Let the domain $O=\mathbb{C}-(-\infty,0)$, the point $z_0 \in O$ and the circle $\gamma=C(0,r<|z_0|)$ in the positive direction. Compute $\int_C \frac{\log z}{z-z_0} dz$.
The answer that my teacher gives us is $2πi \log z_0$. I know that this answer is false according to the question $\int_C \frac{\log z}{z-z_0} dz$ - Cauchy theorem with $z_0$ outside the interior of $\gamma$ and the answer given by Ron Gordon. I think he has used the simple Cauchy formula with the primitive of $\log′ z$. Is there a contradiction at this answer I can mention him? To know that he gives us only the simple solution (he doesn't give us his resolution method).
Yes, you are correct. If $r<|z_0|$ the integral is 0. You can't very easily calculate this one directly, as $\int \dfrac{\log z}{z} dz$ doesn't integrate nicely.
The Cauchy integral formula says that if the function is holomorphic over simply connected subset of the complex plane, then the integral will be 0 over any closed contour on that subset.
The next thing you learn is that if the function fails to be analytic at a single point, then integral of every path that encloses that point is the same. And finally, any contour integral can be evaluated by analyzing just the points where the function fails to be analytic.