vortoca

2022-06-30

If a function is holomorphic in a certain domain, does it mean it necessarily has an antiderivative
Say I have a holomrphic function f(z) in domain D, say $|z|>4$ and my function is not holomorphic for $|z|\le 4$ - it has some poles there. Does it mean there is necessarily an antiderivative for f(z)?
I know I can use Morera's theorem and calculate all the residues inside $|z|\le 4$ and check if their sum is zero. However what seemed more intuitive to me is to prove that since ${\int }_{\gamma }f\left(z\right)dz$ for every $\gamma$ contained in D, depends only on the start and end points, there exists an antiderivative for f(z).
Is my prof correct? Can I conclude that in general, a holomorphic function in a domain D not only has infinitely many derivatives in D, but it has also infinitely many antiderivatives there?

Jayvion Mclaughlin

Expert

Explanation:
No. If $f\left(z\right)=\frac{1}{z}$ (on $\left\{z\in \mathbb{C}\mid |z|>4\right\}$), then f has no antiderivative.
But if f has the additional property that ${\int }_{\gamma }f\left(z\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}z$ dpends only upon the initional and the end point of $\gamma$, then, yes, f has an antiderivative. It does not follow, however, that such an antiderivative will have an antiderivative as well. Take $\frac{1}{{z}^{2}}$, for instance.

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