Chapter 9, Problem 39RE

To determine

To find: $zw$ and $\frac{z}{w}$ of Complex number $z=5\left(\cos {10}^0+i\sin {10}^0\right)$ and $w=\left(\cos {355}^0+i\sin {355}^0\right)$ in the polar form.

Answer to Problem 39RE

The mentioned complex number in the polar form is $zw=5\left(\cos {365}^0+i\sin {365}^0\right)$ and $\frac{z}{w}=5\left(\cos {345}^0-i\sin {345}^0\right)$ .

Explanation of Solution

Given information:

The mentioned complex number is $z=5\left(\cos {10}^0+i\sin {10}^0\right)$ and $w=\left(\cos {355}^0+i\sin {355}^0\right)$ .

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

The trigonometric formula $\cos \left(-\theta \right)=\cos \theta$ and $\sin \left(-\theta \right)=-\sin \theta$ .

Calculation:

Consider the mentioned complex number $z=5\left(\cos {10}^0+i\sin {10}^0\right)$ and $w=\left(\cos {355}^0+i\sin {355}^0\right)$ .

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 $r_1=5$ , $r_2=1$ , ${\theta }_1={10}^0$ and ${\theta }_2={355}^0$ .

Recall the trigonometric formula $\cos \left(-\theta \right)=\cos \theta$ and $\sin \left(-\theta \right)=-\sin \theta$ .

Therefore the mentioned complex number in polar form $zw=5\left(\cos \left({10}^0+{355}^0\right)+i\sin \left({10}^0+{355}^0\right)\right)\Rightarrow zw=5\left(\cos {365}^0+\sin {365}^0\right)$ and $\frac{z}{w}=\frac{5}{1}\left(\cos \left({10}^0-{355}^0\right)+i\sin \left({10}^0-{355}^0\right)\right)\Rightarrow \frac{z}{w}=\left(\cos \left(-{345}^0\right)+i\sin \left(-{345}^0\right)\right)\frac{z}{w}=\left(\cos \left({345}^0\right)-i\sin \left({345}^0\right)\right)$ .

Thus the required Complex number in the polar form is $zw=5\left(\cos {365}^0+i\sin {365}^0\right)$ and $\frac{z}{w}=5\left(\cos {345}^0-i\sin {345}^0\right)$ .