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.
What is your favorite proof that $e^{ix}$ has a period of $2\pi$?
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0great example of how stupid the moderators are. This could have been an interesting discussion – 2018-12-28
3 Answers
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.
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0@Auburn Everything. The map $\theta \to \exp(i \theta)$ wraps the real line around the unit-radius circle. It provides a rigorous way to parameterize points on the circle and measure angles. The real and imaginary parts are the coordinates of those points (and therefore define the values of cosine and sine on real numbers). – 2018-12-28
$ e^{ix} = \cos x + i \sin x \ . $
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2Define (cos x, sin x) to be the rotation image of (1, 0) by x about the origin. They are thus obviously periodic with period 2π. The Maclaurin series for sine and cosine follow from this definition. Define exp(z) by its power series and it follows that $e^{ix}=\cos x+i\sin x$. – 2010-09-02
$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$
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 $2\pi$.
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0Can you explain the last line? You go from $T = 2n\pi$ to $2\pi$ – 2010-09-10