Question: Consider $\gamma_1 = [z_0 - r, z_0 + r]$ and $\gamma_2: [0, \pi] \rightarrow \mathbb{C}$, $\gamma_2(t) = z_0 + re^{it}$. Let $z \notin \gamma_1^* \cup \gamma_2^*$. Evaluate $ \frac{1}{2\pi i} \int_{\gamma_1 * \gamma_2} \frac{1}{(w - z)} \, dw. $
I have some doubts in this solution,
Solution:
If $|z| > 1$, then we are done. So assume that $z$ lies strictly inside the upper-semicircle.
Let $\gamma_3 : [\pi, 2\pi] \rightarrow \mathbb{C}$, $\gamma_3(t) = z_0 + re^{it}$. Then clearly, $ \frac{1}{2\pi i} \int_{\gamma_1 * \gamma_2} \frac{dw}{(w - z)} \, + \frac{1}{2\pi i} \int_{\gamma_3 * \tilde{\gamma}_1 } \frac{dw}{(w - z)}$
$ = \frac{1}{2\pi i} \int_{\gamma_2} \frac{dw}{(w - z)} \, dw + \frac{1}{2\pi i} \int_{\gamma_3} \frac{dw}{(w - z)} \ $
$ = \frac{1}{2\pi i}\int_{C(0;1)} \frac{dw}{(w - z)} $
= 1.
Since $z$ lies in the unbounded component of $\mathbb{C} \setminus (\gamma_3 * \tilde{\gamma}_1)^*$, so $\frac{1}{2\pi i} \int_{\gamma_3 * \tilde{\gamma}_1} \frac{1}{(w - z)^2} \, dw = 0$. Hence, $ \frac{1}{2\pi i} \int_{\gamma_1 * \gamma_2} \frac{dw}{(w - z)} = 1. $
Where $\int_{\gamma_1 * \gamma_3} = \int_{\gamma_1} + \int_{\gamma_3}$ and $\tilde{\gamma}_1$ is the opposite path of $\gamma$
I could not understand the above solution like
What is happening when $|z| > 1$ ?
Why is $\gamma_3$ being constructed in such a way?