Chapter 9, Problem 38RE

To determine

To find: $zw$ and $\frac{z}{w}$ of Complex number $z=2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ and $w=3\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$ in the polar form.

Answer to Problem 38RE

The mentioned complex number in the polar form is $zw=6\left(\cos {360}^0+i\sin {360}^0\right)$ and $\frac{z}{w}=\frac{2}{3}\left(\cos {240}^0+i\sin {240}^0\right)$ .

Explanation of Solution

Given information:

The mentioned complex number $z=2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ and $w=3\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)$ .

Here we can write the mentioned complex number in the degree form as $z=2\left(\cos {300}^0+i\sin {300}^0\right)$ and $w=3\left(\cos {60}^0+i\sin {60}^0\right)$ .

Formula used:

If $z=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)$ and $w=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ then $zw=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)$

Also $\frac{z}{w}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)$ .

Calculation:

Consider the mentioned complex number $z=2\left(\cos {300}^0+i\sin {300}^0\right)$ and $w=3\left(\cos {60}^0+i\sin {60}^0\right)$ .

Recall the above mentioned formula that if $z=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)$ and $w=r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)$ then $zw=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right)$ and $\frac{z}{w}=\frac{r_1}{r_2}\left(\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right)$ .

Here $r_1=2$ , $r_2=3$ , ${\theta }_1={300}^0$ and ${\theta }_2={60}^0$ .

Therefore the mentioned complex number in polar form $zw=6\left(\cos \left({300}^0+{60}^0\right)+i\sin \left({300}^0+{60}^0\right)\right)\Rightarrow zw=6\left(\cos {360}^0+\sin {360}^0\right)$ and $\frac{z}{w}=\frac{2}{3}\left(\cos \left({300}^0-{60}^0\right)+i\sin \left({300}^0-{60}^0\right)\right)\Rightarrow \frac{z}{w}=\frac{2}{3}\left(\cos \left({240}^0\right)+i\sin \left({240}^0\right)\right)$ .

Thus the required Complex number in the polar form is $zw=6\left(\cos {360}^0+i\sin {360}^0\right)$ and $\frac{z}{w}=\frac{2}{3}\left(\cos {240}^0+i\sin {240}^0\right)$ .