Euler gave the most concise (I think so) connection between $e,\,i$ and $\pi$ (called Euler's identity), which states: $$e^{\pi i}+1=0$$ I would like to see some proofs of this. You can mention (I mean, give link to the webpage or article) as many as proof as you can. What I want to ask is, give alternatives to Euler's proof (which used Taylor series).
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Check this: https://www.youtube.com/watch?v=v0YEaeIClKY – irene dovichi Oct 17 '20 at 08:29
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@irenedovichi: It's my favorite mathematics youtube channel. It was a nice video. – Oct 17 '20 at 08:30
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Do these answer your question: 1 2 3 4 – Anindya Prithvi Oct 17 '20 at 09:04
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We start by defining $f=e^z$ as a function that obeys $f'= f$, and its inverse function $f^{-1}=\ln$ such that $z=\ln f$. We find $1=z'=f'(\ln f)'=f(\ln f)' $ and deduce that $\ln f=\int^fw^{-1}dw$ which we can also write as $$ \ln z=\int_1^z\frac{dw}{w} $$ where we have chosen an arbitrary constant of integration as the value of the function at $z= 1 $ in our definition (we could of course start directly from this integral). We can now carry out an integration along the unit circle where $w=\cos\theta +i\sin\theta$ and $dw=(-\sin\theta +i\cos\theta)d\theta$, to get: $$ \ln(-1)=\int_{\arg 1}^{\arg(-1)}\frac{-\sin\theta +i\cos\theta}{\cos\theta +i\sin\theta}d\theta=\int^{\pi}_{0}id\theta=i\pi $$ Thus $$ e^{i\pi}=-1 $$
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