# How do you simplify [sqrt2(cos((7pi)/4)+isin((7pi)/4))]div[sqrt2/2(cos((3pi)/4)+isin((3pi)/4))] and express the result in rectangular form?

Apr 21, 2018

$\left[\sqrt{2} \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right)\right] \div \left[\frac{\sqrt{2}}{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)\right]$

$= \frac{\sqrt{2} \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right)}{\frac{\sqrt{2}}{2} \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)}$

$= \frac{2 \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right)}{\left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)}$

$= \frac{2 {e}^{i \frac{7 \pi}{4}}}{{e}^{i \frac{3 \pi}{4}}}$

$= 2 {e}^{i \left(\frac{7 \pi}{4} - \frac{3 \pi}{4}\right)}$

$= 2 {e}^{i \pi}$

$= 2 \left(\cos \pi + i \sin \pi\right)$

$= 2 \left(- 1 + i \cdot 0\right) = - 2 + i \cdot 0$

Apr 21, 2018

The answer is $- 2 + 0 i$

#### Explanation:

Two divide two complex numbers, we divide their moduli and subtract their arguments.

So our number will have modulus

$\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\frac{1}{2}} = 2$

and argument of

$\frac{7}{4} \pi - \frac{3}{4} \pi = \frac{4}{4} \pi = \pi$

So our number will be

$2 \left(\cos \pi + i \sin \pi\right) = 2 \left(- 1 + 0 i\right) = - 2 + 0 i$