showing that $\int_\gamma f=0$

Question

Let $G \subset \mathbb{C}$ be an open star domain with the vantage point $z_0$, let $f: G \rightarrow\mathbb{C}$ be holomorphic and let $ \gamma: [0,1] \rightarrow G$ be a closed $C^1$-curve. I got to show that $\int_\gamma f=0$ Is it possible and how can I use Cauchy's integral theorem?

Answer 1

Since $G$ is starlit, there exists a primitive $F$ of $f$ on $G$, by the fundamental theorem of calculus the result follows.

$G$ is starlit and $f$ is holomorphic on $G$ $\implies $ there exists a primitive $F$ of $f$ on $G$ can easily be shown.

Since $G$ is open,let $z\in G$ then $\exists \delta >0$ $D_{\delta}(z) \subset G$, further let $h \in D_{\delta}(0)$ you would have to consider two cases, $z_0, z, z+h$ non-colinear and $z_0, z, z+h$ colinear.

To show that the difference $F(z+h)-F(z) = \int_{[\alpha,z]}f $ is defined. This involved CT for triangular contours.

Then you must show that $F$ is differentiable. and you get the result.

Answer 2

$G$ is an open star domain so it is simply connected, $f: G \rightarrow\mathbb{C}$ is holomorphic and $\gamma$ is a closed curve, if it is also rectifiable (finite length) you can apply Cauchy's integral theorem which tells you that $\int_\gamma f=0$.

Answer 3

Yes, you can use Cauchy's Theorem. Some version of CT, anyway.

Various versions of the theorem have various hypotheses, and the point to the problem is to verify that the hypotheses hold. We can't tell you exactly how to do that because we don't know what versions of CT have been proved.

But hint: Since the domain is star-shaped, any closed curve is null-homotopic (homotopic to a constant).