Suppose f is continuous in a region Ω. Prove that any two primitives of f (if they exist) differ by a constant

Question

Suppose $f$ is a complex-valued continuous function in a region $\Omega$. Prove that any two primitives of f (if they exist) differ by a constant. The statement seems to be obvious but I am missing a rigorous concrete proof. Thanks in Advance.....

Answer 1

You need the following thing:

If $f: \Omega \to \mathbb C$ is holomorphic, where $\Omega$ is a domain, then: $f' = 0 \implies f$ is constant.

Outline of proof:

Write $f=u+iv$. That $f' = 0$ gives:

$$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y} = 0$$

Since $\Omega$ is open and connected, the region of $\mathbb R^2$ over which $u$ an $v$ are defined, which is identified with $\Omega$ through the identification of $\mathbb C$ and $\mathbb R^2$, is open connected. Connectedness implies "stairway-wise connectedness", i.e. that any two points in the set can be joined by a "stairway" (a union of arranged segments each parallel to either the $x$-axis or the $y$-axis). Now prove that $u$ and $v$ are constant on any such stairway, hence constant on the whole set. Hint: to do the latter thing, use, for a fixed $y$, the function $g(t) = u(t,y)$, and for a fixed $x$, the function $h(t) = u(x,t)$ (do the same for $v$).