# How do you evaluate  e^( ( pi)/2 i) - e^( ( 4 pi)/3 i) using trigonometric functions?

the answer is $= \frac{1}{2} + i \left(1 + \frac{\sqrt{3}}{2}\right)$
Let $z = {e}^{i \frac{\pi}{2}} - {e}^{4 i \frac{\pi}{3}}$
Use the relation ${e}^{i \theta} = \cos \theta + i \sin \theta$
So ${e}^{i \frac{\pi}{2}} = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = 0 + i = i$
and ${e}^{4 i \frac{\pi}{3}} = \cos \left(\frac{4 \pi}{3}\right) + i \sin \left(\frac{4 \pi}{3}\right) = - \frac{1}{2} - i \frac{\sqrt{3}}{2}$
so $z = i + \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1}{2} + i \left(1 + \frac{\sqrt{3}}{2}\right)$