Since $e^{i2\pi a}=(e^{i2\pi})^a=(1)^a=1$ but $e^{i2\pi a}=\cos(2\pi a)+i\sin(2\pi a)\neq1$, why is that?
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gobears21
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Please use Approach0 to search if your question exists here before asking. – Toby Mak Feb 12 '22 at 07:40
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2What makes you think that complex exponentiation satisfies $(z^a)^b= z^{ab}$? – J. De Ro Feb 12 '22 at 07:41
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2$\sin (2\pi a)=0$ and $\cos (2\pi a)=1$, there is nothing wrong. – sirous Feb 12 '22 at 07:52
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@sirous But the OP says that it should work for any $a$... – Jean Marie Feb 12 '22 at 08:18
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This question and its answers are more precise: https://math.stackexchange.com/q/2582046/305862 – Jean Marie Feb 12 '22 at 08:21