# Prove $e^{i \pi} = -1$ [duplicate]

Possible Duplicate:
How to prove Euler's formula: $\exp(i t)=\cos(t)+i\sin(t)$ ?

I recently heard that $e^{i \pi} = -1$.

WolframAlpha confirmed this for me, however, I don't see how this works.

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## marked as duplicate by Qiaochu YuanMay 4 '11 at 22:27

Do you know the imaginary unit $i$? Do you know how complex exponentiation is defined? – lentic catachresis May 4 '11 at 22:22
Did you read en.wikipedia.org/wiki/Euler's_formula ? – lhf May 4 '11 at 22:23
To answer your (now deleted) question on how to put an equation into LaTeX, you can click on edit (below your question) to see what I did. Some links and references on that can be found here – t.b. May 4 '11 at 22:28
@Theo Buehler, thanks for that. I was trying to use parenthesis instead of braces. I'll remember that next time. – mbreedlove May 5 '11 at 0:24

This identity follows from Euler's Theorem, \begin{align} e^{i \theta} = \cos \theta + i \sin \theta, \end{align} which has many proofs. The one that I like the most is the following (sketched). Define $f(\theta) = e^{-i \theta}(\cos \theta + i \sin \theta)$. Use the quotient rule to show that $f^{\prime}(\theta)= 0$, so $f(\theta)$ is constant in $\theta$. Evaluate $f(0)$ to prove that $f(\theta) = f(0)$ everywhere.
Take $\theta = \pi$ for your claim.