# Question f593a

Aug 25, 2016

$\frac{1}{2} c i s \left(\frac{3 \pi}{8}\right)$

#### Explanation:

Consider the 2 complex numbers.

${z}_{1} = {r}_{1} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right)$

and ${z}_{2} = {r}_{2} \left(\cos {\theta}_{2} + i \sin {\theta}_{2}\right)$

dividing them gives.

$\frac{{z}_{1}}{{z}_{2}} = \frac{{r}_{1} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right)}{{r}_{2} \left(\cos {\theta}_{2} + i \sin {\theta}_{2}\right)}$

multiply numerator and denominator by $\left(\cos {\theta}_{2} - i \sin {\theta}_{2}\right)$

$\frac{{z}_{1}}{{z}_{2}} = \frac{{r}_{1} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right) \left(\cos {\theta}_{2} - i \sin {\theta}_{2}\right)}{{r}_{2} \left(\cos {\theta}_{2} + i \sin {\theta}_{2}\right) \left(\cos {\theta}_{2} - i \sin {\theta}_{2}\right)}$

distribute the brackets.

$= \frac{{r}_{1} \left(\cos {\theta}_{1} \cos {\theta}_{2} + \sin {\theta}_{1} \sin {\theta}_{2} + i \left(\sin {\theta}_{1} \cos {\theta}_{2} - \cos {\theta}_{1} \sin {\theta}_{2}\right)\right)}{{r}_{2} \left({\cos}^{2} {\theta}_{2} + {\sin}^{2} {\theta}_{2}\right)}$

=[(r_1)/(r_2)](cos(theta_1-theta_2)+isin(theta_1-theta_2)........ (A)#

(A) is obtained using the $\textcolor{b l u e}{\text{trigonometric identities}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ and }$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

and of course, the denominator ${\cos}^{2} {\theta}_{2} + {\sin}^{2} {\theta}_{2} = 1$

The result being $\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{| {z}_{1} / \left({z}_{2}\right) | = \frac{{r}_{1}}{{r}_{2}} \text{ and } a r g \left(\frac{{z}_{1}}{{z}_{2}}\right) = {\theta}_{1} - {\theta}_{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\textcolor{b l u e}{\text{----------------------------------------------------------------}}$

Here ${r}_{1} = 5 , {\theta}_{1} = \frac{5 \pi}{8} \text{ and } {r}_{2} = 10 , {\theta}_{2} = \frac{\pi}{4}$

$\Rightarrow \frac{5 c i s \left(\frac{5 \pi}{8}\right)}{10 c i s \left(\frac{\pi}{4}\right)} = \frac{1}{2} c i s \left(\frac{5 \pi}{8} - \frac{\pi}{4}\right) = \frac{1}{2} c i s \left(\frac{3 \pi}{8}\right)$