Chapter 9, Problem 36RE

To determine

To find: $zw$ and $\frac{z}{w}$ of Complex number $z=\left(\cos {205}^0+i\sin {205}^0\right)$ and $w=\left(\cos {85}^0+i\sin {85}^0\right)$ in the polar form.

Answer to Problem 36RE

The mentioned complex number in the polar form is $zw=\cos {290}^0+i\sin {290}^0$ and $\frac{z}{w}=\cos {120}^0+i\sin {120}^0$ .

Explanation of Solution

Given information:

The mentioned complex number is $z=\left(\cos {205}^0+i\sin {205}^0\right)$ and $w=\left(\cos {85}^0+i\sin {85}^0\right)$ .

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 $z=\left(\cos {205}^0+i\sin {205}^0\right)$ and $w=\left(\cos {85}^0+i\sin {85}^0\right)$ .

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 $r_1=1$ , $r_2=1$ , ${\theta }_1={205}^0$ and ${\theta }_2={85}^0$ .

Therefore the mentioned complex number in polar form $zw=\left(\cos \left({205}^0+{85}^0\right)+i\sin \left({205}^0+{85}^0\right)\right)\Rightarrow zw=\left(\cos {290}^0+\sin {290}^0\right)$ and $\frac{z}{w}=\left(\cos \left({205}^0-{85}^0\right)+i\sin \left({205}^0-{85}^0\right)\right)\Rightarrow \frac{z}{w}=\left(\cos \left({120}^0\right)+i\sin \left({120}^0\right)\right)$ .

Thus the required Complex number in the polar form is $zw=\cos {290}^0+i\sin {290}^0$ and $\frac{z}{w}=\cos {120}^0+i\sin {120}^0$ .