I am reading a lecture note on complex analysis, and the following is in it:
Let $p\in\mathbb{C}$ and $r>0$. Let $\gamma$ be the curve in $\mathbb{C}$ with parametrization $ z: [0,2\pi] \to \mathbb{C}$ given by $$ z(t) := p + re^{it} $$ If $z_0\in B(p,r)$, then $$ \int_{\gamma} \frac{1}{z-z_0}\ dz = 2\pi i. $$ If $z_0\in \mathbb{C}\setminus\bar{B}(p,r)$, then $$ \int_{\gamma} \frac{1}{z-z_0}\ dz = 0. $$
I understand the proof included, but it did not give me an intuition as to why this is true. Is there any intuition behind this fact? If we think about complex numbers as vectors in $\mathbb{R}^2$, then $z-z_0$ is the vector from $z_0$ to $z$, so I drew $z-z_0$ as $z$ ranges over $\gamma$. Does this picture explain anything?
