My brother has came up with this equality:
$$ {e}^{i\pi} = -1 \quad\Rightarrow\quad {\left({e}^{i\pi}\right)}^{2} = 1 \quad\Rightarrow\quad \boxed{2\pi i = 0} $$
But that result sounds impossible, so, where is the error?
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1In complex $e^x=e^y$ doesn't imply $x=y$. – Nosrati Mar 11 '17 at 13:50
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1It is not true that, for $z_1,z_2 \in \mathbb{C}$, $$ e^{z_1}=e^{z_2}\implies z_1=z_2 $$ – Olivier Oloa Mar 11 '17 at 13:51
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the exponential function is $2\pi i$-periodic, it means that $$e^x=e^{x+2\pi i},\quad\forall x\in\Bbb C$$ – Masacroso Mar 11 '17 at 14:34