# How do you multiply e^(( 2 pi )/ 3 i) * e^( pi/2 i )  in trigonometric form?

May 15, 2018

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

#### Explanation:

${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$

${e}^{i {\theta}_{1}} \cdot {e}^{i {\theta}_{2}} = = \cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)$

${\theta}_{1} + {\theta}_{2} = \frac{2 \pi}{3} + \frac{\pi}{2} = \frac{7 \pi}{6}$

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

May 15, 2018

The answer is $= = - \frac{\sqrt{3}}{2} + \frac{1}{2} i$

#### Explanation:

Another method .

${i}^{2} = - 1$

Euler's relation

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

Therefore,

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

$= \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(0 + i\right)$

$= \frac{1}{2} i - \frac{\sqrt{3}}{2}$

$= - \frac{\sqrt{3}}{2} + \frac{1}{2} i$