Chapter 9, Problem 11CT

To determine

The quotient $\frac{w}{z}$, where $z=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)$ and $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$, and express the answer in polar form and exponential form.

Answer to Problem 11CT

Solution:

The quotient $\frac{w}{z}$ in polar form is $\frac{3}{2}\left(\cos \frac{33\pi }{20}+i\sin \frac{33\pi }{20}\right)$, and the exponential form is $\frac{3}{2}{e}^{i\cdot \frac{33\pi }{20}}$.

Explanation of Solution

Given information:

$z=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)$ and $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$

Explanation:

Let $z=2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)$ and $w=3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)$.

$\frac{w}{z}=\frac{3\left(\cos \frac{11\pi }{90}+i\sin \frac{11\pi }{90}\right)}{2\left(\cos \frac{17\pi }{36}+i\frac{17\pi }{36}\right)}$

First express the complex numbers in exponential form.

$\frac{w}{z}=\frac{3e^{i\cdot \frac{11\pi }{90}}}{2e^{i\cdot \frac{17\pi }{36}}}$

Thequotient of $z_1=r_1{e}^{i{\theta }_1}$ and $z_2=r_2{e}^{i{\theta }_2}$ is find as $\frac{z_1}{z_2}=\frac{r_1}{r_2}{e}^{i\left({\theta }_1-{\theta }_2\right)}$

$\Rightarrow \frac{3}{2}{e}^{i\cdot \left(\frac{11\pi }{90}-\frac{17\pi }{36}\right)}$

Thus, $\frac{w}{z}=\frac{3}{2}{e}^{i\cdot \left(-\frac{7\pi }{20}\right)}$

$=\frac{3}{2}\left(\cos -\frac{7\pi }{20}+i\sin -\frac{7\pi }{20}\right)$

The argument must be in between $0$ and $2\pi$. So, it is $-\frac{7\pi }{20}+2\pi =\frac{33\pi }{20}$

$=\frac{3}{2}\left(\cos \frac{33\pi }{20}+i\sin \frac{33\pi }{20}\right)$

$=\frac{3}{2}{e}^{i\cdot \frac{33\pi }{20}}$

Therefore, the quotient $\frac{w}{z}$ in polar formis $\frac{3}{2}\left(\cos \frac{33\pi }{20}+i\sin \frac{33\pi }{20}\right)$, and the exponential form is $\frac{3}{2}{e}^{i\cdot \frac{33\pi }{20}}$.