A property of Analytic functions How can I find ALL analytic functions $f : \Bbb{C}\longrightarrow \Bbb{C}$ such that for any $z$, $f(f(z)) = z^4$ (or generally $f(f(z)) = z^r$ for some fixed $r$) ?
 A: You are looking for an entire function with
$$f(f(z)) = z^r$$
for all $z$ and a given $r \in \mathbb{N}$.
First let's treat the case $r = 0$. If $f$ were not constant, $f(\mathbb{C})$ would be a nonempty open set on which $f \equiv 1$, which implies $f \equiv 1$ on all of $\mathbb{C}$. So $f$ must be constant, and $f(z) = z^0$.
Now, for $r > 0$, we obviously must have $f$ non-constant, so there are two possibilities for the behaviour at $\infty$,


*

*$\infty$ is a pole, i.e. $f$ is a polynomial of degree $k \geqslant 1$. Then a simple computation shows that $r = k^2$, and $f(z) = \alpha\cdot z^k$ where $\alpha^{k+1} = 1$. (If the lowest-degree non-zero term of $f$ has degree $m$, $f \circ f$ has a non-zero term of degree $m^2$, hence $f$ must be a homogeneous polynomial. If $f(z) = \alpha z^k$, then $f(f(z)) = \alpha(\alpha z^k)^k = \alpha^{k+1} z^{k^2}$.)

*$\infty$ is an essential singularity. But then $f(\mathbb{C}\setminus D_R)$ omits at most one point of $\mathbb{C}$ for every $R > 0$, hence $f(\mathbb{C}\setminus D_R) \supset \mathbb{C}\setminus D_R$ for all large enough $R$, and $f(f(\mathbb{C}\setminus D_R))$ omits at most one point, hence $f\circ f$ cannot have a pole in $\infty$.
So: There are only solutions for square $r = k^2$, and then all solutions are
$$f_n(z) = e^{2\pi i n/{(k+1)}} z^k.$$
A: $$f'(f(z))f'(z)=rz^{r-1}$$
Then $f'$ only vanishes at $z=0$. Then the function is $z^ae^{g(z)}$ for $g$ analytic.
Hence $$f(f(z))=z^{a^2}e^{ag(z)}e^{g(z^ae^{g(z)})}.$$
Then $a^2=r$. Therefore, if $r$ is not a square the function is not going to be analytic at the origin. Assume $r$ is a square. Then $e^{ag(z)+g(z^ae^{g(z)})}=1$. So, $ag(z)+g(z^ae^{g(z)})=2k\pi i$.
From this we get that $g'(z)+g'(z^ae^{g(z)})\left(az^{a-1}e^{g(z)}+z^ae^{g(z)}g'(z)\right)=0$
From this we get that $g'(0)=0$. I am being lazy to continue typing derivatives. But I think that if you keep differentiating and using that the previous derivatives at $z=0$ are zero, you get that all derivatives of $g$ are zero at $z=0$. So $g$ is constant.
Therefore the function is $f(z)=Cz^a$, from where $z^r=C^{a+1}z^{a^2}$. So, $C$ is an $a+1$ root of unity.
