Use Morera's Theorem to prove the following: Let $\Omega\subset_{\text{open}} \mathbb{C}$,$z_{0}\in\Omega$. Suppose $f$ is continuous in $\Omega$ and analytic in $\Omega \backslash \{z_{0}\}$. Then $f$ is analytic in $\Omega$
I am not really sure how to proceed. Morera's theorem states:
Theorem: Morera If $g$ is a continuous function in the open disc $D$ such that for any triangle $T$ contained in $D$: $$ \oint_{T} g(z) \, dz = 0 $$ Then $g$ is holomorphic in $D$.
$\Omega$ is an open subset of $\mathbb{C}$. So at $z_{0}$ there is some open disk of radius $r$ contained in $\Omega$, $D(z_{0},r)\subset \Omega$. If I can show that Morera's thoerem holds on this disk, then I am done, since holomorphic functions are always analytic - so is $f$ is holomorphic on $D(z_{0},r) \supset \{z_{0}\}$ then it is analytic at $z_{0}$.
I know need only concern myself with triangles $T$ such that one of the vertices is $z_{0}$, since all other cases follow automatically from analyticity (holomorphicity) of $f$ and Cauchy's/Goursat's theorem(s).
However, how can I evaluate these integrals??
It's not true that you need only consider triangles with one vertex $z_0$. Indeed those are not a problem: the integral over a triangle with one vertex $z_0$ is the limit of integrals over triangles that $z_0$ is outside of, which are $0$ by Cauchy.
But you can't use Cauchy/Goursat when $z_0$ is inside your triangle. To handle those, break up such a triangle into three triangles, each of which has $z_0$ at one vertex. The integral over your triangle is the sum of the integrals over the three sub-triangles (as the integrals over the interior edges cancel). And by the previous paragraph, the integrals over those triangles are $0$.