For a continuous function on a complex domain, endpoint independence, zero closed-contour integrals, and existence of a primitive are equivalent

If $f$ has a primitive $F$ on $U$, then $F^{\mathrm{\prime }}=f$ is continuous by the Given, so [L5] applies to every rectifiable contour in $U$ and gives endpoint independence; endpoint independence makes every closed-contour integral zero because the constant contour with the same endpoint has integral $0$.

Assume every closed-contour integral is zero. By [L1] and [L2], fix a basepoint $z_0\in U$; for each $z\in U$ at least one polygonal path in $U$ runs from $z_0$ to $z$. Any two such paths carry the same integral: concatenating one with the reversal of the other is a closed contour, whose integral is by [L3] the difference of the two, and the closed-loop hypothesis makes that difference $0$.

So for each $z\in U$ there is a unique complex number shared by the integrals of $f$ along all polygonal paths in $U$ from $z_0$ to $z$; define $F(z)$ to be that number. This specifies $F$ uniquely from the data of step 1.2, with no path selected and no choice principle used.

For sufficiently small $h$, the segment from $z$ to $z+h$ lies in $U$. By [L3] and [L4], $F(z+h)-F(z)={\int }_0^{1}f(z+th)hdt$, so division by $h\mathrm{\ne }0$ gives an average tending to $f(z)$ by continuity. Thus $F^{\mathrm{\prime }}(z)=f(z)$.

The construction proves that zero closed integrals imply a primitive, completing both directions of the equivalence. On piecewise-$C^1$ contours the componentwise statement agrees with the real vector-field equivalence [L6], whose open and path-connected hypotheses hold by [L1] and [L2].