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I'm working through my Complex Analysis notes independently (because of lockdown mode) and would like to draw a general conclusion from some seemingly disjointed theorems.

One theorem states that functions having an antiderivative are path independent. Another (Cauchy-Goursat) implies path independence of analytic functions on simply connected domains. It is also proven that functions having an antiderivative are analytic. Furthermore, an analytic function is continuous, which, together with path independence, let's us construct an antiderivative. So it would seem analyticity and existence of antiderivative are equivalent. Is this true? Am I missing a subtlety?

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Analyticity is equivalent to the existence of an antiderivative for functions on a simply connected domain. It is not true on arbitrary domains: for instance, the function $f(z)=1/z$ is analytic on $\mathbb{C}\setminus\{0\}$ but does not have an antiderivative (locally it has an antiderivative given by a branch of a logarithm, but you can't define a consistent global antiderivative).

Another formulation is that analyticity is equivalent to locally having an antiderivative (i.e., every point has a neighborhood on which the function has an antiderivative). For the forward direction, just note that since a disk is simply connected, an analytic function has an antiderivative on a small disk around each point of its domain. For the reverse direction, locally having an antiderivative implies being locally analytic, but analyticity is a local property so this just means it is analytic.

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  • $\begingroup$ What do you mean by 'branch of a logarithm'? And how can you prove $f(z)=\frac{1}{z}$ doesn't have a global derivative? Thanks a lot! $\endgroup$ Oct 29, 2020 at 1:39
  • $\begingroup$ @othi: A branch of a function is a restriction to a function to a subset of the Complex plane where it is analytic. Remember logz =ln|z|+iargz is an antiderivative , but only locally, as the argument cannot be defined globally: if you go around a circle, your argument starts at 0, and at the same endpoint becomes $2\pi$ after one loop. $\endgroup$
    – MSIS
    Nov 7, 2021 at 22:09

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