# How do you multiply e^(( 3 pi )/ 8 i) * e^( pi/2 i )  in trigonometric form?

Well, we knkw that ${e}^{i \theta} = \cos \theta + i \sin \theta$
And that ${e}^{i {\theta}_{1}} \cdot {e}^{i {\theta}_{2}} = {e}^{i \left({\theta}_{1} + {\theta}_{2}\right)} = \cos \left({\theta}_{1} + {\theta}_{2}\right) + i \sin \left({\theta}_{1} + {\theta}_{2}\right)$
$\frac{3 \pi}{8} + \frac{\pi}{2} = \frac{7 \pi}{8}$
$\cos \left(\frac{7 \pi}{8}\right) + i \sin \cos \left(\frac{7 \pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2} i \approx 0.92 + 0.38 i$