# Existence of zeroes of holomorphic Functions

I really am struggling with this kind of easy question, i hope you can help me out here.

Let $\Omega \subset \mathbb C$ be a domain, $f : \Omega \rightarrow \mathbb C$ a holomorphic function and $z_0 \in \Omega$ be a zero of $f$.

1) Show that if f is not the constant nullfunction there exists a $k \in \mathbb N$, so that $f^{(n)}(z_0) = 0 \quad \forall n \lt k$ and $f^{(k)}(z_0) \neq 0$

2) Furthermore show that there always exists a $\epsilon \gt 0$ so that for all $0 \lt r \lt \epsilon$ the equation$$\frac{1}{2 \pi i} \int_{\gamma_{B_r(z_0)}} \frac{f´(z)}{f(z)}$$ returns the order of zeroes of f.

• This is a little hard to follow, but you seem to be asking about the multiplicity of the root $z_0$, and the isolation of $z_0$ from other zeroes of $f$ (if any). – hardmath Jan 14 '17 at 22:10
• Use the analyticity of $f$ to write it as a taylor series. Because $f$ is not identically zero, there is at least one coefficient of the taylor series that is not zero... – asdjfksj Jan 14 '17 at 22:12

Hints: for (1), use the identity principle to prove that if all the derivatives at $z_0$ are zero, then $f=0$ (look at the Taylor expansion in some ball around $z_0$). For (2) use that the zeros of any holomorphic function are isolated: write $f(z)=(z-z_0)^{{\rm ord}(f,z_0)}g(z)$, with $g$ holomorphic and $g(z_0)\neq 0$ and then compute that integral.