complex integral: $\oint dz/z$ I'm trying to determine $$\oint \frac{dz}{z}$$ on a closed path $\mathcal C$, where $\mathcal C$ is the circle $|z|=100$ traversed anticlockwise.
What I did was re-write the function as $$\oint \frac{dz}{z-0}$$ and by Cauchy integration formula, it straight away gives $2\pi i$.
Is this method correct? Share a better/correct method and explain my mistakes, if not.

If $f(z)$ is an analytic function inside a simple closed curve $C$,
except for a finite number of isolated singular points $z_i$, $i \in \mathbb{N}$ located inside C, then
\begin{equation}
\oint f(z) dz= 2 \pi i \sum_{j=1}^N r_j
\end{equation}
where $r_j$ are the residue of $f(z)$ at $z = z_j$. This is know as the Cauchy's residue theorem.
In your case  $f(z)=1/z$ and $z_1=0$. If $C$ is a circle of radious R, i.e. $|z| = R$ the residue of $z^{-1}$ is unity and the integral is equal to $2 \pi i$. The Cauchy's residue theorem shows that the correct modification of Cauchy’s Theorem, when $f(z)$ contains one isolated singular point at $z_0 \in D$, is that the integral be proportional to the residue of $f(z)$ at $z_0$, where $0 < |z-z_0|< R$.
