Find all functions $f:\mathbb{C}\to\mathbb{C}$ satisfying $f(\overline{f(z)}) = \overline{z}$. I have the above question and am stuck with it. It seems very general. So far, I've found the solutions $f(z) = z$ and $f(z) = \overline{z}$. But that really was a lucky guess.
My conjecture is that any solution must be a planar isometry, but I can't find a proof nor a counterexample. So far, what I've tried is to write $f(z)$ as $g(z) + z$, but this makes the resulting expression for $g$ more complicated.
I don't need a full answer, but it would be very helpful to have hints suggesting a way forward. Thanks in advance!
 A: You're conjecture is definitely not true as stands. You can produce an uncountable number of such functions as follows:
Let $\psi : \mathbb{H} \to \mathbb{H}$ be any permutation of the upper-half plane. Define $f(z) = z$ for $z \in \mathbb{R}$ and define $f \mid_{\mathbb{H}} = \psi$. Then when $\textrm{Im}(z) < 0$ we let $f(z) = \overline{\psi^{-1}(\overline{z})}$.
A: I'm assuming you mean for $f$ to be analytic? Assuming that you did: observe that if $g(z) = \overline{f(\bar{z})}$, then $g$ is the analytic inverse function of $f$. It follows that $f$ is an entire function which is also a bijection from $\Bbb{C} \to \Bbb{C}$. It is a well known fact that such an $f$ must be a linear polynomial. Assuming $f(z) = az +b$, we compute that $g(z) = \bar{a}z + \bar{b}=  f^{-1}(z) = \frac{1}{a}(z - b)$. Hence $\bar{a} = 1/a$-- in other words, $|a| = 1$-- and $\bar{b} = -b/a$. In other words, $-arg(b) = arg(b) + \pi - arg(a)$: $2arg(b) = arg(a) - \pi$. We conclude that $f$ must be of the form $az + b$ where $|a| = 1$ and $b$ lies on the ray from the origin defined by $arg(b) = \frac{1}{2} (arg(a) - \pi)$. 
