# How do you multiply e^(( 5 pi )/ 3 i) * e^( 3 pi/2 i )  in trigonometric form?

Mar 31, 2018

${e}^{i \frac{7 \pi}{6}}$

#### Explanation:

By Euler's formula:
${e}^{i x} = \cos x + i \sin x$

Hence, the above becomes a FOIL problem:

${e}^{i \frac{5 \pi}{3}} \cdot {e}^{i \frac{3 \pi}{2}}$
$= \left(\cos \left(\frac{5 \pi}{3}\right) + i \sin \left(\frac{5 \pi}{3}\right)\right) \left(\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right)$
$= \cos \left(\frac{5 \pi}{3}\right) \cos \left(\frac{3 \pi}{2}\right) + i \cos \left(\frac{5 \pi}{3}\right) \sin \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{5 \pi}{3}\right) \cos \left(\frac{3 \pi}{2}\right) - \sin \left(\frac{5 \pi}{3}\right) \sin \left(\frac{3 \pi}{2}\right)$

Let's plug in the values, which we know from the unit circle:
$= \frac{1}{2} \cdot 0 + i \frac{1}{2} \cdot - 1 + i \cdot - \frac{\sqrt{3}}{2} \cdot 0 - \left(- 1\right) \cdot \left(- \frac{\sqrt{3}}{2}\right)$
$= - \frac{1}{2} i - \frac{\sqrt{3}}{2}$
Which is clearly on the unit circle! With a little thinking, we realize this is at $\theta = \frac{7 \pi}{6}$.

We can now realize that since this is an exponential, we could have used an exponential rule that we know:
${e}^{a} \cdot {e}^{b} = {e}^{a + b}$
${e}^{i \frac{5 \pi}{3}} \cdot {e}^{i \frac{3 \pi}{2}} = {e}^{i \frac{5 \pi}{3} + i \frac{3 \pi}{2}} = \exp \left(i \left[\frac{5 \pi}{3} + \frac{3 \pi}{2}\right]\right) = \exp \left(i \frac{19 \pi}{6}\right)$
Subtracting off a full rotation ($2 \pi$) from the argument leaves us with
$= {e}^{i \left(19 - 12\right) \frac{\pi}{6}} = {e}^{i \frac{7 \pi}{6}}$
as we found trigonometrically.