Does there exist $g$ s.t $g'=f$?

Question

I have the following homework question:

Let G be the bounded open set shown in gray in this picture, whose boundary consists of eight line segments. The endpoints of those segments are, as shown, the points $-2,-1,-1+4i,1+4i,1,2,2+5i,-2+5i$.

Let $f:\, G\to\mathbb{C}$ be an arbitrary function which is holomorphic in $G$. Does there a function $g:\, G\to\mathbb{C}$ which satisfies $g'(z)=f(z)$ for all $z\in G$ ? one tool that might be useful here is the Identity Theorem.

What I did:

I believe that I can construct such a function, but I am unsure if my construction is correct:

I would take some sequence of points $\{z_{i}\}_{i=1}^{\infty}$, s.t there exist $\{r_{i}\}_{i=1}^{\infty}$ s.t $D(z_{i},r_{i})\subseteq G$ and s.t $$\cup_{i=1}^{\infty}D(z_{i},r_{i})=G$$

I believe that such a sequence of points can be obtained by choosing all the points $x+iy\in G$ s.t $x,y\in\mathbb{Q}$.

Moreover, I think that the points can be arranged s.t $$D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$

We first consider $D(z_{0},r_{0})$ - there is some $g_{0}:\, D(z_{0},r_{0})\to\mathbb{\mathbb{C}}$ s.t $g_{0}'=f$ in $D(z_{0},r_{0})$, since $f$ is holomorphic.

We continue with $z_{1}$and get $\widetilde{g_{1}}$, since $g_{0}'=\widetilde{g_{1}}'$ for all points in $D(z_{0},r_{0})\cap D(z_{1},r_{1})$ then $g_{0}-\widetilde{g_{1}}$ is a constant $c$, and we can pick $g_{1}=\widetilde{g_{1}}+c$. We take $g(z)$ will be eventually defined for every $z\in G$ as one of the $g_{i}$'s.

This construction looks a bit fishy to me, I am not sure about the "moreover" part and even if so, I am not totally convinced I can arrange the constants to fit to get a holomorphic function $g$.

ADDED:

Instead of $$D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$ we can relax the condition to be $$\cup_{i=1}^{r}D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$

Is my construction correct ?

I would also appreciate to see another approach (maybe one that uses the Identity Theorem as suggested)

Answer 1

You don't need so many disks. The region $G$ is the union of three rectangles $R_i$ intersecting in two squares $Q_1=R_1\cap R_2$ and $Q_2=R_2\cap R_3$. In each of the $R_i$ the given function $f$ has a primitive $F_i$, by the most elementary version of Cauchy's theorem. In $Q_1$ the function $F_2-F_1$ has derivative zero, so it is a constant $c_1$, and similarly in $Q_2$ the function $F_3-F_2$ is a constant $c_2$. It follows that the function $$F(z):=\cases{F_1(z)+c_1\quad&$(z\in R_1)$ \cr F_2(z)\quad&$(z\in R_2)$ \cr F_3(z)-c_2\quad&$(z\in R_3)$ \cr}$$ is a well defined primitive of $f$ in $G$.

Of course observing that $G$ is simply connected makes all of the above superfluous. It is a standard theorem of complex analysis that a function $f$ holomorphic in a simply connected (i.e., without holes) domain $G$ has a primitive on $G$.