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

Feb 18, 2017

The answer is $= - \frac{3}{2} + i 3 \frac{\sqrt{3}}{2}$

#### Explanation:

We need Euler's relation

${e}^{i \theta} = \cos \theta + i \sin \theta$

$\cos \left(\frac{2}{3} \pi\right) = - \frac{1}{2}$

$\sin \left(\frac{2}{3} \pi\right) = \frac{\sqrt{3}}{2}$

Therefore,

$\cos \left(\frac{7}{6} \pi\right) + i \sin \left(\frac{7}{6} \pi\right) = {e}^{\frac{7}{6} i \pi}$

$\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = {e}^{i \frac{\pi}{2}}$

So,

$\frac{3 \left(\cos \left(\frac{7}{6} \pi\right) + i \sin \left(\frac{7}{6} \pi\right)\right)}{\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)}$

$= 3 {e}^{\frac{7}{6} i \pi} / {e}^{i \frac{\pi}{2}}$

$= 3 {e}^{i \pi \left(\frac{7}{6} - \frac{1}{2}\right)}$

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

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

$= - \frac{3}{2} + i 3 \frac{\sqrt{3}}{2}$