# How do you multiply e^(( 3pi )/ 2 ) * e^( pi/2 i )  in trigonometric form?

Mar 25, 2018

We have to use $\textcolor{red}{\text{Euler's Identity}}$, that claims that

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

By plugging in $x = \frac{\pi}{2}$ respectively, we get :

${e}^{\textcolor{red}{\frac{\pi}{2}} i} = \cos \textcolor{red}{\frac{\pi}{2}} + i \sin \textcolor{red}{\frac{\pi}{2}} = i$

I am not sure whenever you meant ${e}^{\frac{3 \pi}{2} i}$ or simply ${e}^{\frac{3 \pi}{2}}$. If you meant what you wrote, then you simply have :

${e}^{\frac{3 \pi}{2}} \cdot {e}^{\frac{\pi}{2} i} = i {e}^{\frac{3 \pi}{2}}$

If not, then you can use some basic properties of powers in order to simply things :

${e}^{\frac{3 \pi}{2} i} \cdot {e}^{\frac{\pi}{2} i} = {e}^{\frac{3 \pi + \pi}{2} i} = {e}^{2 \pi i}$

Apply the formula again.

${e}^{\textcolor{red}{2 \pi} i} = \cos \textcolor{red}{2 \pi} + i \sin \textcolor{red}{2 \pi} = 1$.