# How can you use trigonometric functions to simplify  8 e^( ( 3 pi)/2 i )  into a non-exponential complex number?

Apr 16, 2016

$- 8 i$

#### Explanation:

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

So, $8 {e}^{\frac{3 \pi}{2} i} = 8 \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$
$\cos \left(\frac{3 \pi}{2}\right) = 0 \mathmr{and} \sin \left(\frac{3 \pi}{2}\right) = \sin \left(\pi + \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{2}\right) = - 1$

So, $8 {e}^{\frac{3 \pi}{2} i} = - 8 i$.