Chapter 9, Problem 3CRT

a.

To determine

Write polar form of $z$ .

Answer to Problem 3CRT

  $z=2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$

Explanation of Solution

Given information:

  $z=\sqrt{3}-iandw=6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

Calculation:

Consider the point in Cartesian coordinates $\left(\sqrt{3}-i\right)$ the polar form is,

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \theta ={\tan }^{-1}\frac{y}{x}\\ r=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(-i\right)}^2}\\ r=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(-i\right)}^2}\left[\therefore {\left(-i\right)}^2=1\right]\\ r=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2}\\ r=2\\ \theta =\frac{11\pi }{6}\end{array}$

  $z=2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$

Hence, polar form is. $z=2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$

b.

To determine

Find $zwandz/w$ .

Answer to Problem 3CRT

  $zw=12\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $\frac{z}{w}=\frac{1}{3}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$

Explanation of Solution

Given information:

  $z=\sqrt{3}-iandw=6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

Calculation:

Consider

  $z=2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ and $w=6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

Now calculate $zw$ ,

  $\begin{array}{l}zw=\left[2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)\right]\times \left[6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)\right]\\ zw=2\times 6\left(\cos \left(\frac{11\pi }{6}+\frac{5\pi }{12}\right)\right)+i\sin \left(\frac{11\pi }{6}+\frac{5\pi }{12}\right)\\ zw=12\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\end{array}$

Hence, $zw=12\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

Now calculate $z/w$ ,

  $\begin{array}{l}\frac{z}{w}=\frac{2\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)}{6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)}\\ \frac{z}{w}=\frac{1}{3}\left(\cos \frac{11\pi }{6}-\frac{5\pi }{12}\right)+i\sin \left(\frac{11\pi }{6}-\frac{5\pi }{12}\right)\\ \frac{z}{w}=\frac{1}{3}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\end{array}$

Hence, $\frac{z}{w}=\frac{1}{3}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$

c.

To determine

Find $z^{10}$ .

Answer to Problem 3CRT

  $z^{10}=1024\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)$

Explanation of Solution

Given information:

  $z=\sqrt{3}-iandw=6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

Calculation:

  $\begin{array}{l}{z}^n=r^n\left(\cos \left(\theta n\right)+i\sin \left(\theta n\right)\right)\\ z^{10}=2^{10}\cos \left(\frac{11\left(20\right)\pi }{6}+i\sin \left(\frac{11\left(20\right)\pi }{6}\right)\right)\left[\therefore r=2,n=10,\theta =11\pi /6\right]\\ z^{10}=1024\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)\end{array}$

Hence, $z^{10}=1024\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)$

d.

To determine

Find the three cube root of $z$ .

Answer to Problem 3CRT

  ${\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{11\pi }{18}\right)+i\sin \left(\frac{11\pi }{18}\right)\right]$

  ${\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{23\pi }{18}\right)+i\sin \left(\frac{23\pi }{18}\right)\right]$

  ${\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{35\pi }{18}\right)+i\sin \left(\frac{35\pi }{18}\right)\right]$

Explanation of Solution

Given information:

  $z=\sqrt{3}-iandw=6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)$

Calculation:

Apply general form of Demoivre’s Theorem and find the first root,

  $\begin{array}{l}{z}^{1/n}=r^{1/n}\left[\cos \left(\frac{x+k2\pi }{n}\right)+i\sin \left(\frac{x-k2\pi }{n}\right)\right]\\ {\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{11\pi }{18}\right)+i\sin \left(\frac{11\pi }{18}\right)\right]\\ \left[\therefore z=\sqrt{3}-i,r=2,x=11\pi -6,n=3,k=0\right]\end{array}$

Hence, first root is, ${\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{11\pi }{18}\right)+i\sin \left(\frac{11\pi }{18}\right)\right]$ .

Apply general form of Demoivre’s Theorem and find the second root,

  $\begin{array}{l}{z}^{1/n}=r^{1/n}\left[\cos \left(\frac{x+k2\pi }{n}\right)+i\sin \left(\frac{x-k2\pi }{n}\right)\right]\\ {\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{23\pi }{18}\right)+i\sin \left(\frac{23\pi }{18}\right)\right]\\ \left[\therefore z=\sqrt{3}-i,r=2,x=11\pi -6,n=3,k=1\right]\end{array}$

Hence, second root is ${\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{23\pi }{18}\right)+i\sin \left(\frac{23\pi }{18}\right)\right]$ .

Apply general form of Demoivre’s Theorem and find the third root,

  $\begin{array}{l}{z}^{1/n}=r^{1/n}\left[\cos \left(\frac{x+k2\pi }{n}\right)+i\sin \left(\frac{x-k2\pi }{n}\right)\right]\\ {\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{35\pi }{18}\right)+i\sin \left(\frac{35\pi }{18}\right)\right]\\ \left[\therefore z=\sqrt{3}-i,r=2,x=11\pi -6,n=3,k=2\right]\end{array}$

Hence, third root is ${\left(\sqrt{3}-i\right)}^{1/3}=2^{1/3}\left[\cos \left(\frac{35\pi }{18}\right)+i\sin \left(\frac{35\pi }{18}\right)\right]$ .

  $\begin{array}{l}{z}^n=r^n\left(\cos \left(\theta n\right)+i\sin \left(\theta n\right)\right)\\ z^{10}=2^{10}\cos \left(\frac{11\left(20\right)\pi }{6}+i\sin \left(\frac{11\left(20\right)\pi }{6}\right)\right)\left[\therefore r=2,n=10,\theta =11\pi /6\right]\\ z^{10}=1024\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)\end{array}$

Hence, $z^{10}=1024\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)$