Solving $z^z=z$ in Complex Numbers I wanted to find all complex numbers $z\neq0$ such that $z^z=z$. I observed that $z=\pm1$ satisfies the equation. But I had problems when tried to find all the possible solutions since $z^z$ may take more than one value. I attempted this:
$$z=r(\cos\theta+\mathrm i\sin\theta)$$
$$\therefore\log z=\log r+\mathrm i\theta$$
$$z^{z-1}=1$$
$$\therefore (z-1)\log z=2\pi n\mathrm i\qquad n\in\mathbb Z$$
$$\therefore\cases{(r\cos\theta-1)\log r-r\theta\sin\theta=0\\\theta(r\cos\theta-1)+r\log r\sin\theta=2\pi n}$$
But I couldn't go further. I'm also aware that adding an integer multiple of $2\pi$ to $\theta$ may give another possible value for $\log z$. Can anyone help please?
 A: The more I think about this question, the more I like it. The key to it is to have a precise idea of what we’re talking about.
We need an unambiguous definition of the natural logarithm, $\log$. It can be defined as a single-valued function only on a simply-connected domain in $\Bbb C$ that omits the origin. Since we know all the positive real solutions of our equation, namely $z=1$, we might as well omit from the plane the whole nonnegative real axis. Then we may specify that $0<\Im(\log z)<2\pi$, so that this logarithm maps onto the open strip between the real axis and a line parallel to it and $2\pi$ units above.
Now, to the equation $z^z=z$ we apply log and get $z\log z=\log z+2k\pi i$, and so $(z-1)\log z=2k\pi i$. This is really infinitely many equations, one for each integer $k$. The value $k=0$ gives us our known value $z=1$, and if you try it for $k=-1$, you can check that since $\log(-1)=\pi i$, there’s your other known solution. It would be fun to see whether there are other solutions for this value of $k$, but I’m going to bet that each other value of $k$ leads to at least one solution.
I’m posting this incomplete answer, and will look for a value with $k=-2$.
A: We can write
$$
z = z(r,\theta),
$$
so we actually have
$$
z(r,\theta)^{ z(r,\theta) } = z(r,\theta).
$$
Note that
$$
z(r,\theta) = e^{\ln(r) + \mathbf{i} \theta}
= r \cos(\theta) + \mathbf{i} r \sin(\theta),
$$
so we obtain
$$
e^{ \big[ \ln(r) + \mathbf{i} \theta \big]
\big[ r \cos(\theta) + \mathbf{i} r \sin(\theta) \big] }
= e^{\ln(r) + \mathbf{i} \theta},
$$
therefore
$$
\big[ \ln(r) + \mathbf{i} \theta \big]
\big[ r \cos(\theta) + \mathbf{i} r \sin(\theta) \big]
= \ln(r) + \mathbf{i} \big\{ \theta + 2 n \pi \big\}.
$$
So we obtain
$$
\Big[ r \ln(r) \cos(\theta) - r \theta \sin(\theta) - \ln(r) \Big]
+ \mathbf{i} \Big[ r \ln(r) \sin(\theta) + r \theta \cos(\theta)
- \theta - 2 n \pi \Big] = 0.
$$

(added to show step)

We get
$$
\left[
\begin{array}{rcl}
r \ln(r) \cos(\theta) - r \theta \sin(\theta) &=& \ln(r)\\
r \ln(r) \sin(\theta) + r \theta \cos(\theta) &=& \theta + 2 n \pi
\end{array}
\right.
$$
So
$$
\Big[ r \ln(r) \cos(\theta) - r \theta \sin(\theta) \Big]^2
+ \Big[ r \ln(r) \sin(\theta) + r \theta \cos(\theta) \Big]^2
= \ln^2(r) + \big( \theta + 2 n \pi \big)^2
$$
Thus
$$
r^2 \ln^2(r) + r^2 \theta^2 = \ln^2(r) + \big( \theta + 2 n \pi \big)^2
$$

So we obtain

$$
r^2 \ln^2(r) + r^2 \theta^2 = \ln^2(r) + \big( \theta + 2 n \pi \big)^2,
$$
or
$$
\big[ r^2 - 1 \big] \big[ \ln^2(r) + \theta^2 \big]
= \big( \theta + 2 n \pi \big)^2 - \theta^2.
$$
The case $r=1$
We obtain
$$
\big( \theta + 2 n \pi \big)^2 - \theta^2 = 0,
$$
whence
$$
\theta = - n \pi,
$$
Thus
$$
z = \pm 1
$$
The case $r \ne 1$
We obtain
$$
\big[ r^2 - 1 \big] \big[ \ln^2(r) + \theta^2 \big]
= 4 n \pi \Big( \theta + n \pi \Big).
$$
But as
$$
z(r,\theta) = z(r,\theta + 2 k \pi),
$$
so the right part can be positive or negative, while the left part does not change sign.
There are no solutions for the case $r \ne 1$.

So
$$
z^z=z \Rightarrow z = \pm 1,
$$
as the only solutions.
