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I have this question from ahlfors page149:" there is no longer any need to give a separate proof of Theorem 15 in the presence of removable singularities. "

Theorem 15: If $f(z)$ is analytic in $\Omega$, then $$\int_\gamma f(z)dz = 0$$ for every cycle $\gamma$ which is homologous to zero in $\Omega$.

If there is removable singularities in $\Omega$. I wonder how to prove the theorem rigorously.

Here is my attempt.

we only need to prove it when there is one removable singularity.
Suppose $a\in \Omega $ is a removable singularity, which means $$\lim_{z\rightarrow a}(z-a)f(z)=0$$ then $f(z)$ is analytic in the region $\Omega^{'}=\Omega -\{a\}$, then there exists a analytic function $g(z)$ in $\Omega$ which coincides with $f(z)$ in $\Omega'$.

Then apply theorem 15 to $g(z)$, the integral can be written as $$\int_\gamma f(z)dz =\int_\gamma g(z)dz = 0 $$

Is this proof valid? Or how to give a rigorous proof?

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  • $\begingroup$ Where did you explicitly use the fact that $a$ was removable? In concluding that such a $g$ exists? $\endgroup$ Commented May 5, 2023 at 14:03
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    $\begingroup$ Yes, if $a$ is removable, then we can define the value $f(a)= \frac{1}{2\pi i}\int_{C}\frac{f(z)dz}{z-a}$,where $C$ refers to a circle about $a$ and its inside are contained in$ \Omega$ @DerekAllums $\endgroup$
    – F.Liu
    Commented May 5, 2023 at 14:11
  • $\begingroup$ Did you mean Cauchy's theorem? $\endgroup$ Commented May 5, 2023 at 14:18

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The proof looks correct. It would be worth to simply describe it as follows: "After we remove the singularity, $f$ becomes analytic, so theorem 15 holds for $f$."

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  • $\begingroup$ Thanks for the advice! $\endgroup$
    – F.Liu
    Commented May 5, 2023 at 14:14

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