# What is the trigonometric form of 9 e^( (3pi)/2 i ) ?

Nov 27, 2017

$9 \cdot \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

#### Explanation:

Recall Eulers formula:
${e}^{i z} = \cos z + i \sin z$

In this case $z = \frac{3 \pi}{2}$

Therefore $9 {e}^{\frac{3 \pi}{2} i} = 9 \cdot \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$

If you wish to simplify this further, $\cos \left(\frac{3 \pi}{2}\right) = 0$, and $\sin \left(\frac{3 \pi}{2}\right) = - 1$, therefore:

$9 {e}^{\frac{3 \pi}{2} i} = 9 \cdot \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right) = 9 \cdot \left(0 + i \cdot \left(- 1\right)\right) = - 9 i$