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

May 10, 2018

$\sin \left(\frac{11 \pi}{6}\right) + i \sin \left(\frac{11 \pi}{6}\right)$

Explanation:

A number given a $r {e}^{i \theta}$ can be written as $r \left(\cos \theta + i \sin \theta\right)$

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

${r}_{1} {r}_{1} = 1 \cdot 1 = 1$
${\theta}_{1} + {\theta}_{2} = \frac{\pi}{3} + \frac{3 \pi}{2} = \frac{11 \pi}{6}$

$\sin \left(\frac{11 \pi}{6}\right) + i \sin \left(\frac{11 \pi}{6}\right)$

e^((11pi)/6i

May 10, 2018

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

Explanation:

This is another point of view.

By Euler's relation

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

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

${e}^{\frac{\pi}{3} i} = \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{\sqrt{3}}{2} i$

${e}^{\frac{3}{2} \pi i} = \cos \left(\frac{3}{2} \pi\right) + i \sin \left(\frac{3}{2} \pi\right) = 0 - i$

Therefore,

${e}^{\frac{\pi}{3} i} \cdot {e}^{\frac{3}{2} \pi i} = \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) \cdot \left(- I\right)$

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

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