# How do you multiply e^(( pi )/ 4 i) * e^( 3 pi/2 i )  in trigonometric form?

Mar 2, 2018

The answer is $= \frac{\sqrt{2}}{2} \left(1 - i\right)$

#### Explanation:

Apply Euler's Identity

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

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

Therefore,

${e}^{\frac{\pi}{4} i} = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \left(1 + i\right)$

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

So,

$z = {e}^{\frac{\pi}{4} i} \cdot {e}^{\frac{3}{2} \pi i} = \frac{\sqrt{2}}{2} \left(1 + i\right) \cdot \left(- i\right) = \frac{\sqrt{2}}{2} \left(- i - {i}^{2}\right)$

$= \frac{\sqrt{2}}{2} \left(1 - i\right)$

$\text{Verification}$

$z = \left(\cos \phi + i \sin \phi\right) = \frac{\sqrt{2}}{2} \left(1 - i\right)$

$\cos \phi = \frac{\sqrt{2}}{2}$

$\sin \phi = - \frac{\sqrt{2}}{2}$

$\phi = - \frac{\pi}{4}$, $\left[2 \pi\right]$

$z = {e}^{\frac{\pi}{4} i} \cdot {e}^{\frac{3}{2} \pi i} = {e}^{\left(\frac{\pi}{4} + \frac{3}{2} \pi\right) i} = {e}^{\frac{7}{4} \pi i} = {e}^{- \frac{1}{4} \pi i}$