Cauchy integral of $\frac{1}{z}$ over closed curve

Let $\gamma: [a,b] \rightarrow \mathbb{C}$ be a closed curve with $0 \notin \gamma([a,b]).$ Compute the Cauchy integral of $f(z):=\frac{1}{z}$ with regards to $\gamma$.

I have to compute

$$\begin{align} F(z)=\frac{1}{2 \pi i} \int _{\gamma}\frac{f(w)}{w-z}dw=\frac{1}{2 \pi i} \int _{\gamma}\frac{1}{w^2-wz}dw \end{align}$$

but as the curve is arbitrary I don't know how to go on. Can you guys help me?

Solution 1:

We assume that $z \not \in \gamma ([a,b])$, otherwise the integral is not defined. We can then calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{1}{w} \frac{1}{w-z} \, dz$ by the residue formula:

• If $z = 0$ then $\frac{1}{w} \frac{1}{w-z} = \frac{1}{w^2}$ has residue zero.
• If $z \neq 0$ and both $z$ and $0$ are inside the interior of $\gamma$, then $\frac{1}{w}\frac{1}{w-z}$ has two simple poles, with residue $\frac{1}{0-z}$ for the pole at 0 and $\frac{1}{z}$ for the pole at $z$, thus in total we get residue zero. Thus if only one of $0$ or $z$ is in the interior of $\gamma$, then we get respectively the residues $\frac{1}{-z}$ and $\frac{1}{z}$. If both $0$ and $z$ are outside of $\gamma$ then the integral is zero.