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When I was learning about complex exponents, I came through this expression $e^{ix}=\cos(x)+i\sin(x)$.
But I don't understand why $e$ is used here and I don't have any knowledge about geometrical interpretation of $e$.

EDIT: Now after knowing the proof, I'm wondering whether it can be proved geometrically using circles.

Love Invariants
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    Here's the proof https://en.wikipedia.org/wiki/Euler%27s_formula#Proofs . I'm surprise your text doesn't show it. – layabout Jun 19 '18 at 18:26
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    The short explanation is that the taylor series of the two expressions exactly match up. $e^z = 1+z+\frac{z^2}{2}+\frac{z^3}{3!}+\frac{z^4}{4!}+\dots+\frac{z^n}{n!}+\dots$, meanwhile $\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\dots$ and $\sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\dots$ – JMoravitz Jun 19 '18 at 18:27
  • DuncanH, JMoravitz- thanks. I should've seen it. – Love Invariants Jun 19 '18 at 18:29
  • If you search "cos i sin proof" here, you'll find lots of proofs via lots of approaches. This answer includes a geometric visualization of $(1+i\theta/n)^n$ (which becomes $e^{i\theta}$ as $n\to\infty$); I've seen animated versions of that argument around here, but didn't find it with a cursory search. – Blue Jun 19 '18 at 18:52

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