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.

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