0

Let $U\subset\mathbb C$, $f:U\to\mathbb C$ holomorphic and $z_0\in U$ a zero of order $k$ of the function $z\mapsto f(z)-f(z_0)$. Show that there exists a biholomorphic function $\Phi$ of an open neighborhood of $0$ into an open neighborhood of $z_0$ where $f\circ\Phi(w)=f(z_0)+w^k$.

For this problem I am literally clueless since I couldn't relate my first attempt to rewrite $f(z)-f(z_0)=h(z)^k$ with $h$ having the zero $z_0$ of order 1 to the problem itself. So far I couldn't come up with any other approach. Do you have any hints on that?

Christian Ivicevic
  • 3,123

1 Answers1

3

Hint: $h$ is injective in a neighbourhood of $V$ of $z_0$. Consider the inverse function on $h(V)$.

Details:

As you already said, $f(z_0) = 0$ with multiplicity $k$ implies that there is a neighbourhood $\tilde U \subset U$ of $z_0$ and a holomorphic function $h : \tilde U \to \Bbb C$ such that $f(z) = f(z_0) + h(z)^k$ in $\tilde U$. $h$ satisfies $h(z_0) = 0$ and $h'(z_0) \ne 0$, therefore $h$ is injective in a neighbourhood of $z_0$, i.e. $h$ maps a neighbourhood $V$ of $z_0$ bijective to a neighbourhood $W$ of $w=0$. The inverse function $\Phi = h^{-1}: W \to V$ is holomorphic and satisfies $f(\Phi(w)) = f(z_0) + w^k$ for $w \in W$.

Martin R
  • 113,040
  • Is it true that you deduce from $h(z_0)=0$ and $h'(z_0)\neq 0$ that $h$ is locally biholomorphic around $z_0$ and therefore injective? – Christian Ivicevic Jun 05 '16 at 14:09
  • @ChristianIvicevic: Yes. One possible argument uses Rouché and is sketched in this answer http://math.stackexchange.com/a/541858/42969. I will see if I can find a better reference. – Martin R Jun 05 '16 at 14:31
  • So far we haven't discuess Rouché yet - only basic properties of zeros like the ones we both mentioned and your solution is very helpful and only relies on what I may use. The only question left is how to exactly explain why $h$ maps to $W$ around $w=0$? How come that this is not another point $w$? – Christian Ivicevic Jun 05 '16 at 14:40
  • @ChristianIvicevic: Because $h(z_0) = 0$. Using Rouché or the Argument Principle one can show that each value in some neighbourhood of $w_0 = 0$ is taken exactly once in a neighbourhood of $z_0$. – Martin R Jun 05 '16 at 15:20