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What does the ratio of the circumference of a circle to its diameter have to do with the base of the natural logarithm and $\sqrt{-1}$?

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    The answers on [this MO question](http://mathoverflow.net/questions/27126/interpreting-the-famous-five-equation) are relevant.2011-10-12
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    sqrt(-1)=i (from complex number definition (a+i*b)2011-10-12
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    I don't think it has so much to do with "the base of the natural logarithm" as it has with "the natural exponential function".2011-10-12
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    I'm sure this is a duplicate... though my search skills are on the fritz today.2011-10-13
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    If you plot the function $f(x) = e^{ix} = \cos x + i\sin x$ in the complex plane, then every point is of the form $(\cos x, \sin x)$, so the graph of the function is just the unit circle. Then, as you increase $x$ from $0$ to $2 \pi$ you're traversing the unit circle. The statement $e^{i \pi} = -1$ is then just a subtle way of saying that if you start at $(1, 0)$ (i.e. starting at $x = 0$) then if you travel $\pi$ radians around the unit circle, you will end up at $(-1, 0)$. Of course, this doesn't explain why there is a connection between $e^{ix}$ and the unit circle...2014-03-01
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    @nsanger You did explain why there is a connection between $e^{ix}$ and the unit circle, it's because its equal to $\cos x + i \sin x$, and that can easily be shown using Taylor series.2014-03-01
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    What I meant was, it doesn't shed any light on why we might expect $e^{ix} = \cos x + i \sin x$. Whether or not it follows directly from the Taylor series of $\exp(x), \sin x,$ and $\cos x$, these algebraic manipulations don't give a clear intuitive reason for why the connection is there.2014-03-01

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