I encountered this problem as a previous result of an exercise in a text book way before proving Cauchy Theorem, so I think there must be another way to prove it without it.
Show that $\int_{\gamma(0;r)} \frac{1}{z-a} dz = 0$ for $|a|>r>0$
Here is my try using the Fundamental Theorem of Calculus, but I'm not sure if it is the right way to proceed.
Parametrize the circline by $\Gamma_r = re^{it}$ for $t\in \left[0,2\pi \right]$. Now $f(z)=\frac{1}{z-a}$ is holomorphic in the open disc $D(0;r)$, so there exist a function $F$ such that $f=F'$ in the disc (this is the step I think I got wrong). Then
$$\int_{\gamma(0;r)} \frac{1}{z-a} dz =\int_{\gamma(0;r)} F'(z) dz = F(\gamma(2\pi))-F(\gamma(0))=F(r)-F(r)=0$$
Is this proof okay? And if not, can you help me to solve it properly? Thanks in advance!
Well, I would consider using the fact that there exists such an $F$ to be using Cauchy's theorem, since the usual proof of this for arbitrary holomorphic functions on simply connected domains uses Cauchy's theorem. But in this case you can avoid that by just writing down such an $F$ explicitly: $F(z)=\log(z-a)$, for some branch of $\log$ defined in the ball of radius $r$ around $-a$.