Notice that by partial fraction decomposition we have,
$$
\int_{\gamma} \frac{1}{z^2-2z}\ dz = \frac{1}{2}\int_{\gamma} \frac{1}{z}\ dz - \frac{1}{2} \int_{\gamma} \frac{1}{z-2}\ dz.
$$
We know by the Fundamental Theorem of Calculus for Contour Integrals that since $\gamma$ is a closed path and since the principal branch of the log is analytic on some open set containing $\gamma$ then $(1/2) \int_{\gamma}(1/z)\ dz = 0$. For the other part we can see, that in a similar fashion of Problem 2 part (c), we parameterize $\gamma: [0,2\pi] \to \mathbb{C}$ as $ \gamma(\theta)= e^{i\theta} + 2$ a…
