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

Given: The punctured plane $U=\mathbb{C}\setminus \{0\}$ and $f(z)=1\mathrm{/}z$.

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

For dimension at least two, punctured Euclidean space is polygonally connected (For $n\ge 2$, the punctured space ${\mathbb{R}}^n\setminus \{0\}$ is polygonally connected).

Around a positively oriented circle centred at $0$, $\int z^{-1}dz=2\pi i$ (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).

The integral of a continuous complex derivative over every closed rectifiable contour is zero (The integral of a continuous complex derivative over every closed rectifiable contour is zero).

A continuous function on a complex domain has a primitive exactly when its closed-contour integrals vanish (For a continuous function on a complex domain, endpoint independence, zero closed-contour integrals, and existence of a primitive are equivalent).

For complex polynomials $P,Q$, the set where $Q\mathrm{\ne }0$ is open and $P\mathrm{/}Q$ is holomorphic there (Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero).

Complex differentiability at a point implies continuity there (Complex differentiability at a point implies continuity there).