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can someone say what wrong with the bellow stament?

$$ e^{\imath\pi}+1=0\\ e^{\imath\pi}=-1\\ (e^{\imath\pi})^2=(-1)^2\\ e^{2\imath\pi}=1\\ (e^{2\imath\pi})^\imath=1^\imath\\ e^{2\imath^2\pi}=1^\imath\\ e^{-2\pi}=1^\imath $$

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    $\begingroup$ Taking $\sqrt[i]{\dots}$ on each side of the one-before-last equation. You do this after raising each side to the power of $i$. This is the same as $-1=\sqrt[2]{(-1)^2}=\sqrt[2]{1}=1$. $\endgroup$ – barak manos Nov 30 '16 at 15:40
  • $\begingroup$ @Bacon: No, I misinterpreted that stage. OP doesn't take $\sqrt[i]{\dots}$ on each side, but merely replace $i^2$ with $-1$ on the LHS. $\endgroup$ – barak manos Nov 30 '16 at 15:45
  • $\begingroup$ @barakmanos Oh yes, duly noted and I interpreted it quickly, that way too! Deleted my comment, thanks for commenting. $\endgroup$ – Kevin Nov 30 '16 at 15:53
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Nothing is wrong yet. You didn't write $1^i=1$ at the end of the proof, you left it for the reader to assume. In the complex plane, $a^b=\exp(b \log a)$ and the logarithm is multivalued unless you are careful with branch cuts. We have $\log 1=0+2k\pi i$ for $k \in \Bbb Z$. If we take $k=-1, 1^i=\exp(-2\pi)$ and there is no problem.

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