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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.

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    Even if x is complex, $\exp(ix)$ is still periodic.2010-09-02
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    No takers for using the ODE?!2010-09-02
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    Maybe I didn't really frame the question clearly enough, but these are all kind of boring proofs. Perhaps I'd get some more interesting responses on MathOvervlow?2010-09-03
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    I suppose the recognition that e^ix = cosx + isinx from Taylor series is really surprising when you first see it. But I find it inelegant because you're reasoning about an infinite number of terms (granted, they're really simple terms) in order to understand something about a finite number of functions. I kind of assumed there'd be another proof out there that doesn't rely so heavily on Taylor series...2010-09-03
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    great example of how stupid the moderators are. This could have been an interesting discussion2018-12-28

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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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    The first time I ever read that prologue, it gave me goosebumps.2015-03-07
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    real and imaginary parts of exp(iz)?2018-12-28
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    @Auburn Yes, that's correct. I'll edit to clarify.2018-12-28
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    this is a nice way to think about it. I wonder what is the motivation for looking at the real and imaginary parts of e^iz though. What does that have to do with circles and triangles?2018-12-28
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    @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
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$$ e^{ix} = \cos x + i \sin x \ . $$

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    That pushes the question back: how do you define cosine and sine, and then--based on that definition--how do prove that their common period is 2 pi? One would like a definition that makes it relatively easy to prove the important properties of exp, especially that it is an entire analytic function of the complex plane.2010-09-02
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    You're right, but assuming the knowledge of cosine, sine and Euler identity it's an streamlined proof, isn't it? :-)2010-09-02
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    Well, yes, but then it's a tautology. Now you have to prove that the lcm of the periods of sine and cosine equals $2 \pi$!2010-09-02
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    Define (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
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$$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$$

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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    Can you explain the last line? You go from $T = 2n\pi$ to $2\pi$2010-09-10