One can easily show that, given any closed loop $\gamma$ around $a$, then $\int_{\gamma} \frac{1}{z-a} dz=2k\pi i$ with $k \in Z$. I personally used the following argument. Let: $$h(t)=\int _\alpha ^{t} \frac {\gamma'(t')}{\gamma(t')-a} dt $$ then if we let: $$g(t)=e^{-h(t)}(\gamma(t)-a)$$ we have that $g'(t)=0$. Since $g$ is constant, and $\gamma$ is a closed loop, the following equality holds: $$g(\beta)=e^{-h(\beta)}=e^{-h(\alpha)}=e^0=1$$ We can conclude that $h(\beta)=2k\pi i$, $k \in Z$.
This argument is very convincing but it does not allow us to say anything about the nature of $k$. What is the easiest way to show that indeed $k$ is the number of times that $\gamma$ winds around $a$ ? Thanks in advance.
My suggestion:
Consider the formula for line integrals $$ \begin{align*} \int_\gamma f(z)\,dz &=…\text{ (substitution }z=\gamma(t)\, , \,dz=\gamma’\,dt) \\ …&=\int_a^b (f\circ \gamma )(t)\cdot \gamma’\,dt =\int_a^b f(\gamma (t))\cdot \gamma'(t)\,dt \end{align*} $$ and apply it on the path $\gamma(t)= z_0+re^{i t}\, , \,0\leq t\leq 2k\pi$
$$ \begin{align*} \oint_\gamma \dfrac{1}{z-z_0}\,dz&=\int_0^{2k\pi}\dfrac{1}{re^{it}} \cdot ire^{it} \,dt=i\cdot \int_0^{2k\pi}dt=2k\pi i \end{align*} $$