Title: Why is log(i^i) + pi/2 = 0?
Date: 2025-07-11 15:45

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I was watching [Lennart
Augustsson](https://en.wikipedia.org/wiki/Lennart_Augustsson)'s [MicroHs, a
tiny Haskell Compiler](https://www.youtube.com/watch?v=SJwvPEq4Mok) talk,
and saw that he mentioned a pretty neat equality: $$log(i^i) + \frac{\pi}{2} = 0$$,
and punctued it with a casual "it's obvious if you think about it." And even
though I dislike trigonometry and complex numbers, it's indeed pretty trivial,
thanks to [Euler's identity](https://en.wikipedia.org/wiki/Euler%27s_identity),
as usual when dealing with exponentials and complex numbers:

%%
\begin{aligned}
e^{i\pi} = -1 &\iff e^{i\pi} = i^2\\\\
&\iff \log{e^{i\pi}} = \log{i^2}\\\\
&\iff i\pi = 2 \log{i}\\\\
&\iff \pi = 2\frac{\log{i}}{i}\\\\
&\iff \frac{\pi}{2} = \frac{\log{i}}{i}\\\\
&\iff \frac{\pi}{2} = -\log{i^i}\\\\
&\iff log(i^i) + \frac{\pi}{2} = 0\\\\ ■
\end{aligned}
%%

Nothing too complicated indeed, and you can use this to get the value of
$$i^i$$, which is $$e^{-\frac{\pi}{2}}$$.