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If $f(z)$ is analytic in $|z|<1$ and $f'(0)\not =0$ prove that there exists an analytic function $g(z)$ such that $f(z^n)=f(0)+(g(z))^n$ in the nbd. of origin.

Since $f$ is analytic so Taylor's series expansion of $f$ about $z=0$ is $\displaystyle f(z)=\sum_{k=0}^{\infty}a_kz^k$. Also , $f(0)=a_0$. Then $\displaystyle f(z^n)=f(0)+\sum_{k=1}^{\infty}a_kz^{nk}=f(0)+z^nh(z^n)$ , where $h$ is analytic. But from here how I can prove the required result ?

dudas
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  • You have to find $g$ inductively. Let $b_1$ be such that $b_1^n=a_1$. Then define $b_2,b_3,\dots$ inductively so that $g(z)=\sum b_iz^i$. You will need to assume $a_1=f'(0)\neq 0$, presumably. – Thomas Andrews Nov 30 '15 at 15:53
  • And you've misused $g$ in your last paragraph. That is not your $g$, that is some other function. $g(z)$ will be an $n$th root of $f(z^n)-f(0)$ – Thomas Andrews Nov 30 '15 at 15:56
  • Important question: is $n$ fixed? Do we need to prove that there exists $g(z)$ that works for every $n$ or rather that for every $n$ we can find a suitable $g(z)$? – bartgol Nov 30 '15 at 16:18

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The function $f$ can be written in the form $$f(z)=f(0)+z h(z)$$ with $h$ analytic in a neighborhood of $0$, and $h(0)=f'(0)=:c\ne0$. Therefore the function $h$ has an analytic $n$-th root near $0$: $$h(z)=\bigl(p(z)\bigr)^n$$ for some function $p$ which is analytic in a suitable neighborhood $U$ of $0$. We therefore can write $f(z)=f(0)+ z\bigl(p(z)\bigr)^n$, from which we obtain $$f(z^n)=f(0)+ z^n\bigl(p(z^n)\bigr)^n\qquad(z\in U)\ .$$ This shows that $$g(z):=z\>p(z^n)$$ satisfies the claim.

Christian Blatter
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