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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?

PJ Miller
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2 Answers2

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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 $g(z)=z{\sf exp}\bigg({\frac{\tilde{g}(z)}{n}}\bigg)$.

Ewan Delanoy
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  • Why is there $r>0$ such that $\tilde{h}$ is nonzero on the disk $D(r)$? I don't think we know that $\tilde{h}$ is analytic yet... – PJ Miller Oct 23 '13 at 00:09
  • @PJMiller 1. Of course $\tilde{h}$ is analytic on a neighborhood of $0$ since $h$ is and $\tilde{h}(z)=\frac{h(z)}{z^n}$. 2. For the nonzero point, we do not need analyticity, just continuity ; if $\varepsilon=|h(0)|$ there will be a $r > 0$ such that $|h| > \frac{\varepsilon}{2}$ on $D(r)$. – Ewan Delanoy Oct 23 '13 at 05:32
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Since $f$ is holomorphic on the disc $U:=\{z\in\mathbb C:\left|z\right|<R\}$, there exists a sequence $(a_k)$ of complex numbers such that : $$\forall z\in U,\ f(z)=\sum_{k=0}^{+\infty} a_k z^k$$

Let $h:z\mapsto f(z^n)-f(0)$. We have $h(0)=0$. You can write :

$$\forall z\in U,\ h(z)=\sum_{k=1}^{+\infty}a_kz^{nk}=z^n \tilde{h}(z)$$

Where $\tilde{h}:z\mapsto\displaystyle\sum_{k=0}^{+\infty}a_{k+1}z^{nk}$ is holomorphic on $U$ and $\tilde{h}(0)=a_1=f'(0)\neq 0$.

Hence $0$ is a zero of $h$ of order $n$.

Philippe Malot
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