## A Simple Proof of Euler’s Formula

Euler’s formula (well, one of Euler’s formulas) states that $e^{i \theta} = \cos \theta + i \sin \theta.$  The proof I am most familiar with is based on the Maclaurin series for $e^x, \sin x$, and $\cos x$.

I ran across a different, simpler proof recently on Math Stack Exchange, posted by user02138.  It doesn’t give any insight into why the formula is true, but it is very short.

Let $f(\theta) = e^{- i \theta} (\cos \theta + i \sin \theta)$.  Then $f'(\theta) = -i e^{-i \theta} (\cos \theta + i \sin \theta) + e^{- i \theta} (- \sin \theta + i \cos \theta).$  Simplifying, we have $f'(\theta) = e^{- i \theta} ( - i \cos \theta + \sin \theta - \sin \theta + i \cos \theta ) = 0.$  The fact that $f'(\theta) = 0$ implies that $f(\theta) = C$ for some constant C.  Since $f(0) = e^0 (\cos 0 + i \sin 0) = 1,$ $f(\theta) = 1,$ and thus $e^{i \theta} = \cos \theta + i \sin \theta.$

Some of the other answers to “How to prove Euler’s formula” are interesting and worth perusing as well.