Chapter 9, Problem 40RE

To determine

To find: $zw$ and $\frac{z}{w}$ of Complex number $z=4\left(\cos {50}^0+i\sin {50}^0\right)$ and $w=\left(\cos {340}^0+i\sin {340}^0\right)$ in the polar form.

Answer to Problem 40RE

The mentioned complex number in the polar form is $zw=4\left(\cos {390}^0+i\sin {390}^0\right)$ and $\frac{z}{w}=4\left(\cos {290}^0-i\sin {290}^0\right)$ .

Explanation of Solution

Given information:

The mentioned complex number is $z=4\left(\cos {50}^0+i\sin {50}^0\right)$ and $w=\left(\cos {340}^0+i\sin {340}^0\right)$ .

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 $z=4\left(\cos {50}^0+i\sin {50}^0\right)$ and $w=\left(\cos {340}^0+i\sin {340}^0\right)$ .

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 $r_1=4$ , $r_2=1$ , ${\theta }_1={50}^0$ and ${\theta }_2={340}^0$ .

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 $zw=4\left(\cos \left({50}^0+{340}^0\right)+i\sin \left({50}^0+{340}^0\right)\right)\Rightarrow zw=4\left(\cos {390}^0+\sin {390}^0\right)$ and $\frac{z}{w}=\frac{4}{1}\left(\cos \left({50}^0-{390}^0\right)+i\sin \left({50}^0-{390}^0\right)\right)\Rightarrow \frac{z}{w}=4\left(\cos \left(-{290}^0\right)+i\sin \left(-{290}^0\right)\right)\Rightarrow \frac{z}{w}=4\left(\cos \left({290}^0\right)-i\sin \left({290}^0\right)\right)$ .

Thus the required Complex number in the polar form is $zw=4\left(\cos {390}^0+i\sin {390}^0\right)$ and $\frac{z}{w}=4\left(\cos {290}^0-i\sin {290}^0\right)$ .