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I have these two mathematical statements: 1) $e^{i\pi}=-1$ and 2) $\ln(-1)=i\pi$. Now I want a proof of these statements. Can anyone help me proving these statements?

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    $\begingroup$ Are you familiar with Euler's identity? How have you defined the functions involved, and what have you tried? $\endgroup$ – user61527 Nov 23 '13 at 21:26
  • $\begingroup$ Do you know the definitions of the functions involved? $\endgroup$ – Git Gud Nov 23 '13 at 21:28
  • $\begingroup$ You can prove $e^{ix}=\cos(x)+i\sin(x)$ by deriving both sides, then put $x=\pi$ $\endgroup$ – LeeNeverGup Nov 23 '13 at 21:28
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    $\begingroup$ It depends on how $e^z$ was defined to you. $\endgroup$ – Emanuele Paolini Nov 23 '13 at 21:28
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    $\begingroup$ @NorbertWillhelm, it's time for you (and us all) to learn the lesson: WA is wrong lots of times in lots of different subjects. It is, of course, a great site where one can verify what was already done, but to base one's answer on it...tsk,tsk,tsk. Fly by Night is right, and the reason is a little deep if you haven't yet studied complex analysis. $\endgroup$ – DonAntonio Nov 24 '13 at 12:20
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That's straightforward $$e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1$$

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    $\begingroup$ You can put MathJaX code between $$. $\endgroup$ – Ian Mateus Nov 23 '13 at 21:35
  • $\begingroup$ @IanMateus thank you :) I didn't realize that... this is my first answer here ... :D $\endgroup$ – Jekyll Nov 23 '13 at 21:42
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If you take the definition (high school, usually)

$$\text{For}\;\;x\in\Bbb R\;,\;\;e^{ix}:=\cos x+i\sin x\implies e^{\pi i}=\cos\pi+i\sin\pi=-1$$

and now, choosing the branch $\;\{z\in\Bbb C\;;\;\text{Im}\,z\ge 0\}\;$ for the logarithm (in this case, it means that $\;\arg(-1)=\pi\;$ ) , we get

$$\text{Log}\,(-1):=\log|-1|+i\arg(-1)=0+\pi i$$

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