±l,±2,...).
4 Combining these two
sets of values,
log(/
1/2
)
we
find that
= |n +
nm
^
41
On the other hand, 1
1, -logi=.
2
Thus the
lnl
+ i|
of values of log(i
set
= \n + —\m
^ + 2nn
2
) is
the
= 0,±l,±2,...).
(n
1
same
as the set of values of
— logi,
and
we
2
may write log(/
(b)
Note
1/2
)
= |log/.
that
log(i
2 )
= log(-l) = In 1 + {it + 2mt)i = {2n + \)iti
(n
= 0,±l,±2,...)
but that
21ogi'
=2
^ + 2mt
lnl + i'l
= {An + Y)Td
Evidently, then, the set of values of log(/
2 log i. That
2
To
solve the equation logz
2
is
)
= 0,±l,±2,...).
not the same as the set of values of
is,
log(i
7.
2
(/i
= iitll,
)*21ogi.
write exp(logz)
= exp(/7r / 2),
2
or z
= e'*n =
i.
) is the real component of any (analytic) branch of 2 logz, it is harmonic in every domain that does not contain the origin. This can be verified directly by writing
10. Since ln(jc
u{x,y)
+y
= ln(* 2 + y 2 ) and showing
that
^(jc.y) + u (x,y)
= 0.
SECTION 31 1.
Suppose
that
Rez > t
and Rez2 >
zt
0.
Then
= r exp i'0j t
and
z1
-r1 exp i'0 2
where
— it
2
~ it < 0, < — 1
2
,
it
and 2
it ^ — <©,< 2
2
,
,
42
-% <
The fact that
Log(ztz2 )
t
+
2
= Log[(r
x
enables us to write
r2 )exp /(0 t
+
2 )]
= lnfe) + i{Q + x
2)
= (In r + i'0i ) + (In r2 + 1'0 2 ) = Logfo exp i©, ) + Log(r2 exp i0 2 ) {
= LogZ! + Logz2
3.
.
We are asked to show in two different ways that
log
(a)
One way is
= ln
Another way
is
+ iarg
to first
show
(z,
t
to refer to the relation arg
log
(bj
=logz -logZ2
—
= argZ[ - argz2
in Sec.
* 0, z2
- = -logz (z * 0). To do this, we write ]
z
and then
log^ = log^e"'8 j = ln^ + i(-0 + 2imt) = -[lnr + i(0- 2imt)] = -logz, where n
= 0,±1,±2,.
. . .
This enables us to use the relation
log(z z2 ) 1
= logz +logz2 1
and write
f
log
log \ Z2J
= logz, + log \2 = logz,- log z2
0).
7 and write
= (lnl^l+zargz!) - (lnlz2 l+i'argz2 ) = log^ - logz2
that log[
?t
.
= re
j
43 5.
The problem here
is to
verify that
1/n
=exp -UogzJ
z
given that
1/ e
valid
when n = 1,2,...
m
Then, since
integer. z
it is
= e~
z
is
.
To do
(n
= -l,-2,...),
we put m = -n, where n is a negative we may use the relations z' = 11 z and
this,
a positive integer,
1
to write
p(^ =
1,
iog
4
exp(llogz)] = exp(-llogz) = expQlogz).
SECTION 32 1.
In each part below,
(a)
(1
+
)' 1
n=
0,±l,±2,....
= exp[i log(l + 1)] = exp< i lnV2+i(^ + 2n^
=
n — ln2-[ — + 2nn = exp^--^ - 2n^jexp^ln2j. {i
exp|
1
Since n takes on
all integral values, the
term
-2nn here can be
replaced by +2nit.
Thus
(1
(-1) '* 1
(b)
2.
+
/)''
=exp^-| + 2n^exp^ln2j.
= exp^log(-l)J = exp|^[ln 1 + i(n + 2nn]^ = exp[(2n + 1)/].
= exp(/Logi) = exp^ln 1 +
(a)
P. V.
/'
(b)
P.V.
[j(-l - V3/)]
"
/
-|
exp
-n
= exp |3^Log[|(-l - V3/)| = exp = exp(2/r2 )exp(/3^) = -exp(2;r 2 )
3ni\
)ne-i
In
44
P.V.
(c)
(1
- i) 4 =
exp[4/Log(l - 1)]
'
= exp 4iflnV2-/j
= e*[cos(41n V2) + isin(41n V2)] = «*[cos(21n2) + isin(21n2)]. 3.
Since
- 1 + V3i = 2e 2w
(-1 + V3/)
3 '2
=exp |log(-l + V30] = exp|| ln2 +
/|
^Y
+ lnn
= exp[ln(2 3/2 ) + (3n + l)*ri] = 2-^2 exp[(3/i + where n = 0,±1,±2,....
Observe
that
if
n
(-1 + V3i)
5.
odd, 3n+l
l);n']
3/2
©
<
According
it.
according to Sec. 32, that value
exp(
— Logz = exp |
(-1 + V3/)
is
odd; and so
is
even; and this means that
3/2
arise.
Specifically,
=±2V2.
We consider here any nonzero complex number where -it <
3n+l
even, then
is
= -l. On the other hand, if n is exp[(3/i + Y)iti] = 1. So only two distinct values of exp[(3n +
z
in the exponential
to Sec. 8, the principal value of z
Un is
form
Zq
= rQ exp i'0 o
^/r^exp^i— j;
,
and,
is
= ^ (in r + i'0 o ) exp(ln ^/r~) exp^z'
= nfc exp^j
^
These two expressions are evidently the same.
7.
Observe that when c = a + bi
power
c i
is
any fixed complex number, where c *
0,±1,±2,..., the
can be written as
T = exp(clogi) = exp< (a + bi)
]sil
+ i^— + 2nit
— + 2mt \+ia\ — + 2mt Vl .12 )
= exp -b
it
it
(n
= 0,±l,±2,...).
Thus
C I/
and c
it is
clear that
* 0,±1,±2,...
c lz
I
is
l= exp -b\
— + 2nit
multiple-valued unless b
ensures that
c i
is
(»
=
0, or c is real.
multiple- valued even
when b = 0.
Note
= 0,±1,±2,...),
that the restriction
.
SECTION 33 1.
The desired
derivatives can be found
by writing
-'~ k ^
k
e--e- = j_(d —sin = ±f' dz
dz
2i
ie
= Yi\
iz
e-
)
2i\dz
)=
——
+ie
_d—-e -
iz
dz
j
= cosz
2
and
d d(e* + e-*} lfd =- — — cosz = — dz dz\ 2 ) 2\dz
= -[ie -ie ")- = V 2
2.
From
e
k
—d
+
e
_ fe
dz
= -sinz. 2i
i
the expressions
s'mz
= e*-e'*
,
and
cosz
=
k
e
+ e~
iz
2
2i
we see that cosz + ismz
=
e
k
K
+e-'
+ ,
2
3.
Equation
(4), Sec.
33
Zi
k
-e~ k
= ek
.
2
is
2 sin zt cos z2 Interchanging
e
= sinfo + z2 ) + sin^ - Zj ).
and z 2 here and using the 2 cos z sin z2 x
fact that sin z is
we have
= sin(z + z1 ) - sin(zi - z2 ). t
Addition of corresponding sides of these two equations
2(sin zx cos z^
an odd function,
+ cos z
v
sin Zj )
now yields
= 2 sin(z + Z2 ), t
or
sin(z!
+ z2 ) = sin z, cos z2 + cos z
{
4.
sin 22
Differentiating each side of equation (5), Sec. 33, with respect to z v
cos(zt
+z2 ) = cosZ[ cosz2 -sinz
t
sinz2
.
we have
46
7.
(a)
From
2
the identity sin z
+ cos 2 z = 1, we have
—— + —— = —— 2
cos z
ft)
1
cos z
or
'
1
+ tan
z
= sec
z.
cos z
Also,
sin z .
sm
9.
,22
2
cos z
sin z
From
+
,
cos z ,
.
=
sin z
z
1 ,
.
or
,
1
+ cot
z
= esc
z.
sin z
the expression sin z
we find
= sin x cosh y +
1
cos x sinh y,
that
= sin 2 xcosh 2 y + cos 2 x sinh 2 y
2
Isinzl
= sin 2 x(l + sinh 2 y) + (1 - sin 2 jc)sinh 2 y = sin 2 x + sinh 2 y. The expression cos z
on the other hand,
= cos x cosh y + i sin * sinh y,
us that
tells
Icoszl
2
= cos 2 jccosh 2 y + sin 2 jtsinh 2 y = cos 2 *(1 + sinh 2 y) + (1 - cos 2 x) sinh 2 y
= cos 2 x + sinh 2 y.
10.
2
Since sinh y
is
never negative, 2
(a)
Isinzl
it
follows from Exercise 9 that
> sin 2 x,
or
Isinzl
> lsin*l
and that (b)
11.
In this problem
Icoszl
2
> cos 2 x,
or
Icoszl
>
I
cos xl.
we shall use the identities I
sin zl
2
= sin 2 x + sinh 2 y,
I
cos zl
2
= cos 2 x + sinh 2 y.
y
.
47
(a)
Observe that 2
sinh y =lsinzl
2
- sin 2 x < Isinzl 2
and 2
Isinzl
= sin 2 x + (cosh 2 y - 1) = cosh 2 y - (1 - sin 2 x)
= cosh 2 y - cos 2 x < cosh 2 y. Thus 2
sinh y
(b)
2
< cosh 2 y,
lsinhyl
or
On the other hand, 2
sinh y
=
cos zl
I
2
- cos 2 x <
I
cos zl
2
and Icoszl
2
= cos 2 x + (cosh 2 y - 1) = cosh 2 y - (1 - cos 2 x) = cosh 2 y - sin 2 x <, cosh 2 y.
Hence sinh
13.
By writing
/(z)
2
2
y < Icoszl ^ cosh
= sinz
2
y,
lsinhyl< Icoszl^ coshy.
or
= sin * cosh y -i cos* sinh y, we have
=sin(;c- iy)
f(z)
= u(x,y) + iv(x,y),
where u( x,
If the
)
= sin x cosh y
and
Cauchy-Riemann equations wx = vy
=
cos;ccoshy Since coshy
x = — + nn
is
(n
never zero,
it
= 0± 1,±2,...)-
Wj, ,
v(x,y)
= -v,
are to hold,
sin*sinhy =
and
follows from the
= -cosx sinh y
first
it is
easy to see that
0.
of these equations that cosx
Furthermore, since sin x
is
= 0;
that
is,
nonzero for each of these values
2 of x, the second equation tells us that sinhy equations hold only at the points
z
= 0,
or y
= 0. Thus
= - + n7t
the
Cauchy-Riemann
(n
= 0±l,±2,...)-
2
no neighborhood of any point throughout which /
Evidendy, then, there
is
we may conclude that
sinz
is
is
not analytic anywhere.
The function /(z) = cosz = cos(jc -/y) = cos x cosh y + isinx sinh y can be /(z)
= u(x,y) + iv(x,y),
where u(x, y)
and
analytic,
= cos x cosh y
and
v(x, y)
= sin xsinh y.
written as
48
If the
Cauchy-Riemann equations ux = vy uy ~ -vx ,
sin x cosh y
The
first
of these equations
cos nn *0, it follows that equations hold only when
tells
=
16. (a)
is
cos x sinh y
= 0.
= 0, or x = n7t(n = 0,±l,±2,...). Since = y 0. Consequently, the Cauchy-Riemann
us that sin;c
sinhy = 0,
z
So there nowhere
and
hold, then
or
= nn
(n
no neighborhood throughout which /
is
analytic,
and
= 0±l,±2,...).
this
means
that cosz is
analytic.
Use expression
(12), Sec. 33, to write
cos(iz)
= cos(-y + ix) = cosy cosh x — i sin y sinh x
and cos(i'z)
This shows that
(b)
Use expression
= cos(y + ur) = cosycoshx-isinysinh*.
cos(i'z)
= cos(iz)
for all z.
(1 1), Sec. 33, to write
sin(i'z)
= sin(-y + ix) = -sinycosh x - i cosysinhx
and sin(i'z)
= sin(y + ix) - sin y cosh x + cosy sinh jc. i
Evidently, then, the equation sin(zz)
= sin(i'z)
sinycoshx = 0,
Since cosh*
is
sin(/z)
tells
= sin(/z)
= nn
equivalent to the pair of equations
cosysinhx
first
= 0.
of these equations
tells
us that siny
= 0.
= 0,±1,±2,...). Since coshtt = (-1)" ^ 0, the second = So we may conclude that us that sinhx 0, or that x = 0. if and only if z = + inn = nni (n = 0,±1,±2,. .).
Consequently, y equation
never zero, the
is
(n
.
17. Rewriting the equation sinz
= cosh4
as sin;ccoshy
+ icos;tsinhy = cosh 4, we
need to solve the pair of equations
sin x cosh y
= cosh 4,
cos x sinh y
=
see that
we
49
for x
and
any*
y. If
since
cos* =
0.
y
= 0,
the
equation becomes sin*
first
sin*^l and cosh4>l.
= cosh 4, which cannot be
So y*0, and
satisfied
by
the second equation requires that
Thus
x=
~- + n7t
(n
= 0±l,±2,...).
Since
sin^Y + »wj =
the first equation then is
even,
it
becomes
follows that y
= ±4.
The problem here
is to
coshy = cosh 4, which cannot hold when n
(-1)"
Finally, then, the roots of sinz
Z
18.
(-l)",
=
(f
+ 2n7C ±4i
odd. If n
are
= 0±l,±2,...).
(n
]
find all roots of the equation cosz
equation as cosxcoshy-/sin;tsinhy
= cosh 4
is
We
= 2.
= 2. Thus we need to solve
cos*coshy =
2,
sinjrsinhy
since cos x = 2 x and y. We note that y * second in the pair of equations to be solved for
if
y=
start
by writing
that
the pair of equations
= and
0,
that is impossible.
So
the
us that sin;c = 0, or that x = nn (n = 0±l,±2,...). The first equation then tells us that (-l)"coshy = 2; and, since coshy is always positive, n must be even. That is, x = 2nn {n = 0±1,±2,...). But this means that -1 coshy = 2, or y = cosh 2 Consequently, the roots of the given equation are tells
.
z
To express cosh" y
= cosh"
1
2, or
1
coshy =
2, 2.
= 2n^ + /cosh _1 2
which has two values, This
tells
(e
y
us that e
y
y
+ e~ - 4;
apply the quadratic formula to obtain e y the observation that
at this alternative
(2-V3)(2 + V3)
form of the
z
way, we begin with
and, rewriting this as
= 2 ± V3,
= ln
2 + V3
we arrive
in a different
= 0±1,±2,...).
f-4(e y ) + l = 0,
we may
ln(2-V3) = ln
(n
or y
!
2 + V3
= ln(2 ± V3).
Finally, with
= -ln(2 + V3),
roots:
= 2n;r±tln(2 + V3)
(n
= 0±l,±2,...).
e
50
SECTION 34 1.
To find
we
the derivatives of sinhz and coshz,
d(e -e' z
d
.
,
z
)
d
1
write
,
.
z
+ e' z
_ ZN
e
_,. x
__
and
d + d _d coshz= _^__j = __ (e!+
2
3.
z
(
,
Identity (7), Sec. 33, is sin z
that
2 i'
2
sinh z
\
1
+ cos 2 z = 1.
sin(z'z)
we find
z
e
,
z2
by
iz 2
identities that
we have
2
z
z
-e
.
,
=sinlu
.
& here and using the identities
= coshz,
= 1.
cos^ + Zj) = cosz^oszj -sinzjsinzj. Replacing Zj by iZj and cosf/^ +z2 )] = cos(/z )cos(/z2 )-sin(iz )sin(iz2 ). The same 1
were used just above then lead
1
to
+ z2 ) = coshzj cosh^ + sinl^ sinhz2.
cosh(z1
6.
e
or 2
here,
)=
cos(iz)
cosh z - sinh z Identity (6), Sec. 33, is
,
Replacing z by
= isinhz and
+ cosh 2 z = 1,
,
We wish to show that lsinh.*l
in
(a)
two
cosh*
different ways.
Identity (12), Sec. 34, is lcoshzl this tells
us that sinh
2
= sinh 2 x + cos 2 y. Thus
x < Icoshzl
2
2
2
2 ,
or sinh x\< Icoshzl. I
2
Icoshzl -sinh
On
2
x>0; and
the other hand, since
= (cosh x - 1) + cos y = cosh x - (1 - cos y) = cosh x - sin 2 y, we know 2 2 2 2 Icoshzl -cosh x<0. Consequently, lcoshzl
(b)
Exercise U(b), Sec. 33, recalling that cosi'z
7.
(a)
2
Observe
tells
2
2
2
us that lsinhyl
= coshz and
iz
= ~y + ix, we
obtain the desired inequalities.
that
sinh(z +
m) =
that
=
=
=
= -sinhz.
— 51
(b)
Also,
cosh(z +
(c)
From
7ti)
parts (a)
=
and
u/tanh(z
=
L
f
we find
(b),
= + /n) _
— = ——— = -
1
'-z—
-
= -coshz.
that
—— + sinh(z
+ ?n) = -sinhz = -
cosh(z
9.
—
7n)
—
sinhz
-coshz
coshz
=
tanhz.
The zeros of the hyperbolic tangent function t
.
=
tanhz
sinhz
coshz are the
same
tanhz are the zeros of coshz, or z
15.
(a)
Observe
which are z = nm (n =
as the zeros of sinhz,
that,
since sinhz
to solve the pair
= ^-j + rucji
=/
x = 0,
= 0,±1,±2,...).
can be written as sinh x cos y + /cosh xsiny
= i, we need
of equations
sinh x cos y
If
(n
The singularities of
0,±1,±2,...).
= 0, cosh x sin y = 1.
the second of these equations
becomes siny = l; and so y =
— + 2nn it
2 (n
= 0,±l,±2,...). Hence
(n
If
(/i
x*0,
the
first
equation
= 0,±l,±2,...). The second
Rewriting cosh z
=^
as
cosh x cos y 4- i sinh x sin y
,
,
sinhxsiny
= 0.
see that
y=
or
= l. But there
= — we 2
2
= 0,
we have no additional roots
2 the pair of equations
coshxcosy = —
cosy
that
then becomes (-l)"coshjr
value of x satisfying this equation, and
(b)
requires
= 0,±l,±2,...).
is
of sinhz
—vmt 7T
no nonzero
= i.
x and y must satisfy
52
x = 0,
If
cosy =
1
—
.
the second equation
Thus y = cos
1
%
1
— = ±— + 2nn
z
If
x
becomes all x.
0, the
=
Thus no additional
(n
first
= 0,±1,±2,...), and
equation becomes
this
means
= {ln±^7ti
second equation
(-1)" cosh x
and the
satisfied
is
But
tells
this
roots of
(n
us that y
= nn
(n
that
= 0,±l,±2,...).
= 0,±1,±2,...). The first then
equation in x has no solution since cosh*
coshz =
—
>1
for
are obtained.
2
16. Let us rewrite
coshz = -2 as cosh* cos ;y
+ isinhjcsin;y = -2. The problem
evidently to
is
solve the pair of equations
cosh x cos y If is
If
* = 0,
the second equation
Since cosn^
is satisfied
sinh x sin y
and the
- 0.
first
reduces to cos y
no roots of coshz = -2 arise. find from the second equation that siny = 0, or
no y satisfying
x*Q, we
= -2,
= -2.
Since there
this equation,
= (-l)",
equation can hold only
it
follows from the
when n
z
is
first
1
2
x = cosh
-1
2.
+ (2n + l)m 1
= ±ln(2 + V3) + (2n + l)ni
= 0,±l,±2,...).
= -2. But
this
Consequently,
(n
Recalling from the solution of Exercise 18, Sec 33, that cosh" 2 these roots can also be written as
z
(n
equation that (-i)"coshx
odd, in which case
= cosh"
y-nn
= 0,±1,±2,...).
= ±ln(2 + V3), we note that
(n
=
0,±1,±2,...).
.
Chapter 4 SECTION 37
2.
jQ^J*=j(£-l)<*^
(a)
Te
dt
=
V3 —T = —ircos—^.•f + -1 = — + 1
|v 2
Since le~**l= e~
fcj
3
2/L
L 2/
o
bx
3
6 fe
IV" T"*
udt =
lira
L
The problem here
is to
\e
*•
mB
e-
Mde =
r
27T
1
*
*
when
m*n,
when
m = n.
we write 2*
/
and observe that when
=
J
e
Me~ M dO = 2wje'^'de
m#n, 2ic
i(m-n)B
i(m — n)
When
1
= - lim(l - e - bt ) = - when Re z > 0.
Jj=0
o
this,
4'
verify that ?s?
To do
4
J
/
-;
we find that
,
e""
3.
,i
*sin—
-
m = n,
1
1
i(m-n)
i(jn-n)
= 0.
/becomes 2*
/= \dd = 2n; and the verification
4.
First
is
complete.
of all, 7*
J But
7t
e
(I+,)x
etc
=
x
J
e cos * dtc
+
J e * sin x dx. 1"
also,
,(!+<>
l
+
I
e*e'"-l l
+i
-e*-l l-i l + i l-i
l
+ e*
+
e" A—+ —
i
-
x
x
Equating the real parts and then the imaginary parts of these two expressions,
je'cosx dx = - ^
+e
jVsin;c_* =
and
+ e*
2
o
Consider the function w(t)
= e" and observe that
lie
2tc
jw(t)dt=je
it
dt
=i-i=o.
=
i
i
K Since \w(c)(2n - 0)| =\e \2n c in the interval
l
we find that
= 2%
for every real
number c,
it is
clear that there
is
no number
< t < 2n such that lie
jw(t)dt=w(c)(2n-0).
(a)
Suppose Thus
that w(t) is even. It is straightforward to a
a
a
show
a
and
that u(t)
v(f)
must be even.
a
jw(t)dt=ju(t)dt+ijv(t)dt=2ju(t)dt + 2ijv(t)dt
a
a
= 2 j u(t)dt +
(b)
z
J
a
v(t)dt
Suppose, on the other hand, that w(t) a
J
odd.
is
a
w(t)dt
=2
J
It
w(t)dt.
follows that u(i) and
v(f) are
odd, and so
(n
= 0,1,2,...),
a
= ju(t)dt + J v(t)dt =
+ / = 0.
1
Consider the functions
1
P„(x)
where -1 < x <
1.
K
Since
2 x + i^l-x cosd = V* 2 + (1 -
it
"
_____
= ~j(x + iVl - x 2 cos ej d6
2
)cos
2
d<^x
2
+(l-
follows that 1
T.
P„(*)|<-f Lc
|
+ iVl-x 2 cos0
X
^<-fj0 =
l.
2 )
= 1,
55
SECTION 38 1.
(a)
Start
by writing
—a
I
=
J
-a
w(-t)dt
=
-b
The
substitution
r = -t
a
-b
two
integrals
b
on the right then yields
b
b
= -)u(T)dT-ijv(T)dz = ju(t)dr+ij v(t)dr = jw(r)dt. b
That
j v(-t)dt.
i
-b
in each of these
a
I
J
-a
ui-t)dt +
b
a
a
a
is,
-a
b
jw(-t)dt = jw(r)dt. -b
(b)
a
Start with b
1
b
b
= jw(t)dt = ju(t)dt+ijv(t)dt a
a
and then make the substitution
t
=
a
each of the integrals on the
is
P
I
P
a
That
P
= ju[(T)]'(T)di:+ ijv[(T)]' (r)dr = a
J
{x)dx.
a
is,
b
P f
jw(t)dt=jw[(/)(x)]<}> (x)dx.
3.
The slope of the
line
through the points (a, a) and (fi,b) in the ft plane
b-a
m=— p-a So
the equation of that line
.
is
b-a
t-a = -
p-a
,
(t-a).
is
right.
The
result
+
56 Solving this equation for
t,
one can rewrite
b-a
t
Since
t
=
If
B-a
as
aB-ba T+-^
B-a
.
0(f), then,
...
4.
=—
it
Z(t) = z[^(T)] where zit) ,
b-a
aB-ba
B-a
B-a
= x(t) + iy(t) and
Z(T)
t
= 0(t), then
= 4^(i:)] + /y[0(T)].
Hence
Z'(t)
=
=
5.
If
w(t)
at
x'MtWW + iy'miWW
wcd = z'[
vim]
= f[z(t)] and
= u(x,y) + iv(x,y),
f(z)
w(f)
The chain
=
+ i-f
ax
z(t)
= x(t) + iy(t), we have
= u[x(t),y(t)] + Hx(t),y(t)].
rule tells us that
du — =
,
ux
x+u y y
.
,
and
dv — =
,
Vj
(
x'
,
+ vy y',
and so
w'(0 = In
(ux x'
+ uy y') + i(yx x' + vy y').
view of the Cauchy-Riemann equations ux = vy and uy w'(t)
= -v,,
then,
= (ux x' - vx y') + i(yxx' + uj) = (ux + ivx )(*' + iy').
That is,
w'(0
when
f
=f
-
= {ux [x(t),y(t)] + ivMt)Mt)])[x'(t) + //(f)] = f'[z(t)]z'(t)
SECTION 40 1.
(a)
C be the semicircle z = 2e'°
Let
(0
<
-2
<
n),
2
x
shown below.
Then
+e
= 2/
(b)
Now let C be the semicircle
This
is
the
same
= 2e i0
z
f£±2
2.
(a)
The
arc
is
C:z = l + e
i0
2jt
Jc
= 2i(-i + 2n-i-7t) = 4 + 2m.
+
f JC
the value here being the
(z
- \)dz =
z
= 2e i0
(0 <
< 2k).
In this case,
^-^-dz = Am,
sum of the
7
values of the integrals in parts (a) and
(n< 6<2n). Then Ijc
.''29
w w ne J(l + « - l)fe <*0 = ije dO = i 2/
=
Thus
2k
C denote the entire circle
Finally, let
(n
as part (a), except for the limits of integration.
.'9
(c)
= 2j'(j + # + 1) = -4 + 2m.
I(^_^) = I(1_1
)
=
.
(b).
58 (b)
Here C: z
= x (0 < x < 2). Then
Jc
3.
(z-l)dz = J(jt-l)<£c =
In this problem, the path
C is the sum of the paths Q C C 2,
,
The function
to
= 0.
-jc
3
,
and
C4
that are
be integrated around the closed path C is /(z) = ne Kt We observe that and find the values of the integrals along the individual legs of the + .
c = c + C2 + C3 i
Q
square C. (i)
Since C,
is
z
=x
(0
< x < 1),
^^ = ^^ = ^-1. i
Jc
1
o
(ii)
C
Since
2 is
z
= 1 + iy (0 < y < 1), i
(Hi)
C
Since
3 is
= (1 - x) +
z
1
(0
i
< x < 1),
i
i
o
(ivj
Since
C4 is
z
= i(l - y) (0 < y < 1), i
= J *»*
i
4^^'HMy = ^JV^y = -2. o
Finally, then, since
we find
shown below.
that l
\c
ne*~ dz
= 4{e* -1).
The path
C is the sum of the paths 3 Q:z = x + ix (-1<*<0)
3
C2 :z = x + ix (0
and
Using
/(z)
= lonC,
and
=[
f(z)dz
/(z)
f(z)
= 4y = 4x 3 on C2
,
we have 1
jc
f(z)dz
+JVJ C2
= Jl(l + i3x 2 )<£c + -1
The contour z(fc)
C
has
4j^ +
t + [*t +
some parametric
+ i3* 2 )
1
-1-10
= Wlli + i[* 3
(l
o
1
= Jit + 3/jA; 2 dbc +
3
J4*
5
12ijA: dtc
6
2i [* ]J :
representation z
= 1 + i + 1 + 2/= 2 + 3/.
= z(t) (a
z(a)
= z and x
= Z2. Then
Jc
= \z\t)dt = [ Z (0]* = z(b) - z(a) = z2 -zv a
To
integrate the branch
around the circle C: z
Let z
=e
C
= e ,e
(0
be the positively
,e
(0
<
<
2;r), write
oriented
circle
1*1=1,
with parametric representation
< B < lit), and let m and n be integers. Then
lc
m z z"dz
But we know from Exercise
=
m
i6
j
{e
)
n
ie
(e-
)
e
i£ dd = i je'<"
+1)
3, Sec. 37, that
lK
je
Me- Md0-- [0 [lit
when
when
m*n, m-n.
VM
60 Consequently,
m+1 * n
when
frt\fc-j° lc
m + l = n.
\2m when
8.
Note that
C is
suggests the parametric representation C:z
With
2 2 x +y =
the right-hand half of the circle
that representation,
-
So, on C,
4.
* = ^4-y 2
2
= ^4-y +iy (-2 < y < 2),
to
.
This
be used here.
we have
dy
2
(
2
I= + ^y 4%
= j(-y + y)dy + ij\t
iW
-2 2
2
^
4
..2
= 4i J -2
= 4/ [sin"
10. Let
(a)
C
be the
f
circle
=
z
1
(1)
- sin"
1
(-1)]
= 4i
= z + Re ie (-Jt<6<
= Am.
n)
-^Rie dd = i]dd = 2m. ie
f
-K
(b)
When n = ±1,±2,...,
=
—
m (e *
- e-'" *) = / 1
— 2R°
.
sinn/r
=
n
11. In this case,
where a
is
any
real
number other than
zero, the
with a instead of n, yield the result
a~ l
f Jc a
(z-Zo)
dz
=
i—-sin(a^) a
same
steps as in Exercise 10(b),
61 12.
The function f(z)
(a)
representation
is
continuous on a smooth arc C, which has a parametric
z = z(t) (a
Exercise 1(b), Sec. 38, enables us to write p
b
\f[z(t)]z'(t)dt
=f
/[Z(T)k'[0(T)]0'(T)
a
a
where Z(T) = Z[0(T)]
But expression
(14),
Sec 38,
tells
(«
us that
zWt)W'(t) = Z'(t); and so P
*
jf[z(t)]z\t)dt =jf[Z(r)]Z'(t)dr.
f(z) is piecewise continuous on C. Since C can be broken up into a finite chain of smooth arcs on which f(z) is continuous, the identity obtained in part (a) remains valid.
Suppose that
(b)
C is any contour and that
SECTION 41 1.
Let
C be the arc of the circle
\z\=
2 shown below.
o
Without evaluating the
2
integral, let us find
x
an upper bound for j
note that if z
is
a point on C, 2
|z
Thus
-l|>|lz
2
l-l|
= |lzl 2 -l| = l4-ll = 3.
jc ~T~J
To do
this,
we
<
z
62
Also, the length of
2.
The path
C is
^(47r)
=
So, taking
n.
C is as shown in the figure below.
M = ^ and L = n
The midpoint of
,
we find that
C is clearly the closest point on
V2 C to the origin. The distance of that midpoint from the origin is clearly -y-,
the length of
being V2.
Hence
if
z
is
any point on C, Id >
V2 —
This means
.
that, for
such a point
4 Izl
Consequendy, by taking
M - 4 and L = V2 ||c
3.
The contour
To
^|
,
we have
S ML = 4V2.
C is the closed triangular path shown below.
find an upper
bound
for
\jc
z
\e
(e
-
- z)dz
z
l I
\e
\
,
we let
z be a point on
+ \z\ = e x + jx 2 +y 2
.
C and observe that
C
63 x
But e <
1
less than or
Hence, by
x < 0, and
since
equal to 4. Thus
writing
le*
M = 5 and L =
- zl < 5 when 12, we have (e
Note
that if lzl=
R (R > 2),
^x 2 + y 2
the distance
z
of the point z from the origin
on C. The length of
is
C is
then
2
l2z -ll<2lzl
2
2 +l = 2/? +l
and
4
lz
2 2 2 + 5z + 41 = lz + II lz + 41 >
2 Izl |
2
-1
1
Izl
-4 = (tf 2 - 1)(/? 2 - 4). |
1
Thus 2
2z 4
z
when
lzl=
R
(R >
2
-l
2 +5z +4
4
z
+ 5z 2 +4l
Since the length of
2).
2
2z -l
W+5z
I,
2
+4
2* 2 + l
I2z -ll
teR(2/?
^
CR
2
2
2
(R -l)(R2 -4)
is
^R,
then,
+ 1)
R2
/?l
2
~(/? -l)(/? -4)
R 2 J} and
it is
Here
CR
is
the positively oriented circle
Logz
Izl
lln/? + /0l
=R(R>
<
ln/?+l0l 2 ?
since
-n
Q
is,
1).
If
< "
R2 )
R tends to infinity.
clear that the value of the integral tends to zero as
z
^r
is
a point on
CR
,
then
+ ln/g
R2
of course, 2nR. Consequently, by taking
M = ———
and
L = 2kR,
always
evidently 12.
-z)dz < ML = 60.
z
is
we see
that
Logz
n + k\R R
< ML = 2n
dz z
*
Since it
iim it
+ \xiR _ =
lira
1//? — — = u, -
follows that J c*
Let
Cp
z
be the positively oriented circle \z\=p (0
suppose that /(z)
is
analytic in the disk Id
<
shown
in the figure below,
and
1.
y
We let
= exp^-
bounded disk in that disk.
observe that
Izl
< 1,
We if
z
is
there
is
V
1
\
Jppx
we
note
that, since
are asked to find an upper
a point on
Cp
jc
inasmuch as
M
is
Cp is z-
bound
/(z)
m
M such that for
= \z- m \\az)\<
I
z
M
2np, we may conclude
m f(z)dz
limf
that
<^j=2np = 27tMjp.
independent of p, U2 z-
it
follows that
f(z)dz
|
is
,
U2
Since the length of the path
exp^-i-j
a nonnegative constant
\z-
that,
1
^ log z j = exp|^- ^ (In r + id) =
of the power function here; and
Note
I' /a
represent any particular branch
z
-1/2
I
= 0.
(r
> 0,
a < < a + 2ft)
continuous on the closed l/(z)l^
m f{z)dz
M for each point z .
To do
this,
we
— 65
SECTION 43 1.
The function
z" (n
=
0,1,2,...)
has the antiderivative z"
Jc
I
n+1
^
_»+l
n
.
to
/(n
+ 1) everywhere
in the finite
a point z2 , 1
+1
/i
+l
n + 1^
2
1
'
i/2
2
"r
fa;
x
Zl
.
2.
C from a point z
any contour
plane. Consequently, for
+1
z
e*
—e
e
=
j
it
+1
+
l
it
i
it
>r+2i
cosl
(b)
—
l
= 2S
J
=
~
U-2)
«
= —lie
i„(f + ,)=:
e
—e
e
I
2/
= « + -. e
3
4
1
1
1
4
4
0.
4 i
1
3.
= ±1,±2,...) always has an antiderivative = z Zq. So, by the theorem in Sec. 42,
Note the function (z -Zq)"' (n does not contain the point
L(z-zo
5.
C
for
any closed contour
Let
C denote any contour from
real axis.
any domain that
r dz = l
does not pass through
that
in
Zq.
z = -1 to z = 1 that, except for This exercise asks us to evaluate the integral
its
end
points, lies
above the
t
I
=
jz'dz, -i
where
z'
denotes the principal branch
z'
= exp(/Logz)
(lzl>0,-^< Argz<
it).
66
An z
cannot be used since the branch But the integrand can be replaced by the branch
antiderivative of this branch
= -1.
z'
since
can
it
= exp(z'logz)
not even defined at
is
—
lzl>0,-
agrees with the integrand along C. Using an antiderivative of this
new
branch,
we
now write _('+!
*
-
i+l
[(l)
+1
1
_ _£_("^(i+l)(Inl+iO) _ i +
j
(
(i+lXlnl+i*)"! J
-
^ j~
£
('>D'ogl
£
(M)\og(-D
+e = _J_ h _ g-*^*) - ^ ' l+ i + V
j
1 i
.
1
L
1-/
d-o.
SECTION 46 2.
The contours
C
t
and
C2
are as
shown
in the figure below.
In each of the cases below, the singularities of the integrand lie outside
so the integrand
is
analytic
C
x
or inside
on the contours and between them. Consequendy,
imdz=if(z)dz.
C2
;
and
67
(a)
When
f(z)
=
(b)
When
f(z)
=
/(z)
=
(c)
(a)
When
-
the singularities are the points z
z .
—
+2 .
the singularities are at z
,
=
= 2nn (n = 0,±1,±2,...).
sm(z 1 2)
-, the singularities are at
z
= 2nm
(n
= 0,±1,±2,.
In order to derive the integration formula in question, around the closed rectangular path shown below.
Since the lower horizontal leg e~
l*
along that leg
is
represented by z
we
.
.).
integrate the function e
= x (-a < x < a),
the integral of
is
xl
je-
xl
dx = 2je- dx.
-a
Since the opposite direction of the upper horizontal leg has parametric representation z
= x + bi (-a £x
-]e'
{x+bi)1
dx = -e
the integral of e'
bl
xl
\e
iUx
e
dx
=
1*
along the upper leg
V
is
2
cos2bxdx + ie" \e
j
xl
sm2bxdx,
or simply u 2
-2/ jy* Since the right-hand vertical leg e~'*
along
is
2
cos2fct<£t.
represented by z
= a + iy
it is
O
je-
B
(a+iyf
idy
=
!«-* je
yl
e-
i2ay
dy.
(0
< y < b),
the integral of
68 Finally, since the opposite direction of the left-hand vertical leg has the representation
z
= -a + iy
(0
< y < b),
the integral of e~ l * along that vertical leg is b
b
-fe
H - a+iy)1 idy = -ie- al je y *e i2ay dy.
According to the Cauchy-Goursat theorem, then, a
a
b
xl
b 2\e~ dx - 2e \je-
V_
xl
b
cos2bxdx + ie** \e
y \-'lla
a nay > dy - ie~ je y *e dy = 0;
f :
and
this
X
reduces to
4 xl
]e~
(b)
We now
let
xl
coslbxdx = e~ b \e~ dx + e
a -»
°°
- (a
^ bl) ]e yZ sinlaydy.
in the final equation in part (a), keeping in
mind
the
known
integration formula
xl
\e' dx and the fact that
(al+bl)
e-
yl
\e
sinlaydy < e
- (a2+bl)
yl
je
dy ->
as
a ->
oo.
The result is
xl
e
cos2bxdx = ^-e- b2
(b>0).
2
6.
We let C denote the entire boundary of the semicircular region appearing below. It is made up of the leg C, from the origin to the point z = 1, the semicircular arc C2 that is shown, and the leg C from z = -1 to the origin. Thus C = C + C + C 2 3
t
y
3
.
69
We also let
f(z) be a continuous function that
=
by writing /(0)
is
defined on this closed semicircular region
and using the branch
/(z)
of the multiple-valued function z
m .
= V7e ie/2
r
The problem here
is to
C
>o,-f<*
evaluate the integral of f(z)
by evaluating the integrals along the individual paths C„ C ,and C3 and then 2 adding the results. Li each case, we write a parametric representation for the path (or a related one) and then use it to evaluate the integral along the particular path. around
(i)
C
z
:
v
= re i0 (0
JCi /(z)^=jVF.lJr
(ii)
C2
:
z
=1
I
B •
e'
(0
f(z)dz
<6<
ie
ie
2
dd = ije^ n d0 = if^e na
o
(Hi)
-Cy
z
= re ix
(0
<
r
1).
*
'2
]
L3i
o
<
[V^=|.
Then
it).
= ]e wn
=
Jo
- 1) = - 1 (1 + /).
= 3
3
Then
mdz = -\_ c f{z)dz = -)^e *'\-\)dr = i\^dr = i\-r vll i
\c5
o
The desired
L3
o
Jo
=j 1
f(z)dz + j
Cj
f(z)dz+i f(z)dz = l-h+i) + "-J
3
The Cauchy-Goursat theorem does not apply since f(z)
is
3
= h 3
0.
not analytic at the origin, or even
defined on the negative imaginary axis.
SECTION 48 In this problem,
3
result is
jc f(z)dz
1.
=-/.
\
*" J
we let C denote the square contour shown in the figure below. y 2i
c i
-2,
2
r
-2/
Jt
6 (a)
(b)
f
Jc
Jc
.
,.
,
.
=2m \e~
f
^2
z (z
+ 8)
2z +
1
r
coshz
,
tan(z/2)
f
Jc
Jc z
Jc— * r
=
dz
z \
^
= 2m(-i) = 2n.
,
.
+ 8>
7
z-0
coshz
d3
2tb* f
,
tan(z/2)
r
2
U;
8.
L2J z= _ 1/2
Jc b^sr* "IT
,
iJ^L + Lz
-(_i/2) r
=
I
,
4j
2
=
60,1,1
4
= 0. f(0)
Jz=0
2«f
.
Z=*o
= 2«i"|
4-sec
2
U
Let
(aj
1
= »rsec 2 f
C denote the positively oriented circle
The Cauchy
dz
c
+4
\z
—
I
when
-2
- il = 2, shown below.
J
z _2/)(z +
r
20
l/(z + 2i) dz
Jc
= 2m\
= 2m
z-2/
z
+ 2ij z=2i
Applying the extended form of the Cauchy integral formula,
f
Jc z 2 (
<2.
integral formula enables us to write
dz
JCrz 2
(b)
—2
& + 4)
2
_ f ~ Jc
2
dz ffc (z
_ 2/)
2 (
z
+ 2i)
2
1/ U + 2Q _ rf l/(z 1+1 ~ J c (z-2i) (* -
Z
=
we have
2£ 1!
1
(z
+ 2i)
71 3.
Let
C be the positively oriented circle
I
z
I
= 3 and consider the function ,
g(w)=\
We wish to find g(w) when w = 2
dz
and when
\w\
>3
(lwl*3).
(see the figure below).
y
Cm )(
i
w = 2/3
V
x
We observe that #(2)
= Jc
_2~
On the other hand, when
5.
z Iwl
>
3, the
= 2 ^'[ 2z2 " z ~ 2 ]
is
/ is analytic inside inside C, then
f£&*»2«rk)
= 2^1(4) = 8m.
Cauchy-Goursat theorem
Suppose that a function not on C. If z
z=2
and
tells
us that g(w)
and on a simple closed contour
f^>*
/ft)* =
f
= 0.
C and that z
is
^
Thus f
J^
The Cauchy-Goursat theorem to C,
7.
f\z)dz = f /(g)
tells
us that this last equation
each side of the equation being
ie
C
f
it),
and
— dz=[ -^-dz = 2m\e —
Jc z
is
also valid
when
zq is
exterior
0.
be the unit circle z = e (-n£d< Cauchy integral formula reveals that Let
'
Jc z
let
at ]
J * =0
a denote any
= 2m.
real constant.
The
On the other hand,
the stated parametric representation for
—dz = ]^~Qie w de =
\c Z
-It
K
It
=i
J
jexp[a(cos0 + isin 6)]d6
/
e
-K
C gives us
e
acos
V
flSin
e
acose d0 = ije [cos(asm 6) + isin(asin 6)]dd -It
-ft
K
It
= - je acos0 sin(asin 8)dd + ije acos9 cos(asin 6)dd. -It
-1t
Equating these two different expressions for the integral It
— e
r I
az
c z
dz,
we have
It
- je
acose
sin(asin d)dd
-It
+ i je acosB cos(asin d)dd = 2m. -It
Then, by equating the imaginary parts on each side of this
last
equation,
we see that
it
je
tt<:ose
cos(asmd)dd =
27f,
-« and, since the integrand here
is
even,
je
(a)
The binomial formula enables us
We
cos(asin 6)dd
=
Jt.
to write
note that the highest power of z appearing under the derivative
differentiating
(b)
acose
it
n times brings
it
down
to z".
So
Pn (z)
is
2
is
z
",
a polynomial of degree
and n.
We let C denote any positively oriented simple closed contour surrounding a fized point z.
The Cauchy
integral formula for derivatives tells us that
rtf-tf^fSfzlLj, dz"
Hence
the polynomials
K '
Pn (z)
Im^is-z)^
1
in part (a) can be written
(„
= 0,1,2,...).
73 (c)
Note
that 2
C?
-l)"
_
(j-1)"(j +
n+l
(s-iy
+ S-l
1)"
(s-iy
Referring to the final result in part (b), then,
we have
Also, since 2
(s (j
- 1)"
+ ir
+1
=
-
(s
+ 1)" _
(*
+ l)"
(s
+1
5
- 1)"
+1
'
we have
2" 2to* j c
9.
+i
s
We are asked to show that 1
rf(s)ds
— m Jc (t ni (s-z)
(a)
In view of the expression for /' (z) in the
/'(z + Az)-/'(z)
_
1
lemma,
1
r
2ml
Az
*
3
f(s)ds
1
(.y-z-Az)
2
(s-z)
2
Az
2(s-z)-Az 2ni[(s-z- Az) (s-z) 1
f
Then
/'(z + Az)-/'(z)
Az
1
f
f
/(j)
7ri|(5-z)
3
_
1 i
r
2^4 1
2(j-z)-Az 2 2 (s-z-Az) (s-z) 3(j-z)Az-2(Az)
(s-z)
2
^' 2 3 /( J f.c — 2iTi 2^/J(,-z-Az) (5-z) f
"
3
74 (b)
We must show that r
3(.s-z)Az-2(Az)
2
2 (3DIAzl+2IAzl )M < jf(s)ds
c(s-z-Azf(s-z?
Now D,d,M, and L
(rf-IAzl)V
The
are as in the statement of the exercise in the text.
triangle
inequality tells us that
I3(j
- z)Az - 2(Az) 2 l< 31* - zl
I
Azl
+ 21 Azl 2 < 3D! Azl + 21 Azl 2
we know from the verification of the Is - z - Azl > d -I Azl> 0; and this means that
Also,
I
(s
expression for /' (z) in the
- z - Az) 2 (s - z) 3 > (d l
.
1
Azl )
2
lemma
3 d > 0.
This gives the desired inequality.
(c)
If
we let Az
tend to
in the inequality obtained in part (b)
we find that
This, together with the result in part (a), yields the desided expression for /"(z).
that
75
Chapter 5 SECTION 52
We are asked to show in two ways that the sequence
1.
(-1Y
z„=-2 + A-f
(n
= l,2,...)
(n
= l,2,...)
n
One way is
converges to -2.
=-2
xn
Another way
is to
2.
is
0, respectively,
=
observe that \z„- (-2)1
~ (- 2)| < e
z„
where n
two sequences
y„=t-9-
and
-2 and
of real numbers converge to 51.
to note that the
if z„
= - 2 + i^-
(n
= 1, 2,.
.
.
n
),
to
apply the theorem in Sec.
Thus for each e>0,
n>n
whenever
,
1
any positive integer such that n >
Observe that
and then
then
/•„=lzJ=i/4 +
— -»2.
But, since
2fl
the sequence
3.
Suppose \zn
that
0„
=Argz2„-»^
(n
limzB
= 1,2,...)
.
is,
for each
= Argz^,
(n
e >0, there
is
a positive integer
In view of the inequality (see Sec. 4)
lzn
it
2 „_,
= 1,2,...),
does not converge.
= z. That
- z\< £ whenever n > n
and
follows that llzj-lzlke whenever
-zl>llzj-idl,
n>n
Q
.
That
is,
limlzj=lzl.
such that
=
76 4.
The summation formula found
in the
example
when
If
we put
z
=
<
where
jr(re'*)"
r
< 1,
52 can be written
in Sec.
lzl
the left-hand side
becomes
= jr rV" e = j^V cosnfl +
u=l
n=l
n=l
r
n
sinn0;
n=l
and the right-hand side takes the form
re
w
- re ie
1
w 2 - re~ w _ re -r _ ie 2 w ie ~ 1 - r(e + e~ ) + r ~ 1 - re~
rcos
1
'
1
- r 2 + irsin
- 2rcos + r 2
Thus
^ > r Zl
.
-
.
_
—
rcos0-r 2
rsin0
.
—
cosnd + iy r smnd = T+ T 2 l-2rcos0 + r l-2rcos0 + r 2 t& .
i
Equating the real parts on each side here and then the imaginary parts, summation formulas
E„r
Q= cosn0
B=1
where
6.
Suppose
<
r
—
rcosd-r 2 -r l-2rcos0 + r 2
< 1. These formulas
that
^z
B
=5.
clearly hold
To show
that
V> r
j
and
«
^
when
—
a=smnd •
rsin0
appeal to the theorem in Sec. 52. First of
r
=
^z„ =S, we
5>„=X
all,
write zn
=xn +iy„,
5>
B
=7.
OO
Then, since
^(— y„) = —Y,
it
follows that
n=l
jU = 2ta - % = 2>„ + )
—
«*(->.)]
T 2
arrive at the
,
too.
we note that
and
we
l-2rcos0 + r
n=l
n=l
.
x - iY = j.
S=
X + iY
and
77 8.
Suppose
£z„ = S
that
and
^w„ = T.
In order to use the theorem in Sec. 52,
we write
b=1
z»
=*,+%> S = X + iY
and
w„=un
2>b = * f,yH =Y
and
5>„ = t/,
+ivn
,
T = U + iV.
Now B=l
B=l
|>
B
=
V.
B=l
Since
2(^+"„) = X + t/
it
X^+vJ^+y,
and
follows that
J,[(xn +u„) + i(yn +v„)] =
X + U + i(Y+ V).
That is,
+%) + ("„ + J'vJ] = X + iY+(t/ + /V), B=l
or
B=l
SECTION 54 2
1.
Replace z by z in the
known series ln
~
coshz =
X
z (Izl<
~)
to get «>
cosh(z
Then, multiplying through
this last
2 )
=
2
_4b
Z
(lzl< ~).
equation by z, **
zcosh(z
2 )
= ]£
z
we have the desired result:
4n+1
tS(2n)!
(Izl<~).
78 2.
Replacing z by z -
(b)
1
in the
known expansion : z
n
= 5>7
e
"
n=0
;
we have
~Z
(lzk°o).
„, n!
n=0
So e
^rt =e £(£zi)l
(ld
r.=0
3.
We want to find the Maclaurin series for the function
4
z
To do
this,
we first replace
z by -(z
z
z
+9
9
1
l
4
/
9) in the
+ (z 4 /9)'
known expansion
r-^ly as well as
Then,
if
its
Ozl
condition of validity, to get
we multiply through this
last
equation by
|-,
we have the desired expansion:
n
2
6.
Replacing z by z in the representation
2n+l
sinz
z
=
(lzk~),
Zo
(2n + D!
we have 4b+2
sin(z
2
) = £(-1)"
*j
"„
Qz\<
oo).
Since the coefficient of z" in the Maclaurin series for a function f(z)
shows
The function
4n)
(0)
=
/
and
—^— has a singularity at z =
1.
<2B+1)
(0)
So
l-z \z
- < V2, i\
/
(0)/n!, this
that
f 7.
(n)
is
=
(n
the Taylor series about z
=i
is
= 0,1,2,...).
valid
when
as indicated in the figure below.
y
i
1
To
\V2
J
we start by writing
find the series,
This suggests that
i
1
1
l-z
(l-i)-(z-O
we replace z by
(z
l-i
l-(z-/)/(l-0"
- 1) / (1 - i)in the known expansion
(ld
and then multiply through by
:•
*
n=0
The desired Taylor series
is
then obtained:
(z-iT.
i-z
9
The
identity sinh(z
+ ni) = -sinhz sinhz
So, if we replace z
sa-o*
and the periodicity of sinhz, with period 2 m,
= -sinh(z +
by z - ni
in the
(lz-/I
ni)
known
= -sinh(z - ni).
representation
tell
us that
80
we find
and then multiply through by -1,
sinhz
13.
Suppose
<
that
<
Izl
4.
—
= -V i£_JEL! (2» + D!
S
<
Then
that
Iz / 41
(
< 1, and we can use
the
|
Z
known expansion
7^- = 2y
4z-z
2
(IzKl).
< Izl < 4,
To be specific, when
1
_ m\< ~).
111-lym ^fz =y£_ 4z~W ^4 fl+l
!_£
4z
= J_ + 4z
= _L + y. y^_ ~4" ^4" +1
4Z
+2
'
B
4
SECTION 56 1.
We may use the expansion
™ 2n+l
sin
to see that
when 0<
Izl
%.
3.
Suppose
(UI< ~>
< °°,
jn
that 1
-S
( ir
and
y
(-!>•
1 ,
recall the
,
f
(-ir
Maclaurin series representation
= Z^ T^'2 1-z
( |Z|<1 )-
±! „=o
This enables us to write
1 1
+Z
1
^
'r-iiRT-i^ — Z n=0 v
1h
„=o z
Z
Replacing n by n -
1
1
in this last series
and then noting
that
81
we arrive at
the desired expansion:
v
1
4.
The
singularities
of the function /(z)
there are Laurent series in
=
are at the points z
powers of z for the domains
=
Y z"
(lzl< 1)
Z
„ =2
„_
1
=
and
and z = 1
<»
1.
Hence
(see the
figure below).
To
find the series
when
<
\z\<
Z
Z
„=0
A
2
As
for the
domain
< lzl<
1
/(z)=
5.
(a)
The Maclaurin
<»,
1,
recall that
Z
n=0
note that
II / zl
=
"7|(z")
series for the function
Z
Z
< 1 and write
=
"7'I^17I)
and write
z+
~%^~%7-
l
is
valid
when
z-1
\z\<
1.
To
find
it,
we
recall
the Maclaurin series representation
= TIT 5>"
(ld<1)
and write
for
1-z z
+l
fziA
~
-<« +
•°
"=1
1
»izr 12 «=0
<-<
- »X »=0
11=1
-
-X^ n=0
1
- X*' «=0
(b)
To
find the Laurent series for the
Maclaurin series for
same function when
—— that was used in part (a). 1
-z
<
Since
1+ZJ n= i)\ZJ z
l,
n =oZ
1 here,
n =0
we
recall the
we may write
Z
z
(l
1
The function /(z) = z(l
the figure below.
Hence
=
and z = ±z, as indicated
there is a Laurent series representation for the
and also one for the domain
To
has isolated singularities at z
+ z2 )
1
°°,
find each of these Laurent series,
which
is
exterior to the circle lzl=
domain
For the domain
/(*>
we recall the Maclaurin series representation
Z
1
+Z
2
n=0
< Id < 1, we have
= ~'7~~t = 1
t HT = tc-DV-
Z B=0
On the other hand, when
B=0
1
1
= -+ £(-1)
V"
1
= £(-l)
2n+l
n+1
z
Z
Z
11
1
^ r__l_V _ ^ (-If _ v (-1)" n=0
£
n=l
+1
.2n+l £~
'
2
z In this second expansion,
1
1.
(lzl< 1
in
we have used
1 the fact that (-1)""
= (-l) fl-1 (-l) 2 = (-1)" +1
.
1).
83 8.
(a)
Let a denote a real number, where -1
< a < 1.
Recalling that
j— = |>"
(IzKl)
»=0
enables us to write
_a
a
_v
1 i
fl
z-a"7"l-(a/z)"^7Tr
'
or
a
(b)
Putting z
= e'° on each
^ a"
= 2j—
(lal
side of the final result in part (a),
a
n
e
we have
-inB
But
a e
ie
-a
(cos0-a)-isin0
(cos0-a) + i'sin0 (cos0-a)-i'sin0
acos0-a 2 -iasin0 l-2acos0 + a 2
and
XflV" 9 n=l
cosn6-i^a n smne. b=1
n=l
Consequendy,
2,a"cosnd =
when -1 < a <
10. (a)
-—
and
r-
l-2acosd + a 2
Ya" smn9 = r ~ l-2acosd + a 2
1.
Let z be any fixed complex number and plane. The function
C the unit circle w = e'*
{-it
<
^ n)
in the
w
/(w) = exp 2v has the one singularity w = in the C, as shown in the figure below.
w
w>
plane. That singularity
is,
of course, interior to
.
w
Now
the function
plane
f(w) has a Laurent
According to expression
series representation in the
domain
I<
<
(5), Sec. 55, then,
exp
(0
[II
where the coefficients J„(z) are
exp
Using the parametric representation expression for
Jn (z)
J" i z) -
1
r
That
|
T7; 2m 'J -
dw
w = e'*
= 0,±l,±2,...).
(n
(-n <$
for C, let us rewrite
as follows:
exp
=
]
Wn+l
2m Jc
this
w——
K
2 v~ v JC» e
*
- iV'd0 =
1
2m
|
exp[izsin ]e~
m *d
_
is,
K J„ (z)
The
last
=
1
^
J
exp[-i
- z sin 0)]
(n
= 0,±l,±2,...).
(h
= 0,±1,±2,...).
expression for /„ (z)in part (a) can be written as *
= ^i(^)
1
|[cos(n0-zsin0)-ism(n0-zsin0)]d0 T~ 2;r J
\cos(n-zsin 2tc j„
=
— 2% 1
*
2 cos(«0 1
i
- zsin )d
f sin (n-z sin )d
2?r j
i
2k
85 That
is,
*
= — J cos(«0 - zsin 1
J„(z)
(n
= 0,±l,±2,...).
iti
11.
fa)
The function f(z) unit circle
is
C.z- e'*
analytic in
(-it
<
<
some annular domain centered
it) is
at the origin;
and the
contained in that domain, as shown below.
For each point z in the annular domain, there
n=0
is
n=l
a Laurent series representation
<
where
(n — ?T
-*?T
and
— 7F
Substituting these values of a„ and b„ into the series,
or
/(a.J-J/e-w.J-lJ/,,.,
—
we then have
= 0,1,2,...)
86 (b)
Put z
= e' e
in the final result in part (a) to get
or
/(«")
If
k(0)
= ^- J/(e* )d* +
= Ref(e' e ),
w(0)
1£ J/(^)cos[n(0 -
then, equating the real parts
=
—
J«(0)
on each side of this
last
equation yields
+— 2 J"(0)cos[n(0-0)]
SECTION 60 1.
Differentiating each side of the representation
1-z we find
n=0
that
U
z;
az „ =0
n=0
az
„ =1
n=0
Another differentiation gives
(i
2.
— z;
"Z n=0
Replace z by
as well as in
1 / (1
its
„=o
- z) on
«z
B=1
n=0
each side of the Maclaurin series representation (Exercise
1)
condition of validity. This yields the Laurent series representation
4 = £<=!>>ZI>
«
87 3.
Since the function /(z) valid in the
To
= 1/z
has a singular point at z
Taylor series about z
=2
is
1111 2 + (z-2)
z
it
its
open disk \z-2\<2, as indicated in the figure below.
find that series, write
to see that
= 0,
2
l
+ (z-2)/2
can be obtained by replacing z by -(z - 2) / 2 in the
known expansion
1
(lzl<
1).
Specifically,
l_ly|-
(Z-2) T (Iz-2I<2),
z
ifA.
Z
R-0
2
J
or
(lz-2l<2).
Differentiating this series term
z
by term, we have
- 2r = E^tH" + J J = l^n(z y l
7
1
-~l
l)(z
- 2)"
(lz
- 2I< 2).
n=0
Thus
-| 4.
(
_ 1)>+1)
^y
Consider the function defined by the equations
V-l f(z) = 1
when
z
* 0,
when z = 0.
(lz-2l<2).
When
z
* 0,
/(z) has the power series representation '
/(*) =
2
\
3
+ £ + l. + £. + .^ -1 = 1+ ,
l
z
2!
1!
Since this representation clearly holds
z —z +— 2!
3!
when
z
=
too,
it is
+••
3!
actually valid for all z.
Hence/
is entire.
Let
C
point
be a contour lying
w=1
in the
w = z,
to a point
as
open disk \w -
shown
w According to Theorem
1
II
<1
in the
w
plane that extends from the
in the figure below.
plane
in Sec. 59,
we can integrate the Taylor series
1
representation
(lw-u
w
„=0
term by term along the contour C. Thus
dw "
n=0
f
— = [— =
n=0
But
[Logw] z = Logz-Logl = Logz 1
and n+l
4
J
(w-l)"=J(w-l)"rfw =
n+
l
(z-1) Jl
n+l
n+l
Hence
(-D
^n + and, since (-1)"
-1
^
l
= (-ly-^-l) 2 = (-1)" +1 Logz =
,
this result
X Lii— («-!)" n=l
(Iz-lkl);
n becomes
(Iz-lkl).
— 89
SECTION 61 1.
The
singularities
of the function /(*)
= -^L-
are at z
=
find the Laurent series for/that is valid in the punctured disk
,
± L The problem
here
is to
below.
y i
°
\
x
/
4
-i
We begin by recalling the Maclaurin series representations
2!
1!
Qz\<
3!
oo)
and 1
l-z which enable us
= l + z + z 2 + z 3 +---
Oz\
to write
e
z
= l + z + ^-z + -z 3 +2
2
(\z\< oo)
6
and 1
(\z\
Multiplying these last two series term by term,
2 l
z
-
+\ ,
we have
the Maclaurin series representation
= l + z + ±z 2 +*z 3 +2
-z
6 2
-z
3
-
4
z +-
5
1.2
which
is
valid
side of the
when
\z\<
The desired Laurent
1.
series is then obtained
by multiplying each
above representation by -:
1
z(z'
+ l)
1 = -+l--z
z
,
2
5
2
z
6
+••
(0
90 4.
We know the Laurent series representation
1117
1 2
3
z sinhz
from Example tells
6
z
Expression
2, Sec. 61.
-
of
us that the coefficient
(0
Z
360
z
(3), Sec. 55, for
the coefficients b„ in a Laurent series
in this series can be written
z
dz
2m where
C is the circle
lzl=
1,
taken counterclockwise. Since bx
'
6.
The problem here
is to
is
= - \,
then,
6
c z sinhz
use mathematical induction to verify the differentiation formula
[/(^(z)]
The formula
lc z sinhz
(n
"=E Af \z) g
(n)
when n = 1
clearly true
[/(z)s(z)]'
k
k)
(n
(z)
since in that case
it
= l,2,...).
becomes
= /(z)s'(z) + f(z)g(z).
We now assume that the formula is true when We start by writing true when n = m +
n=
m
and show how, as a consequence,
1.
UWz)]
(m)
(m+1)
={[f(z)g(z)]'}
=mz)g'(z)f
=1 *=0
m)
(m)
,
=[f(z)g
(z)
+ nz)g(z)]
l(m)
+[f(z)g(z)]
(t)
r / (^
(m - t+i)
/
(^)+s
(m - t+1)
(t)
(z)^
(z)
V* /
= /(z)*
(m+1)
(z)
+
£ *=1
'm^
(
m
N
/W( z )^('» +1 -^(z) +
n+1)
/
('
(z)^(z).
it is
But
m
frrC
ml
+ 1)! kl(m + l-k)l
ml
(ro
|
k
,
k-l)
(*-l)!(m-* + l)!
kl(m-k)l
'm + V <
k
>
and so
inz)g(z)r
+i)
n+i
=nz)g°
y
\z)+
z
m
or
[/(^(z)]
The
/(z)
is
Write g(z)
It is
m +r f\z)t«-
,
k
\z).
an entire function represented by a series of the form
f(z)
(a)
=iT
now complete.
desired verification is
We are given that
(m+i)
= z + a2 z 2 +ajz 3 +-'-
(lzl<°°).
= f[f(z)] and observe that
straightforward to
show
that
/(z) = /'[/(z)]/'(z),
= /" [/(z)][/'(z)] 2 + f'[f(z)]f" (z),
g"(z)
and
3
g"\z)
= /'"[/(z)][/'(z)] + 2f(z)f"(z)f"[f(z)] +
f
t/(z)]/' (z)/" (z) + /'[/(*)]/"'(*)
Thus
S(0)
= 0,
^'(0)
= 1,
g"(0)=4a2
,
and
g'"(0)
= 12(a22 + a3 ),
and so /[/(z)]
= z + 20.Z 2 + 2(4 + a,)z 3 +-
•
(lzl< °°).
•
—
)
.
92 Proceeding formally,
(b)
we have 2
/[/(*)]
= f(z) + fljC/Cz)] + «,[/(z)]
3
+- •
•
= (z + fljZ 2 + OjZ 3 +•••) + (<*2Z 2 + 2^z 3 +-
2^ +
=z+
2(a*
+ a3>z 3 +-
• •
)
+ (fl3z 3 +•
• •
• •
Since
(c)
sinz
=z-
—
+• • •= z
3!
+ Oz 2 + f - - V+-
=
the result in part (a), with Oj
and a3 =
•
(IzKoo),
•
6)
V
-—
,
tells
us that
6
sin (sinz)
8.
= z-— z 3 +---
(lzl< ~).
We need to find the first four nonzero coefficients in the Maclaurin series representation
— coshz
This representation
is
(\z\
= E^z" t^n\
2
\
numbers
valid in the stated disk since the zeros of coshz are the
The series (/i = 0,±l,±2,...)» the ones nearest to the origin being z = ±^-i2 J contains only even powers of z since coshz is an even function; that is, E2tt+l =0 z
= \— + n7t\i
(n
U
= 0, 1, 2,
. .
.
).
To
find the series,
coshz =
1
+— + — + —+•••= 2!
into
1.
The
4!
l
+ -z
series
2
+—z 24
2
6!
4
+"+—z 720 6
(lzl< °°)
result is
,
=1 coshz or
we divide the
1
2
z
2
+
— 5
24
4
61
z
6
z
720
+••
I
V
n>
i^ lzl<—
(i
2
\ I,
93 1
1
,
=1
2
z
coshz
2!
+—z 4
61
4!
6!
5
6
f.
z +•••
I
.
lzl<
V
—n 2
Since
_!_ = £ coshz this tells
+
^ 2!
+ Jl z 4 + JSl z «+... 4!
f|
6!
V
us that
£0=1,
E2 = -1, £4 = 5,
and
E
6
=-61.
z |
2
.
94
Chapter 6 SECTION 64 1.
(a)
Let us write
_i- = I.-i- = I(i- z + z z
+z
The residue
(b)
z
at
z
1
+z
2
3 2 -z + ...) = i-l + z -z +...
z
(0
z
= 0, which is
the coefficient of -, z
is
clearly
1.
We may use the expansion = l-
cosz
2!
4!
1
1
6!
to write
ZC
m
1
osH = z^l--. ? + -. 7 --. 7 + 1
1
1
111111
"\
= z--.- + -.-3---
...J
7
-f.
(0
The residue
(c)
at z
= 0,
or coefficient of -, z
is
now seen to be -— 2
Observe that
z-sinz
1.
1
= -(z-smz) = -
z
.
.
Since the coefficient of
-
T4
3!
5!
(0).
*'j
r
z
z
T2
5!
3!
in this Laurent series
is
0, the residue at z
=
is 0.
z
(d)
Write
cotz 4
z
and
cosz
1 4
'
.
sinz
z
recall that
cosz
=1 1
zT
2
2!
and
_ ~
T4
+z ,
4!
,
=1
T z
2
2
+
-T z
4
24
(lzl<
~)
95 Dividing the series for sinz into the one for cosz,
cosz
z
z
1
we find that
3
Thus cotz 4
z
Note z
_
I
"z
4
fl
z
U
3
+
T5
11 11 + '-)-7~37~V~z -
that the condition of validity for this series is
= nit(n = 0,±l,±2,...).
now evident that
It is
4
due
(e)
to the fact that sinz
has residue
z
(0
—45-.
at z
<&<*).
=
when
= 0.
Recall that
=z+
sinhz
|y
^+
+
... (|
z
and 1
= l + z + z 2 + -.i
(lzl<~).
1-Z There
is
a Laurent series for the function
sinhz
that is valid for
and
1-z
1.
To
1
find
(
.
it,
we
first
1
^
multiply the Maclaurin series for sinhz
2
1
z + -z +
1
6
120
3
,
5
z +•
:
3
z
1
+-z
5
+•
5
z +-
7
z
+ tz o
3
+•••
(0
—
.
96
We then see that sinhz 4
z
(l-z
)
z
In each part,
(a)
To
1
3
•••
47—^ —z (1
at
z
=
Jc
—^
dz,
is
— 6
J
C denotes the positively oriented circle
evaluate
(0
6 z
z
This shows that the residue of
2.
—17 + -•- +
=
2
lzl= 3.
we need the residue of the integrand at
z
= 0. From
the Laurent series
exp(-z) 2
_ —
1
^
1
_2
z
2
+
1!
^
we see that the
—
3
z
z
2!
1 ) +...=_ 2
™P^ dz = 2n -l i
•
jc
11
,
z
2
valid for
1
)
= -2jti
11
1
—+ 1
1
1
1!
2!
3!
z
°°, tells
A
.
.
,
(0
3!
.
exp^jdz, we must
Z
= Z2 + — +
is
2!
,
+...
find the residue of the
= 0. The Laurent series
,
which
z
1!
z
1
+
-1. Thus
is
Likewise, to evaluate the integral integrand at z
z
J
1
1
,
3!
required residue
lc
(c)
,
+
1
4!
11
1 2
z
us that the needed residue
is
— 6
.
Hence
—
—
z
97 z -=
r
(d)
As
for the integral
JC 7*l
z
+1 —dz, we need the two residues — 9r -2z z+1 z -2z
z
2
one
at z
=
and one z
at z
z(z-2)'
z-2)
—
z+
To
.
{
2A
z
z
2
2
(z-2) + 3
1
%
1 2
-
that the coefficient of
z-2
2
+ (z-2)
when
z-2j
2V
(z-2)
l
+ (z- 2)/2
2
|
z-2
2\
- 21 < 2, and
note that the coefficient of
in this product
z-2 is
—3
.
Finally, then,
by the residue theorem,
Jc
In each part of this problem,
(a)
If/(z)
-,
l-z
1 _2 / when
-
2
z
-2z
V
2
2)
C is the positively oriented circle
lzl= 2.
then
:
—4=-—-;
=-
\zj
z
in this last
1
z-2
valid
z) l-(z/2)
= 2, we write
obtain the residue at z
1
is
can be found by writing
z
z(z-2)
which
=
when 0
valid
is
2 z)
2
\
product
z
+1
f
is
+1
= 2. The residue at
z(z-2)
which
of
1.
-z
This
z
1-z
tells
us that
f(z)dz
r
= -- v l+z 3 +z 6 +••• y
= 2mRes <=°
z
=--— z
= 2ni(-l) = -2 m. \ f(-) \z) z
z
2
-
(b)
When
f(z)
1
= 1
+z
>
2
we have
Thus
= 2mRes 4"/f- ] = 2m(0) = 0.
f
(c)
If
/(z)
= -,
it
follows that
f ic
Let
~ /[-) = -• 2
Z
C denote the circle
f{z)dz
lzl=
VZ/
= 2m Res -^/f - ] = 2/ri(l) = 2m. z z =o
1,
z
z
=
T—
The Maclaurin
(b)
Referring to the Maclaurin series for e
Now
the
-
\ZJ
taken counterclockwise.
(a)
series e
Evidently, then,
Z
(lzl< °°) enables us to write
in this series occurs
z
once again,
when
let
n-k = -l,
us write
or k
= n + l.
So, by the residue
z
theorem,
l> expf-ljz = 2m-i—
Jc
The
vz/
final result in part (a) thus reduces to
(«
+ l)!
(/j
= 0,1,2,...).
99 5.
We
are given
two polynomials P(z)
= a +a z + a2 z 2 +--- + a„z''
<2(z)
= b + blZ + b2 z 2 + --- + bm z m
(a„*0)
l
and
(bm
* 0),
where m>.n + 2. It is
show
straightforward to
z
'"" 2
pQ 1 z> -
J_ 2
"
that
Qz
fl
+ a z "" 3 + a2* m
V" +
2(1 / z)
~4
i
1
fri*"*"
+
+
V"
2
w """ 2
+ fl^ + +K •
•
•
1
• • •
Observe that the numerator here is, in fact, a polynomial since m -n - 2 > 0. Also, since bm * 0, the quotient of these polynomials is represented by a series of the form 2
do+diZ + dzZ + ••-. That is, P(l/z)
1
1
and
L —r we see that
z
P(\lz)
that
C is
^
.
rt
z
2
(0
= 0.
2(1 /z) all
of the zeros of Q(z)
positively oriented.
plane except at the zeros of <2(z), obtained that
P(z) dz ,c
j
,
has residue
Suppose now that assume
j
Q(z)
lie inside
Since P(z)/Q(z)
is
a simple closed contour C, and analytic everywhere in the finite
follows from the theorem in Sec. 64 and the residue just
it
P(l/z) = 2mRes J_2 = 2^*0 = 0. z=0 z 'ea/z)
C is negatively oriented, this result is still true since then
If
Piz) J__
icQ {z)
P(z)
f
J-c
dz
= 0.
Q(z)
SECTION 65 1.
(a)
From
the expansion
(Izl<~), 1!
we
2!
3!
see that
ii + ,n =z+1+ ^,^11 _._ + _._
zexp|
_j
..
(0
100
-
The principal part of zexpl
at the isolated singular point z
I
=
is,
then,
z
+ 2!'z and z
(b)
The
=
is
+ "'
3!Y
;
an essential singular point of that function.
isolated singular point of
is at 1
involves powers of z 2
z
z
+z
+ 1, we begin by
= -1.
Since the principal part at z
= -1
observing that
= ( 2 + 1) 2 - 2z - 1 = (z + 1) 2 - 2(z + 1) + 1.
This enables us to write
e
_ (z+lf
Since the principal part
is
,
The
point z
t
The
1
z+1 the point z
,
+1
is
= -1
a (simple) pole.
is
the isolated singular point of £ -£ ) and z
we can write
^- — + — -•••1 = 1- — + — -•••
1
3!
z\,
5!
(0
5!
3!
J
principal part here is evidently 0,
and so z =
is
a removable singular point of the
smz
.
function
(d)
+!
m
=
z
-
—^— z
sinz
The
!)
z+1
1+z
(cj
-% +
.
isolated singular point of
cos z is
z
= 0.
Since
z 4
2
cosz
1
—I
the principal part is
(
z
z
I
2!
4!
.
= I_± + £l_... ;
This means that z
z
=
(0
4!
2!
COSZ is
a (simple) pole of
(e)
Upon
writing
—
^—-j
(2
=
— z)
isolated singular point z
(of order 3).
.
z
z
we
find that the principal part of
(2-z)
(z-2)
=2
is
simply the function
itself.
That point
is
3
at its
evidently a pole
..
101 (a)
The singular point is z = 0. Since
1-coshz
(
1
°°,
6
we have
4!
2!
V
when
,4
_2
1- !+£.+£.+£_+.. 6!
Here the singular point
l-exp(2z)_
is
also z
= 0.
(
2z
I
1!
1
= _2_ J__2^ ll'z
when
°°,
we have
The
3
Since
2
V
_1
and
4!
3!
2j_
2
exp(2z)
= e 2(z -V=e
1
+
js
z
2V 2V
3
2 73
2!
2!*z
m=3
ex P^ z ^ (z-1)
singular point of
6!
2
2
B=
"\
5!
_j__2^_2^
3!'z
IT*"
4!
3
=
— 4 3
3!
(c)
4!
z
m = l and B = —— = -— 2!
(ft)
2\
J
= \ jhe Taylor series 2
2(z-l)
l
2 (z-l)
2
3
.
2 (z-l)
2!
1!
3
(Izl< oo)
3!
enables us to write the Laurent series
exp(2z)
(z-1)
Thus
2
m = 2 and
Since / is analytic at Zq,
f(z)
Let
2
_^
+— L(z-lr
1!
1 •
z-1
2
2
2!
3!
2 2 + — + -J-(z-l)+-
(0
2
= e 2 — = 2e 2
it
= f(z
.
has a Taylor series representation
)+^-(z-z
g be defined by means of the equation
)
+ ^^-(z-z
2
)
+-
(Iz-ZoK^,).
102 (a)
Suppose that /(z
g(z)
*0. Then
)
1
=
/<0 + ^(*-*) + ^U-*o)
Z-Zn
This shows that
(b)
£ has a simple pole
at Zq
Suppose, on the other hand, that /(z
)
+-
(0
°;
2!
1!
2
,
with residue /(z
l
).
= 0. Then
2!
1!
(0<\z-z \
is
just 0, the point z
=
is
a removable singular
point of g.
4.
Write the function 3_2
=
/(Z)
(a>0)
(?W
as
/
_ =
(z)
4Kz) /
-n3
3 3_2
where
0(z)
=v
8a z
.
(z
Since the only singularity of 0(z)
z
*M(Z -
Hz) = ««0 + about z
is at
= -ai,
ai)
+
n3
-
+ «T
0(z) has a Taylor series representation
^(z- ai? +
(lz-ai'l<2a)
= ai. Thus
/(z)
1
=
(z-m)
3
0(ai) + ^rr^Cz
- fl +
Now straightforward differentiation reveals
^ (z) =
—
—
4
3
16a /z-8fl z ;
(z
„4
+ ai) ,
2
- ai) +
(0
that
16a (z) = ,
.
and
(z
2!
1!
2
3
(z
(z
-4fl/z-a
+ ai)
3
2 )
Consequently,
Q(ai)
= -a 2 i,
0'(a/)
= ™, and
"(ai)--i.
This enables us to write
/(Z)
The
=
(z-aif [~
a2i
principal part of /at the point z
ill
- ai
z
~
fll) (Z m)2 + " (z f ~ ~i " ']
= ai
is,
(0
< lz - ml < 2a).
then,
all 2 (z - ai)
a (z
2 i
- ai) 3
'
SECTION 67 1.
(a)
The function f(z) where that
(b)
If
z
(/>(z)
=1
is
+2 z~l
Z
= z 2 +2, and
has an isolated singular point at z
observing that 0(z)
a pole of order
analytic
Writing /(z)
and nonzero
at z
=
z-1 = 1, we see
m = 1 and that the residue there is 5 = 0(1) = 3.
we write f(z)
we
Hz)
=
zrf,
lz + 1
where
0(z)
z- -
=
8'
1
see that z
point,
= --
is
a singular point of /. Since 0(z)
/has a pole of order
m=3
there.
The
B _ r(-V2) _
residue
is
analytic and nonzero at that
is
3 16'
2!
fc)
is
= 1.
The function expz z has poles of order
2
expz
+ ;r
2
m = 1 at the two points B
exp TO
i
2 to
and the one
at z
(z-/n)(z +
= -ni
is
z
7ri)
= ±m. The residue at
_ -1 _ ~ In 2 to i
'
z
=
to is
104 1/4
2.
(a)
Write the function /(z)
=
(I
z
/(z)
Hz) =z ^7. z+1
The function 0(z)
where
is
zl
+1
> 0,
< are z < 2 it)
= z m =e 4
analytic throughout
its
as
(lzl>0,0
domain of
definition, indicated in the
figure below.
•
Branch cut
o
-1
Also,
jl
/
J'osH) Ja«i+<*) it it 1+i , n = (-1) n i/4 _ =e 4 = £4 = e iwM = cos— + zsin— = —f=-*0. .
.
<9(-l)
4
4
This shows that the function /has a pole of order being
5 = 0(-l) =
1
V2
m = 1 at z = -1, the residue there
+/
V2'
(b)
Write the function /(z)
From
= Logz 2 2 (z + D
as
/(z) has a pole of order
this, it is clear that
differentiation then reveals that
Res
Logz & 2
(z
+ l)
2
=
=
tf
+ 2i 8
m=2
at z
=
i.
Straightforward
105 (c)
Write the function .1/2
/(*) =
2
(ld>0,0
2
+D
(z
as .1/2
f(z)
= (z-i)
=
where
2
(z
2
+
Since
f
W
2(z
+ /) 3
and
r 1/2 - *-''*/4 -
_i_
_L
7
Res
f 2
*=<~(z
(a)
,-i/2
_
1/2 1
„„2
+ l)
j*/4
—
= ft'(t>
_
1
' .
'
1
8V2"
We wish to evaluate the integral 3
+2
3z ,c
where
C is the circle
the singularities z
- 21 = 2 taken in the counterclockwise direction. That circle and ± 3i of the integrand are shown in the figure just below.
\z
= 1,
dz,
2
(z-l)(z +9)
,
Observe that the point z and that
= 1, which
is
the only singularity inside C,
the integrand
3
3z + 2 Res2 (z-l)(z +9)
3z z
3
2
+2 +9
1
2
According to the residue theorem, then, 3z
3
+2
l z-\Z^9) dz = 1M {l) =ni i
-
is
a simple pole of
106 (b)
C is changed to be the positively oriented circle
Let us redo part (a) when in the figure below.
= 1, ± 3/
In this case, all three singularities z
already
know from
Izl
= 4, shown
of the integrand are interior to C.
We
part (a) that 3
3z + 2 Resz * =1 (z-l)(z +9)
It is,
moreover, straightforward 3
to
+2 (z-l)(z' + 9) 3z
Res <= 3i
1
2
show that 3
+2 (z-l)(z + 3/). 3z
15 + 49/
12
and 3
3z
Resz =-3,-
3
+2
_ (z
The residue theorem now
+9 ) tells
+2 _ i) (z _ 3/ ) (z
+2 _ /l 5 dz = 2m\- + 2 } c( z -l)(z +9) V2
4.
(a)
Let
C
,
15 + 49/
12
denote the positively oriented circle r
integral
dz
-rJc z (z
12
j=-3i
us that
3
3z
f
15-49/
3z
— has singularities at
+ 4) y
z
=
Izl
+
15-49/"\
"
= 2, and
=6, m.
note that the integrand of the
and z = - 4. (See the figure below.)
.
107
To
find the residue of the integrand at z
= 0, we recall the expansion
(lzl
and write
_
1
1
1 3
z\z + 4)
4z
_l
_n-3
+ (z/4)
Now the coefficient of -
here occurs
when n =
2,
and
(0
we see that
z 1
1
z\z + 4)
64
Res *=°
,
Consequendy,
Jc
(b)
Let us replace the path at
-2 and with radius
z\z + 4)
C in part 3.
It is
(a)
*=°
-4,
by the positively oriented
^
us that z
0(-4) = -1 /
64.
= -4
centered
— 64
z\z + 4)
_
(z)
Tw-T+Ty tells
+ 21 = 3,
we write
1
This
circle \z
that
Re S
the residue at
32
shown below.
We already know from part (a)
To find
\64)
is
...
.
where
a simple pole of the integrand and that the residue there
Consequendy,
J<^ 5
1
Hz)= 7-
(z+4)
{64
64J
is
108
cosh 7tz dz
f
:
Let us evaluate the integral
J^ C
TIT Z^Z
—+ *
7T"» ij1
= 0,±i of
All three isolated singularities z
C is
wnere
the positively oriented circle lzl= 2.
1
the integrand are interior to C.
residues are
=— — +1 z(r+l)
_
s
z=o
_
cosh nz
cosh nz
Res
Res
=1,
5
z
—+
cosh 7cz =
2
z(z
l)
.
= cosh tfz z(z + i)_
1
.~2'
and
—
_ cosh nz = cosh tez Res 2 +l)
*=-<"z(z
_ 1 .~2'
z(z-i)
z=-i
Consequendy, coshflzdz
r
=
,c z(z
6.
In each part of this problem,
(a)
It is
straightforward to
f„\ = lf/(z) This function
2
/,
„ = 2m
+l)
1
+—1 +—1 2
V
\
.
.
= 4m.
2,
C denotes the positively oriented circle
show
Oz + 2)
lzl= 3.
that
/n_ + 2z) ZllJ-zd-^ + Sz)2
2
(3
1
111611
z(z-l)(2z + 5)'
-y /^-j has a simple pole at
z
= 0, and
—
+ 2)2 dz = 2/ri Res *=° Icz(z-l)(2z + 5) (3z
(b)
,
= 9ni.
= 2/ri <2,
z/J
Likewise,
w
if/(z)= '
The function
z\l-3z) (l+z)(l + 2z
\
)
Jc(i + z )(l
)
*-3 2 -*-M-l z
U;
has a simple pole at z
Z(1 " 3z) f
±en
4
+ 2z
4 )
= 2;riRes *=°
z(z + l)(z
= 0, and we
V
1
4
+2)
find here that
= 27ri| -- = -3;ri. |
The desired
109 (c)
Finally,
zV"
The
point z
=
=
+ z3 )
.
The
residue
is
0'(O),
where
\z
z
^ =7f
1 Jl -jf -
a pole of order 2 of
is
±f
then
-
+ z3
1
'
Since (l
+z
3
z
)e
-e l 3z 2
\z)
the value of
0'(O)
f
'c 1
+ z3
So
is 1.
dz
3\Z
= 2m Res
7 fil)
=2jti
W =2m
-
SECTION 69 1.
(a)
Write
esc z
=
= -^4, sinz
where
/?(z)
= 1 and
= sin = 0,
and
tf'(0)
q(z)
= sin z.
q(z)
Since
p(0) z
-
= l*0,
^(0)
must be a simple pole of cscz, with residue P(0)
=- = 1
q'(0)
(b)
= cos0 = l*0,
From Exercise
2, Sec. 61,
CSCZ
we know that
^
= - + — Z+ z
1
1
.
3!
1
1
T 2
L(30
3 z +-
1
Since the coefficient of residue being
1.
-
here
is 1, it
(0
it).
5!.
follows that z
=
is
a simple pole of cscz, the
(a)
Write
z- sinhz = 2
p(z) 6
^)'
z sinhz
^) = *- sinh *
and
tf(z)
=z
2
sinhz.
Since
p(Ki)
it
= m*0,
and
(m) = j? * 0,
follows that
Rc „ *=*«'
(b)
q(m) = 0,
z-sinhz 2
=
z sinhz
p(ni) q'(,ni)
= m__i_ n1 k
Write exp(zr)
p(z)
=
6
sinhz It is
~q(zj'
^= )
exP(*')
?(z)
= sinhz.
easy to see that
«-*»
sinhz
r
q\7ti)
*=-*/
s jnhz
P
g'(-»0
Evidently, then,
Res
exp(
£
+ Res
expU0 = _ «Pff*) + CTpW*) 2
sinhz
(a>
sinhz
2
Write
/(z)
= ^7T'
where
P(z)
= z and
o(z)
= cosz.
Observe that
+ n«:j = «(f .2
(n
= 0,±l,±2,...).
Also, for the stated values of n,
p\- + n7c\ = - + n7t*
and
^| + «^ = -sin^| + n^ =
(-l)
fl+1
^0.
Ill
So
the function f(z)
=
has poles of order
cosz
m = 1 at each of the points r
z„=-+nn ~"
(n
2
The corresponding
(b)
= 0,±l,±2,...).
residues are
Write
tanh ^
Bothp and q
=
where
= sinhz and
p(z)
q(z)
= coshz.
are entire, and the zeros of q are (Sec. 34)
=
z
I
^- + nn nw
2
In addition to the fact that
J ~+
= 0,±l,±2,...)
= 0, we see that
jf
(
(n
\i
f
J
P
i[j
+ nn
)*)
= sinh
= icosnn = /(-!)" *
(^ +
and
= Sinh
+
(f '
*'((f
So
the points z
=
+
(«
= 0,±1,±2,...)
residue in each case being
p\
[
+ n?n =
— + nn
}i
I
)
* °-
are poles of order
m=1
of tanhz, the
Let
C be the positively oriented circle
shown just below.
lzl= 2,
TO
(a)
To
evaluate the integral
tanz
and
Jc
= £^£l
tan z dz ,
72
we write the integrand as
where
)
recall that the zeros of cos z
p(z)
are z
namely z = ±tt/2, are interior tanz interior to C. Observe that
zeros,
Res tanz = z=t/2
n(nll)
=-1
' ,
x
= sin z and
= -^ + n;r to C,
and
tf(tf/2)
q(z)
(n
= cosz,
= 0,±1,±2,...). Only two
and they are the isolated
p(-x/2) _ Res tanz _ „ qX-npi)
_
•
of those
singularities of
L
Hence tanz*fe
f
(fe)
The problem here
is to
= 27ri(-l-l) = -47ri.
evaluate the integral
.
Jc sinh2z
To do
this,
we
write the
integrand as
— — = £Q \
sinh2z
Now
sinh2z
=
t
where
p(z)
= l and
g(z)
= sinh2z.
q(z)
when 2z = nm (n = 0,±1,±2,...),
or
when
—
(/z
= 0,±l,±2,...).
2
Three of these zeros of sinh2z, namely singularities
show
and±-^-> are inside
of the integrand that need to be considered here.
that
Res *=o
_J_ = £W = _J_ = I2' sinh2z
q'(0)
2cosh0
C
and are the isolated
It is
straightforward to
113
p(m/2) _
1
Res
1
q'(m/2)
*=»/2sinh2z
1
1
2cosrc
2cosh(;ri)
2'
and
= PtE!2± =
1
Res z =-«v2
S inh2z
1
2cosh(-;rz)
q'(-jti/2)
1
=
2cos(-;r)
=
*
2'
Thus
Jc
5.
CN
The simple closed contour
Within
Q,
is
the function -=
z
z
To
find the residue at z 2, Sec. 61, and write
1
z sinz
U
s inh2z
as
shown
-to.
2
2
in the figure below.
has isolated singularities at
smz
=
and
z
= ±nn
(n
= l,2,...,AT).
= 0, we recall the Laurent series for
- + —z + = -t-cscz = -ri 2 2 1
1
z
z
1
-= + 3
z
1
1
6 z
+
1
1
3!
\z
1
3
z +-
z
(3!)
5!
1
z+-
2
(3!)
1
1
cscz that was found in Exercise
5!_
(0
114
This
tells
—
-=4 z smz
us that
1
Res «=°
As for the points
z
=
has a pole of order 3 at z
= ±nn
(n
= 1,2,
—
= ^tt, ~Y~ z smz q(z)
. .
.
,
z
i
and
that
1
smz
6
N), write
where
p(z)
= 1 and
q(z)
= z 2 sinz.
Since
p(±rut)
it
= 1*0,
q(±nn) -
q\±nn) = n 2 n 2 cosnit = (-l)"n z 7F z
and
0,
56
0,
follows that
1
Res
*=±»* z sinz
(-1)"
(-1)"
1
2
(-1)"«V
~~
(-1)"
nV
So, by the residue theorem,
cfe J,c"
Rewriting
this
=
2m 6
z sinz
equation in the form
A (-1)" il and recalling from Exercise tends to infinity,
n
+1
2
=
2
points
4i'-kz sinz
7, Sec. 41, that the value
of the integral here tends
to zero as
N
we arrive at the desired summation formula:
it
C
dz
f 2
12
f (-ir The path
g
tt
here
is
±2 and ±2 +
n
1
2
2 = n
12
the positively oriented boundary of the rectangle with vertices at the i.
The problem
is to
evaluate the integral
dz c (z 2 -l) 2
+3'
115
The
polynomial
isolated singularities of the integrand are the zeros of the
= (z 2 -l) 2 +3.
q(z)
Setting this polynomial equal to zero
the property z also the
2
= l±V3i.
It is
2
and solving for z
,
we find that any
zero z of q(z) has
straightforward to find the two square roots of
two square roots of 1-V3i. These are the four zeros of
q(z).
1+ V3i and
Only two of those
zeros,
z,.^-.^*L
d*+L,
and
he inside C. They are shown in the figure below.
To
and -Zq
find the residues at z
_
1 2
(z -l)
2
P(z) ,
+3
This polynomial q(z)
,
we write the
integrand of the integral to be evaluated as
where
= 1 and
p(z)
q(z)
= (z - 1) +3.
q(z)
is,
of course, the
same q(z)
p and q are analytic at Zq and that p(z 2 q\z) = 4z(z - 1) and hence that ?'(*<,)
We may conclude, then, that
z
)
* 0.
as above; hence q(z
Finally,
it is
)
= 0.
Note, too, that
straightforward to
show
that
= 4z (4 - 1) = -2V6 + 6V2/ * 0. is
a simple pole of the integrand, with residue
P(z
)
q'(z
)
_
1
-2V6+6V2i'
Similar results are to be found at the singular point -Iq.
To be
specific,
q X-Zo ) = -q '(z„ ) = -7(^) = 2V6 + 6V2i * 0, the residue of the integrand at -Zq being
q'i-Zo)
2V6 + 6V2/*
it is
easy to see that
(
Finally,
by the residue theorem,
^
We
Z
2
2
- 1) + 3
are given that /(z)
1
= 2ml
f
\-2sf6 + 6V2i
= l/[?(z)] 2 where ? ,
is
1
+
=
2V6 + 6V2i J
analytic at z*,
__ 2^2
= 0, and m = l at
?(z
These conditions on 4 tell us that q has a zero of order #(z) = (z - Zq)^(z), where £ is a function that is analytic and nonzero
)
?'(z
v
at z
;
+ 2g'(z
).
and
So /has a pole of order 2
=
-^-T
at z
,
= (z -z
q'(z)
Then, by setting z
=z
)g'(z)
+ g(z) and
in these last
q'(z
-M^.
we know that
)#(z),
= (z-z
—^.
and
Res/(z) = 0'(zo ) =
But, since q(z)
=
where
,
)
?"(z)
= (z-z
two equations, we find
= g(z
and
)
?"(z
)
,,
)g (z)
that
= 2s'(z
).
Consequently, our expression for the residue of /at z can be put in the desired form: Q
Res/(z) = - q "^\
(a)
To
find the residue of the function esc 2 z at z
csc z
~ r
<
sn2
»
.
= 0, we write
wnere
^(z)
= sinz.
W(z)J Since q
is entire,
#(0)
= 0, and
q'(0)
= 1 * 0,
the result in Exercise 7 tells us that
2 Rescsc z = - q
[
3
= 0.
Hence
this enables
us to write
f(z)
)*0.
—
(b)
The residue of the function
^-—j
(z
+z
=
at z
can be obtained by writing
)
1
-,
{z
Inasmuch asqis
+z
2
)
-
q(z)
= z + z2
[q(z)f
entire, q(0)=0,
Res
where
and q'(0)
=-j£<°>_ = - 2
1
(z+z
= 1*0, we know from Exercise 7
1 )
[«W
.
that
.
118
Chapter 7 SECTION 72
1.
To
—
evaluate the integral
we
-, J— x +1
integrate the function f(z)
closed contour
= Z
n
shown below, where
R>L
y c.
We see that
J
_s x
2
+l
J
^
2
z
+l
where
5 = Res —r—— = Res «'
+1
z
(z-f)(z +
Z + l'_U
Thus
f
f
Now if z is a point on Q, 2
lz
+ ll>llzl 2 -ll=rt 2
-l;
and so
E.
R 2 -l
J,
-»0
as
—dx
n =—
i_J_ R2
Finally, then
1 J _
dx 1 —7 = * + 2
1
1 tt>
or
:
Jjc
2
+1
2
2i
+1
around the simple
-
119 °°
2.
The
integral
dx
— — (x +1) 5
J J
can be evaluated using the function
rrj"
_((jc
—+ 1
2
+ l)
5-
and the same
1)
Here
1.
2+
= 27riB,
i\2 fc( z 2 + i) .2
,
1
B = Res
*"<~V + 1)
Since
.
2
Hz)
1
.
= 12—^ ^' + l) 2 TT (z-i) 2
.
where
2
,
x
0( z )
=
B = '(i) = -7-
that
,
— 1
U+
(z
we readily find
=
(z
simple closed contour as in Exercise
where
f(z)
and so
4i
fife
J „0t 2 ?
If
z
is
a point on
CR we know
+ l) 1)
(*
2
^(z J<
2
from Exercise
,
2
lz
2
+l)
2
1 that
+ ll>i? 2 -l;
thus
b The
desired result
<—inR
dz az
t
2
(z
is,
2 + l)
We begin the evaluation of the fourth roots of -1,
_%
dx
i(x
[
2
+i)
2
f*
~>0
R-*°°.
as
r
or
"2'
dx
%_
JcTTif'T
by finding the zeros of the polynomial z*
and noting that two of them are below the real
shown below, where R>1,
two roots
k
~ €jit/* _
1
1
+
"V2
V2
and J3X/4
+ 1, which are
+1
consider the simple closed contour the
,2
,
then,
°r
3.
J? 3
t=
iJT/4„iJt/2
1
.
/
V
1
i
axis.
In fact, if
we
that contour encloses only
r
120
Now
where
R - Res— — ^
and
= Res—
2?,
*=*2
The method of Theorem 2
in Sec.
69
tells
us that
z,
Z
+1
and z1 are simple poles of
— :
z that
= -7^=rr-4zj 4
311(1
= -T' =TT-4 4z z
fi
2
3
Zi
since
4 z,
= -1 and
zt
= -1.
2
Furthermore,
B +B2 =-j(z +z2 ) = l
l
2
~ 2V2"
Hence t
dx
}R x*
_ n
+ l~j2
dz
t
Jc,
z*
+ i'
Since
J,
c«z
4
+l
we have °°t
dx
n
iy + l"V2'
°r
°r
dx
L
4
n
+ l"2V2-
and
+1
121
— —x—dx— —— 2
r
We
wish
to evaluate the integral
=
=
2
.
We
use the simple closed contour
l(x +l)(x +4)
shown below, where R>2.
must find the residues of the function /(z) =
We
—
,
:
z
= i and
z
= 2i. They
at its
+ l)(z + 4)
(z
are
B = Res/(z) = l
(z
+ /)(z 2 + 4).
6/
and fi2
=Res/(z) = *=2<
(z (7
2 Z
3/'
+l)(z + 2z). ?=2i
Thus 2
*A
f
2
2
+ l)(x' K(x +l)(x ijtf
+ 4)
+fJJc. c
z dz
+
+ 4)
or
(x R
z
is
a point on 2
lz
CR
,
2 n r z dz = --f 2 J +l)(x +4) 3 c(z 3 + lXz 2 + 4)
x dx
J
If
—
2
* 2
then
+ ll^llzl 2 -ll = /? 2 -l
and
2
lz
2
+4l2>lld -4l
= /? 2 -4.
Consequendy, 2
z 2
(z
and
+ l)(z 2 +4)
2
-»0 as
2
(R -l)(R -4)
we may conclude that x dx i(jf"
+l)(^ + 4)
n 3
dx J
2
(jt
2
+ l)(* + 4)
it
6
simple poles
'
122
x dx— ———5 — —5- can be evaluated with the aid of the function {(x + 9)(x + 4) f
5.
The
5
integral f
2
2
2
f(z)
2
2
+9)(z +4)
(z
2
R > 3.
and the simple closed contour shown below, where
y
We start by writing 2
x dx
z dz
+ 9){x 2 + 4) 2
J {x 2 R
Jct Z 2 (
+ 9)(^ + 4) 2
where z
2
—
A = Res Ti—r~n «-* (z + 9)(z +4) z
z
and
z
5,
z
= Res—z
2
5
=
+9)(z
(z
z
=-.
+4r
Now 3
t 2
1
(z
To
find
+ 3i)(z + 4)
2
50/' J?=3<
B2 we write ,
2
__0(z)
z 2
(z
2
+9)(z + 4)
x
._,
-,
2
(z-2/)
where
,
,
0(z)
:
Z"
=
2
(z
+ 9)(z + 2/)
2
Then 2?2
This
tells
13
= 0'(2/) =
200/
us that A J
R
2
x dx 2
(x +9)(x
2
+4)
z
=JL_fJ
2
100
a 2
2
a\2' + 9)(z.2a +4) .
:
Finally, since 2
3
z dz 1,c
7r/?
2
2
*(z +9)(z +4)
we find
2
2
j -»
2
as
(R -9)(R -4)
R -> 00,
that
*
2
d!x
2 2 z f 0c +9)(;c (* +
*
7T
+ 4)
2
or
100
1
U
2
etc
+9)(a:
2
+4)
2
200'
.
.
7.
In order to
show that
P we introduce
V
xdx f
= -—
2
'^(x + l)(x 2 + 2x + 2)
5'
the function
2
2
+ l)(z +2z + 2)
(z
and the simple closed contour shown below.
y
/ x
*
>
X
R Observe that the singularities of /(z) are z
= -1 - i
lower half plane. Also,
in the
if
at
i,
z„=-l + i and
their conjugates
R > V2, we see that
K
jf(x)dx + j f(z)dz = 2m(B
+B
l
),
where 2?
*
=Res/(z)=
z
«-«o
(z
= _JL + _L-I
+l)(z-z
)
io
io
Z=Zo
and
= _L_I-
B =Resf(z) =
1'
l
(z
+ /)(z +2z + 2)
10
5
Evidently, then,
£<&
f
4U
2
2
+ 1)(jc + 2x + 2)
zrfz zdz
= _£_f 5
2
2
^(z + l)(z + 2z + 2)
Since
Z*fe
r
h as
2 z
(z (z
R -»
+ l)(z l)( z + 2z + 2)
°°, this
means
zdz
r
2 i
~
(z (z
z 2
+ l)(z-z
)(z-z^)
that ,.
hm
r
—
xdx ;
2
2
*-*~L(x +l)(x + 2x + 2) This
is
the desired result.
=
— n 5
2
(tf
-!)(/? -V2)
2
->0
124 8.
The problem here closed contour
There
is
establish the integration formula
is to
shown below, where R >
1
= 3
z
when R > dz
JclZ 3
where the
.
1
dz
r
3
Jc«
+l
According
.
z
=r
(0
<
<
r
using the simple
that is interior
,
dz
JC2Z 3
+1
+1
iicn namely z - e
to the residue theorem,
r
.
,
+1
z=Zo
3
Z
+l
legs of the closed contour are as indicated in the figure.
representation z
=
-
1.
only one singularity of the function f(z)
to the closed contour
—
J
Since
Q
has parametric
R),
^z = f dr kz + l~{ r3 + l' f
3
and, since
-C2
can be represented by z J
dz
f
JcJZ 3
dz-
r
J-c2Z 3
+i
= re ann
_
+1
(0
,K*n J2 e dr
*f
J(re
,2,c/3
3
R),
_
n ,/3
Jr 3 + 1'
+l
)
f
Furthermore,
5S.V + 1
3z
2-
~
3*
i2 *' 3
Consequently,
V
U
+l
3^
2ff/3
Jca2 >
+r
But 1
2^i?
R -l
3
<. 3
•h z + i
>
3
n as R->.
This gives us the desired result, with the variable of integration r instead of x
2
dr f
JV + 1 »
t
A
3(e
a *n
-e
m
2ni
l
-e
)
3(e
—e
)
:
n
2%
3sm(2^/3)
3\3
— 125
m
and n be integers, where Let formula
(a)
x
2n
m < n.
The problem here
-dx = —esc In \
[—5i
<
+1
+1
The zeros of the polynomial z
(-l)
1/(2B)
.(2k
2n
- -1.
Since
+ i)n (k
In clear that the zeros of z
it is
2n
+1
in the
derive the integration
In
occur when z
=exp
is to
= 0,1,2,.. .,2n-l),
upper half plane are
(*
and that there are none on the
(b)
With the aid of Theorem 2
real axis.
in Sec. 69,
2m
Res-^— = <-»V + l
^
.
Putting
a=
2m + l
we can
n,
= 0,l,2,...,n-l)
c
we find
2m
that
,
= J_ c *(— )+> \_ z ''2n-l 2wc, 2n .
(*
,
1
= 0,1,2... .,n-l).
write
In
2(m-n)+l
" eX
|. (2fc
+ l)7r(2m- 2w + i)l
T
In
J
[\(2* + l)(2m + l);rl '—
= expl
^-
i-i
,
exp[-/(2fc
.
N = + l);r] ,
Thus -2m
Res*=<*z
In
(2t+l)o 2
"
+l
view of the identity (see Exercise n-l
k=0
(*
In
10, Sec. 7)
n
i
1
~
= 0,1,2
/i-l).
126 then,
—m
,2m
n-l
f ^z
2n
+
n-l
l
n
m
e
n
(c)
l-e i2an l-e :2a
e
iitm+l)
*-l -e
e
Consider the path shown below, where
n n
e'
-e
-ia
,a
nsma
R>1.
y c.
The residue theorem
x
r
tells
us that
2m
z
r
2m
54
+
r 2m
^T' +1
or
}R x Observe
2n
+l
that if z is a point
2m lz
l
k
nsina
CR
on
= /?2m
z
2n
+l
then
,
2n
and
\z
+ \\>R 2n
-I.
Consequently,
1
W'+l dz
R2m
-2»
R 2a -l
2(n-m)-l
->0;
-2n
"
/?
1-
1 2«
and the desired integration formula follows.
10.
The problem here
is to
evaluate the integral
dx 2
2 Kx -a) + l] 2
Ji
where a
is
any
real
number.
- ia
n
2i
a
e
'
We do this by following the steps below.
m n
e
i2an
ia
'
e
-l
-e~
ia
127 (a)
Let us
first
find the four zeros of the polynomial
= (z 2 -a) 2 + l.
q(z)
Solving the equation q(z)
=
for z
2 ,
we
obtain z
2
=a±i. Thus two
a + i, and the other two are the square roots of two square roots of a + i are the numbers
the square roots of 5, Sec. 9,
the
Zq
where
A = 4c? + 1.
=-j=>(^A + a+rtA-a)
Since (±z
2
=z\
)
-z
and
=a + i = a-
the
i,
a-
of the zeros are
By
i.
Exercise
,
two square roots of a - i, are
evidently
The four zeros of q(z) tells
us that z
-z
and
z
.
just obtained are located in the plane in the figure below,
and -Zo
lie
above the real axis and
that the other
two zeros
lie
which
below
it.
y •
• Zo
X
(b)
•
•
-Zo
Zo
Let q(z) denote the polynomial in part (a) and define the function ;
f(z)
1
=
[q(z)f
which becomes the integrand developed specific,
q\z) 2'(z
we
be evaluated when z = x. The method To be is a pole of order 2 of /.
note that q
is
and
entire
recall
from part
(a) that q(Zo)
= 0.
Furthermore,
= 4z(z 2 - a) and z\=a + i, as pointed out above in part (a). Consequently, ) = 4z„ (Zq -a) = 4iz * 0. The exercise just mentioned, together with the relations
zl=a + i
and
1
+ a1 = A2
g"(Zo)
WW As
in the integral to
in Exercise 7, Sec. 69, reveals that z
for the point
-z
,
=
,
also enables us to write the residue
12z
2
- 4a _
2
3zp
2
3
(4iz
q'(-z)
16iz z
)
we observe
-a _
+ i)-a a — i 16i(a + i)zo' a-i 3(a
that
= -q\z)
and
of /at
q'\-z)
= q'Xz).
Zq'.
2
__
a — /(2a + 3)
16A
2
Zo
Since q(-z
)
=
and #'(-z
2 off. Moreover,
if
B2
_
)
= -q'do) = 4i'z
0, the point
-Zq
also a pole of order
is
denotes the residue there,
_
_
3
g»frb)
f
[^'(-^,)]
_ -5 =
1
3
3
[
.
x
l^'(^o)]
Thus
B.+B2 = B.-B.= 2r'Im5. = —Vim 1
l
1
-g + f(2fl 2 +3)
1
8A2/
We now R >\z
\
Zn
integrate /(z) around the simple closed path in the figure below,
and
CR
where
denotes the semicircular portion of the path. The residue theorem
us that R
jf(x)dx +
= 27ri(A + B2 ),
f(z)dz
or
dx
Im
I;
In order to
show
-a + i(2a 2 +3)
dz 2
•
[
that
dx lim
f
*^~ Jc«
2
= 0,
[?(z)]
we start with the observation that the polynomial
q(z) can be factored into the form
q (z) = (z - Zq )(z + Zo )(z ~ Zo )(z + z
lz±Zol>llzl-IZoll
= /?-lzol
and
\z±Zq\>\
)•
Izl-lz^l
I
= R-\z
\.
tells
a
129 This enables us to see that
\q(z)\
4
^
(/?-lz„l)
CR
when z is on
Thus
.
1
(R-\z for such points,
and
\)'
we arrive at the inequality nR 8
8
'
(tf-IZoO
which
tells
us that the value of this integral does, indeed, tend to
as
R
tends to
<».
Consequently,
-q + /(2a 2 + 3) n Im 4A 2 ^0
dx P.V.
2
j J
_ .[(*
But the integrand here
is
-a) 2 +l]
2
even, and
z
Im
-a + i(2a +3)
= Im V2
Zo
-a + i(2a 2 + 3)
-jA + a
- i-yjA-a
VA + a +/VA-a VA + a -i^A-a
So, the desired result is
k where
2
2
-a) + l]
2
=^[(2a
2
+ 3)VA^ + .VA^],
A = Vfl 2 + 1.
SECTION 74 r
1.
The problem here
is to
evaluate the integral
—
=
_*.(*+ this,
we
introduce the function f(z)
=
— (z
inside the simple closed contour
course, in the lower half plane.
+a
)(z
— X*+&)
cosx dx 2
—n~' whose +b
shown below, where
where a > b > 0. To do
singularities ai
and bi
lie
)
R > a. The
other singularities are, of
130 According to the residue theorem, a
e
jj? + a
2
dx
)(x
2
J/(z)e
ft
= 2»XA + ^).
£fe
+b 2 -+ / r.
where
= Res[/(zK<] =
B,
*=<«
v 2 +b 2 ) + ai)(z
-
.
(z
2a(*
.
:=oi
2
-a 2 )/
and z
52
=Res[/(z)e'
]
e-»
=
2b(a
=bi
2
-b 2 )i-
That is,
n 2 2 a -b
e"dx
Is
b
(e
_£1 - \Az)e*dz,
b
\
a
or
(x
Now,
if
z
is
2
+ a 2 )(x 2 + b 2 )
a point on
CR
iz
a
2
-b
Rejf(z)e k dz.
2
,
l/(z)l<
md\e
MR fi
where
M» = *
— (R
it
i
=
5-
=
-a 2 )(R 2 -b 2 )
nR (R
2
-a
2
)(R
2
-b 2 )
-»
as
R -» 00.
cosjtd!*
This problem
2 (jc
is to
+ a 2 )(jc 2 + fc 2 )
a
2
-& 2
evaluate the integral
—
=—
- has the singularities ±1; and so z +1 contour shown below, where R > 1.
/(z)
2
follows that
r
2.
=
\=e~ y <1. Hence
ReJc/(z)^&|<|jCj[ /(z)e'Vz <>MB nR = So
-a\
( -b
cos*dfr
y
(a>b>0). l
a
ft
f
COS — =
we may
fl.y
dx,
where a > 0.
The function
integrate around the simple closed
<
131
We start
with
)-1-—dx+ X +1
-R
jf(z)e
iai
dz
= 2mB,
C,
where UK = ^-: B = Res[f(z)e ] 1 J
z+i
*="
2i
Hence lOX
\—^dx = ne-
a
-
\f{z)e
iai
dz,
or
cosax
_
,, _ = ^ a -Re /(zy
,
r
.
,„ T
,
Since
L-
\f{z)\
we know
that
icR
Re
\f{z)e^dz
R 2 -V
and so cos ax
J x2
dx = ne~
+l
That is, r
cosax —5
,
J* 2 +l ox
4.
To
^
XS evaluate the integral
\
I
2
X dx,
x +3
2
z
where
z,
= V3i. The
=—
z
Zl
(a>0).
2
we first introduce the function
+3
point zt lies above the
fi&P^M. -
it
(z-zOiz-zJ x
where
axis,
and
zi lies
Hz) =
below
,
Z-Zt
it.
If
we write
.
132
we see that
zx
is
a simple pole of the function f(z)e i2z and that the corresponding residue
V3/exp(-2V3) _ exp(-2V3) B = (j>(z ) = l
1
2V3i
2
Now consider the simple closed contour shown in
the figure below,
where
R > VJ
y
Integrating f(z)e'
2i
around the closed contour, R
we have
ilx
jf—dx = 2mB
l
-if(z)e
i2i
dz.
Thus
j
Now, when
z
is
-^T^dx = Im(2^)-Im jcJ(z)e i2i dz.
l/(z)l<
and
so,
by
CR
a point on
M
R,
,
where
M =—
>
R
as
R -» °°;
limit (1), Sec. 74,
limf f(z)e
iu
R-*~JC,
dz
= 0.
Consequendy, since
Jmj J(z)e i2t dz\<\j nz)e iU dz c c we
arrive at the result
|^
+3
or
j™H.,fr = fexp(-2V3).
„*+3
J
2
is
133
The
integral to
be evaluated
is
* f^"* dx, where a > x +4
f
L /(z)
= ~Tj7^' and b y computing the fourth »
^=V2e te/4 =l + both
and
i
z2
below the
tell
we find that the singularities
shown below, where
in Sec.
-
f
x
3
e
tax
7+4 *
+
k f(z)eiKdz = 27n'^ + ^
where giaZ
5 - Res^g* 4 1
t-tt
z
z
+4
4z
a(1+,)
' '
i
-
g
=
3
4
4
= ggH ~ 4
and
o7 — Kes — z=z
—=
=
lZ * + 4
=
=
4
4z*
4
4
Since
27n'(A+£2 ) =
we are now able
.
e
+e
= i7te
flu; \
cos a,
to write
r
J
R > V2
.
The
69 for finding residues
us that R
define the function
other two
real axis.
The residue theorem and the method of Theorem 2 poles
-4,
We
= V2V 3 *' 4 = V2V*' V*' 2 = (1 + i)i = -1 + i
the simple closed contour
lie inside
singularities lie
roots of
0.
x smax
r
-^r^pfr = « "cosa-Im J .
/Uy*.
at
simple
— 134 Furthermore,
if z is
l/(z)l
and
this
means
CR
a point on
<
then
,
according to limit
ehZdz
(1), Sec. 74.
7 J
\
* \fCR
->
R -> <»;
as
2
^ox
me
dZ -»
as
/?
->
°°,
3
x sinax — —ax = ne x +4 ,
3
consider the simple closed contour
.
cos a
(a>0).
* Jf
(x
2
+ l)(* 2 +9)
Its singularities in
-
iaz
Finally, then,
In order to evaluate the integral
2
-
4
that
R
f(z)=
MR = —R—
where
^ L fW
8.
— R 3
MR
,
we
introduce here the function
the upper half plane are
R > 3.
shown below, where
Since
Res[/(z)e*]=
z
1
'f2 (z+0(r+9)
16e
and 3
Res[/(z)e
iz
«= 3 ' 1
the residue
theorem
tells
x
t
,(
3
] J
-z e*
= 71z (z
+ l)(z + 3i)
Z=3i
3 »
us that
e
a
dx
(^ + l)(^ + 9)
+
/'i
r
Jc/^
=
H-I?
16
or
r
I6e
j:
sin x dx
n
(9
\
t
16e
3
'
i
and
3i,
and
we
135
Now if z is a point on CR \f{z)\
So, in
view of limit
(1),
Im and
this
means "r
,
then
MR*
-
R
l(x
The Cauchy
R-> oo.
Sec. 74,
J
\c
(z)e*dz\
< |
J
\c
->0
(z)e*dz
as /?-> oo;
that
2
+\){x +9)
8e{e
2
i
x smxdx
)'
°
principal value of the integral
T
it
°f
J
(x
2
+ l)(x +9)~ 2
(9
I6e \e
nx
f
J
function /(z)
as
n
2
(R -l)(R 2 -9)
_jc^nxdx_ _ n_(9__\ 2
—
M* = —
where
f x +4* + 5
can
2
|,
e foun(j
2
^ih
^
^
1
=
r +4z + 5
and the simple closed contour shown below, where
Using the quadratic formula to solve the equation z 2 singularities at the points zl
=-2 +
i
and zx
+4z + 5 = 0, we
=-2-i. Thus
f(z)
=
interior to the closed
The residue theorem
contour and z
tells
x
is
below
the real axis.
us that
where
S = Res and so
smxdx 2me -Imf r = Im l J „ x + Ax + 5 L(zi-^i). r I
,Zl
—5
k
f(z)e dz,
R > V5.
find that
——
3
(z-zjiz-zj is
f
,
/ has
where
z^
136 or
%
sinxdx
r
Now, if z is a point on Q, l/(z)l
lz
then \e \=e~ y
where
M
„
.
,
f
< 1 and
=
R
Hence
Im
f
/(z)^|<|f
M nR = —
<
R
(/?-V5) and
To
as
R -» oo,
we may conclude that .
10.
-»
2
find the
Cauchy
7
V
J J
~—
sin;t<£t
2
7
=
x + 4x + 5 'a
#
„
.
sm2
-
e
principal value of the improper integral
^
the function /(z)
=
z+l r + 4z + 5
same simple closed contour as
—
z+1
,
x + 1 ^ cosx
f
J
.
x
2
+4x + 5
dx,
we
where z = -2 + /, and z = -2 - 1, and x
(z-zJiz-zO
x
in Exercise 9. In this case,
where
(z
5 = Res
+ l)e*
(Zt+l)g
_
(z-^Xz-zi)
fel
=
(-l
(z-zi)
+ Qg-2
''
2ez
Thus
or
r
Finally,
we
(x
observe that
\f(z)\
+ Dcosjc
if
where
z
is
.
tt,
a point on
Ms =
v
.
Cs
,
shall use
r
then
R+l
R+l
(R-izMR-m
(/?-V5)
-» 2
as
R -» oo.
the
137 Limit
(1),
Sec. 74, then
us that
tells
RcJCa /(z)«*
as /?
->
oo,
and so
_,, pv
12.
(aj
r -
Since the function /(z) integral
(* + ^ —
l)cos;t
n,
_
r^ = — (sin 2 -cos
:
= exp(i'z 2 )
,
is entire,
.
2).
the Cauchy-Goursat theorem tells us that
around the positively oriented boundary of the sector
has value zero.
A parametric
The closed path
is
its
0£r
shown below.
from the origin to the point segment from the origin to the point
representation of the horizontal line segment
R is z = x (0
for the
A
je
ixl
dx
+
lll
jc
e
dz-e
iK
rl
dr
'*\e-
=
or
)e
ixl
dx
rl
= e "'*]e- dr-\ i
e
c
By
e dz.
equating real parts and then imaginary parts on each side of this last equation,
see that
1*1
R 2
r jcos(x )dx = -j=je~ *dr-Rej
C*
e*dz
o
and R
|
2
fsin(jc )dbt
1
R
= -H
we
138 (b)
A parametric representation for the arc CR
is
it/* izl
e
jc
Since
.iR
z
cosZ0
1
dz= \e*
= 1 and
iB
'
\
ie
'Rie
it
d9 = iR\e-
(0
<
<
7r / 4).
Hence
RlA * 26 iRla" ie i6 e e de.
follows that
ic/4
,
Jc
e*dz
Then, by making the substitution (3), Sec. 74, of Jordan's inequality,
je-*
lsia26
= 2d in this we find that
2
(c)
= Re ie
a/4
a
= 1,
\e
z
dd.
last integral
2¥~4R
-»
as
and referring
to the
form
R -» oo.
In view of the result in part (b) and the integration formula
z
o
it
follows from the last two equations in part (a) that
2
]cos(x )dx =
^^
and
2
Jsin(*
)d!*
=
SECTION 77 1.
The main problem here
is to
-
derive the integration formula
cos(ax)
- cos(foc) 2
using the indented contour
shown below. y
,
dx = —(p - a)
(a>0,b>0),
139
Applying the Cauchy-Goursat theorem to the function
/(*)
=
we have f J
M
f(z)dz+[ f(z)dz+l f(z)dz+i f(z)dz = 0, JC JL *
«C.
1
or
Since
^
have parametric representations
and
i0 L :z = re = r(p
we can see
- L2 :z = re" = -r(p
and
l
that
^
^ /<«)& + j f(z)dz = jj(z)dz-l_J{z)dz = \
S
wr
&
ibr
~e
dr +
Li
e
— iar
~e
\
—ifer
dr
- (e* + e"*') (g^+g^ ) -5 J cos(ar) - cos(fcr) -5 dr. dr = „ r
f
J
,
,
2J
Thus
p
In order to find the limit of the first integral
fiz)
=
\ z
^
this
2!
+ •••
we see that
lim
z
f
p—»0 JCp
(/az) (
=
|
3!
right here as
N
3
(iaz) |
1!
/(a-fc)
From
2
iaz |
on the
(1H I
p -» 0, we write
ibz(ibz)\(ibz)\ ••• H
1
1!
2!
h
3!
(0
is
f(z)dz
a simple pole of f{z), with restdue~Ik = i(a - b). Thus
= -B
7ti
= -i(a - b)m = %{a - b).
140
As
CR
for the limit of the value of the second integral as
R -» «>, we note that if z is a point on
then
,
rt
^
\e"*\+\e
M
e-v
\
+ e -* ^1 + 1
2
Consequently,
<—t kR = R2
now
It is
p -»
clear that letting
—R -*0
as
and /?-><» yields
^7 cos(ar)-cos(frr )L J dr= n(b-a) j
2J
o
This
is
the desired integration formula, with the variable of integration r instead of x.
Observe
that
when a -
and b =
2, that result
7 1 - cos(2;t)
3
J
But
cos(2jc)
A
=
n-
= 1 - 2 sin 2 x, and we arrive at
2
*
I
2.
,
dx
becomes
Let us derive the integration formula
r
x fa
|(7TIF' where x"
= exp(aln x) when * > 0. _
whose branch cut is the path shown below.
=
(l-a)TT
,
(
4cos(a,/2 )
- 1
'
We shall integrate the function
z"
exp(alogz)
.
_
3;^
origin and the negative imaginary axis, around the simple closed
p Branch cut
By Cauchy's residue
f JL
f(z)dz+i f{z)dz+[ f(z)dz+i f(z)dz = 2mResf(z). JL JC
f
f(z)dz+
i
That
theorem,
z
i=i
p
is,
f
f(z)dz
= 2mResf(z)~
f(z)dz-
f
f(z)dz.
f
Since
L
l
:z
= re i0 =r(p
-I2 :z = re = -r(p
the left-hand side of this last equation can *
be written R
a(lnr+i"0)
«
o(lnr+/)r)
*
a
= J1 71— 2 (r
-4-
2
dr + «"*
1)
a
\ ,i 2 J(r +1)
= (1 +
*
a
f
/+
'J(r 2
. 2
dr.
1)
Also,
Res/(z) =
where
(z)
= (z
the point z
=
i
+
2
'
being a pole of order 2 of the function f(z). Straightforward differentiation
reveals that
'(-\
0'(z)
a(z
_ = e-(a-l)logz L
+ i)-2z
(z
+
3
142
and from
this it
follows that
-iaKll Res f(z) = -ie
fl-a
We now have
Once we show
that
limf f(z)dz =
and
lira f JC.
p-*0 iC„
f(z)dz
= 0,
we arrive at the desired result:
„a _r J
o(^+l)
The
first
2
^_
_/i
JaKll
_\ — a)
_e
e
+ e m* iaic
2
l
of the above limits
is
e
-iaKll
^(1 - a) \
—iaKll
shown by
(l-a)ff
4
e
ia*'2
+e
-ia*l2
AcQS ( a7t / 2 )
writing a+l
f(z)dz
I and noting
that the last
(l-p 2 )
term tends to
as
np = 2r
p -»
Tip
(1-p
2 )
a+1>
since
_
1
\jc
3.
last
mdz
2
(tf
-l)
2
""
2
(R -l)
as /?-»<» since 3
term here tends to
The problem here
is to
1
2
1
|1-
- a > 0.
2
n jr
i,
2
by
for the second limit,
derive the integration formulas
-yxln* "(Mxlnx f l
As
Ra R4
and the
0.
x +l
_ .
. ,
2
6
7f
VI Vjc
J* 2 + l
,
7t
V3
integrating the function
H
,
f(r )
z
m logz e^logz _ 2
z
+l ~
z
2
+l
lzl>0,
—
—
143 around the contour shown in Exercise
f(z)dz +
f
f(z)dz
f
2.
As was
= 2mResf(z) -
the case in that exercise,
f(z)dz -
f
f(z)dz.
f
Since
/W-M z-i =
the point z
is
i
=
where
z+i
a simple pole of f(z), with residue
Res/(z) = 0(O = ye'*/6
.
The parametric representations
L :z = re i0 =r(p
-L :z = re = -r(p
1
can be used to write
and
J.
=
f
^.J^l^fr.
Thus
Vrlnr
^ „ J^TT^ + f
r2 I p
By equating
Ifrlnr+in Vr ^
gl /3
(
1
real parts
m^dr +
1 ; p J
r
>
+i
dr=
2
;r
T
e
/^z-jc mdz.
-i c
on each side of this equation, we have
cos(;r / 3)
l^fdr - nsin{7t
1
3)\-^-dr =
-—
sin(;r / 6)
-Ref /(zVfe-Ref /(*)&; and equating imaginary parts yields * 3/~~ i
sin(;r / 3)
f-^-f J
r+1
* flfr
3/~"
= + *rcos(w / 3) \——dr 2 J
r
+l
2
—
cos(;r / 6)
2
•Imf /(z)
Now
sin(^/3)
= ^, cos(^/3) = ^-, z
show
sin(7r/6)
Z
= ^-, cos(^/6) = 2
p->0 Jc, '
and
lim f /(z)
— 2
that
limf f(z)dz =
/(z)
= 0.
and
it is
routine to
144 Thus
3f V7lnr
2
r
*
2
W3
r
J
r
+l
2
W 2
2 ;r
.
+l
4
That is,
_
3
W3 2
l
2
2
1
2
2
Solving these simultaneous equations for
n2
_ ~
T*
2
4
/j
and 72
we
,
arrive at the desired integration
formulas.
4.
Let us use the function v
=
/(z)
and the contour
in Exercise
f
2
to 2
3
j
{x +l
f
n I* n lzl>0,--
f,
—
-5
J
8
Q
2
,
dtc
'
2
=n 0.
x +l
around the closed path shown in Exercise
f(z)dz + j
i
V
} Inx
and
2
Integrating /(z)
(logz)
show that
n ——-<& = — (lnx)
2
-2— Z +1
2,
we have
nz)dz = 2mResAz)-[ f(z)dz-\
f(z)dz.
Since
/W-M Z-i the point z
=/
is
where
z+i
a simple pole of /(z) and the residue
Res/(z) = 0(0 = *=•'
^
=
(lnl
2/
is
+ t-;r/2 ) 2 2/
= _*1
'
8i
Also, the parametric representations
L :z = re i0 =r{p
and
-L :z = re 2
iic
= -r(p
2
enable us to write
J,/W*"f$£*
y^,]^f
and
P
P
Since
p
p
p
then,
p
p
p
Equating real parts on each side of this equation,
p
we have
p
and equating imaginary parts yields r
,
27c\^-dr = lmi f(z)dz-lm[
f(z)dz.
p It is
straightforward to
show
lim
f
that
f(z)dz
=
and
lim
Hence
and
Finally,
inasmuch as (see Exercise
1,
Sec. 72),
r
dr 2
Jr + 1
we arrive at the desired integration
formulas.
_ = % 2'
f
f(z)dz =
0.
dr
.
146 5.
Here we evaluate the integral (x
We
a>b>0.
dx, where J
+ a)(x + b)
consider the
function
expl
.1/3
(z
+ a)(z + b)
(z
^logz
+ a)(z + fc)
and the simple closed contour shown below, which is similar to the one used in Sec. 77. The numbers p and R are small and large enough, respectively, so that the points z = -a and
= -b
z
are between the circles.
A parametric (p
<
r
<
R),
representation for the upper edge of the branch cut from
and so the value of the integral of /along that edge
r f
o
exp
(ln/-
_LL :
(r + a)(r
+ iO)
^
r
+ b)
=
A representation for the lower edge from value of the integral of / along that edge from
£
f
{( r
p
R
+ a)(r + b)
to is
to
p
R
is
z
L3
\ dr
= -e i2 *n
to the residue theorem, then,
R
is
z
= re i0
^
= re n * (p
is
\
i(r + a)(r + b)
According
to
is
*exp -(lnr + /2;r) f
p
dr.
R).
Hence
the
147
where
exp|j(lna +
exp|jlog(-a)j
3=Res/(z)
-a+b
gM3
/;r)J
a-b
^
a-b
and
1
exp |log(-t)
B2 =Resf(z) =
exp
(\nb + in)
n
Mb — a b
i" -
-b + a
Z=-b
-b + a
Consequently,
2«-^) . J/(^, J/( ^=
^
•
V7
(\-e™)\
--
dr
^
Now
L
f(z)dz
(a-p)(b-p)
(a-p)(b-p)
and
/(z)& <
V* (R-a)(R-b)
D= 2nR
—
2*/?
2
— •-
1
n >0
(R-a)(R-b) MR 2
as
„
Hence
2me * /3 e/a~-Mb) - 2" -
,
r
o
(r + fl)(r
+ ft)
3
(1
e''
^
n(Ma-Mb) sin(;r/3)(a-fc)
*)
(
V*
'
'*/3
/3
(e'*
_ n{Ma~-Mb) = 7~~ ^/3~
Replacing the variable of integration r here by x,
{(x + a)(x + b)
2^
e™
i
f
fc^n V3
-y(«-*)
-
-Mb) - *)
2ar
V^-Vfr
V3~"
a-fc
'
we have the desired result: Ma'-Mb
a-b
(a>b>0).
148 6.
(a)
Let us
first
use the branch
exp^--logz
-i/2
z z'
2
+l
z
and the indented path shown below
lzl>0,--^
+l
to evaluate the
improper integral
P Branch cut
Cauchy's residue theorem
f
M
tells
f(z)dz+i f(z)dz + -' (
-*
us that
f '^2
f(z)dz+l f(z)dz = 2mResf(z), •'Cp Z=i
or
f
/(z)
+J
f(z)dz
= 2mResf(z)-
f
/(z)rfz-
f
/(z)
Since
I^-.z^re'
=r(p
and
-l^.z-re^ = -r(p
we may write dr 1)
Thus
149
Now the point
Res f{z) = J=<
z
=i
is
evidently a simple pole of f(z), with residue
m
z
exp -I^inl + f.5
exp[-|logij
—. + z~
2i
i
2i
2iVV2
2i
Furthermore,
no
Jp
= 72^ n -TvTT 1-p Vp(i-p .
L
f(z)dz
it
2
2
as
p->0
)
and
\\c
Finally, then,
k4r
m
we
is
the
->
as /?->oo.
have (1
which
n
same
'
VKr 2 + l)
V2
as
r
_
dbc
it
J-vW+1) ~V2' To
evaluate the improper integral
J[ Q
-1/2
-j=
—
^x(x 2 + l)
,
we now
use the branch
expf-^-logz z
z
(kl>0,0
+l
and the simple closed contour shown in the figure below, which Sec. 77.
C„ and
We stipulate that p < 1 CR .
and
R > 1,
is
similar to Fig.
so that the singularities z
= ±i
99
in
are between
—
1
Since a parametric representation for the upper edge of the branch cut from
z
= re' (p£r<
R), the value of the integral of /along that edge
— 1
«exp „_.. r
(lnr + i'O)
A representation for the lower edge from
p
to is
value of the integral of/ along that edge from
-f
R
2
2
R
to
z
is
p
-U = -e-*
2
R
= re' 1 " (p
is
dr
=
f
_i
dr.
1
where
exp -i-flnl-H-*
exp --log/
-1/2
2l
B =Res/(z) = [
z
+i
2/
-;«/4
2
2/
2i
and
^ .-1/2
52 = Res/(z) = *=-'
That
exp
z-
4
exp
iogH)
—
.3* If, , lnl + i
2V
1
-2i
2i
-2i
J
„Vr(r +1)
J
J
Since
VP(i-p
2 )
2k Jp l-p 2
and 2;cR
2;r
-»0
-i3*r/4
2
is,
In p
is
is
Hence, by the residue theorem, R
R
+ l)
~(\nr + iliz) L
to
„
V^(r
«exp
p
as
oo,
so the
151
we now find that i
,
J J V^(r + l) 2
dr
—
—e
e
=n
it
=n
When
x, instead of r, is
+e
e
e
= Kco&
urvr
used as the variable of integration here,
we have
the desired
result:
_
dx
7
it
SECTION 78 1.
Write 2
dd _ r ]5 + 4sin0~-k r
C is the positively
where
J z -z~ \
5+4
= -il 2
? J o
2.
5
is
1.
The
quadratic formula tells us that the
far right here are z
a simple pole interior to C; and the point z
= -i/2 and z - -2i. The point
= -2i
is
w-u
M artier'— + 4sin0 *=--42z +5iz-2j
To evaluate
2
L4z + 5i_L_ 1/2
exterior to C.
Thus
-wfi)-^. 3 U»7
the definite integral in question, write
f
dd
Jl + sin
_ 2
r
dz
1
0~Jc 1
+
(7
-7 2i
where
~^2z + 5iz-2'
iz
oriented unit circle \z\=
on the
dz
r
2
l
singular points of the integrand
z
dz_
1
C is the positively oriented unit circle
-i\ 2
'
z
i
_
t
4izdz
~Jc z *_6 z 2 + r
) \z\=
1.
This circle
is
shown below.
r
.
152 Solving the equation (z
2
2
)
- 6(z 2 ) + 1 =
find that the zeros of the pol ynomial z
Those zeros
are, then,
4
for z
2
with the aid of the quadratic formula,
2
- 6z + 1
numbers z such
are the
that z
2
= 3 ± 2V2
= ±^3 + 2-^2 and z = ±V3-2>/2. The first two of these zeros are and the second two are inside of it. So the singularities of the
z
exterior to the circle,
integrand in our contour integral are
z
;
=V3-2V2
means
indicated in the figure. This
and
z1
=-zv
that
where
By
= Res—4 *-* z
_
4iz 2
-6z + l
4izt
/
4^-12^
zf-3
(3
-2V2) -3
2-V2
and
_
4/z
B2 = Res
4
-4/z,
2
3
-4z + 12z
-6z + l
*=-*.z
1
1
zf-3
2V2"
Since
2k V2
\
i
2^(A-H52 ) = 2^|--^j = ±|.-l| = V2^ the desired result is
J l
7.
Let
+ sin 2
C be the positively oriented unit circle
lzl=
In view of the binomial formula (Sec. 3)
1.
In
Jsin
2"
BdO =
IJ
2
2
we
sin
2"
ddd = ± 1
(
z-z
J 2M
i^/c|(T)
2 " +i
2 " +1
(-l)"i
ft
^ *
2i
z2
-l\2«
dz iz
" -z (
"
1)iz
2
2 " +1
"^
(-1)"/ J c (-l)
dz
153
Now each of these last integrals has value zero except when 1
jc
z dz
k = n:
= 2m.
Consequently,
(2w) !(-!)"
1
jsm 2n ddd = 2
"
2»+i
2m =
(2n)!
2
(n! )2
2
2fl
7T.
(«!)
SECTION 80 5.
We are given a function / that is contour C, and we assume inside C,
where each zk
The object here
To do
this,
is to
is
show
we consider the
that
analytic inside
of multiplicity
zf'iz)
/(z)
is
mk z 1
z-zk zg'(z)
Since the term
— —
.
= l,2,...,n)
(See the figure below.)
kth zero and start with the fact that
= (z-zk ) m >g(z),
analytic and nonzero at zk
_ —
mt
(&
that
f(z)
where g(z)
and on a positively oriented simple closed
/has no zeros on C. Also, / has » zeros zt
zg\z) — _ g(z)
m
k
.
From
this, it is
(z-zk ) + mk zk
zg\z) _
_ -WlH
1
z-zt
straightforward to
,
zg\z)
g(z)
g(z)
here has a Taylor series representation at zk ,
g(z) has a simple pole at zk and that
/(«)
An application of the residue theorem now yields
the desired result.
, 1
show
that
m zk k
.
z-zk zf'iz)
it
follows that
Hz)
.
154 6.
(a)
To detennine lzl=
1,
number of zeros of the polynomial z 6 - 5z 4 + z 3 - 2z
the
inside the circle
we write /(z)
= -5z 4
and
= z 6 +z 3 -2z.
g(z)
We then observe that when z is on the circle, l/(z)l
=5
6
and
ls(z)l
+ lzl 3 +2lzl = 4.
Since \f(z)\>\g(z)\ on the circle and since /(z) has 4 zeros, counting multiplicities, inside
it,
the theorem in Sec.
80
tells is that
the
sum
6
Az) + g(z) = z -5z*+z 3 -2z also has four zeros, counting multiplicities, inside the circle.
(b)
Let us write the polynomial 2z
=9
/(z)
Observe
that
and
when z is on
l/(z)l
=9
4
- 2z 3 + 2z 2 - 2z + 9
g( z )
sum
/(z) + g(z), where
= 2z 4 -2z 3 + 2z 2 -2z.
the circle lzl=
1,
4
and
as the
l#(z)l<2lzl +2lzl
3
+ 2ld 2 +2ld = 8.
Since l/(z)l>l#(z)l on the circle and since /(z) has no zeros inside /(z) + g(z)
7.
Let
4
2
3
= 2z - 2z + 2z - 2z + 9
C denote the circle
I
zl
sum
=2
The polynomial z + 3z 3 + 6 can be /(z)
On
the
has no zeros there either.
4
(a)
it,
= 3z 3
written as the
and
$(z)
sum of the polynomials
= z 4 +6.
C,
l/(z)l
= 3lzl 3 = 24
and
l^(z)l
= lz 4 +6l^izl 4 + 6 = 22.
Since l/(z)l>l#(z)l on C and /(z) has 3 zeros, counting multiplicities, inside C, follows that the original polynomial has 3 zeros, counting multiplicities, inside C. 4
(b)
The polynomial z - 2z 3 + 9z 2 + z - 1 can be /(z)
On
= 9z 2
and
g(z)
written as the
sum of the polynomials
= z 4 -2z 3 +z-l.
C,
|/( z )|
= 9lzl 2 = 36 and
U(z)l=lz
4
3 -2z + z-ll
it
155
C
and /(z) has 2 zeros, counting multiplicities, inside C, the original polynomial has 2 zeros, counting multiplicities, inside C.
Since l/(z)l >\g(z)\ on follows that
2
3
5 The polynomial z + 3z + z + 1 can be
(c)
/(z)
On
= z5
and
$(z)
written as the
sum of the polynomials
= 3z 3 +z 2 + l.
C, 5
l/(z)l=lzl
= 32
and
= l3z 3 +z 2 +ll<3lzl 3 +lzl 2 +l = 29.
I#(z)l
Since \f(z)\>\g(z)\ on C and /(z) has 5 zeros, counting multiplicities, inside C, follows that the original polynomial has 5 zeros, counting multiplicities, inside C.
10.
it
The problem here
is to
give an alternative proof of the fact that any polynomial
P(z)
where n >
1,
= a + a,z +
+ a^z"" + aH z" 1
• • •
(a.
* 0),
has precisely n zeros, counting multiplicities. Without loss of generality,
may take an =l
it
we
since
r
P(z)
= a„
Ea.
+ BL z + ... + EB=L^ +z ^
Let
/(z)
Then
let
= zB
and
g(z)
= a +o z + - + an _ z"' 1
l
z
is
.
R be so large that /?>l + la + lo + ---+laB _
If
1
l
a point on the circle C: lzl= R,
1
l.
l
1
we find that -1
lg(z)l fS la
l
+ l^llzl + ••• + la^llzl" = la + iaJR + ••• + la^ltf-
1
l
1
< Iflol/?"- + Ifl!!/?" 1
-1
+ ••• + t^.J/?"
-1
= (la + laj + - + l
= l/(z)l.
Since /(z) has precisely n zeros, counting multiplicities, inside arbitrarily large, the desired result follows.
C and since R
can be made
-
156
SECTION 82 1.
The
singularities of the function
3
2s
5-4 are the fourth roots of 4.
They
are readily found to be
or
V2, See the figure below, where
-V2,
and
y > -4l and R > V2 +
-V2*.
y.
Y+iR
The
function
—-4
2?V'
=—
e
s
has simple poles at the points
j
= V2,
jt
= V2/,
j,
= - V2,
and
*3
and
V
X Res^FW] = 2 Res ^il 2
3
=
f
2
2
•
+
= cosh V2f + cos V2f.
o„3*.r
2j£
2
=
157
Suppose
now that s
\s\=\y+Re'°\
It
is
CR
a point on
,
and observe
that
ie \s\=\y+Re \>\y-R\ = R-y>«j2.
and
follows that
\2s
3 \
= 2\s\ 3 <2(R+yf
and s
I
4
- 41 > lid 4 -4I> (/? - y) 4 - 4 > 0.
Consequendy,
lF(*)l<
(J?+ y)
^ (/?-y)
.
4
'
->0as/?->
-4
This ensures that
/(/)= cosh V2r + cos V2f.
2.
The polynomials
in the denominator of
2s
F(s) (s
have zeros at s
= -1 and
s
= -1 ± 2i.
e"F(s)
t
= -1 + 2/'. The
-2 + 2s + 5)
Let us, then, write
e*(2j-2)
= (j
where j
+ l)(s
2
+ IXj-^X^-J!)
points -1, su and s{ are evidently simple poles of e"F(s) with the
following residues:
i_ ^=Resfe"F(j)] L J
g
Jt
z=-l
52 = Res[*"F(,)l =
B =Res[e«F(s)]= WJ L 3
~,
^'^^ (J,
+1X51-5.)
(2^-2) -Si)
" g <(2
^- 2)
Sl e '(2 Sl
= fi - 1
-2)
(5,+lX^-J!)
W<,
158 It is
easy to see that
i2t
— —e -t + e -t ,
f e i2 '-e- i2
Now let s
+
n
'
+ e- i2
''
=
2
2/
v
e
'
be any point on the semicircle shown below, where y >
and R >
-\/5
+ y.
Since
\s\=\y +
we find
Re
ie
\
and
i6
\s\=\y+ Re \>\y- R\=
R- y> V5,
that
I2j-2I<2IjI+2<2(/?+7) + 2,
\s
+ l\>\\s\-\\>(R-y)-l>0,
and 2
\s
+2s + 51=1 s - Sl \\s - s,\ > (\s\-\sj) 2 > \{R - 7) 2 - V5
f > 0.
Thus
I
and
m
^=1
=
<
2
Is
+ II \s + 2j + 51
we may conclude
+2 2 [(* - y) - !][(/? - 7) - V5]
.
Y)
that
/(/)
= e"'(sin It + cos It - 1).
as
R -> oo,
159 4.
The function „2
s
—a 2 (a>0)
(r+a 2 ) 2 has singularities at s
= ±ai. So we consider the
simple closed contour shown below, where
Y>0 and R>a+y.
Upon
writing
F(s)
yv
=
.
(s
we
see that
F(j).
is
- a/) 2
where
0(s)
analytic and nonzero at s
Furthermore, F(s)
=
a pole of order 2 of F(j); and
at points
=-
= ai. Hence
where F(s)
we know from
where
is analytic.
expression
Res [e"F(s)] + Res [e^Fis)] = 2Re[e
sQ is a pole of order
(2), Sec.
ia,
(fc
1
+ b2 t)]
m=2
Consequently, s
is
of
also
82, that
,
and b2 are the coefficients in the principal part
- ai
s
of F(s) at
ai.
(s
- ai) 2
These coefficients are readily found with the aid of the
Taylor series for
(s)
about s
=
ai:
1 (s
- ai)
(s)
1
= (s
- ai)
first
two terms
in the
160
=
It is
(ai)
^)
straightforward to
0'(ai)
r + 7^- + -
show
,
that
(ai)
(0
= 1 / 2 and
\ai)
= 0, and we
find that
2a).
^=
and
b2 =l/2. Hence
Res [e*F(sj\ + Res [e"F(s)] = 2Re
^(j'j = tcosat.
We can, then, conclude that f{t)
= tcosat
( fl
provided that F(s) satisfies the desired boundedness condition. z
is
a point on
CR
this
2
\z
means
for that condition,
when
,
\z\=\y+Re and
As
> 0),
ie
\£y+R = R+y
and
ie
\z\=\y+Re \>\y-R\=
R- y > a;
that
-a 2 \<\z\ 2 +a 2 <(R+y) 2 + a 2
and
2
\z
+a 2 \>\\ z 2 -a 2 \>(R-y) 2 -a 2 \
>0.
Hence
6.
We are given „.
which has isolated
singularities at the points
-n = °>
_ (2n-l)^
„ *.=
2
This function has the property F(s)
f(t)
To
sinh(xs)
.
_
.
311(1
,f
=
J"
= ~ (2n-l)n.
,
(n
*
„ = l,2,...). ,
2
F(s), and so
= Res [e*F(sj\ + J^Res [«*F(j)] + Res [e"F(j)]j.
find the residue at j
= 0, we write
3 _ xy + (xy) / 3 + 2 2 2 j cosh* 5 (l + * /2! + ...)"
sinh(xy)
!
• • •
+ jcV/6 + — s + s /2 + -
oc
0
3
{
2
.
161 Division of series then reveals that s
is
a simple pole of F(s), with residue x; and,
according to expression (3), Sec. 82,
= Res F(s) = x. Res le^Fis)] L 1 s=s
s=s„
As
= 1,2, ...), we write
for the residues of F(s) at the singular points a; (n
=
F(s )
= s 2 cosh j
= sinh(jcs) and
where
-
.
We note that .
.
.
—^—
(2n-l);a
.
= isin^
_
5*0
,
and
„ = 0;
v
q(sn )
furthermore, since q'(s)
we find
„
that
?W .
=
(2n-l)V.
.
ism
(2w-1)tf
.(2n-l)
= -,i
-r
4
2
^V.smnttcos— jt
'='.
1
7r
Sec. 82,
(4),
J
art
,
n\
( sm\nn-.
now gives
—
2 2 n = v(2n-l) ^ (-l)"/*0.
2) is
'
,
4
a simple pole of F(s), and
(2n-l)
?'(*.)
Res
2
2
-costt7rsin—
2
V
view of Theorem 2 in Sec. 69, then, sn
Expression
-J.
J
.(2n-l)
In
= 2 j cosh j + j 2 sinh a,
J
2
us
He. L[e"F(s)]J - 2 Re(-± 2 '='"»
.
[;r
8
—
-^ (2n-l)
(-1)"
2
i^^expTifi^l] 2
—
(2n-l)*ct
= —?•— ^—rsinn 1 (2«-l) 2 .
sin
-
2
2
L
—
(2n-l)nt cos-
2
JJ
,
162
Summing all of the above
-x
f(t)
7.
residues,
+— —
we
arrive at the final result:
-j sm
2_,
cos-
—
The function F(s)
=
1 -
JCOSh(5
where
it is
agreed that the branch cut of s
=
isolated singularities at s
in
= 1,2,
The point
...).
1/2
.ycosh(.y
and dividing expression
)
does not
and when cosh(j
l/2
)
a simple pole of F(s), as
is
this last
denominator into
1.
is
s
along the negative real axis, has
or at the points s„
seen by writing
+ s 2 /2 + s 3 I2A+-
Sec. 82, tells us that
for the other singularities,
F(s)
=
^
l.
we write
where
p(s)
= 1 and
q(s)
= scosh(s m ).
Now p(sn )
= 1*0 and
q(sn )
= 0;
also, since
q\s) = ±s
it is
straightforward to
x
4
(*„)
show
m sinh( 5
1/2
+ cosh(* 1/2 ),
)
that
(2n-l);r n\ (2n-V)7t = -^-^-sinl(nit - = ±J±-^L(-Vf±^ .
I
= _( 2n Jj n
In fact, the residue is found to be
Resfe" F(s)] = Res F(s) =
As
lie
= 0,
4l + (j 1/2 ) 2 /2! + (/ /2 ) 4 /4!+-]
)
(3),
s
m
'
1/2
1;
and
1
,
163
So each point sn
is
a simple pole of F(s), and
pw = i.i=«l
ResF(J)=
#
Consequently, according to expression
= Res \e"F(s)] J L
e'"'
'=',
K 2n-l
CO
(3), Sec. 82,
Res F(s) = - £-^-exp ;r 2n -
(2n-l)Vf (n
•
= l,2,...).
Finally, then,
/(/)
= Res [e
st
F(s)]
£ Res
+
[**F(j)1 ,
or
(2n-l)Vf nZi 2n-l
Here we are given the function
.
.
_
coth(^y / 2) s
2
cosh(^y/2)
_
~(/ + l)sinh(^s/2)'
+l
which has the property F(s) = F(J).
We
consider
first
the singularities at s
= ±i. Upon
writing
F(j)\
u where
=
\
jl /•
s-i
we
find that, since 0(i)
3(b), Sec. 65];
the (/i
residue
and the same of e"F(s)
= 0,±l,±2,...),
F(s ) =
(s
= 0,
£yl q(s)
the point is true
is
or at the points s
where
p(s)
+ i)sinh(ns f 2)
a removable singularity of F(s) [see Exercise
i is
of the point -i.
The other
0.
—
coshers/ 2)-
=
= 2ni
= coshf
—
singularities
= 0,±1,±2,...). To
(n
\2
At each of these
and q(s) = J
J
2
(s
points,
it
occur when
follows that
nsl2 = nni
find the residues,
+ l)sinh[
—
\2
we write
and note
that
p(2ni)
= cosh(nm) = cos(nn) = (-1)" *
and
q{2ni)
= 0.
Furthermore, since
q\s) =
2
(s
+
1)|
cosh^H + 2,sinh(-0 j
we have q\2ni)
= {-An 2 + l)-^cosh(n^i) = (-4n 2 + l)-cos(n^) = ^
/r(4w
2
- 1)
(-1)"*0.
2
Thus *= 2 »<-
Expressions (3) and (4) in Sec. 82
now tell us
Res[e"F(5)l
(«
n An 2 -I
q\2ni)
= 0,±1,±2,...).
that
= ResF( y) = 1
and
A_
Res[e"F(j)]+ Res \e"F(sj\ = 2 Re e i2 "'( 2
is,
(«
n An 2 -\
n An -I The desired function of t
cos 2nt
= 1,2,...).
then,
4^cos2/if
2
*r*\
The function F(V) K '
where
it is
= 1,2,
.
.
.)•
sinh(x? s
2
sinhjxs
The point 1/2
)
m smh(s
)
_
s
J sinh(^
s
=
=
m
is
)
(0<*<1),
1/2
)
does not
and when
sinh(j
l/2
)
lie
along the negative real axis, has
= 0,
a pole of order 2 of F(s), as
m + (xs U2 1 3! + (xs m [s + {s m f / 3! + (s m 3
xs 2
«
1 '2
2
agreed that the branch cut of s
isolated singularities at in
-
)
1 '2
)
5
)
/ 5!
+•
or at the points is
seen by
first
s
= -n 2 it2
writing
x + x sl 6 + xV 1 120 +• 3
• •
5
1 5! +•••]
~
s
2
+ s 3 1 6 TP 1 120 +•••
•
—
-
165 and dividing the series in the denominator into the series in the numerator. The result sinh(xy 2
•
.
1/2
)
/
rsinhCs In
view of expression
1/2 '2 1
\
_ —
1
1
3 =X x— -(x X ~X) -x] ~T 2 + ~T\
6
s
)
— 1
h'
is
(0<\s\<
2 7T ).
s
Sec. 82, then,
(1),
= Ux'-x) + xt = \x{x % - 1) + xt. Res [e"F(s)] 1 1 "°°
As
6
for the singularities s
F(s)
=
4t q(s)
Observe that p(-n
2
- -n 2 n2
n2 )*0
= sinh(^ 1/2 ) and
2
2
q{-n n
and
= 2s
easy to see that q'(-n
= 1,2, ...), we write
(n
where
q'is)
it is
6
sinh(5
1/2
)
n 2 ) * 0. So
2
= Q.
)
q(s)
= s 2 sinh(j 1/2 ).
Also, since
+ ijj 1/2 cosh(j 1/2 ),
the points s
= -n 2 ^2
(n
= 1,2, ...),
are simple poles
and
of
2sinh(xs >=-« 2 jt j
£
™
(j)
1/2
1/2
cosh(5
= —5*
1/2
.1-1
)
+1
— «
(-1)"
2
)
5
sinn;cc
(/i
= 1,2,...).
(n
= l,2,...).
Thus, in view of expression (3), Sec. 82,
=
B+1
4 ^f-*"" V ^nnx 2
(-1)
Res
le"F(s)]
f(t)
= Rcj[e^F(s)] + ±Re
'
•
Finally, since
we arrive
at the
AyF(
expression
fit)
—n
2v(-ir',,v,, 1 g " sin
= ~x(x 2 -l) + xt 6
10.
S )],
The function
F00=—
**
*
-3
jr
— jsinhj
5
has isolated singularities at the points s
=
and
j„
= nm,
s„
= -nm
(n
= 1,2,.
.
.).
166
Now jsinhj
3
= 5i 5|
s + —s +---)
)
6
and division of this
= s 2 + —s 4 +-
(0
oo),
6
series into 1 reveals that
in
1
(0<\s\<7t).
This shows that F(s) has a removable singularity at s
have a removable singularity
there;
and so
s Res \e 'F(s)] = L
J
s=s
To
find the residue of F(s) at s n
^0) = and observe
-—
= nm
where
p(.y)
(n
= 1,2,.
.
.),
we write
= sinhj-j and
= -nm * 0,
q(nm) = 0,
and
Consequently, F(.s) has a simple pole at s B
Res F(.) =
4^
=
tf'(n;n)
Res
0.
=
2 ,y
sinh.s
that
p{nni)
Since F(j)
s
Evidently, then, e 'F(s) must also
.
= F(s ), st
\e F(s)]
2 2 n+l q'(nm) = n 7t (-l) *0.
and
,
= -rrar nV(-l) n+1
^
0»
= U,...).
n;r
the points sn are also simple poles of F(s); and
+ Res \e"F(sj\ = 2 Re
(
l
T ie inn = 2Re
Ml
=2
we may write
(-1)"
(icos/i7#-sin/Mtt)
sinnnt. nit
Hence the desired
result is
/(0 = Res [e"F(sj\ +
f (Res [e*F(j)l + Res
or
-smnm.
[^'F(.y)l},
7
167 11.
We
consider here the function „.
.
F(s)
a>
where
and cd*G)„
= 0,
5
Because the
first
=
sinh(xj)
s(s
+co)coshs
2
—
=
h
(n
= ±tui,
5
— and
(0
.
= 1,2,...). The singularities ^
= ±6)J
(n
of F(j) are at
= l,2,...).
term in the Maclaurin series for sinh(xs)
xs,
is
it is
easy to see that s
=
a removable singularity of e"F(s) and that
= 0.
Res[«*F(j)l
To find
= coi, we write
the residue of F(s) at s
cv
x
F(5)
from which
= ^^: s — coi
follows that s
it
rv \ Res F(s) = t>
Since fjs)
= F(J),
=
^/
1
1
s=s
u where
coi is
-\
=
(0)i)
;
,
+ cm)coshs
simple pole and
——
sinh(jcflw)
/sin ox
coilcoi cosh(o)i)
-2co cos o
j
for the residues at s
= 2^-
sinh(xy)'
then,
j
F(s)
=s{s
= 2Re|" j Res [e*F(5)l Res \e"F(s)]+ l l As
v
jl
(s)
= ©„/
where
Si
L-2tf)
(n
= 1,2,.
p(j)
.
.),
"^
iVl = 2
—
J
2fiJ
-
cos©
we put
= sinh(*.y) and
sinfitt =
2
sin G)f COSfl)
F(s) in the form
q(s)
= (s 3 + G) 2s)coshs.
q(s)
Now
p(fl)„i)
= swh(x(Q„i) =
q'(s)
we find
i
sin
fi)„.x
*
and
q(0)„i)
= (s 3 + 0) 2 s)sinhs + (3s 2 +
= 0. 2
fiJ
Also, since
)cosh s,
that
q'(o)J)
= (-coli + Q) 2 G)J)smh(o)n i) = -Q)„ (m 2 -
Hence we have a simple pole
at s
= an i,
q'(o)n i)
with residue
-con (a)
2
- G) 2 )sm G)n
=
sin oar sin g» 2
G)
cosg>
is
168 Consequently,
Resfe* F(s)]+ Res \e"F(s)] = 2Re
2
= 2-
2
-G)n ((Q -G)n )sin(Dn
But
sin o)„
n;r - -j
= sin(
j
= (-l)" +l and this means ,
1
>
-
2
0)n (a}
2 -Q)„)smo) n
that
2^^>^*
Res[e"F(*)l + Res f g 'T(j)1 =
j-*.!
smw.xsinco.t
2
6>_
a)
sin6)"'
-co
2
Finally,
/(f)
That
= Res [e"F(s)] + (Res [ e "F(*)] + Res
[e"F(*)]J +
J)f Res
is,
_
sinffit*sin6tt [
CO
COSO)
2
y
^
sina)„;csinfly 0) n
© - a:
[e"F(*)]
+ Res
[e"F(s)]\
