17.2

766 CHAPTER 17 Functions of a Complex Variable

RemarksMany of the properties of the real system hold in the complex number system, but there are some remarkable differences as well. For example, we cannot compare two complex numbers z1 � x1 � iy1, y1 � 0, and z2 � x2 � iy2, y2 � 0, by means of inequalities. In other words, state-ments such as z1 z2 and z2 � z1 have no meaning except in the case when the two numbers z1 and z2 are real. We can, however, compare the absolute values of two complex numbers. Thus, if z1 � 3 � 4i and z2 � 5 � i, then |z1| � 5 and |z2| � !26, and consequently |z1| |z2|. This last inequality means that the point (3, 4) is closer to the origin than is the point (5, �1).

In Problems 1–26, write the given number in the form a � ib.

1. 2i 3 � 3i 2 � 5i

2. 3i 5 � i 4 � 7i 3 � 10i 2 � 9

3. i 8 4. i 11

5. (5 � 9i) � (2 � 4i) 6. 3(4 � i) � 3(5 � 2i)

7. i (5 � 7i) 8. i (4 � i) � 4i(1 � 2i)

9. (2 � 3i)(4 � i) 10. (12 � 1

4i)(23 � 5

3i)

11. (2 � 3i)2 12. (1 � i)3

13. 2

i 14.

i

1 1 i

15. 2 2 4i

3 1 5i 16.

10 2 5i

6 1 2i

17. 13 2 i2 12 1 3i2

1 1 i 18.

11 1 i2 11 2 2i2

12 1 i2 14 2 3i2

19. 15 2 4i2 2 13 1 7i2

14 1 2i2 1 12 2 3i2 20.

14 1 5i2 1 2i3

12 1 i22

21. i (1 � i)(2 � i)(2 � 6i) 22. (1 � i )2(1 � i)3

23. (3 � 6i) � (4 � i )(3 � 5i ) � 1

2 2 i

24. 12 1 3i2 a 2 2 i

1 1 2ib2

25. a i

3 2 ib a 1

2 1 3ib 26.

1

11 1 i2 11 2 2i2 11 1 3i2

In Problems 27–32, let z � x � iy. Find the indicated expression.

27. Re(1/z) 28. Re(z 2) 29. Im(2z � 4z � 4i) 30. Im( z

2 � z 2) 31. Zz � 1 � 3iZ 32. Zz � 5z Z

In Problems 33–36, use Definition 17.1.2 to find a complex number z satisfying the given equation.

33. 2z � i(2 � 9i) 34. z � 2z � 7 � 6i � 0 35. z 2 � i 36. z

2 � 4z

In Problems 37 and 38, determine which complex number is closer to the origin.

37. 10 � 8i, 11 � 6i 38. 12 � 1

4 i, 23 � 1

6 i 39. Prove that |z1 � z2| is the distance between the points z1 and

z2 in the complex plane. 40. Show for all complex numbers z on the circle x 2 � y 2 � 4 that

Zz � 6 � 8iZ � 12.

17.1 Exercises Answers to selected odd-numbered problems begin on page ANS-000.

17.2 Powers and Roots

Introduction Recall from calculus that a point (x, y) in rectangular coordinates can also be expressed in terms of polar coordinates (r, u). We shall see in this section that the ability to express a complex number z in terms of r and u greatly facilitates finding powers and roots of z.

Polar Form Rectangular coordinates (x, y) and polar coordinates (r, u) are related by the equa-tions x � r cos u and y � r sin u (see Section 14.1). Thus a nonzero complex number z � x � iy can be written as z � (r cos u) � i(r sin u) or

z � r (cos u � i sin u). (1)

We say that (1) is the polar form of the complex number z. We see from FIGURE 17.2.1 that the polar coordinate r can be interpreted as the distance from the origin to the point (x, y). In other words, we adopt the convention that r is never negative so that we can take r to be the modulus FIGURE 17.2.1 Polar coordinates

y

x

rθ

θ

θ

z = x + iy

r sin

r cos

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17.2 Powers and Roots 767

of z; that is, r � Zz Z. The angle u of inclination of the vector z measured in radians from the positive real axis is positive when measured counterclockwise and negative when measured clockwise. The angle u is called an argument of z and is written u � arg z. From Figure 17.2.1 we see that an argument of a complex number must satisfy the equation tan u � y/x. The solutions of this equation are not unique, since if u0 is an argument of z, then necessarily the angles u0 2p, u0 4p, … , are also arguments. The argument of a complex number in the interval �p u � p is called the principal argument of z and is denoted by Arg z. For example, Arg(i) � p/2.

■ EXAMPLE 1 A Complex Number in Polar Form

Express 1 � 23i in polar form.

Solution With x � 1 and y � �23, we obtain r � Zz Z � #1122 1 122322 � 2. Now since the point (1, �23) lies in the fourth quadrant, we can take the solution of tan u � �23/1 � �23 to be u � arg z � 5p/3. It follows from (1) that a polar form of the number is

z 5 2 acos 5p

31 i sin

5p

3b .

As we see in FIGURE 17.2.2, the argument of 1 � 23i that lies in the interval (�p, p], the principal argument of z, is Arg z � �p/3. Thus, an alternative polar form of the complex number is

z 5 2 ccos a 2p

3b 1 i sin a 2

p

3b d .

Multiplication and Division The polar form of a complex number is especially convenient to use when multiplying or dividing two complex numbers. Suppose

z1 � r1(cos u1 � i sin u1) and z2 � r2(cos u2 � i sin u2),

where u1 and u2 are any arguments of z1 and z2, respectively. Then

z1z2 � r1r2[(cos u1 cos u2 � sin u1 sin u2) � i(sin u1 cos u2 � cos u1 sin u2)] (2)

and for z2 � 0,

z1

z25

r1

r2 [(cos u1 cos u2 � sin u1 sin u2) � i(sin u1 cos u2 � cos u1 sin u2)]. (3)

From the addition formulas from trigonometry, (2) and (3) can be rewritten, in turn, as

z1z2 � r1r2 [cos(u1 � u2) � i sin(u1 � u2)] (4)

and z1

z25

r1

r2 [cos(u1 � u2) � i sin(u1 � u2)]. (5)

Inspection of (4) and (5) shows that

Zz1z2 Z 5 Zz1 Z Zz2 Z , 2 z1

z22 5 Zz1 Z

Zz2 Z , (6)

and arg(z1z2) � arg z1 � arg z2, argaz1

z2b � arg z1 � arg z2. (7)

■ EXAMPLE 2 Argument of a Product and of a QuotientWe have seen that Arg z1 � p/2 for z1 � i. In Example 1 we saw that Arg z2 � �p/3 for z2 � 1 � 23i. Thus, for

z1z2 5 i11 2 23i 2 5 23 1 i and z1

z25

i

1 2 23i5 2

23

41

1

4 i

FIGURE 17.2.2 Two arguments of

z � 1 � 23i in Example 1

y

x

1 – √

– /3π

5 /3π

3i



it follows from (7) that

arg1z1z22 5p

22 p

35p

6 and argaz1

z2b 5 p

22 a2p

3b 5 5p

6.

In Example 2 we used the principal arguments of z1 and z2 and obtained arg(z1z2) � Arg(z1z2) and arg(z1/z2) � Arg(z1/z2). It should be observed, however, that this was a coincidence. Although (7) is true for any arguments of z1 and z2, it is not true, in general, that Arg(z1z2) � Arg z1 � Arg z2 and Arg(z1/z2) � Arg z1 � Arg z2. See Problem 39 in Exercises 17.2.

Powers of z We can find integer powers of the complex number z from the results in (4) and (5). For example, if z � r (cos u � i sin u), then with z1 � z and z2 � z, (4) gives

z 2 � r 2[cos (u � u) � i sin (u � u)] � r 2(cos 2u � i sin 2u).

Since z 3 � z2z, it follows that

z3 � r3(cos 3u � i sin 3u).

Moreover, since arg(1) � 0, it follows from (5) that

1

z2 � z�2 � r�2[cos(�2u) � i sin(�2u)].

Continuing in this manner, we obtain a formula for the nth power of z for any integer n:

z n � r n(cos nu � i sin nu). (8)

■ EXAMPLE 3 Power of a Complex Number

Compute z 3 for z � 1 � 23i.

Solution In Example 1 we saw that

z 5 2 c cos a2p3b 1 i sin a2p

3b d .

Hence from (8) with r � 2, u � �p/3, and n � 3, we get

11 2 23i23 5 23 c cos a3a2p3bb 1 i sin a3a2p

3bb d

� 8[cos(�p) � i sin(�p)] � �8.

DeMoivre’s Formula When z � cos u � i sin u, we have Zz Z � r � 1 and so (8) yields

(cos u � i sin u)n � cos nu � i sin nu. (9)

This last result is known as DeMoivre’s formula and is useful in deriving certain trigonometric identities.

Roots A number w is said to be an nth root of a nonzero complex number z if w n � z . If we let w � r(cos f � i sin f) and z � r (cos u � i sin u) be the polar forms of w and z , then in view of (8) , w n � z becomes

rn(cos nf � i sin nf) � r (cos u � i sin u).

From this we conclude that rn � r or r � r1/n and

cos nf � i sin nf � cos u � i sin u.

By equating the real and imaginary parts, we get from this equation

cos nf � cos u and sin nf � sin u.


These equalities imply that nf � u � 2kp, where k is an integer. Thus,

f 5u 1 2kp

n.

As k takes on the successive integer values k � 0, 1, 2, …, n � 1, we obtain n distinct roots with the same modulus but different arguments. But for k � n we obtain the same roots because the sine and cosine are 2p-periodic. To see this, suppose k � n � m, where m � 0, 1, 2, …. Then

f 5u 1 21n 1 m2p

n5u 1 2mp

n1 2p

and so sin f 5 sin au 1 2mpn

b , cos f 5 cos au 1 2mpn

b .

We summarize this result. The nth roots of a nonzero complex number z � r (cos u � i sin u) are given by

wk 5 r1>n c cos au 1 2kpn

b 1 i sin au 1 2kpn

b d , (10)

where k � 0, 1, 2, …, n � 1.

■ EXAMPLE 4 Roots of a Complex NumberFind the three cube roots of z � i.

Solution With r � 1, u � arg z � p /2, the polar form of the given number is z � cos(p/2) � i sin(p/2). From (10) with n � 3 we obtain

wk 5 1121>3 ccos ap>2 1 2kp

3b 1 i sin ap>2 1 2kp

3b d , k 5 0, 1, 2.

Hence, the three roots are

k 5 0, w0 5 cos p

61 i sin

p

6523

21

1

2 i

k 5 1, w1 5 cos 5p

61 i sin

5p

65 2

23

21

1

2 i

k 5 2, w2 5 cos 3p

21 i sin

3p

25 2i.

The root w of a complex number z obtained by using the principal argument of z with k � 0 is

sometimes called the principal nth root of z. In Example 4, since Arg(i) � p/2, w0 � 23/2 � (1/2)i is the principal third root of i.

Since the roots given by (8) have the same modulus, the n roots of a nonzero complex number z lie on a circle of radius r1/n centered at the origin in the complex plane. Moreover, since the difference between the arguments of any two successive roots is 2p/n, the nth roots of z are equally spaced on this circle. FIGURE 17.2.3 shows the three roots of i equally spaced on a unit circle; the angle between roots (vectors) wk and wk � 1 is 2p/3.

As the next example will show, the roots of a complex number do not have to be “nice” numbers as in Example 3.

■ EXAMPLE 5 Roots of a Complex NumberFind the four fourth roots of z � 1 � i.

Solution In this case, r � 22 and u � arg z � p/4. From (10) with n � 4, we obtain

wk 5 12221>4 ccos ap>4 1 2kp

4b 1 i sin ap>4 1 2kp

4b d , k 5 0, 1, 2, 3.

FIGURE 17.2.3 Three cube roots of i

y

x

w2

w1 w0

17.2 Powers and Roots 769



The roots, rounded to four decimal places, are

k 5 0, w0 5 122 21>4 ccos p

161 i sin

p

16d 5 1.0696 1 0.2127i

k 5 1, w1 5 12221>4 ccos 9p

161 i sin

9p

16d 5 20.2127 1 1.0696i

k 5 2, w2 5 12221>4 ccos 17p

161 i sin

17p

16d 5 21.0696 2 0.2127i

k 5 3, w3 5 12221>4 ccos 25p

161 i sin

25p

16d 5 0.2127 2 1.0696i.

In Problems 1–10, write the given complex number in polar form.

1. 2 2. �10

3. �3i 4. 6i

5. 1 � i 6. 5 � 5i

7. �23 � i 8. �2 � 223i

9. 3

21 1 i 10.

12

23 1 i

In Problems 11–14, write the number given in polar form in the form a � ib.

11. z 5 5 acos 7p

61 i sin

7p

6b

12. z 5 8 22 acos 11p

41 i sin

11p

4b

13. z 5 6 acos p

81 i sin

p

8b

14. z 5 10 acos p

51 i sin

p

5b

In Problems 15 and 16, find z1z2 and z1/z2. Write the number in the form a � ib.

15. z1 5 2 acos p

81 i sin

p

8b , z2 5 4 acos

3p

81 i sin

3p

8b

16. z1 5 22 acos p

41 i sin

p

4b ,

z2 5 23 acos p

121 i sin

p

12b

In Problems 17–20, write each complex number in polar form. Then use either (4) or (5) to obtain a polar form of the given number. Write the polar form in the form a � ib.

17. (3 � 3i)(5 � 523i) 18. (4 � 4i)(�1 � i)

19. 2i

2 2 2i 20.

22 1 26i

21 1 23i

In Problems 21–26, use (8) to compute the indicated power.

21. (1 � 23i)9 22. (2 � 2i)5

23. (12 � 1

2i)10 24. 1222 1 26i24

25. acos p

81 i sin

p

8b12

26. c23 acos 2p

91 i sin

2p

9b d 6

In Problems 27–32, use (10) to compute all roots. Sketch these roots on an appropriate circle centered at the origin.

27. (8)1/3 28. (1)1/8

29. (i)1/2 30. (�1 � i)1/3

31. (�1 � 23i)1/2 32. (�1 � 23i)1/4

In Problems 33 and 34, find all solutions of the given equation.

33. z4 � 1 � 0 34. z8 � 2z4 � 1 � 0

In Problems 35 and 36, express the given complex number first in polar form and then in the form a � ib.

35. acos p

91 i sin

p

9b12 c2 acos

p

61 i sin

p

6b d 5

36.

c8 acos 3p

81 i sin

3p

8b d 3

c2 acos p

161 i sin

p

16b d 10

37. Use the result (cos u � i sin u)2 � cos 2u � i sin 2u to find trigonometric identities for cos 2u and sin 2u.

38. Use the result (cos u � i sin u)3 � cos 3u � i sin 3u to find trigonometric identities for cos 3u and sin 3u.

17.2 Exercises Answers to selected odd-numbered problems begin on page ANS-000.


17.2

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