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I read through and pretty much understand most of Goursat's Theorem in $\textit{Complex Analysis}$ by Gamelin. The theorem states that if $f(z)$ is a complex-valued function on a domain $D$ such that $f'(z_0)$ exists at each point $z_0 \in D$, then $f(z)$ is analytic on $D$. Here the requirement that $f'(z)$ be continuous is redundant. This section only had one problem on p. 124: Find an application for Goursat's Theorem in which it is not patently clear by other means that the function in question is analytic. Seems kind of vague to me, and I've been struggling with this for quite a bit. I would appreciate some input on how we should proceed.

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Maybe something like: Let $f(z)$ be a solution of some (complex) differential equation $y'=g(z,y)$. $f'(z)$ exists since it's a solution. Thus it's analytic. So solutions are automatically analytic. –  Bill Cook Oct 23 '11 at 1:28
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How about the local inverse theorem ? take a holomorphic map with no explicit inverse such as exp(z) + z + a ; by the holomorphic version of the local inverse theorem it admits a local inverse near z=1 for instance. that inverse is then analytic.

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