$F(x)=f'(x)$ in complex analysis conditions

Question

could use some help, cant find a straight forward definition.

Let there be a complex function $f(x):\omega \to \mathbb C$

What are the conditions for $f(x)$ and $\omega $, so that $f(x)$ would have a pre derivative function $F(x)$?

$F'(x)=f(x)$

Answer 1

The existence of an antiderivative (or primitive) for holomorphic functions is subjected to conditions which are similar to those regulating exactness of differential forms. As stated in the comments above, there is a genera theorem:

Theorem. Let $f:\Omega \subseteq \mathbf{C}\longrightarrow \mathbf{C}$ a holomorphic function defined on an open subset of the complex plane. Then $f$ has an antiderivative $F$ defined on $\Omega$ if and only if the integral

$$\int_C f(\zeta)d\zeta$$

vanishes for every closed curve $C\subseteq \Omega$.

As remarked above, however, this vanishing condition is pretty hard to evaluate; clearly it holds for simply connected opens, so holomorphic functions on such sets have a global antiderivative. But if the domain is not simply connected there are no shortcuts.

In general, what you can say is that an holomorphic function $f$ defined on some open set $\Omega\subseteq \mathbf{C}$ has a local antiderivative: namely, for every $p\in \Omega$ there exists an open neighborhood $U_p$ in which $f$ has a holomorphic antiderivative.

Think about $f(\zeta)=1/\zeta$ defined in the punctured plane $\mathbf{C}\setminus \{0\}$. It is holomorphic and it is easy to see that path integrals along circles enclosing the origin don't vanish. Thus $f$ does not have a global antiderivative, but indeed resticting to any open disk $D$ such that $0\notin D$ allows to determine a holomorphic branch of $\log z$ over $D$, which is a local antiderivative for $f$ in $D$.