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?
 A: 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.

A: If, as you say, you're not supposed to use the Cauchy theorem, then I guess you're supposed to use the parametrization of the curve.
For a given $\gamma: [a,b] \ni t \mapsto \gamma(t) \in \mathbb C$:
$$ F_\gamma(z) = \frac{1}{2\pi i}\int_\gamma \frac{1}{w(w-z)} dw = \frac{1}{2\pi i}\int_a^b \frac{1}{\gamma(t) \big(\gamma(t)-z\big)}\frac{d\gamma(t)}{dt}dt$$
You cannot actualy calculate this integral unless you knwo what the curve is, but given a parametrization, it can be done for any curve, closed or not.
