Is it true that a complex function has a global antiderivative if and only if it integrates to zero over every closed curve?

I am somehow thinking that these properties must be equivalent, unfortunately I do not know a theorem that says it:

$f$ has a global antiderivative iff the line integral $\int_{\gamma}f$ over every closed curve in the domain of f is zero? Is this correct?

1 Answer

Let $\Omega$ be the domain of $f$ (an open set). If $f$ has a global antiderivative, an application of the fundamental theorem of Calculus along with the definition of path integrals shows that $\int_\gamma f(z) \,dz = 0$ for every closed curve $\gamma$ in $\Omega$.

Now assume that $\int_\gamma f(z) \,dz = 0$ for every closed curve $\gamma$ in $\Omega$. Assume, without loss of generality, that $\Omega$ is connected. (Otherwise we can treat each component of $\Omega$ individually.) Fix $z_0 \in \Omega$ and define $$F(z) = \int_{\gamma(z)} f(\zeta) \,d\zeta$$

where $\gamma(z)$ is a path from $z_0$ to $z$. This function is well-defined regardless of the path we choose because of the initial hypothesis. We claim that $F' = f$.

Fix $z \in \Omega$. There is a neighborhood $D(z, r) \subset \Omega$ for some $r > 0$. Let $w \in D'(z, r)$. We have:

$$\frac{F(w) - F(z)}{w - z} = \frac{1}{w-z}\int_{[z, w]} f(\zeta) \,d\zeta$$

Hence:

$$\left|\frac{F(w) - F(z)}{w - z} - f(z)\right| \le \frac{1}{w-z}\int_{[z, w]} \left|f(\zeta) - f(z)\right| \,d\zeta$$

Fix $\epsilon > 0$. By the continuity of $f$, we can make $r$ small enough so that $\forall \zeta \in [z, w] : \left|f(\zeta) - f(z)\right| < \epsilon$. Thus:

$$\left|\frac{F(w) - F(z)}{w - z} - f(z)\right| \le \epsilon$$

Since $z$ and $\epsilon$ are arbitrary, it follows that $F' = f$.

• hey can you show the first direction more precisely please? Nov 17, 2021 at 16:02
• where you used the fact that domain is connected Sep 24, 2022 at 11:05