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?
[Math] Solving $z^z=z$ in Complex Numbers
complex numbersexponentiation
Best Answer
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 $k=-2$.