# How do you multiply e^(( 5 pi )/ 4 i) * e^( pi/2 i )  in trigonometric form?

Apr 13, 2016

${e}^{\left(5 \frac{\pi}{4}\right) i} \cdot {e}^{\left(\frac{\pi}{2}\right) i} = - \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}$

#### Explanation:

A complex number can be written in polar form in two ways - either as $r \cdot {e}^{i \theta}$ or as $r \cos \theta + i r \sin \theta$.

Hence,

${e}^{\left(5 \frac{\pi}{4}\right) i} = \cos \left(5 \frac{\pi}{4}\right) + i \sin \left(5 \frac{\pi}{4}\right)$ and

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

Hence, ${e}^{\left(5 \frac{\pi}{4}\right) i} \cdot {e}^{\left(\frac{\pi}{2}\right) i}$

= $\left(\cos \left(5 \frac{\pi}{4}\right) + i \sin \left(5 \frac{\pi}{4}\right)\right) \cdot \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

= $\left(\cos \left(5 \frac{\pi}{4}\right) + i \sin \left(5 \frac{\pi}{4}\right)\right) \cdot \left(0 + i\right)$

as $\cos \left(\frac{\pi}{2}\right) = 0$ and $\sin \left(\frac{\pi}{2}\right) = 1$

= $\left(i \cos \left(5 \frac{\pi}{4}\right) + {i}^{2} \sin \left(5 \frac{\pi}{4}\right)\right)$

= $- \sin \left(5 \frac{\pi}{4}\right) + i \cos \left(5 \frac{\pi}{4}\right)$

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