The Cauchy-Goursat Theorem

1. Mar 19, 2013

Bachelier

The Cauchy-Goursat Theorem states:

Let $f$ be holomorphic in a simply connected domain D. If C is a simple closed contour that lies in D, then
$\int_C f(z) \mathrm{d}t = 0$ ​

Now if $f$ is holo just on $|C| \bigcup \ int(C)$ (i.e $f$ holo only on the contour and inside of it, if we take $z_0 \in \mathbb{C}$ can we deduce from the theorem that

$\int_C \frac{f'(z)}{z-z_0} \mathrm{d}t = \int_C \frac{f(z)}{(z-z_0)^2} \mathrm{d}t = 0$ whether $z_0 \in \left[\ |C| \bigcup \ int(C) \right]$ or not

Since both equal $0$.

Also where does $z$ must belong to for the theorem to apply?

2. Mar 19, 2013

Bachelier

I think I got it. I cannot apply Cauchy inside because of the singularity problem.

Now my problem is, even though we don't know if $f$ is holo outside by the givens, can we still apply Cauchy outside $C$ and claim that

$$\int_c{f(z)} \mathrm{d}t = 0$$​

Last edited: Mar 19, 2013