# Rigorous proof of cauchy's theorem in the presence of removable singularities

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?

• Where did you explicitly use the fact that $a$ was removable? In concluding that such a $g$ exists? Commented May 5, 2023 at 14:03
• 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 Commented May 5, 2023 at 14:11
• Did you mean Cauchy's theorem? Commented May 5, 2023 at 14:18

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 $$f$$."