Complex curveintegral, singularities inside the set

Question

I've just begun doing complex curve integrals using Cauchys theorem, and there is one thing i dont really understand.

The exercise if following:

Compute $\int_\gamma \frac{dz}{z}$ where $\gamma$ is the unit circle oriented counterclockwise.

My solution is using Cauchys theorem:

$f(a)=\frac{1}{2\pi i}\int_\gamma\frac{f(z)}{z-a}dz$ if i know that im working with a open set $G$

I know that $f(a)=1$ so that gets me that $\int_\gamma\frac{f(z)}{z-a}dz=2\pi i$ which is the correct answer without taking into account the singularity.

My question is, how does the singularity point $0$ plays its role here? Am i calculating this wrong, or how do i use the Theroem when there is singularities within the set $G$?

Answer 1

In your example, $f(z)=1$ is a constant function. Notice that $f$ is the numerator of the integrand.