$\int_C \frac{\log z}{z-z_0} dz$ - Cauchy theorem with $z_0$ outside the interior of $\gamma$ 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 $\int_C \frac{\log z}{z-z_0} dz$. 
How do I solve the problem with the Cauchy theorem knowing that the point $z_0$ is outside the interior of $\gamma$? Do I have to use the primitive of $\log z$, i.e. $\log' z = \frac{1}{z}$?
Thank you to bear in mind that I have not yet seen the Residue theorem.
 A: To evaluate the contour integral, we use a keyhole contour to avoid the branch cut.  Consider the integral
$$\oint_{C'} dz \frac{\log{z}}{z-z_0} $$
where $C'$ is the keyhole contour.  The contour integral is then
$$\int_C dz \frac{\log{z}}{z-z_0} + e^{i \pi} \int_1^{\epsilon} dx \frac{\log{x}+i \pi}{-x-z_0}+ i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{\log{\epsilon}+i \phi}{\epsilon e^{i \phi} - z_0}+e^{-i \pi} \int_{\epsilon}^1 dx \frac{\log{x}-i \pi}{-x-z_0}$$
The third integral vanishes as $\epsilon \to 0$.  We can combine the second and fourth integrals to further simplify:
$$\int_C dz \frac{\log{z}}{z-z_0} - i 2 \pi \int_0^1 \frac{dx}{x+z_0} $$
By Cauchy's theorem, the integral over $C'$ is zero.  Thus, when $z_0$ is exterior to the unit circle $C$:

$$\int_C dz \frac{\log{z}}{z-z_0} = i 2 \pi \log{\left (1+\frac1{z_0} \right )} $$

ADDENDUM
From the above, it should be simple to evaluate the integral when $z_0$ is inside the unit circle.  By the residue theorem, when $z_0$ is interior to $C$,
$$\int_C dz \frac{\log{z}}{z-z_0} = i 2 \pi \log{\left (1+z_0 \right )} $$
