Chapter 9.7, Problem 23E

To determine

To find $\frac{z_1}{z_2}$ and express the result in rectangular form.

Answer to Problem 23E

The value of $\frac{z_1}{z_2}$ is $8\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$ and the rectangular form of the $\frac{z_1}{z_2}$ is $-4-4\sqrt{3}i$

Explanation of Solution

Given information:

The equation is

  $z_1=4\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\text{ and }{z}_2=\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$

Calculation:

The quotient can be obtained as

  $z_1=4\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\text{ and }{z}_2=\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$

First find the modulus and amplitude of the quotient.

  $\begin{array}{l}r=\frac{r_1}{r_2}\left[r=\text{modulus}\right]\\ r\text{ =}\frac{4}{\frac{1}{2}}\left[\text{since, }{r}_1=4\text{ and }{r}_2=\frac{1}{2}\right]\\ r=\frac{4\times 2}{1}\\ r=8\end{array}$

And

  $\begin{array}{l}\theta ={\theta }_1-{\theta }_2\left[\theta =\text{amplitude}\right]\\ \theta \text{ =}\frac{5\pi }{3}-\frac{\pi }{3}\left[\text{since, }{\theta }_1=\frac{5\pi }{3}\text{ and }{\theta }_2=\frac{\pi }{3}\right]\\ \theta \text{ =}\frac{5\pi -\pi }{3}\\ \theta =\frac{4\pi }{3}\end{array}$

The value of $\frac{z_1}{z_2}$ is $8\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$

Now find therectangular form of the $\frac{z_1}{z_2}$ .

  $\begin{array}{l}8\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)=8\left(\left(-\frac{1}{2}\right)+i\left(-\frac{\sqrt{3}}{2}\right)\right)\left[\text{since, }\cos \left(\frac{4\pi }{3}\right)=-\frac{1}{2}\text{ and }\sin \left(\frac{4\pi }{3}\right)=-\frac{\sqrt{3}}{2}\right]\\ =8\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\\ \text{ =}-4-4\sqrt{3}i\end{array}$

The rectangular form of the quotient is $-4-4\sqrt{3}i$

Hence,

The value of $\frac{z_1}{z_2}$ is $8\left(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}\right)$ and therectangular form of the $\frac{z_1}{z_2}$ is $-4-4\sqrt{3}i$