$m \in (0,1]$
$e^{-im\pi}=(e^{i(-\pi)})^m=(\cos(-\pi)+i\sin(-\pi))^m=(-1)^m$
Can someone confirm if this is correct?
I noticed it can be $(-1)^{-m}$ which was a bit weird.
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1Same thing, but $(-1)^m$ looks nicer. – André Nicolas Mar 15 '16 at 00:34
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@AndréNicolas it is also a bit weird how $(-1)^m \neq -1$.... – snowman Mar 15 '16 at 00:35
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1Why is that wierd ? . $(-1)^2 = 1$ – Shailesh Mar 15 '16 at 00:35
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@Shailesh but m is not above 1. but I think I see how it is not equal to -1 now because sometimes it is not defined, for example when m=1/2. – snowman Mar 15 '16 at 00:37
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I'm not so sure that $(-1)^m$ is a good notation when $m$ is not an integer. See here for a discussion: http://math.stackexchange.com/a/317546/169852 – Mar 15 '16 at 00:45
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No, try $m=\frac12$. We have $e^{-i\pi/2}=-i$, but taking the principal branch of the complex exponential, $\left(e^{-i\pi}\right)^{1/2}=(-1)^{1/2}=i$.
Chris Culter
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You could fix the equation to read either $e^{im\pi}=(-1)^m$ or $e^{-im\pi}=(-1)^{-m}=1/(-1)^m$. – Chris Culter Mar 15 '16 at 00:44