Chapter I, Problem 29E

Write the number in polar form with argument between 0 and 2π.

29. $z=\sqrt{3}+i,w=1+\sqrt{3}i$

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 29E

The polar form of the complex number $zw$ is $zw=4\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$.

The polar form of the complex number $\frac{z}{w}$ is $\frac{z}{w}=\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)$.

The polar form of the complex number $\frac{1}{z}$ is $\frac{1}{z}=\frac{1}{2}\left[\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right]$.

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 $z=\sqrt{3}+i\text{and}w=1+\sqrt{3}i$.

Rewrite the complex number $z=\sqrt{3}+i\text{and}w=1+\sqrt{3}i$ 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 $z=\sqrt{3}+i$.

Obtain the argument of the complex number $z=\sqrt{3}+i$.

$\tan \theta =\frac{1}{\sqrt{3}}$

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

Obtain the modulus of the complex number $z=\sqrt{3}+i$.

$\begin{array}{c}r=\left|\sqrt{3}+i\right|\\ =\sqrt{{\sqrt{3}}^2+{\left(1\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

Thus, the value of $r=2$.

Thus, the polar form of the complex number $z=\sqrt{3}+i$ is $2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$.

Similarly, obtain the polar form of the complex number $w=1+\sqrt{3}i$.

The argument of argument of the complex number $w=1+\sqrt{3}i$ is $\theta ={\tan }^{-1}\left(\sqrt{3}\right)=\frac{\pi }{3}$

The modulus of the complex number $w=1+\sqrt{3}i$ is,

$\begin{array}{c}r=\left|1+\sqrt{3}i\right|\\ =\sqrt{{\sqrt{3}}^2+{\left(1\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

Thus, the polar form of the complex number $w=1+\sqrt{3}i$ is $2\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$.

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

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

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

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

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

$\begin{array}{c}\frac{z}{w}=\frac{2}{2}\left[\cos \left(\frac{\pi }{6}-\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{6}-\frac{\pi }{3}\right)\right]\\ =\cos \left(-\frac{\pi }{6}\right)+i\sin \left(-\frac{\pi }{6}\right)\end{array}$

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

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

$\frac{1}{z}=\frac{1}{2}\left[\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right]$

Thus, the polar form of the complex number $\frac{1}{z}$ is $\frac{1}{z}=\frac{1}{2}\left[\cos \frac{\pi }{6}-i\sin \frac{\pi }{6}\right]$.