What is your favorite proof that $e^{ix}$ has a period of $2\pi$? [closed]

as a function of a real variable, apparently. Part of the freedom in choosing a proof is that you get to choose what definition of $e^{ix}$ to start from -- do you use a differential equation? a power series? a definition in terms of trig functions? Another bit of freedom is that you get to choose what definition of $\pi$ to start from.

Solution 1:

My favorite has always been Walter Rudin's proof in the prologue to his "Real and Complex Analysis" (2nd Ed.). Here's a sketch:

• Define $\exp$ in terms of the power series.

• By manipulating the series, deduce that $\exp$ is a homomorphism from the additive group to the group of complex units.

• Show it satisfies the usual first order ODE.

• Define $\cos z$ and $\sin z$ as the real and imaginary parts of $\exp(iz)$, respectively.

• Define $\pi$ as twice the smallest positive real root of $\cos$.

• Deduce that $\exp( i \pi / 2) = i$.

• By multiplying, conclude that $2 \pi i$ is a period of $\exp$.

• Show, by means of the preceding properties, that no smaller period exists.

Solution 2:

$$ e^{ix} = \cos x + i \sin x \ . $$

Solution 3:

$$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$$

We have to find $T$ for which $e^{iT} = 1$

$$\rightarrow cos(T) + isin(T) = 1$$

$$\rightarrow sin(T) = 0$$ for all $$T = 2n\pi , n = 0,1,2,3...$$

So, period is $2\pi$.