# How can you use trigonometric functions to simplify  16 e^( ( 5 pi)/6 i )  into a non-exponential complex number?

$16 \left[\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right]$
Euler's theorem states : ${e}^{i \theta} = \cos \theta + i \sin \theta$
Here $\theta = \frac{5 \pi}{6}$
hence  16e^((5pi)/6 i = 16 [ cos((5pi)/6) + isin((5pi)/6) ]