Is it true that $e^{\frac{i}{2}\pi}+e^{-\frac{i}{2}\pi}=0$?

Question

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.

Answer 1

By Euler's formula, $$ e^{\frac{i\pi}{2}}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i $$ and $$ e^{\frac{-i\pi}{2}}=\cos\Big(-\frac{\pi}{2}\Big)+i\sin\Big(-\frac{\pi}{2}\Big)=-i $$ hence the sum is zero.

Answer 2

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 $$e^{i\pi/2}+e^{-i\pi/2}= \cos (\frac \pi 2) +i\sin (\frac \pi 2)+ \cos (-\frac \pi 2)+i\sin (-\frac \pi 2)=i\Big( \sin(\frac \pi 2)+\sin(- \frac \pi 2)\Big)=i(1-1)=0$$

Answer 3

$e^{i \theta}=\cos ( \theta) +i \sin (\theta) $

Answer 4

Using Euler's formula we have that:$\cos(\pi/2)+i\sin(\pi/2)+\cos(\pi/2)-i\sin(\pi/2) =2\cos(\pi/2)=0$

Answer 5

Multiply both members with $e^{\frac{i\pi}{2}}\neq 0$ to obtain the famous Euler relation: $$ e^{i\pi}+1=0$$