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Suppose $f(z)$ is analytic on $A=\{z: 0<|z|<\delta\}$ with an essential singularity at 0 such that its Laurent expansion is

$$ \sum_{n=-\infty}^\infty a_nz^n $$ Ahlfors in his book "Complex Analysis" seems to say that $f(z)-\frac{a_{-1}}{z}$ need not be the derivative of an analytic function on $A$.

Does anyone know an example? If such example exists, I suppose that $$ \int_C f(z)\,dz $$ need not be equal to $2\pi i a_{-1}.$ Am I right?

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up vote 1 down vote accepted

I couldn't find any such claim in Ahlfors when I browsed through my edition.

Anyway, a convergent Laurent series converges uniformly on any compact subannulus, in your example on any $r \le |r| \le R$ if $r > 0$ and $R < \delta$. Hence the series can be integrated termwise and if $C$ is a closed simple curve within the annulus of convergence, then $$\int f(z)\,dz = 2\pi i\,a_{-1}.$$

Furthermore, if $g(z) = f(z) - a_{-1}z^{-1}$, then $$G(z) = \sum_{n \neq -1} \frac{a_n z^{n+1}}{n+1}$$ is a Laurent series with the same annulus of convergence as $g$ and (at least) $f$, and $G' = g$.

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You seem to be right. Suprisingly most books avoid this case: Rudin, Stein-Sharkachi, Ahlfors. That is why I was in doubt. –  TCL Nov 7 '12 at 14:52
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