Chapter I, Problem 32E

To determine

To find: The polar form of $zw,\frac{z}{w}\text{and}\frac{1}{z}$ by putting z and w in polar form.

Answer to Problem 32E

The polar form of the complex number $zw$ is $24\sqrt{2}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$.

The polar form of the complex number $\frac{z}{w}$ is $\frac{4\sqrt{2}}{3}\left(\cos \left(-\frac{13\pi }{12}\right)+i\sin \left(-\frac{13\pi }{12}\right)\right)$.

The polar form of the complex number $\frac{1}{z}$ is $\frac{1}{z}=\frac{1}{8}\left[\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right]$.

Explanation of Solution

Formula used:

Let $z_1=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)$ and $z_2=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ polar form of the two complex number then,

$z_1{z}_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]$, $\frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right]\text{where}{z}_2\ne 0$ and $\frac{1}{z_1}=\frac{1}{r_1}\left[\cos {\theta }_1-i\sin {\theta }_1\right]$

Calculation:

It is given that $z=4\left(\sqrt{3}+i\right)\text{and}w=-3-3i$.

Rewrite the complex number $z=4\left(\sqrt{3}+i\right)\text{and}w=-3-3i$ in polar form.

The polar form of the complex number $z=a+bi$ is $z=r\left(\cos \theta +i\sin \theta \right)$ where $r=\left|z\right|=\sqrt{a^2+b^2}$  and $\tan \theta =\frac{b}{a}$.

Consider the complex number $z=4\left(\sqrt{3}+i\right)$.

Obtain the argument of the complex number $z=4\left(\sqrt{3}+i\right)$.

$\begin{array}{c}\tan \theta =\frac{4}{4\sqrt{3}}\\ =\frac{1}{\sqrt{3}}\end{array}$

Thus, the argument of argument of the complex number $z=4\left(\sqrt{3}+i\right)$ is $\theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}$

Obtain the modulus of the complex number $z=4\left(\sqrt{3}+i\right)$.

$\begin{array}{c}r=\left|4\left(\sqrt{3}+i\right)\right|\\ =\left|4\sqrt{3}+4\right|\\ =\sqrt{{\left(4\sqrt{3}\right)}^2+{\left(4\right)}^2}\\ =\sqrt{64}\\ =8\end{array}$

Thus, the value of $r=8$.

Thus, the polar form of the complex number $z=4\left(\sqrt{3}+i\right)$ is $8\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$.

Similarly obtain the polar form of the complex number $w=-3-3i$.

The argument of argument of the complex number $w=-3-3i$ is,

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-3}{-3}\right)\\ ={\tan }^{-1}\left(1\right)\\ =\frac{5\pi }{4}\end{array}$

The modulus of the complex number $w=-3-3i$ is,

$\begin{array}{c}r=\left|-3-3i\right|\\ =\sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}\\ =\sqrt{18}\\ =3\sqrt{2}\end{array}$

Thus, the polar form of the complex number $w=-3-3i$ is $3\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$.

Use the above formula to obtain the polar of the complex number $zw,\frac{z}{w}\text{and}\frac{1}{z}$ respectively.

Here, $z=8\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ and $w=3\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$.

$\begin{array}{c}zw=8\left(3\sqrt{2}\right)\left[\cos \left(\frac{\pi }{6}+\frac{5\pi }{4}\right)+i\sin \left(\frac{\pi }{6}+\frac{5\pi }{4}\right)\right]\\ =24\sqrt{2}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\end{array}$

Thus, the polar form of the complex number $zw$ is $24\sqrt{2}\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)$.

Compute the polar form of the complex number $\frac{z}{w}$.

$\begin{array}{l}\frac{z}{w}=\frac{8}{3\sqrt{2}}\left[\cos \left(\frac{\pi }{6}-\frac{5\pi }{4}\right)+i\sin \left(\frac{\pi }{6}-\frac{5\pi }{4}\right)\right]\\ =\frac{4\sqrt{2}}{3}\left(\cos \left(-\frac{13\pi }{12}\right)+i\sin \left(-\frac{13\pi }{12}\right)\right)\end{array}$

Thus, the polar form of the complex number $\frac{z}{w}$ is $\frac{4\sqrt{2}}{3}\left(\cos \left(-\frac{13\pi }{12}\right)+i\sin \left(-\frac{13\pi }{12}\right)\right)$.

Obtain the polar form of the complex number $\frac{1}{z}$.

$\frac{1}{z}=\frac{1}{8}\left[\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right]$

Thus, the polar form of the complex number $\frac{1}{z}$ is $\frac{1}{z}=\frac{1}{8}\left[\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right]$.