Do holomorphic functions have primitive? Let $f:\Omega\to \mathbb C$ be called holomorphic on $\Omega$ if it is complex-differentiable on $\Omega$.Define primitive of $F$ on $\Omega$ to be a function such that $F'(z)=f(z)$ for all $z\in \Omega$.We have shown in a theorem that if a continuous function has a primtive  then $\int\limits_{\gamma} f(z)dz=0$ for any closed curve $\gamma$.We also know by Cauchy's theorem that if $f$ is holomorphic on $\Omega$ then $\int\limits_{\gamma}f(z)dz=0$ for any closed curve $\gamma$.This leads me to ask the question whether any holomorphic function on $\Omega$ have a primitive.I want a proof of this fact or a counterexample that this may not be true.Can someone help me?
 A: There is a very deep connection between the shape of $\Omega$ and the existence of primitives on $\Omega$. For now, let's assume that $\Omega$ is connected. Then it can be shown that every holomorphic function $\Omega\to\mathbb C$ has a primitive if and only if $\Omega$ is simply connected. Meaning that $\Omega$ has no holes. Geometrically speaking, if $\Omega$ has a hole, then we could integrate a function along a closed path winding around that hole. And such integrals are not necessarily $0$, contrary to your statement. The integral of a holomorphic function along a closed path is only guaranteed to be $0$ if the path encloses no holes. And conversely, if a path winds around a hole, then we can guarantee the existence of a holomorphic function whose integral along that path is not $0$.
Note that Cauchy's theorem is only true on simply connected domains.
There is also a more general version of Cauchy's theorem: if we can continuously deform a closed path (without tearing or glueing) to obtain a second path, then the integrals of a holomorphic function along those paths are equal. Now if a path can be deformed to just a point, then the integral along that path must be $0$. All paths which can be deformed that way must yield an integral of $0$ for this reason. But if a path winds around a hole, it can't be deformed to a point because the hole is in the way. And for this reason, the integrals along such paths need not vanish. And consequently, a holomorphic function on such a domain with a hole need not have a primitive.
A: Depends on the geometry of $\Omega$. If $\Omega = \bf C\setminus {0}$, the function $f(z)=1/z$ has no primitive. The integral of $f$ over the circle is $2i\pi \not =0$. But if every closed curve satisfies $\int _C f(z) dz=0$, then $f$ has a primitive
