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?