# How can you use trigonometric functions to simplify  4 e^( ( 5 pi)/4 i )  into a non-exponential complex number?

The Moivre formula tells us that ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$.
Apply this here : $4 {e}^{i \frac{5 \pi}{4}} = 4 \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)$
On the trigonometric circle, $\frac{5 \pi}{4} = \frac{- 3 \pi}{4}$. Knowing that $\cos \left(\frac{- 3 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$ and $\sin \left(\frac{- 3 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$, we can say that $4 {e}^{i \frac{5 \pi}{4}} = 4 \left(- \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = - 2 \sqrt{2} - 2 i \sqrt{2}$.