Show that the complex integral of $\frac{1}{z-z_0}$ on $γ_R=Re^{it}$, $ t∈[π,2π]$ with $R>0$, $Im(z_0)<0$ and $z_0∈C$ is equal to $π i$ for ${R\rightarrow \infty}$
My attempt: I was able to show that $π$ is greater than the integral by using the ML inequallity which I don't believe is useful at all. In addition I couldln't find anything when I used the cauchy formula.
Recall Cauchy's formula, for $f$ holomorphic, $\gamma$ some arc and $z_0\in \mathbb{C}$ in the interior of $\gamma$ : $$f^{(n)}(z_0)=\frac{n!}{2i\pi}\oint_\gamma\frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\text{d}\zeta.$$
You're in the case where $f$ is constant : $f(z)=1$, and $n=0$. Since you want $R\to+\infty$, you can assume that $R>|z_0|$, and thus apply the formula to get what you want, by noting that you're only integrating over half of the circle : divide the integral over $[0,2\pi]$ into two integrals, and show that both are equal to what you are looking for, using some basic change of variables in one of the two.