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, θ). We shall see in this section that the ability to express a complex number z in terms of r and θ greatly facilitates finding powers and roots of z.

Polar Form

Rectangular coordinates (x, y) and polar coordinates (r, θ) are related by the equations x = r cos θ and y = r sin θ (see Section 14.1). Thus a nonzero complex number z = x + iy can be written as z = (r cos θ) + i(r sin θ) or

(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 of z; that is, r = . The angle θ 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 θ is called an argument of z and is written θ = arg z. From Figure 17.2.1 we see that an argument of a complex number must satisfy the equation tan θ = y/x. The solutions of this equation are not unique, since if θ0 is an argument of z, then necessarily the angles θ0 ± 2π, θ0 ± 4π, . . ., are also arguments. The argument of a complex number in the interval −π < θ ≤ π is called the principal argument of z and is denoted by Arg z. For example, Arg(i) = π/2.

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point on the center labeled z = x plus i y, and an arrow, r, starting from the origin points at it. A dashed line of height r sine theta and a base on the horizontal axis measuring r cos theta form a right-angled triangle with the arrow. The angle formed by the hypotenuse and base of the triangle is labeled theta.

FIGURE 17.2.1 Polar coordinates

EXAMPLE 1 A Complex Number in Polar Form

Express 1 − i in polar form.

SOLUTION

With x = 1 and y = −, we obtain r = = = 2. Now since the point (1, −) lies in the fourth quadrant, we can take the solution of tan θ = −/1 = − to be θ = arg z = 5π/3. It follows from (1) that a polar form of the number is

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

≡

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point in the fourth quadrant labeled 1 minus under root (3 i), and an arrow starting from the origin points at it. The positive and negative angles formed by the arrow and the horizontal axis are labeled 5 pi over 3 and minus pi over 3, respectively.

FIGURE 17.2.2 Two arguments of z = 1 − i in Example 1

Multiplication and Division

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

where θ1 and θ2 are any arguments of z1 and z2, respectively. Then

(2)

and for z2 ≠ 0,

(3)

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

(4)

and (5)

Inspection of (4) and (5) shows that

(6)

and (7)

EXAMPLE 2 Argument of a Product and of a Quotient

We have seen that Arg z1 = π/2 for z1 = i. In Example 1 we saw that Arg z2 = −π/3 for z2 = 1 − i. Thus, for

it follows from (7) that

≡

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.

Integer 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 θ + i sin θ), then with z1 = z and z2 = z, (4) gives

z2 = r2[cos (θ + θ) + i sin (θ + θ)] = r2(cos 2θ + i sin 2θ).

Since z3 = z2z, it follows that

z3 = r3(cos 3θ + i sin 3θ).

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

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

(8)

EXAMPLE 3 Power of a Complex Number

Compute z3 for z = 1 − i.

SOLUTION

In Example 1 we saw that

Hence from (8) with r = 2, θ = −π/3, and n = 3, we get

≡

De Moivre’s Formula

When z = cos θ + i sin θ, we have = r = 1 and so (8) yields

(9)

The last result that links complex numbers and trigonometry is called de Moivre’s formula after the French mathematician Abraham de Moivre (1667–1754). This formula is useful in deriving certain trigonometric identities. See Problems 37 and 38 in Exercises 17.2.

Roots

A number w is said to be an nth root of a nonzero complex number z if wn = z. If we let w = ρ(cos ϕ + i sin ϕ) and z = r(cos θ + i sin θ) be the polar forms of w and z , then in view of (8) , wn = z becomes

ρn(cos nϕ + i sin nϕ) = r(cos θ + i sin θ).

From this we conclude that ρn = r or ρ = r1/n and

cos nϕ + i sin nϕ = cos θ + i sin θ.

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

cos nϕ = cos θ     and     sin nϕ = sin θ.

These equalities imply that nϕ = θ + 2kπ, where k is an integer. Thus,

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 2π-periodic. To see this, suppose k = n + m, where m = 0, 1, 2, . . . . Then

and so .

We summarize this result. The n nth roots of a nonzero complex number z = r(cos θ + i sin θ) are given by

(10)

where k = 0, 1, 2, . . ., n − 1.

EXAMPLE 4 Roots of a Complex Number

Find the three cube roots of z = i.

SOLUTION

With r = 1, θ = arg z = π/2, the polar form of the given number is z = cos(π/2) + i sin(π/2). From (10) with n = 3 we obtain

Hence, the three roots are

≡

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) = π/2, w0 = + (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 2π/n, the nth roots of z are equally spaced on this circle. FIGURE 17.2.3 shows the three cube roots of i equally spaced on a unit circle; the angle between roots (vectors) wk and wk + 1 is 2π/3.

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows two diagonal arrows moving away from the origin in the first quadrant and the second quadrant, respectively, and another diagonal arrow on the negative vertical axis. The arrows are labeled w subscript 0, w subscript 1, and w subscript 2, respectively, and are enclosed in a circle.

FIGURE 17.2.3 Three cube roots of i

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 Number

Find the four fourth roots of z = 1 + i.

SOLUTION

In this case, r = and θ = arg z = π/4. From (10) with n = 4, we obtain

The roots, rounded to four decimal places, are

≡

17.2 Exercises Answers to selected odd-numbered problems begin on page ANS-44.

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. − + i
  8. −2 − 2i

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

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


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.

  1. (3 − 3i)(5 + 5i)
  2. (4 + 4i)(−1 + i)

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

  1. (1 + i)9
  2. (2 − 2i)5
  3. ( + i)10

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

  1. (8)1/3
  2. (1)1/8
  3. (i)1/2
  4. (−1 + i)1/3
  5. (−1 + i)1/2
  6. (−1 − i)1/4

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

  1. z4 + 1 = 0
  2. 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.

  3. Use the result (cos θ + i sin θ)2 = cos 2θ + i sin 2θ to find trigonometric identities for cos 2θ and sin 2θ.
  4. Use the result (cos θ + i sin θ)3 = cos 3θ + i sin 3θ to find trigonometric identities for cos 3θ and sin 3θ.
  5. (a) If z1 = −1 and z2 = 5i, verify that

    Arg(z1z2) ≠ Arg(z1) + Arg(z2).

    (b) If z1 = −1 and z2 = −5i, verify that

    Arg(z1/z2) ≠ Arg(z1) − Arg(z2).

  6. For the complex numbers given in Problem 39, verify in both parts (a) and (b) that

    arg(z1z2) = arg(z1) + arg(z2)

    and arg = arg(z1) − arg(z2).