Trigonometric Form of Complex Numbers

The Complex Plane

Each complex number in standard form, $ z = x + yi $ , corresponds to an ordered pair of real numbers $ (x, y) $ and thus to a point in a Cartesian coordinate system, referred to as the complex plane. The x-axis in this system is referred to as the real axis, and the y-axis as the imaginary axis.

Figure 29-1

EXAMPLE 29.1 Show $ 4 + 2i $ , -2i, and -3 - i in a complex plane.

The points are represented geometrically by $ (4,2) $ , $ (0,-2) $ , and $ (-3,-1) $ .

Figure 29-2

Trigonometric Form of Complex Numbers

If a polar coordinate system is superimposed on the Cartesian coordinate system, then the relationships $ x = r $ and $ y = r $ hold. Thus every complex number z can be written in trigonometric form:

\[ \begin{aligned}z&=r\cos\theta+ir\sin\theta\\&=r(\cos\theta+i\sin\theta)\end{aligned} \]

This form is sometimes abbreviated as $ z = r $ . The standard form $ z = x + yi $ is referred to as rectangular form. Since the polar coordinates of a point are not unique, there are an infinite number of equivalent trigonometric forms of a complex number. The relationships among x, y, z, r, and $ $ are shown in Fig. 29-3.

Figure 29-3

EXAMPLE 29.2 Write $ 5( + i) $ in rectangular form.

\[ 5\Big(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\Big)=5(0+1i)=0+5i \]

Modulus and Argument of a Complex Number

In writing a complex number in trigonometric form, the quantity r is normally chosen positive. Then, since $ r^{2}=x{2}+y^{2} $ , r represents the distance of the complex number from the origin, and is referred to as the modulus (sometimes called the absolute value) of the complex number. The absolute value notation is used, thus:

\[ |z|=r=\sqrt{x^{2}+y^{2}} \]

The quantity $ $ is referred to as the argument of the complex number. Unless otherwise specified, $ $ is normally chosen so that $ 0 < 2$ .

EXAMPLE 29.3 Write $ z = -6 + 6i $ in trigonometric form and state the modulus and argument for z, choosing $ 0 < 2$ .

$ -6 + 6i $ corresponds to the geometric point $ (-6, 6) $ . Since x = -6 and y = 6, $ r = |z| = = = 6 $ and $ = = -1 $ . Since $ (-6, 6) $ is in quadrant II, it follows that $ = $ . Thus, in trigonometric form, $ z = 6( + i) $ . The modulus of z is $ 6 $ and the argument of z is $ $ . (Note that other, equally valid, arguments for z can be obtained by adding integer multiples of $ 2$ to the argument $ 3/4 $ .)

Products and Quotients of Complex Numbers

Let $ z_{1}=r_{1}({1}+i{1}) $ and $ z_{2}=r_{2}({2}+i{2}) $ be complex numbers in trigonometric form. Then (assuming $ z_{2} $ )

\[ z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]\quad and\quad\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})] \]

DeMoivre’s Theorem

DeMoivre’s Theorem on powers of complex numbers: Let $ z = r(+ i) $ be a complex number in trigonometric form. Then for any nonnegative integer n,

\[ z^{n}=r^{n}(\cos n\theta+i\sin n\theta) \]

Theorem on nth Roots of Complex Numbers

If $ z = r(+ i) $ is any nonzero complex number and if n is any positive integer, then z has exactly n different nth roots $ w_{0}, w_{1}, , w_{n-1} $ . These roots are given by

\[ w_{k}=\sqrt[n]{r}\bigg(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n}\bigg) \]

for $ k = 0, 1, , n - 1 $ . The roots are symmetrically placed and equally spaced around a circle in the complex plane of radius $ $ and center the origin.

EXAMPLE 29.4 (a) Write i in trigonometric form; (b) find the two square roots of i.

1. Since $ i = 0 + 1i $ corresponds to the ordered pair (0, 1), $ r = = 1 $ and $ = /2 $ . Thus

\[ i=1[\cos(\pi/2)+i\sin(\pi/2)]. \]

2. Since n = 2, r = 1, and $ = /2 $ , the two square roots are given by

\[ w_{k}=\sqrt{1}\Big(\cos\frac{\pi/2+2\pi k}{2}+i\sin\frac{\pi/2+2\pi k}{2}\Big) \]

for k = 0, 1. Thus

\[ w_{0}=1\Big(\cos\frac{\pi/2}{2}+i\sin\frac{\pi/2}{2}\Big)=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \]

\[ w_{1}=1\Big(\cos\frac{\pi/2+2\pi}{2}+i\sin\frac{\pi/2+2\pi}{2}\Big)=\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \]

Polar Form of Complex Numbers

In advanced courses, it is shown that

\[ e^{i\theta}=\cos\theta+i\sin\theta \]

Then any complex number can be written as

\[ z=r(\cos\theta+i\sin\theta)=re^{i\theta} \]

Here, unless otherwise specified, $ $ is normally chosen between $ -$ and $ $ . $ e^{i} $ obeys the standard properties for exponents, hence:

For $ z = re^{i} $ , $ z_{1} = r_{1}e^{i{1}} $ , $ z{2} = r_{2}e^{i_{2}} $ , the previous formulas can be written:

\[ z_{1}z_{2}=r_{1}r_{2}e^{i(\theta_{1}+\theta_{2})}\qquad\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}e^{i(\theta_{1}-\theta_{2})} \]

\[ z^{n}=r^{n}e^{i n\theta}\left(\mathrm{D e M o i v r e^{\prime}s~t h e o r e m}\right) \]

The n nth roots of z = re iθ are given by w k = √[n/re i(θ+2πk)/n for k = 0, 1, . . . , n − 1 (nth roots theorem).

SOLVED PROBLEMS

29.1. Write in rectangular (standard) form:

\[ \left(\mathrm{d}\right)20\Big[\cos\Big(\tan^{-1}\frac{3}{4}\Big)+i\sin\Big(\tan^{-1}\frac{3}{4}\Big)\Big] \]

1. $ 4( + i ) = 4(1 + 0i) = 4 $

\[ (b)3\bigg(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\bigg)=3\bigg(\frac{\sqrt{3}}{2}+i\bigg(\frac{1}{2}\bigg)\bigg)=\frac{3\sqrt{3}+3i}{2} \]

4. Let $ u = ^{-1} $ . Then $ u = $ , $ - < u < $ . It follows that

\[ \cos\left(\tan^{-1}\frac{3}{4}\right)=\cos u=\frac{4}{5}\qquad and\qquad\sin\left(\tan^{-1}\frac{3}{4}\right)=\sin u=\frac{3}{5} \]

Hence

\[ 20\Big[\cos\Big(\tan^{-1}\frac{3}{4}\Big)+i\sin\Big(\tan^{-1}\frac{3}{4}\Big)\Big]=20\Big[\frac{4}{5}+i\frac{3}{5}\Big]=16+12i \]

29.2. Write in trigonometric form: (a) -8; (b) 3i; (c) $ 4 + 4i $ ; (d) $ -3 - 3i $ ; (e) 6 - 8i.

1. $ -8 = -8 + 0i $ corresponds to the geometric point $ (-8,0) $ . Since x = -8 and y = 0,

\[ r=\sqrt{x^{2}+y^{2}}=\sqrt{(-8)^{2}+0^{2}}=8\qquad and\qquad\tan\theta=\frac{0}{-8}=0. \]

Since \((-8,0)\) is on the negative \(x\)-axis, \(\theta=\pi\). It follows that \(-8=8(\cos\pi+i\sin\pi)\).

2. $ 3i = 0 + 3i $ corresponds to the geometric point $ (0,3) $ . Since x = 0 and y = 3,

\[ r=\sqrt{x^{2}+y^{2}}=\sqrt{0^{2}+3^{2}}=3\qquad and\qquad\tan\theta=\frac{3}{0}is undefined. \]

Since (0, 3) is on the positive y-axis, \(\theta = \frac{\pi}{2}\). It follows that \(3i = 3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)\).

3. $ 4 + 4i $ corresponds to the geometric point $ (4, 4) $ . Since x = 4 and $ y = 4 $ ,

\[ r=\sqrt{x^{2}+y^{2}}=\sqrt{4^{2}+(4\sqrt{3})^{2}}=8\qquad and\qquad\tan\theta=\frac{4\sqrt{3}}{4}=\sqrt{3}. \]

Since $ (4, 4) $ is in quadrant I, $ = $ . It follows that $ 4 + 4i = 8( + i) $ .

4. $ -3-3i $ corresponds to the geometric point $ (-3,-3) $ . Since $ x=y=-3 $ ,

\[ r=\sqrt{x^{2}+y^{2}}=\sqrt{(-3\sqrt{2})^{2}+(-3\sqrt{2})^{2}}=6\qquad and\qquad\tan\theta=\frac{-3\sqrt{2}}{-3\sqrt{2}}=1. \]

Since $ (-3,-3) $ is in quadrant III, $ = $ . It follows that $ -3-3i=6(+i) $ .

5. 6 - 8i corresponds to the geometric point (6, -8). Since x = 6 and y = -8,

\[ r=\sqrt{x^{2}+y^{2}}=\sqrt{6^{2}+(-8)^{2}}=10\qquad and\qquad\tan\theta=\frac{-8}{6}=-\frac{4}{3}. \]

Since $ (6, -8) $ is in quadrant IV, $ $ may be chosen as $ ^{-1}(-) $ . However, since this is a negative angle, the requirement that $ 0 < 2$ yields the alternative argument $ = 2+ ^{-1}(-) $ . With this argument, $ 6 - 8i = 10(+ i) $ .

29.3. Let $ z_{1}=r_{1}({1}+i{1}) $ and $ z_{2}=r_{2}({2}+i{2}) $ be complex numbers in trigonometric form. Assuming $ z_{2} $ , prove:

\[ \begin{aligned}(a)z_{1}z_{2}&=r_{1}r_{2}\left[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})\right];(b)\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})].\\ (a)z_{1}z_{2}&=r_{1}(\cos\theta_{1}+i\sin\theta_{1})r_{2}(\cos\theta_{2}+i\sin\theta_{2})\\&=r_{1}r_{2}(\cos\theta_{1}+i\sin\theta_{1})(\cos\theta_{2}+i\sin\theta_{2})\\&=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}+i\sin\theta_{2}\cos\theta_{1}+i\sin\theta_{1}\cos\theta_{2}+i^{2}\sin\theta_{1}\sin\theta_{2})\quad by FOIL\end{aligned} \]

In this expression, use $ i^{2} = -1 $ and combine real and imaginary terms:

\[ \begin{aligned}z_{1}z_{2}&=r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}+i\sin\theta_{2}\cos\theta_{1}+i\sin\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2})\\&=r_{1}r_{2}[(\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2})+i(\sin\theta_{2}\cos\theta_{1}+\sin\theta_{1}\cos\theta_{2})]\end{aligned} \]

The quantities in parentheses are recognized as $ ({1} + {2}) $ and $ ({1} + {2}) $ , respectively, from the sum formulas for cosine and sine. Hence

\[ z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})] \]

\[ \left(\mathbf{b}\right)\frac{z_{1}}{z_{2}}=\frac{r_{1}(\cos\theta_{1}+i\sin\theta_{1})}{r_{2}(\cos\theta_{2}+i\sin\theta_{2})}=\frac{r_{1}}{r_{2}}\frac{cos\theta_{1}+i\sin\theta_{1}}{\cos\theta_{2}+i\sin\theta_{2}} \]

In this expression, multiply numerator and denominator by $ {2}-i{2} $ , the conjugate of the denominator, then use $ i^{2}=-1 $ and combine real and imaginary terms:

\[ \begin{aligned}\frac{z_{1}}{z_{2}}&=\frac{r_{1}}{r_{2}}\frac{(\cos\theta_{1}+i\sin\theta_{1})(\cos\theta_{2}-i\sin\theta_{2})}{(\cos\theta_{2}+i\sin\theta_{2})(\cos\theta_{2}-i\sin\theta_{2})}\\&=\frac{r_{1}}{r_{2}}\frac{\cos\theta_{1}\cos\theta_{2}-i\cos\theta_{1}\sin\theta_{2}+i\sin\theta_{1}\cos\theta_{2}-i^{2}\sin\theta_{1}\sin\theta_{2}}{\cos^{2}\theta_{2}-i^{2}\sin^{2}\theta_{2}}\\&=\frac{r_{1}}{r_{2}}\frac{(\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2})+i(\sin\theta_{1}\cos\theta_{2}-\cos\theta_{1}\sin\theta_{2})}{\cos^{2}\theta_{2}+\sin^{2}\theta_{2}}\end{aligned} \]

The quantities in parentheses are recognized as $ ({1}-{2}) $ and $ ({1}-{2}) $ , respectively, from the difference formulas for cosine and sine, while $ ^{2}_{2}+{2}_{2}=1 $ is from the Pythagorean identity. Hence

\[ \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}[\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})] \]

29.4. Let $ z_{1}=40(+i) $ and $ z_{2}=5(+i) $ . Find $ z_{1}z_{2} $ and $ $ .

\[ \begin{aligned}z_{1}z_{2}&=40\Big(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}\Big)5\Big(\cos\frac{3\pi}{5}+i\sin\frac{3\pi}{5}\Big)\quad&\frac{z_{1}}{z_{2}}&=\frac{40}{5}\Big[\cos\Big(\frac{4\pi}{5}-\frac{3\pi}{5}\Big)+i\sin\Big(\frac{4\pi}{5}-\frac{3\pi}{5}\Big)\Big]\\ &=40(5)\Big[\cos\Big(\frac{4\pi}{5}+\frac{3\pi}{5}\Big)+i\sin\Big(\frac{4\pi}{5}+\frac{3\pi}{5}\Big)\Big]\quad&=8\Big(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5}\Big)\\ &=200\Big(\cos\frac{7\pi}{5}+i\sin\frac{7\pi}{5}\Big)\\ \end{aligned} \]

29.5. Let $ z_{1}=24i $ and $ z_{2}=4-4i $ . Convert to trigonometric form and find $ z_{1}z_{2} $ and $ $ in trigonometric and in rectangular form.

In trigonometric form:

\[ z_{1}=0+24i=24\Big(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\Big)\qquad and\qquad z_{2}=8\Big(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\Big) \]

Hence

\[ \begin{aligned}z_{1}z_{2}&=24\Big(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\Big)8\Big(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\Big)\quad&\frac{z_{1}}{z_{2}}&=\frac{24}{8}\Big[\cos\Big(\frac{\pi}{2}-\frac{11\pi}{6}\Big)+i\sin\Big(\frac{\pi}{2}-\frac{11\pi}{6}\Big)\Big]\\ &=24(8)\Big[\cos\Big(\frac{\pi}{2}+\frac{11\pi}{6}\Big)+i\sin\Big(\frac{\pi}{2}+\frac{11\pi}{6}\Big)\Big]\quad&=3\Big[\cos\Big(-\frac{4\pi}{3}\Big)+i\sin\Big(-\frac{4\pi}{3}\Big)\Big]\\ &=192\Big(\cos\frac{7\pi}{3}+i\sin\frac{7\pi}{3}\Big)\\ \end{aligned} \]

To satisfy the requirement that $ 0 < 2$ , subtract $ 2$ from the first argument and add $ 2$ to the second. Thus

\[ z_{1}z_{2}=192\Big[\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\Big]\qquad and\qquad\frac{z_{1}}{z_{2}}=3\Big[\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\Big] \]

In rectangular form:

\[ z_{1}z_{2}=192\Big(\frac{1}{2}+i\frac{\sqrt{3}}{2}\Big)=96+96i\sqrt{3}\qquad and\qquad\frac{z_{1}}{z_{2}}=3\Big(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\Big)=-\frac{3}{2}+i\frac{3\sqrt{3}}{2} \]

29.6. Prove DeMoivre’s theorem for n = 2 and n = 3

Choose $ z_{1}=z_{2}=z=r(+i) $ . Then

\[ \begin{aligned}z^{2}&=zz=r(\cos\theta+i\sin\theta)r(\cos\theta+i\sin\theta)\quad&z^{3}&=z^{2}z=r^{2}(\cos2\theta+i\sin2\theta)r(\cos\theta+i\sin\theta)\\ &=r^{2}[\cos\left(\theta+\theta\right)+i\sin(\theta+\theta)]\quad&=r^{2}r[\cos\left(2\theta+\theta\right)+i\sin\left(2\theta+\theta\right)]\\ &=r^{2}(\cos2\theta+i\sin2\theta)\quad&=r^{3}(\cos3\theta+i\sin3\theta)\\ \end{aligned} \]

Note: Similar proofs can be given easily for n = 4, n = 5, and so on. These suggest the validity of DeMoivre’s theorem for arbitrary integer n. A complete proof for arbitrary integral n requires the principle of mathematical induction (Chapter 42).

29.7. Apply DeMoivre’s theorem to find (a) $ ^{5} $ ; (b) $ (-1+i)^{6} $

\[ \left(a\right)\left[2\Big(\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}\Big)\right]^{5}=2^{5}\Big(\cos\frac{5\pi}{9}+i\sin\frac{5\pi}{9}\Big)=32\Big(\cos\frac{5\pi}{9}+i\sin\frac{5\pi}{9}\Big) \]

2. First write $ -1 + i $ in trigonometric form as $ ( + i) $ . Then apply DeMoivre’s theorem to obtain

\[ (-1+i)^{6}=\left[\sqrt{2}\Big(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\Big)\right]^{6}=(\sqrt{2})^{6}\Big(\cos\frac{9\pi}{2}+i\sin\frac{9\pi}{2}\Big)=8(0+1i)=8i \]

29.8. Show that any complex number $ w_{k} = ( + i) $ , for nonnegative integral k, is an nth root of the complex number $ z = r(+ i) $ .

Apply DeMoivre’s theorem to $ w_{k} $ :

\[ \begin{aligned}w_{k}^{n}&=\left[\bigvee^{n}\lceil r\Big(\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n}\Big)\right]^{n}=(\bigvee^{n}\lceil r\big\rceil^{n}[\cos(\theta+2\pi k)+i\sin(\theta+2\pi k)]\\&=r(\cos\theta+i\sin\theta)\\ \end{aligned} \]

The last equality follows from the periodicity of the sine and cosine functions. Hence $ w_{k} $ is an nth root of z.

29.9. Find the four fourth roots of $ 5( + i ) $

Applying the theorem on nth roots with n = 4, r = 5, and $ = 3 $ , the four fourth roots are given by

\[ w_{k}=\sqrt[4]{5}\Big(\cos\frac{3+2\pi k}{4}+i\sin\frac{3+2\pi k}{4}\Big) \]

for k = 0, 1, 2, 3. Thus

\[ w_{0}=\sqrt[4]{5}\Big(\cos\frac{3}{4}+i\sin\frac{3}{4}\Big)\quad w_{1}=\sqrt[4]{5}\Big(\cos\frac{3+2\pi}{4}+i\sin\frac{3+2\pi}{4}\Big) \]

\[ w_{2}=\sqrt[4]{5}\Big(\cos\frac{3+4\pi}{4}+i\sin\frac{3+4\pi}{4}\Big)\qquad w_{3}=\sqrt[4]{5}\Big(\cos\frac{3+6\pi}{4}+i\sin\frac{3+6\pi}{4}\Big) \]

29.10. (a) Find the three cube roots of -27i; (b) sketch these numbers in a complex plane.

1. First write -27i in trigonometric form as $ 27( + i) $ . Applying the theorem on nth roots with n = 3, r = 27, and $ = $ , the three cube roots are given by

\[ w_{k}=\sqrt[3]{27}\Big(\cos\frac{3\pi/2+2\pi k}{3}+i\sin\frac{3\pi/2+2\pi k}{3}\Big) \]

for k = 0, 1, 2. Thus

\[ w_{0}=\sqrt[3]{27}\Big(\cos\frac{3\pi/2}{3}+i\sin\frac{3\pi/2}{3}\Big)=3\Big(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\Big)=3(0+i1)=3i \]

\[ w_{1}=\sqrt[3]{27}\Big(\cos\frac{3\pi/2+2\pi}{3}+i\sin\frac{3\pi/2+2\pi}{3}\Big)=3\Big(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\Big)=3\Big(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\Big)=-\frac{3\sqrt{3}}{2}-\frac{3}{2}i \]

\[ w_{2}=\sqrt[3]{27}\Big(\cos\frac{3\pi/2+4\pi}{3}+i\sin\frac{3\pi/2+4\pi}{3}\Big)=3\Big(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\Big)=3\Big(\frac{\sqrt{3}}{2}-i\frac{1}{2}\Big)=\frac{3\sqrt{3}}{2}-\frac{3}{2}i \]

2. All three cube roots have magnitude 3, and hence lie on the circle of radius 3 with center the origin (see Fig. 29-4).

Figure 29-4

Note that since the arguments differ by $ 2/3 $ , the three cube roots are symmetrically placed and equally spaced around the circle.

29.11. Find all complex solutions of $ x^{6} + 64 = 0 $

Since $ x^{6} + 64 = 0 $ is equivalent to $ x^{6} = -64 $ , the solutions are the six complex sixth roots of -64.

Write -64 in trigonometric form as $ 64(+ i) $ . Applying the theorem on nth roots with n = 6, r = 64, and $ = $ , the six sixth roots are given by

\[ w_{k}=\sqrt[6]{64}\Big(\cos\frac{\pi+2\pi k}{6}+i\sin\frac{\pi+2\pi k}{6}\Big) \]

for k = 0, 1, 2, 3, 4, 5. Thus the six complex solutions of $ x^{6} + 64 = 0 $ are:

\[ w_{0}=\sqrt[6]{64}\Big(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\Big)=2\Big(\frac{\sqrt{3}}{2}+i\frac{1}{2}\Big)=\sqrt{3}+i \]

\[ w_{1}=\sqrt[6]{64}\Big(\cos\frac{3\pi}{6}+i\sin\frac{3\pi}{6}\Big)=2(0+i1)=2i \]

\[ w_{2}=\sqrt[6]{64}\Big(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\Big)=2\Big(-\frac{\sqrt{3}}{2}+i\frac{1}{2}\Big)=-\sqrt{3}+i \]

\[ w_{3}=\sqrt[6]{64}\Big(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}\Big)=2\Big(-\frac{\sqrt{3}}{2}-i\frac{1}{2}\Big)=-\sqrt{3}-i \]

\[ w_{4}=\sqrt[6]{64}\Big(\cos\frac{9\pi}{6}+i\sin\frac{9\pi}{6}\Big)=2(0-i1)=-2i \]

\[ w_{5}=\sqrt[6]{64}\Big(\cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6}\Big)=2\Big(\frac{\sqrt{3}}{2}-i\frac{1}{2}\Big)=\sqrt{3}-i \]

29.12. (a) Write $ 3e^{i(/3)} $ in rectangular (standard) form; (b) write 6 - 6i in polar form.

\[ 3e^{i(\pi/3)}=3\Big(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\Big)=3\Big(\frac{1}{2}+\frac{\sqrt{3}}{2}i\Big)=\frac{3}{2}+\frac{3\sqrt{3}}{2}i \]

2. 6 - 6i corresponds to the geometric point $ (6, -6) $ . Since x = 6 and y = -6,

\[ r=\sqrt{x^{2}+y^{2}}=\sqrt{6^{2}+(-6)^{2}}=6\sqrt{2}\quad and\quad\tan\theta=\frac{-6}{6}=-1 \]

Since $ (6,-6) $ is in quadrant IV, $ = - $ . It follows that $ 6 - 6i = 6e^{-i/4} $ .

29.13. For $ z_{1}=12e^{i(5/6)} $ , $ z_{2}=3e^{i/3} $ , find (a) $ z_{1}z_{2} $ ; (b) $ $ .

1. $ z_{1}z_{2}=(12e^{i(5/6)})(3e{i/3})=36e^{{i(5/6+/3)}=36e}{i(7/6)} $ . If $ $ is to be chosen between $ -$ and $ $ , then write

\[ 36e^{i(7\pi/6)}=36e^{-i(5\pi/6)} \]

\[ \mathrm{(b)}\frac{z_{1}}{z_{2}}=\frac{12e^{i(5\pi/6)}}{3e^{i\pi/3}}=4e^{i(5\pi/6-\pi/3)}=4e^{i(\pi/2)} \]

29.14. For $ z = -1 + i $ , find $ z^{3} $ in (a) polar and (b) rectangular (standard) form.

1. First write $ -1 + i $ in polar form as $ 2e^{i(2/3)} $ . Then apply DeMoivre’s theorem to obtain

\[ (-1+i\sqrt{3})^{3}=[2e^{i(2\pi/3)}]^{3}=2^{3}e^{2\pi i}=8e^{2\pi i} \]

2. In standard form $ 8e^{2i}=8(+i)=8 $ .

29.15. Find the three cube roots of $ 64e^{i(5/4)} $ .

Applying the theorem on nth roots with n = 3, r = 64, and $ = 5/4 $ , the three cube roots are given by $ w_{k} = e^{i(5/4 + 2k)/3} $ , for k = 0, 1, 2. Thus

\[ w_{0}=\sqrt[3]{64}e^{i(5\pi/4)/3}=4e^{i(5\pi/12)} \]

\[ w_{1}=\sqrt[3]{64}e^{i(5\pi/4+2\pi)/3}=4e^{i(13\pi/4)/3}=4e^{i(13\pi/12)} \]

\[ w_{2}=\sqrt[3]{64}e^{i(5\pi/4+4\pi)/3}=4e^{i(21\pi/4)/3}=4e^{i(7\pi/4)} \]

29.16. Show that $ e^{i} + 1 = 0 $ .

\[ e^{i\pi}+1=\cos\pi+i\sin\pi+1=-1+0i+1=0 \]

SUPPLEMENTARY PROBLEMS

29.17. Let $ z_{1}=8(+i) $ and $ z_{2}=+i $ . Find $ z_{1}z_{2} $ and $ $ .

\[ Ans.\quad z_{1}z_{2}=-4+4i\sqrt{3},\frac{z_{1}}{z_{2}}=8\Big(\cos\frac{2\pi}{9}+i\sin\frac{2\pi}{9}\Big) \]

29.18. Write -12, -8i, 2 - 2i, and $ - + i $ in trigonometric form.

29.19. Use the results of the previous problem to find (a) $ (-8i)(2-2i) $ ; (b) $ $ ; (c) $ (2-2i)^{3} $ .

\[ Ans.\quad(a)-16-16i;(b)-2+2i\sqrt{3};(c)-16-16i \]

29.20. Prove DeMoivre’s theorem for the cases \(n=0, n=1\), and \(n=4\).

29.21. Show that every complex number $ z = r(+ i) $ has exactly n different complex nth roots, for n an integer greater than 1. [Hint: Set $ w = s(+ i) $ and consider the solutions of the equation $ w^{n} = z $ .]

29.22. Find the two square roots of $ -1 + i $ .

\[ Ans.\quad\frac{\sqrt{2}}{2}+i\frac{\sqrt{6}}{2},-\frac{\sqrt{2}}{2}-i\frac{\sqrt{6}}{2} \]

29.23. (a) Find the three complex cube roots of 1. (b) Find the four complex fourth roots of −1.

\[ Ans.\quad(a)1,-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2};(b)\frac{1+i}{\sqrt{2}},\frac{-1+i}{\sqrt{2}},\frac{-1-i}{\sqrt{2}},\frac{1-i}{\sqrt{2}} \]

29.24. (a) Write $ 12e^{i(3/4)} $ in rectangular (standard) form; (b) Write 5i in polar form.

\[ Ans.\quad(a)-6\sqrt{2}+6i\sqrt{2};(b)5e^{i\pi/2} \]

29.25. For $ z_{1}=20e^{i(4/3)} $ , $ z_{2}=2e^{i/2} $ , find (a) $ z_{1}z_{2} $ ; (b) $ $ .

Ans. (a) $ 40e^{i(11/6)} $ or $ 40e^{-i/6} $ ; (b) $ 10e^{i(5/6)} $

29.26. For z = 1 - i, find $ z^{5} $ in (a) polar and (b) rectangular (standard) form.

Ans. (a) $ 4e^{i(3/4)}; $ (b) $ -4 + 4i $

29.27. Find the four fourth roots of $ 81e^{i(2/3)} $

\[ Ans.\quad3e^{i\pi/6}=\frac{3\sqrt{3}+3i}{2},3e^{i(2\pi/3)}=\frac{-3+3i\sqrt{3}}{2},3e^{i(7\pi/6)}=\frac{-3\sqrt{3}-3i}{2},3e^{i(5\pi/3)}=\frac{3-3i\sqrt{3}}{2} \]