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I'm working on this problem

If $f(z)$ is analytic in $|z|<1$ and $f′(0)\neq 0$ prove that there exists an analytic function $g(z)$ such$f(z^n)=f(0)+g(z)^n$ in a neighborhood of $0$.

There is a solution in this thread If $f$ is analytic in $|z|<1$ then prove that $f(z^n)=f(0)+(g(z))^n$. I don't understand one detail.

"There exists an analytic function $h$ s.t. $f(z)=f(0)+zh(z)$ in a neighborhood of $0$ and $h(0)=f'(0)\neq 0$. Then $h$ has an analytic $n$-th root near $0$, i.e. $h(z)=p(z)^n$."

How can we define the analytic $n$-th root? I mean an $n$-th root can only be defined on the region that does not intersect with the negative real axis, right? In this case we are not sure about that, like what if $h(z)=-1$?

T C
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  • It's the difference between $i^2=-1$ and $i=\sqrt{-1}$. You should be able to write out a power series for $p$ with coefficients to be determined, raise that series to the power $n$ and then equate coefficients with the power series for $h$. –  May 26 '20 at 01:14
  • so what you're saying is, in any neighborhood of $0$, no matter what the image through $h$ is, we can always define its $n$-th root? – T C May 26 '20 at 01:20
  • Check out this answer https://math.stackexchange.com/a/2206865/180579 for the $n=2$ case. –  May 26 '20 at 01:28
  • Since $h$ converges in neighborhood of 0, so must $p$. –  May 26 '20 at 01:48
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    if $h(0) \ne 0$ there is a small neighborhood of zero where $h(w) \ne 0$ this means (a branch of) $\log h$ is well defined and analytic in that neighborhood - take any primitive of $h'/h$) , hence any root of any order of $h$ is well defined (any complex power of $h$ for that matter $h^s=e^{s\log h}$) – Conrad May 26 '20 at 01:50

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