# Proof of Cauchy integral theorem on any simply connected set

I am trying to find a reference for a proof of Cauchy integral theorem, ie the fact that given a simply connected open subset $$U$$ of the complex plane, a rectifiable loop $$\gamma$$ contained in $$U$$ and a function $$f$$ holomorphic on $$U$$, then :

$$\int_\gamma f(z)dz=0$$

Is it only true for specific (convex, starred) simply-connected subsets ? I've found proof for a disk or a keyhole from "Goursat's Lemma" in Elias M. Stein and Rami Shakarchi book on complex analysis, but I've struggled to find one that applies to any (simply connected) set. Does it require advanced knowledge on other topics, regarding homotopy for example ?

• It's not really clear what you mean by "any set." If the set isn't simply connected, then either it isn't path connected (in which case $\gamma$ makes no sense as it might live in two disjoint sets), or it has holes, in which case $f$ might analytically extend to a function with a singularity in one of the holes, so the residue theorem would apply instead. Otherwise, as long as you can embed a simply connected set in your "any set" that also contains your curve $\gamma$, it should be fine. May 20, 2020 at 21:38
• Yes, sorry, I was talking about any simply connected subset, without any convexity or additional conditions. May 20, 2020 at 21:44
• This is essentially in Stein and Shakarchi's book: Section 5 of ch.3 proves that if $\gamma_0$ and $\gamma_1$ are homotopic in a domain $\Omega$ on which $f$ is holomorphic then $\int_{\gamma_0} f(z)dz = \int_{\gamma_1} f(z)dz$. But if $\Omega$ is simply-connected, any closed path $\gamma$ is homotopic to the constant path $c(t)=z_0=\gamma(0)$, and clearly $\int_{c} f(z)dz =0$, so $\int_{\gamma}f(z)dz =0$. Appendix B of that book also discusses simply-connected domains in $\mathbb C$, proving, e.g., that a bounded domain D is sim-conn. if and only if $\mathbb C_{\infty}$\D is connected. May 5, 2023 at 10:17