TITLE: Holomorphic in small disk with polynomial power
QUESTION [7 upvotes]: Let $f(z)$ be holomorphic in $|z|<R$, $f'(z)\neq 0$, and $n>0$ is an integer. Show that there exists $r>0$ and $g(z)$ holomorphic in $|z|<r$ such that $f(z^n)=f(0)+g(z)^n$.

The local mapping theorem says that if $f(z)-f(0)$ has a zero of order $n$ at $0$, then we can choose $r$ so that there exists $\epsilon$ such that for all $a$ with $|a-f(0)|<\epsilon$, the equation $f(z)=a$ has exactly $n$ roots in the disk $|z|<r$.
Here, $f(z^n)-f(0)$ has a zero of order at least $1$, since $f(0)-f(0)=0$, but do we know exactly what order it has?

REPLY [4 votes]: This is a complement to Philippe Malot’s answer.
There is a $r >0$ such that $\tilde{h}$ is nonzero on the disk 
$D(r)=\lbrace z | |z| \leq r\rbrace$.
As Wikipedia explains here, we can then construct a holomorphic logarithm $\tilde{g}$ of $\tilde{h}$ on $D(r)$, so that ${\sf exp}({\tilde{g}})=\tilde{h}$ on $D(r)$.
You may then take $g(z)=z{\sf exp}\bigg({\frac{\tilde{g}(z)}{n}}\bigg)$.