TITLE: Does $z^i=i^z$ have any solutions, beside $z=i$?
QUESTION [12 upvotes]: Does this equation have any solutions: 

$$\large{z^i=i^z}$$ 

Putting polar form of $z$ is better for LHS, But rectangular form is suitable for RHS !
What to do? Thanks!

REPLY [15 votes]: $$
\frac1z\log\left(\frac1z\right)=\frac1i\log\left(\frac1i\right)=-\frac\pi2
$$
Thus, there are many solutions, one for each branch of the Lambert W function:
$$
z=e^{-\mathrm{W}\left(-\pi/2\right)}
$$
One solution is Exp[-LambertW[0,-Pi/2]] which is -i.
Another is Exp[-LambertW[-1,-Pi/2]] which is i.
However, another is N[Exp[-LambertW[1,-Pi/2]],20] which is
1.0213233161306520062 - 4.8683538060775645979 i
Of course, the value of $z^i$ depends on which branch of $\log(z)$ is used. In the computation above, I have taken $\log(i)=\pi i/2$, so we have a clear definition of
$$
i^z=e^{\pi iz/2}
$$
However, $\log(z)$ is determined only mod $2\pi i$ and that affects $z^i$ by a factor of $e^{2\pi k}$ for $k\in\mathbb{Z}$.
For example, for $z=1.0213233161306520062-4.8683538060775645979\,i$, we use $\log(z)=1.60429091344801115852-7.64719227612459292313\,i$.