The equation $x^y = y^x$. The following equation has been of interest for a long time. I'm bringing this to your attention again:

$$x^y = y^x$$

How do you attack to find all real and imaginary solutions for this equation? or 

What is the nicest solution that you have seen?
The standard way I have seen
to solve $x^y = y^x$
is to define
$\frac{y}{x} = r$.

Then $y = xr$,
so
$x^{xr} = (xr)^x$
or
$x^r = xr$
or
$x^{r-1} = r$
or
$x = r^{1/(r-1)}$.

From this,
$y 
=xr
= r^{1+1/(r-1)}
= r^{r/(r-1)}
$.

Any value you put in for $r$
(except $r=1$)
gives you $x$ and $y$
that satisfy $x^y = y^x$,
and conversely.

For example,
if you start with
$r = i$,
$\frac{1}{r-1}
=\frac{1}{i-1}
=\frac{-i-1}{(-i-1)(i-1)}
=\frac{-i-1}{2}
$
and
$\frac{r}{r-1}
=\frac{i}{i-1}
=\frac{i(-i-1)}{(-i-1)(i-1)}
=\frac{1-i}{2}
$.
Then
$x = 
i^{(-i-1)/2}
$
and
$y
=i^{(1-i)/2}
$.
(I'll let you work the
standard form for $x$ and $y$.)
y=-x*W(-(ln(x)/x)/ln(x)

W(x) is the Lambert W function