Simple Holomorphic Functions Question

[gauss mouse]

I have the following statement:
Let A\subseteq \mathbb{C} be open and let f\colon A \to \mathbb{C} be holomorphicic (in A). Suppose that D(z_0,R)\subseteq A. Then
f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k\ \forall z\in D(z_0,R), where
a_k=\displaystyle\frac{1}{2\pi i}\int_{|\zeta-z_0|=r}\frac{f(\zeta)}{(\zeta-z_0)^{k+1}}d\zeta and 0<r<R.

My problem is that it now seems, plugging in z_0, that f(z_0)=0 and since this can be done for all z_0\in A, we have f(z_0)=0 for all z_0 \in A, which is absurd. Can anybody tell me what's going on here?

Sorry I haven't formatted this in the usual coursework question way but I don't think it would suit it.

 

[Dick]

I have the following statement:
Let A\subseteq \mathbb{C} be open and let f\colon A \to \mathbb{C} be holomorphicic (in A). Suppose that D(z_0,R)\subseteq A. Then
f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k\ \forall z\in D(z_0,R), where
a_k=\displaystyle\frac{1}{2\pi i}\int_{|\zeta-z_0|=r}\frac{f(\zeta)}{(\zeta-z_0)^{k+1}}d\zeta and 0<r<R.

My problem is that it now seems, plugging in z_0, that f(z_0)=0 and since this can be done for all z_0\in A, we have f(z_0)=0 for all z_0 \in A, which is absurd. Can anybody tell me what's going on here?

Sorry I haven't formatted this in the usual coursework question way but I don't think it would suit it.

The summation starts from k=0. So f(z_0)=a_0 (z-z_0)^0=a_0.

 

[gauss mouse]

0^0=1. Of course. Thanks.