Is it true that $e^{\frac{i}{2}\pi}+e^{-\frac{i}{2}\pi}=0$? As in the title. Is it true that:
$$e^{\frac{i}{2}\pi}+e^{-\frac{i}{2}\pi}=0$$
And if it indeed is, how could one prove it? Maybe it's a silly question, but - frankly I'm kinda new to complex exponentials. 

Use Euler's formula: $$e^{i\theta}=\cos \theta + i\sin \theta$$
I think the easiest proof of this is with Taylor series, shown here: How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?
so $$e^{i\pi/2}+e^{-i\pi/2}= \cos (\frac \pi 2) +i\sin (\frac \pi 2)+ \cos (-\frac \pi 2)+i\sin (-\frac \pi 2)=i\Big( \sin(\frac \pi 2)+\sin(- \frac \pi 2)\Big)=i(1-1)=0$$ 
