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I saw a twitter post (but forgot to save the link) with a guy looking confused, captioned something like:
\begin{equation}\label{eqn:FunkyExponents:20}
\Abs{e^{i \pi}} = \Abs{\pi^{e i}} = \Abs{ i^{\pi e}} = 1.
\end{equation}
EDIT: this is the picture I saw posted on twitter.
If this is true, then the arguments of each of the absolutes are complex numbers on the unit circle. I suspect I’d seen that before, but forgot, so naturally, I had to verify for myself.
First, for \( \pi^{e i} \), we have
\begin{equation}\label{eqn:FunkyExponents:40}
\begin{aligned}
\pi^{e i}
&= \lr{ e^{\ln \pi} }^{e i} \\
&= \cos\lr{ e \ln \pi} + i \sin\lr{ e \ln \pi},
\end{aligned}
\end{equation}
and for \( i^{\pi e} \), we have
\begin{equation}\label{eqn:FunkyExponents:60}
\begin{aligned}
i^{\pi e}
&= \lr{ e^{i \pi/2}}^{\pi e} \\
&= e^{i e \pi^2/2}\\
&= \cos \lr{ e \pi^2/2} + i \sin \lr{ e \pi^2/2}.
\end{aligned}
\end{equation}
Sure enough, this is true. As it happens, two of these special values are nearly equal, because
\begin{equation}\label{eqn:FunkyExponents:80}
\begin{aligned}
e^{\pi} &= 23.1407 \\
\pi^e &= 22.4592,
\end{aligned}
\end{equation}
so \( e^{\pi i} \approx \pi^{e i} \). We can see this visually if we plot the three points, as done in fig. 1.
fig. 1. The three points.