I begin in complex analysis. I have the following formulation for the Cauchy integral:
If $\gamma$ is the boundary of a compact $K$ (not reduced to a single point) included in an open set $D$ of the complex plane and $f$ a holomorphic function on $D$.
Then
$$
\int_\gamma f(z)dz=0 \qquad (1)
$$
Moreover if $a$ is inside $K$, we have
$$
\int_\gamma \frac{f(z)dz}{z-a} = 2i\pi f(a).
$$
Ok. Now if I add the assumption that $D$ is simply connected and $a$ is not inside $K$ then since $\frac{f(z)}{z-a}$ is holomorphic on $K$ and the integral of the form $\frac{f(z)dz}{z-a}$ along $\gamma$ is equal to the integral along any homotopic curve to $\gamma$. Hence :
$$
\int_\gamma \frac{f(z)dz}{z-a} = 0.
$$
Can you confirm me that we need the simply connected assumption to have this result ?
edit: I think I don't need the simply connected assumption in this case since I can just apply the Cauchy theorem (equation $(1)$) to the holomorphic function $\frac{f(z)}{z-a}$ on $D\setminus \{a\}$, the compact being $K$ not containing $a$ and its boundary is the closed curve $\gamma$. I conclude it is zero.
Best Answer
If $\gamma$ is a simple closed curve then $K$ is a topological disk and $\gamma$ is homotopic to a point by a homotopy staying in $K$. In particular, if $a$ does not belong to $K$ the second integral is $0$, whatever $D$ is.