There is an entire function $g$ such that $f(z)=g\left(z^{n}\right)$. Let $f$ be an entire function and $\xi=e^{\frac{2\pi i}{n}}$ for some $n\in \mathbb{N}$. Suppose that $f\left(\xi z\right)=f(z)$ for all $z\in \mathbb{C}$. Show that  there is a entire function $g$ such that $f(z)=g\left(z^{n}\right)$ for all $z \in \mathbb{C}$.
Remark: I have tried to express $g$ in terms of $f\left(\xi z\right)$ but my attempts have been unsuccessful. 
 A: Using power series is cheating because it makes it too easy. Say
$$f(z)=\sum_{k=1}^\infty a_k z^k.$$
Now $f(\xi z)=f(z)$, with uniqueness of the power series coefficients, shows that $$\xi^ka_k=a_k$$for all $k$. If $k$ is not a multiple of $n$ then $\xi^k\ne1$, so $a_k=0$. Hence $$f(z)=\sum_{j=0}^\infty a_{jn}z^{jn}=g(z^n)$$if $$g(z)=\sum_{j=0}^\infty a_{jn}z^j.$$
A: I should look back at the original problem more. Since $f(\xi z) = f(z)$, it holds that $f(\xi^k \sqrt[n]z) = f(\sqrt[n]z)$, and hence the formula I gave in the comment for $g(z)$ reduces to just
$$g(z) = f(\sqrt[n]z)$$
You need the invariance condition $f(\xi z) = f(z)$ to show that this definition is independent of which branch of the n-th root you use. As already mentioned, for $z \ne 0$, $g$ is holomorphic at $z$ because $f$ and $\sqrt[n]z$ both are, for some branch of n-th root. At $0$, you will need to pull out the Cauchy-Riemann conditions.
A: For any $a\neq 0$, we can define a local branch of $\log$ in a small ball $U$ around $a$ and define $$g(z)=f\left(e^{\frac{\log z}{n}}\right)$$ for $z\in U$.  Furthermore, the function $g$ is actually independent of the branch of $\log z$ chosen, since any different branch just multiplies $e^{\frac{\log z}{n}}$ by some power of $\xi$.  Thus these local definitions of $g$ actually give a single well-defined function on all of $\mathbb{C}\setminus\{0\}$.  Since any of the local definitions are clearly holomorphic, $g$ is holomorphic on all of $\mathbb{C}\setminus\{0\}$.  It is also clear that $g(z^n)=f(z)$ for any $z\neq 0$.  Finally, the singularity at $0$ is removable, since $g$ is bounded in any ball around $0$ (if $0<|z|<r$, then $|g(z)|\leq\sup \{|f(w)|:|w|\leq \sqrt[n]{r}\}$).  Thus $g$ extends to a holomorphic function on all of $\mathbb{C}$, and the identity $g(z^n)=f(z)$ also holds at $z=0$ by continuity.
