complex-analysis
Suppose $\Omega$ be an open subset of $\mathbb{C}$, and let $S$ be a finite union of line segments in $\Omega$. Also, suppose we have a function $f$ which is continuous in $\Omega$ and holomorphic in $\Omega \setminus S$. We want to show that $f$ is holomorphic in $\Omega$.
I'm thinking of using Morera and the symmetry principle, but I'm confused on what would happen to the possible intersections and in the 'corners' where the line segments meet.
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 $0$.
I hope it helps you :). I am going to use a general theorem of Cauchy: If $C$ is a closed simple curve such that $f$ is holomorphic over it and its interior then $\int_Cf(z)dz=0$.
We take the triangle such that it intersects the circle $S^1$ and break it with $\alpha$ and $\beta$ where
$\alpha$ is the part of the triangle above the circle and joined with a circular curve ($e^{i\theta}\delta$ where $\delta>1$ and $\theta\in[\theta_1,\theta_2]$), $\beta$ is the part of the triangle below the circle joined with a circular curve ($e^{i\theta}\epsilon$ where $0<\epsilon<1$ and $\theta\in[\theta_2,\theta_1]$). Then if we integrate
$$\int_\alpha fdz+\int_\beta fdz=\int_{l_1} fdz+\int_{l_2} fdz+\int_{l_3} fdz+\int_{l_4} fdz+\int_{l_5} fdz +\int_{\theta_1}^{\theta_2}f(e^{i\theta}\delta)-f(e^{i\theta}\epsilon)d\theta$$
where $1>\epsilon>0$ tends to 1 and $\delta>1$ tends to 1 and $l_1,l_2,l_3,l_4,l_5$ are the lines of the triangle. Then since $f$ is holomorphic in $\mathbb{C}\backslash S^1$, using the theorem of Cauchy, then
$$\int_\alpha fdz+\int_\beta fdz=0.$$
Since $f$ is continuous
$$\lim_{\delta,\epsilon\rightarrow 1} \int_{\theta_1}^{\theta_2}f(e^{i\theta}\delta)-f(e^{i\theta}\epsilon)d\theta=0.$$
So finally
$$\int_T fdz=\lim_{\delta,\epsilon\rightarrow 1} \left(\int_\alpha fdz+\int_\beta fdz\right)=0.$$
Best Answer
First a preliminary: the theorem is still true if the line segments are not contained in $\Omega$, and it's also easier to prove this way. So that's what we're doing here.
We can do this via induction over the number $n$ of line segments. For $n=0$ the function is holomorphic by assumption. So we can go right to the induction step, that is, assume the claim is true for $n$ line segments, and consider the case of $n+1$ segments. Let $L_i,i=1,\dots,n+1$ be the line segments whose union is $S$. Consider just the line segment $L_{n+1}$. Extend this segment to a whole line, which cuts $\Omega$ in two halves, $\Omega_1$ and $\Omega_2$. On both halves we can apply the induction hypothesis, because $f$ is continuous on $\Omega_{1/2}$ and holomorphic on $\Omega_{1/2}\backslash\bigcup_{i=1}^n L_i$, which is just an open set with $n$ line segments removed. So $f$ is holomorphic on both $\Omega_1$ and $\Omega_2$. The question remains wether it is on the line dividing the two. We use Morera for this. Morera says that if a function on an open set is continuous and all integrals along triangle contours vanish, then the function is holomorphic. The function is continuous by assumption. Now take any triangle contour $\Delta$ and consider
$$\int_\Delta f(w)\mathrm dw.$$
If the contour is contained in either $\Omega_1$ or $\Omega_2$, then this integral vanishes by the Cauchy integral theorem. If it intersects the line, make two cuts, both parallel to $L_{n+1}$ at a distance of $\varepsilon>0$ to $L_{n+1}$. One cut on each side. You can recover up to two contours from this, a triangle and/or a quadrilateral. The integrals along these contours vanish because they're completely contained in one of the halves of $\Omega$, on which $f$ is holomorphic.
Now use the continuity of $f$ to show that as $\varepsilon\to0$, the sum of the integrals along the two contours converges to the integral along the original triangle, making that $0$. I leave this to you. Now Morera applies and the function is holomorphic.