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

May 19, 2016

${e}^{\frac{2 \pi}{3} i} \cdot {e}^{\frac{3 \pi}{2} i} = \frac{\sqrt{3}}{2} + \frac{1}{2} i$

#### Explanation:

$a + i b$ can be written in trigonometric form $r {e}^{i \theta} = r \cos \theta + i r \sin \theta = r \left(\cos \theta + i \sin \theta\right)$,
where $r = \sqrt{{a}^{2} + {b}^{2}}$, but here in given question $r = 1$

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

= ${e}^{\left(\frac{4 \pi}{6} + \frac{9 \pi}{6}\right) i} = {e}^{13 \frac{\pi}{6} i}$

= $\left(\cos \left(13 \frac{\pi}{6}\right) + i \sin \left(13 \frac{\pi}{6}\right)\right)$

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

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

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