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.