SECTION 8.3
POLAR FORM AND DEMOIVRE’S THEOREM
483
8.3 POLAR FORM AND DEMOIVRE’S THEOREM At this point you can add, subtract, multiply, and divide complex numbers. However, there is still one basic procedure that is missing from the algebra of complex numbers. To see this, consider the problem of finding the square root of a complex number such as i. When you use the four basic operations (addition, subtraction, multiplication, and division), there seems to be no reason to guess that
Figure 8.6 Imaginary axis
(a, b) r
i
b
θ a
0
Real axis
Complex Number: a + bi Rectangular Form: (a, b) Polar Form: (r, θ )
1i . That is, 2
1i
2 i. 2
To work effectively with powers and roots of complex numbers, it is helpful to use a polar representation for complex numbers, as shown in Figure 8.6. Specifically, if a bi is a nonzero complex number, then let be the angle from the positive x-axis to the radial line passing through the point (a, b) and let r be the modulus of a bi. So, a r cos ,
b r sin ,
and
r a 2 b2
and you have a bi r cos r sin i from which the following polar form of a complex number is obtained.
Definition of Polar Form of a Complex Number
The polar form of the nonzero complex number z a bi is given by z rcos i sin where a r cos , b r sin , r a 2 b2, and tan ba. The number r is the modulus of z and is the argument of z.
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COMPLEX VECTOR SPACES REMARK:
The polar form of z 0 is given by z 0cos i sin where is any
angle. Because there are infinitely many choices for the argument, the polar form of a complex number is not unique. Normally, the values of that lie between and are used, though on occasion it is convenient to use other values. The value of that satisfies the inequality < ≤
Principal argument
is called the principal argument and is denoted by Arg(z). Two nonzero complex numbers in polar form are equal if and only if they have the same modulus and the same principal argument.
EXAMPLE 1
Finding the Polar Form of a Complex Number Find the polar form of each of the complex numbers. (Use the principal argument.) (a) 1 i
Solution
(b) 2 3i
(c) i
(a) Because a 1 and b 1, then r 2 1 2 12 2, which implies that r 2. From a r cos and b r sin , cos
2 a 1 2 r 2
and
sin
2 b 1 . 2 r 2
So, 4 and
4 i sin 4 .
z 2 cos
(b) Because a 2 and b 3, then r 2 2 2 32 13, which implies that r 13. So, cos
a 2 13 r
and
sin
b 3 13 r
and it follows that arctan32. So, the polar form is
z 13 cos arctan
3 3 i sin arctan 2 2
13 cos0.98 i sin0.98 . (c) Because a 0 and b 1, it follows that r 1 and 2, so
z 1 cos
i sin . 2 2
The polar forms derived in parts (a), (b), and (c) are depicted graphically in Figure 8.7.
SECTION 8.3
POLAR FORM AND DEMOIVRE’S THEOREM
485
Figure 8.7 Imaginary axis
Imaginary axis
Imaginary axis 4 3 Real axis
θ −1
1
z=1−i
EXAMPLE 2
2 cos − π + i sin − π 4 4
[ ( )
( )]
1
z=i
2
−2
(a) z =
z = 2 + 3i
θ
θ 1
Real axis
Real axis
2
(b) z = 13[cos(0.98) + i sin(0.98)]
(c) z = 1 cos π + i sin π 2 2
(
)
Converting from Polar to Standard Form Express the complex number in standard form.
3 i sin 3
z 8 cos Solution
Because cos 3 12 and sin3 32, you can obtain the standard form
3 i sin 3 8 12 i
z 8 cos
3
2
4 43i.
The polar form adapts nicely to multiplication and division of complex numbers. Suppose you are given two complex numbers in polar form z1 r1cos 1 i sin 1
and
z2 r2cos 2 i sin 2 .
Then the product of z1 and z2 is given by z1 z 2 r1 r2cos 1 i sin 1cos 2 i sin 2 r1r2 cos 1 cos 2 sin 1 sin 2 i cos 1 sin 2 sin 1 cos 2 . Using the trigonometric identities cos1 2 cos 1 cos 2 sin 1 sin 2 and sin1 2 sin 1 cos 2 cos 1 sin 2 you have z1z 2 r1r2 cos1 2 i sin1 2 . This establishes the first part of the following theorem. The proof of the second part is left to you. (See Exercise 63.)
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COMPLEX VECTOR SPACES
Theorem 8.4 Product and Quotient of Two Complex Numbers
Given two complex numbers in polar form z1 r1cos 1 i sin 1 and z2 r 2cos 2 i sin 2 the product and quotient of the numbers are as follows. z1z2 r1r2 cos1 2 i sin1 2
Product
z1 r1 z 2 r2 cos1 2 i sin1 2 , z2 0
Quotient
This theorem says that to multiply two complex numbers in polar form, multiply moduli and add arguments, and to divide two complex numbers, divide moduli and subtract arguments. (See Figure 8.8.) Figure 8.8 Imaginary axis
z1z2
Imaginary axis
z2
z1
θ1 + θ 2 r 2 r1r2
θ 2 r1
r2
z1
θ1
r1
θ1
θ2
Real axis
r1 r2
z1 z2
θ1 − θ 2 Real axis
To multiply z1 and z2: Multiply moduli and add arguments.
EXAMPLE 3
z2
To divide z1 by z2: Divide moduli and add arguments.
Multiplying and Dividing in Polar Form Determine z1z2 and z1z2 for the complex numbers
z1 5 cos Solution
i sin 4 4
and z2
1 cos i sin . 3 6 6
Because you are given the polar forms of z 1 and z 2, you can apply Theorem 8.4 as follows. multiply
z1z2 5
5
5
3 cos 4 6 i sin 4 6 3 cos 12 i sin 12 1
5
add
add divide
z1 5 cos i sin z2 13 4 6 4 6
subtract
subtract
15cos 12 i sin 12
SECTION 8.3 REMARK:
POLAR FORM AND DEMOIVRE’S THEOREM
487
Try performing the multiplication and division in Example 3 using the stan-
dard forms z1
52 52 3 1 i and z2 i. 6 6 2 2
DeMoivre’s Theorem The final topic in this section involves procedures for finding powers and roots of complex numbers. Repeated use of multiplication in the polar form yields z r cos i sin z 2 r cos i sin r cos i sin r 2cos 2 i sin 2 z3 r cos i sin r 2 cos 2 i sin 2 r 3cos 3 i sin 3. Similarly, z4 r4cos 4 i sin 4 z 5 r 5cos 5 i sin 5. This pattern leads to the following important theorem, named after the French mathematician Abraham DeMoivre (1667–1754). You are asked to prove this theorem in Chapter Review Exercise 71. If z r cos i sin and n is any positive integer, then
Theorem 8.5 DeMoivre’s Theorem
zn r ncos n i sin n.
EXAMPLE 4
Raising a Complex Number to an Integer Power Find 1 3i
12
Solution
and write the result in standard form.
First convert to polar form. For 1 3i, r 12 3 2 2 and tan
3
1
3
which implies that 23. So,
1 3i 2 cos
2 2 i sin . 3 3
By DeMoivre’s Theorem, 2 2 12 i sin 3 3 12(2) 12(2) cos i sin 3 3
1 3i12 2cos
212
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COMPLEX VECTOR SPACES
4096cos 8 i sin 8 4096 1 i 0) 4096.
Recall that a consequence of the Fundamental Theorem of Algebra is that a polynomial of degree n has n zeros in the complex number system. So, a polynomial like px x6 1 has six zeros, and in this case you can find the six zeros by factoring and using the quadratic formula. x6 1 x3 1x3 1 x 1x 2 x 1x 1x 2 x 1 Consequently, the zeros are x ± 1,
x
1 ± 3i , 2
and
x
1 ± 3i . 2
Each of these numbers is called a sixth root of 1. In general, the nth root of a complex number is defined as follows.
Definition of n th Root of a Complex Number
The complex number w a bi is an nth root of the complex number z if z w n a bin . DeMoivre’s Theorem is useful in determining roots of complex numbers. To see how this is done, let w be an nth root of z, where w scos i sin
and
z rcos i cos .
Then, by DeMoivre’s Theorem you have w n s ncos n i sin n, and because w n z, it follows that s n cos n i sin n r cos i sin . Now, because the right and left sides of this equation represent equal complex numbers, you n r and equate principal can equate moduli to obtain s n r which implies that s arguments to conclude that and n must differ by a multiple of 2. Note that r is a n r is also a positive real number. Consequently, for some positive real number and so s integer k, n 2k, which implies that
2k . n
Finally, substituting this value for into the polar form of w produces the result stated in the following theorem.
SECTION 8.3
Theorem 8.6 n th Roots of a Complex Number
POLAR FORM AND DEMOIVRE’S THEOREM
489
For any positive integer n, the complex number z r cos i sin has exactly n distinct roots. These n roots are given by
n2k i sin n2k
n r cos
where k 0, 1, 2, . . . , n 1. Note that when k exceeds n 1, the roots begin to repeat. For instance, if k n, the angle is
Figure 8.9
REMARK: Imaginary axis
n
2 n 2 n n
2π n r
2π n
which yields the same value for the sine and cosine as k 0. Real axis
nth Roots of a Complex Number
The formula for the nth roots of a complex number has a nice geometric interpretation, as shown in Figure 8.9. Note that because the nth roots all have the same modulus (length) n r, they will lie on a circle of radius n r with center at the origin. Furthermore, the n roots are equally spaced along the circle, because successive nth roots have arguments that differ by 2n. You have already found the sixth roots of 1 by factoring and the quadratic formula. Try solving the same problem using Theorem 8.6 to see if you get the roots shown in Figure 8.10. When Theorem 8.6 is applied to the real number 1, the nth roots are given a special name—the nth roots of unity. Figure 8.10 Imaginary axis
−1 + 3i 2 2
−1
1+ 3 i 2 2
1
1− 3 −1 − 3i i 2 2 2 2 6th Roots of Unity
Real axis
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CHAPTER 8
COMPLEX VECTOR SPACES
EXAMPLE 5
Finding the nth Roots of a Complex Number Determine the fourth roots of i.
Solution
In polar form, you can write i as
i 1 cos
i sin 2 2
so that r 1, 2. Then, by applying Theorem 8.6, you have
42 2k4 i sin42 2k4
4 i14 1 cos
cos
8 k2 i sin8 k2.
Setting k 0, 1, 2, and 3 you obtain the four roots z1 cos
i sin 8 8
z 2 cos
5 5 i sin 8 8
z3 cos
9 9 i sin 8 8
z4 cos
13 13 i sin 8 8
as shown in Figure 8.11.
In Figure 8.11 note that when each of the four angles, 8, 58, 98, and 138 is multiplied by 4, the result is of the form 2 2k.
REMARK:
Figure 8.11
cos 5π + i sin 5π 8 8
Imaginary axis
cos π + i sin π 8 8 Real axis
cos 9π + i sin 9π 8 8 cos 13π + i sin 13π 8 8
SECTION 8.3
❑
SECTION 8.3
1. Imaginary
2.
25. 7cos 0 i sin 0
Real axis
3
3i
1
−2
−1
2 − 2i
3.
Imaginary axis
4.
Imaginary axis
3 2 1
−6
1 −1
1
Real axis
2
In Exercises 5–16, represent the complex number graphically, and give the polar form of the number. (Use the principle argument.) 5. 2 2i
6. 3 i
7. 21 3 i
8.
9. 6i
5 2
3 i
11. 7
12. 2i
13. 1 6i
14. 22 i
19.
18. 5 cos
5 5 3 cos i sin 2 3 3
21. 3.75 cos
23. 4 cos
i sin 4 4
3 3 i sin 2 2
4 cos 2 i sin 2 6 cos 4 i sin 4 3
29. 0.5cos i sin 0.5cos i sin
30.
3cos 3 i sin 3 3 cos
31.
2[cos(23) i sin(23)] 4[cos(29) i sin(29)]
32.
cos(53) i sin(53) cos i sin
33.
12[cos(3) i sin(3)] 3[cos(6) i sin(6)]
34.
9[cos(34) i sin(34)] 5[cos( 4) i sin( 4)]
37. 1 i
16. 4 2i
i sin 2 2
28.
1
35. 1 i 4
In Exercises 17–26, represent the complex number graphically, and give the standard form of the number. 17. 2 cos
2 2 i sin 3 3
In Exercises 35–44, use DeMoivre’s Theorem to find the indicated powers of the given complex number. Express the result in standard form.
10. 4
15. 3 i
2
Real axis −2 −3
3cos 3 i sin 3 4cos 6 i sin 6
1 + 3i
3
−6 −5 −4 −3 −2
Real axis
1
27. 2
−1
26. 6cos i sin
In Exercises 27–34, perform the indicated operation and leave the result in polar form.
Imaginary axis
axis
2
491
EXERCISES
In Exercises 1–4, express the complex number in polar form.
1
EXERCISES
20.
3 3 i sin 4 4
7 7 3 cos i sin 4 4 4
i sin 6 6
5 5 i sin 6 6
22. 8 cos
24. 6 cos
36. 2 2i 6
38. 3 i 7
10
39. 1 3i
3
5 5 i sin 6 6
8
41.
3cos
43.
2cos 2 i sin 2
40.
5cos 9 i sin 9
42.
cos
4
5 5 i sin 4 4
44. 5 cos
3
10
3 3 i sin 2 2
4
In Exercises 45–56, (a) use DeMoivre’s Theorem to find the indicated roots, (b) represent each of the roots graphically, and (c) express each of the roots in standard form. 45. Square roots: 16 cos i sin 3 3
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CHAPTER 8
COMPLEX VECTOR SPACES
46. Square roots: 9 cos
2 2 i sin 3 3
47. Fourth roots: 16 cos
4 4 i sin 3 3
5 5 i sin 48. Fifth roots: 32 cos 6 6
49. Square roots: 25i 51. Cube roots:
66. Show that the negative of z rcos i sin is
z r cos i sin .
50. Fourth roots: 625i
125 1 3i 2
52. Cube roots: 42 1 i 53. Cube roots: 8
54. Fourth roots: i
55. Fourth roots: 1
56. Cube roots: 1000
In Exercises 57–62, find all the solutions to the equation and represent your solutions graphically. 57. 59. 61. 63.
x4 i 0 58. x 3 1 0 5 60. x 4 81 0 x 243 0 3 62. x 4 i 0 x 64i 0 Given two complex numbers z 1 r1cos 1 i sin 1 and z 2 r2cos 2 i sin 2 with z 2 0 prove that z1 r1 z r cos 1 2 i sin 1 2 . 2
2
64. Show that the complex conjugate of z rcos i sin is z r cos i sin . 65. Use the polar form of z and z in Exercise 64 to find each of the following. (a) zz (b) zz, z 0
67. Writing
(a) Let z rcos i sin 2 cos
i sin . Sketch z, 6 6
iz, and zi in the complex plane. (b) What is the geometric effect of multiplying a complex number z by i? What is the geometric effect of dividing z by i? 68. Calculus Recall that the Maclaurin series for e x, sin x, and cos x are ex 1 x
x 2 x3 x 4 . . . 2! 3! 4!
sin x x
x 3 x5 x 7 . . . 3! 5! 7!
cos x 1
x 2 x 4 x6 . . . . 2! 4! 6!
(a) Substitute x i into the series for e x and show that ei cos i sin . (b) Show that any complex number z a bi can be expressed in polar form as z rei. (c) Prove that if z rei, then z rei. (d) Prove the amazing formula ei 1.
