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I was messing around with Euler's formula by changing up the values $e$ and $\pi$, and found the following trend:

$$a^{i\frac{\Large \pi}{\Large \log(a)}} = -1$$

It seems to work for all positive real numbers. Is this true? I haven't learned about the identity in school so I apologize if this is trivial or common sense. Thanks!

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    $\begingroup$ The case $a=1$ is problematic. $\endgroup$ – Michael Hardy Jan 1 '17 at 21:19
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By definition, $a^z=e^{z\log a}$. So, for any $\lambda$, $$ a^{i\lambda}=e^{\lambda\log a}. $$ To get $-1$, we need $\lambda\log a=i(2k+1)\pi$ for $k\in \mathbb N$. In particular, for $k=0$, we get $\lambda=i\pi/\log a$. So, as you say, $$ a^{\dfrac{i\pi}{\log a}}=-1. $$

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  • $\begingroup$ If $a=1$, we should probably avoid the principal branch. $\endgroup$ – Simply Beautiful Art Jan 1 '17 at 21:23
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    $\begingroup$ Good point. Kind of overkill, though :) $\endgroup$ – Martin Argerami Jan 1 '17 at 21:35
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Hint:

For $a>0$ and $a\ne 1$ we have $a^{1/\log a}=e$

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