FALSE: every continuous complex-valued function on a convex domain has a primitive

Given: The whole complex plane $U=\mathbb{C}$, the function $f(z)=\bar{z}$, and the positively oriented unit circle $\gamma (t)=\exp (it)$.

Complex conjugation preserves differences and modulus, and $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$ (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive).

The whole Euclidean plane is convex, since every segment between two of its points remains in the plane (A convex subset of ${\mathbb{R}}^m$ contains every line segment between two of its points).

The integral of $z^{-1}$ around the positively oriented unit circle is $2\pi i$ (On a positively oriented circle about a, the integral of (z-a)^m is zero for every integer m except -1, and is 2 pi i for m=-1).

A primitive $F$ of $f$ is holomorphic and satisfies $F^{\mathrm{\prime }}=f$ (A primitive of a complex function on an open set).

If $F$ is holomorphic, $F^{\mathrm{\prime }}$ is continuous, and $\gamma$ is closed and rectifiable, then ${\int }_{\gamma }{F}^{\mathrm{\prime }}=0$ (The integral of a continuous complex derivative over every closed rectifiable contour is zero).

The Euclidean plane ${\mathbb{R}}^2$ is connected (${\mathbb{R}}^n$ is polygonally connected, connected, locally path-connected and locally connected).

A complex domain is a nonempty connected open subset of $\mathbb{C}$ (A complex domain is a nonempty connected open subset of $\mathbb{C}$).