Sunday, August 26, 2018

algebra precalculus - Prove $e^{i pi} = -1$











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



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


Answer



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.


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