I am wondering if this is true. i did a proof but i'm not sure about it.
take $D \subseteq \mathbb C$ to be a domain and let $f: D \to \mathbb C$ be a holomorphic on $D$. then $f$ is analytic on $D$ so if $a \in D$ then we can find a disk $D(a,r)$ and a power series $\sum_{k=0}^{\infty} a_k (z – a)^k$ converging to $f(z)$ for every $z$ in $D(a,r)$. take $F(z) = \sum_{k=0}^{\infty} (c_k/(k+1)) (z-a)^{k+1}$ then $F$ is holomorphic on $D(a,r)$ and $F'(z) = f(z)$ for every $z \in D(a,r)$ so in particular $F'(a) = f(a)$.
So if I define $F$ on $D$ such that $F(z)$ is equal to that series over the corresponding disk around each point in $D$, don't i get a primitive for $f$?
thanks
Best Answer
Since the holomorphic function $f(z) = 1/z$ fails to have a holomorphic primitive on the punctured plane $D = \mathbf{C} \setminus\{0\}$, there must be a flaw somewhere in the proposed argument. That somewhere is at the end, the implicit assertion that the power series (plural) defining $F$ in a neighborhood of an arbitrary point "patch together" globally.
For further reading, the umbrella for this type of consideration is analytic continuation. If you're algebraically-minded, sheaf cohomology is a powerful, sophisticated bookkeeping device "designed" almost expressly for this type of investigation.