Homology version of Cauchy implies the star shaped version

Question

Theorem 1(Cauchy Theorem for star shaped domains)

Let $\Omega \subseteq \mathbb{C}$, be a star shaped domain. If $f$ is holomorphic on $\Omega$, then it possesses a primitive and $\int_{\gamma}f(z)dz=0$, for every closed curve $\gamma$ in $\Omega$.

Theorem 2(Cauchy Theorem homology version)

Let $\Omega \subseteq \mathbb{C}$ be an open set and let $f$ be holomorphic on $\Omega$. If $\Gamma$ is a cycle contained in $\Omega$, such that $\Gamma$ is 0-homologous, then $\int_{\Gamma}f(z)dz=0$.

I know that the second Theorem is the more general one. So there is an exercise that asks us to formally derive the Star shaped version of Cauchy's theorem by using the homology version. Since I am new to the homology version of Cauchy Theorem, I am not sure if what I did was correct.

My attempt: A cycle $\Gamma$ in $\Omega$ is called 0-homologous if $Ind_{\Gamma}(a)=0 \forall a \notin \Omega $. If $\Gamma=\gamma$, where $\gamma$ is some star shaped curve, then $Ind_{\Gamma}(a)=0 \forall a \notin \Omega$, and thus by the homology version we get $\int_{\gamma} f(z)dz=0$.

Question: Is there a better way of doing this? (I kind of have the feeling that my argumentation is somewhat handwavy). How do I argue for the existence of a primitive?

Answer 1

(For some unknown reason — probably because I am using the Tor browser — "Editing is currently forbidden" and I am unable to edit my old answer. So, I make a new one.)

Let $c$ be the star center of $\Omega$. For any $a\notin\Omega$, let $L$ be the half-line $\{t(a-c): t\ge0\}$. Since $\Omega$ is star-shaped, $(a+L)\cap\Omega=\emptyset$. In particular, for any closed curve $\gamma$ whose image $[\gamma]$ lies inside $\Omega$, we have $(a+L)\cap[\gamma]=\emptyset$.

Take $L$ as the branch cut of the argument function. Since $(a+L)\cap[\gamma]=\emptyset$, the curve $\gamma-a$ does not cross $L$. Therefore $\operatorname{Ind}_{\gamma-a}(0)=0$. Hence $\operatorname{Ind}_\gamma(a)=0$.

Alternatively, since $a+L$ is unbounded, $a$ lies inside an unbounded connected component of $\mathbb C\setminus[\gamma]$. Therefore $\operatorname{Ind}_\gamma(a)=0$.

(Since "Editing is currently forbidden", I cannot comment or reply to any comment. Sorry.)