Proposition: Suppose I have two open sets $\Omega_1,\Omega_2 \subset \mathbb{C}$ such that $\Omega_1 \cap \Omega_2$ is nonempty and connected. Let $f: \Omega_1 \cup \Omega_2 \to \mathbb{C}$ is analytic in $\Omega_1 \cup \Omega_2$. If it has primitives $F_i: \Omega_i \to \mathbb{C}$ in $\Omega_i$ for $i = 1,2$, then $f$ has a primitive $F: \Omega_1 \cup \Omega_2 \to \mathbb{C}$ in $\Omega_1 \cup \Omega_2$.
I don't understand why $\Omega_1 \cap \Omega_2$ has to be connected here and how should I use this assumption in the proof. Thanks in advance.
When they are disjoint the result is trivial. Problem arises when they intersect.
Let $F'=f$ in $\Omega_1$ and $G'=f$ in $\Omega_2$. Then $F-G$ has derivative $f-f=0$ on $\Omega_1 \cap \Omega_2$. Since $\Omega_1 \cap \Omega_2$ is connected we can conclude that $F-G$ is a constant $c$. Now it follows $H(z)= G(z)$ in $\Omega_2$ and $H(z)=F(z)-c$ in $\Omega_1$ gives you a primitive on $\Omega_1 \cup \Omega_2$