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Simple Holomorphic Functions Question

  1. Jan 29, 2012 #1
    I have the following statement:
    Let [itex]A\subseteq \mathbb{C}[/itex] be open and let [itex]f\colon A \to \mathbb{C}[/itex] be holomorphicic (in [itex]A[/itex]). Suppose that [itex]D(z_0,R)\subseteq A.[/itex] Then
    [itex]f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k\ \forall z\in D(z_0,R),[/itex] where
    [itex]a_k=\displaystyle\frac{1}{2\pi i}\int_{|\zeta-z_0|=r}\frac{f(\zeta)}{(\zeta-z_0)^{k+1}}d\zeta[/itex] and [itex]0<r<R[/itex].

    My problem is that it now seems, plugging in [itex]z_0[/itex], that [itex]f(z_0)=0[/itex] and since this can be done for all [itex]z_0\in A[/itex], we have [itex]f(z_0)=0[/itex] for all [itex]z_0 \in A,[/itex] 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.
     
  2. jcsd
  3. Jan 29, 2012 #2

    Dick

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    The summation starts from k=0. So [itex]f(z_0)=a_0 (z-z_0)^0=a_0[/itex].
     
  4. Jan 29, 2012 #3
    [itex]0^0=1[/itex]. Of course. Thanks.
     
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