show univalent $g(z)=z+c_{n+1}z^{n+1}+c_{2n+1}z^{2n+1}+c_{3n+1}z^{3n+1}+\cdots$ can be written $g(z)=\sqrt[n]{f(z^n)}$ with $f$ univalent Show that the univalent function $$g(z)=z+c_{n+1}z^{n+1}+c_{2n+1}z^{2n+1}+c_{3n+1}z^{3n+1}+\cdots$$ defined in $D$ can be written in the form $$g(z)=\sqrt[n]{f(z^n)}$$ where $f(z)$ is univalent in D and $f'(0)=1$. It is very easy to see $f(z)=z(1+c_{n+1}z^{1}+c_{2n+1}z^{2}+c_{3n+1}z^{3}+\cdots)^n$satisfies the equation(and converges in $D$), and $f'(0)=1$, but I have trouble showing that $f$is univalent. Any help would be appreciated. Thank you.

You have $g(z)=zh(z^n)$ and $f(z)=zh(z)^n$ (for the evident $h$). Now if $f(z)=f(w)$ (for some $z,w$), i.e. $zh(z)^n=wh(w)^n$ then $z^{1/n}h(z)=w^{1/n}h(z)$ for _some_ choice of $z^{1/n},w^{1/n}$, i.e. $g(z^{1/n})=g(w^{1/n})$, so $z^{1/n}=w^{1/n}$ and thus $z=w$.