Antiderivatives and Path Independence

We prove $(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(1)$, and along the way establish the displayed formula. The technique throughout is direct computation via the chain rule and the single-variable Fundamental Theorem of Calculus, reducing a complex-variable statement to real-variable calculus applied to real and imaginary parts.

Step 1: $(1)\Rightarrow(2)$, with the explicit formula. Suppose $F$ is a primitive of $f$ on $D$: $F$ is holomorphic on $D$ and $F'=f$ everywhere on $D$. Let $\gamma:[a,b]\to D$ be a contour, first assumed $C^1$ (no corners); we handle corners below. Define $g(t) = F(\gamma(t))$ for $t\in[a,b]$. Since $F$ is holomorphic and $\gamma$ is a real-differentiable path with values in $D$, the complex chain rule applies: $F$ is complex-differentiable at each $\gamma(t)$, and $\gamma$ is differentiable at $t$, so

(This chain rule is justified exactly as in the real case: write $F(\gamma(t+h)) - F(\gamma(t)) = F'(\gamma(t))(\gamma(t+h)-\gamma(t)) + o(|\gamma(t+h)-\gamma(t)|)$ using complex differentiability of $F$, divide by $h$, and let $h \to 0$, using continuity and differentiability of $\gamma$.) The function $g:[a,b]\to \mathbb C$ is $C^1$ since $g' = f(\gamma(\cdot))\gamma'(\cdot)$ is continuous ($f$ continuous, $\gamma'$ continuous). Write $g(t) = u(t) + iv(t)$ with $u,v$ real-valued $C^1$ functions on $[a,b]$. By the real Fundamental Theorem of Calculus applied separately to $u$ and $v$,

Hence

where $p = \gamma(a)$, $q=\gamma(b)$. If $\gamma$ is only piecewise $C^1$, with corners at $a = t_0 < t_1 < \cdots < t_n = b$, apply the computation above on each subinterval $[t_{k-1}, t_k]$ (on which $\gamma$ restricts to a $C^1$ contour) to get

a telescoping sum equal to $F(\gamma(t_n)) - F(\gamma(t_0)) = F(q)-F(p)$. This proves the displayed formula, and in particular the right-hand side depends only on $p,q$ (through $F$), not on the specific contour $\gamma$ joining them: this is exactly path independence, statement (2).

Step 2: $(2)\Rightarrow(3)$. Let $\gamma$ be any closed contour in $D$, say $\gamma:[a,b]\to D$ with $\gamma(a) = \gamma(b) = p$. Let $\gamma_0$ denote the constant contour at $p$ (i.e. $\gamma_0(t) \equiv p$ on, say, $[a,b]$); this is a trivial contour from $p$ to $p$ lying in $D$, and clearly $\int_{\gamma_0} f\,dz = 0$ since $\gamma_0' \equiv 0$. Both $\gamma$ and $\gamma_0$ are contours in $D$ from $p$ to $p$. By path independence (2), applied with initial point $p$ and terminal point $p$,

This is exactly statement (3).

Step 3: $(3)\Rightarrow(1)$. Assume every closed contour integral of $f$ in $D$ vanishes. We construct a primitive directly, mimicking the construction of an antiderivative via $F(x) = \int_{x_0}^x f$ from single-variable calculus. Fix a basepoint. Since $D$ is open and connected, and open connected subsets of $\mathbb{C}$ are path-connected by piecewise-linear (in particular piecewise $C^1$) paths — a standard fact: the set of points reachable from a fixed $z_0\in D$ by a polygonal path in $D$ is open (since $D$ is open, any small disk around a reachable point is still reachable by extending the polygonal path) and its complement in $D$ is also open by the same argument, so by connectedness of $D$ this set is all of $D$ — we may fix $z_0 \in D$ and, for each $z \in D$, choose a contour $\gamma_z$ in $D$ from $z_0$ to $z$ (a finite polygonal path suffices). Define

Well-definedness. If $\gamma_z, \tilde\gamma_z$ are two contours in $D$ from $z_0$ to $z$, form the closed contour $\Gamma$ obtained by traversing $\gamma_z$ then $\tilde\gamma_z$ reversed (concatenation of piecewise-$C^1$ paths is again piecewise $C^1$, and reversal negates the integral: $\int_{-\sigma} f\,dz = -\int_\sigma f\,dz$, directly from the substitution $t \mapsto a+b-t$ in the definition of the contour integral). Then

by hypothesis (3), so $\int_{\gamma_z} f\,dz = \int_{\tilde\gamma_z} f\,dz$. Hence $F(z)$ does not depend on the choice of contour, and $F$ is well-defined on $D$. Differentiability and $F' = f$. Fix $z_1 \in D$. Since $D$ is open, choose $r>0$ with the disk $B(z_1, r) \subset D$. For $h \in \mathbb{C}$ with $0 < |h| < r$, let $\sigma_h$ be the straight segment from $z_1$ to $z_1 + h$; this lies inside $B(z_1,r) \subset D$. Concatenating $\gamma_{z_1}$ (from $z_0$ to $z_1$) with $\sigma_h$ gives a contour in $D$ from $z_0$ to $z_1+h$, so by well-definedness of $F$,

Now write, using $\int_{\sigma_h} dw = h$ (since $\sigma_h(t) = z_1 + th$, $t\in[0,1]$, gives $\int_0^1 h\,dt = h$),

Since $f$ is continuous at $z_1$, given $\varepsilon>0$ there is $\delta \in (0,r)$ such that $|f(w) - f(z_1)| < \varepsilon$ whenever $|w - z_1| < \delta$. For $|h| < \delta$, every point $w$ on $\sigma_h$ satisfies $|w - z_1| \le |h| < \delta$, so by the standard estimate $\left|\int_\sigma g\right| \le (\sup_\sigma |g|)\cdot \mathrm{length}(\sigma)$,

Hence

Since $\varepsilon>0$ was arbitrary, $\displaystyle\lim_{h\to 0} \frac{F(z_1+h)-F(z_1)}{h} = f(z_1)$, i.e. $F$ is complex-differentiable at $z_1$ with $F'(z_1) = f(z_1)$. Since $z_1 \in D$ was arbitrary, $F$ is holomorphic on all of $D$ with $F' = f$, i.e. $F$ is a primitive of $f$. This proves (1).

Conclusion. We have shown $(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(1)$, so all three statements are equivalent, and Step 1 additionally established the fundamental-theorem formula whenever a primitive exists. $\blacksquare$

By Cauchy's Theorem (in its homotopy or homology form for simply connected domains), if $D$ is simply connected and $f$ is holomorphic on $D$, then $\oint_\gamma f\,dz = 0$ for every closed contour $\gamma$ in $D$. This is exactly hypothesis (3) of the Path Independence ⇔ Existence of a Primitive theorem, whose proof (Step 3, not depending on this corollary) already shows $(3)\Rightarrow(1)$. Hence $f$ has a primitive on $D$, and the remaining conclusions follow from the same theorem. $\blacksquare$