# Holomorphic function has primitive

Let $$\mathcal{A}$$ be the family of open disks and open squares in the complex plane. Let $$f:D_1 \cup D_2 \to \mathbb{C}$$ be a holomorphic function where $$D_1, D_2 \in \mathcal{A}$$. Prove that $$f$$ has a primitive in $$D_1 \cup D_2$$. The problem here is that all the theorems we were taught that guarantee the existence of primitive require at least an open and connected set. But $$D_1 \cup D_2$$ is not connected unless $$D_1 \cap D_2 \neq \emptyset$$.

Now the proof I came up with (but I'm not really sure about) is the following:

• If $$D_1 \cap D_2 \neq \emptyset$$ then $$D_1 \cup D_2$$ is open, connected and star-shaped. Since $$f$$ is holomorphic we know that it has a primitive (this is a known theorem)

• If $$D_1 \cap D_2 = \emptyset$$ then $$f|_{D_1}$$ is holomorphic and $$D_1$$ is open connected and star-shaped so $$f$$ has a primitive on $$D_1$$. The same holds on $$D_2$$ and we can construct a primitive on $$D_1 \cup D_2$$ by "combining" the primitives on $$D_1$$ and $$D_2$$.

Is this correct? If not, how can I prove it? Thank you.

• Is the union of two disks star-shaped? Your second argument is fine but could do with some justification about why "combination" works (since you don't seem sure). You can certainly define a primitive candidate $F$ piecewise by using the primitives on each disk. You can then always compute $F'(z)$ by restricting to a sufficiently small open nbhd of $z$ (lying entirely inside one of the disks). Jun 15, 2021 at 21:34
• @preferred_anon You're right. The union of open discs is not always star-shaped! Is there a fix for that? Jun 15, 2021 at 21:40
• Of course there is - but you should think about it first :-) Jun 15, 2021 at 21:41
• @preferred_anon I could use a similar combining argument for $D_1 \setminus (D_1\cap D_2)$, $D_1 \cap D_2$ and $D_2 \setminus (D_1 \cap D_2)$ since all of these sets are star-shaped. Apart form $D_1 \cap D_2$ though I don't know for sure that the others are open. Jun 15, 2021 at 21:48
• Let $D_1$, $D_2$ be the circles of radius a little more than 1 centred at $1$ and $-1$. Choose $f(z) = 0$. On $D_1$ my primitive is $F_1(z) = 1$. On $D_2$ my primitive is $F_2(z) = 2$. What value does my primitive take at $0$? Alternatively, on $D_1 \cap D_2$ I choose the primitive $F_3(z) = 3$. Is this holomorphic? Jun 15, 2021 at 21:51

1. $$D_1 \cap D_2 = \emptyset$$. Let $$F_1: D_1 \to \mathbb{C}$$, $$F_2: D_2 \to \mathbb{C}$$ be the two given primitives of $$f$$. Claim: A primitive of $$f$$ on $$D_1 \cup D_2$$ is
$$F(z) = \cases{F_1(z) & z \in D_1 \\F_2(z) & z \in D_2 }.$$
Proof: $$F$$ is well-defined since $$D_1 \cap D_2 = \emptyset$$. Let $$z \in D_1 \cup D_2$$. $$z \in D_1$$ without loss of generality. We compute $$F'(z)$$: Since $$D_1$$ is open, there is some $$\varepsilon > 0$$ such that $$B_{z}(\varepsilon) \subset D_1$$. Then $$\forall w \in B_z(\varepsilon)$$, $$F(w) = F_1(w)$$. Therefore $$F'(z) = F_1'(z) = f(z)$$. Similarly for $$D_2$$.
1. $$D_1 \cap D_2$$. In this case $$D_1 \cup D_2$$ is simply connected so it should come as no surprise that $$f$$ has an antiderivative. Let $$F_1, F_2$$ be as above. On the connected open set $$D_1 \cap D_2$$, $$F_1'(z) = F_2'(z)$$. Therefore $$F_1(z) - F_2(z) = C$$ throughout $$D_1 \cap D_2$$. Claim: A primitive of $$f$$ on $$D_1 \cup D_2$$ is
$$F(z) = \cases{F_1(z) - C & z \in D_1 \\F_2(z) & z \in D_2 }.$$ Proof: $$F$$ is well defined by the previous argument. Just like before, at any point of $$D_1 \cup D_2$$, we can restrict $$F$$ to a small enough ball that $$F$$ is equal to one of the branches. Then $$F'(z) = f(z)$$ as required.