# How can you use trigonometric functions to simplify  8 e^( ( 7 pi)/4 i )  into a non-exponential complex number?

Dec 28, 2015

$8 \cdot {e}^{\frac{7 \pi}{4} \cdot i} = 4 \pi - 4 \pi \cdot i$

#### Explanation:

If we have a complex number in trigonometric form:

$z = | z | \cdot {e}^{i \cdot \varphi}$,

then according to de Moivre formula we can write, that:

$| z | \cdot {e}^{i \cdot \varphi} = | z | \cdot \left(\cos \varphi + i \cdot \sin \varphi\right)$

If we assume that the module $| z |$ is equal to $1$, then the formula simplifies to:

${e}^{i \cdot \varphi} = \left(\cos \varphi + i \cdot \sin \varphi\right)$

So the expression $8 \cdot {e}^{\frac{7 \pi}{4} \cdot i}$ can be written as:

$8 \cdot \left(\cos \left(\frac{7 \pi}{4}\right) + i \cdot \sin \left(\frac{7 \pi}{4}\right)\right) =$

$8 \cdot \left[\cos \left(2 \pi - \frac{\pi}{4}\right) + i \cdot \sin \left(2 \pi - \frac{\pi}{4}\right)\right] =$

$8 \cdot \left[\cos \left(\frac{\pi}{4}\right) - i \cdot \sin \left(\frac{\pi}{4}\right)\right] = 8 \cdot \left[\frac{\sqrt{2}}{2} - i \cdot \frac{\sqrt{2}}{2}\right] =$

$4 \sqrt{2} - 4 \sqrt{2} \cdot i$