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I know hat $\sqrt{z}$ is a multivalued function with a branch point at $z=0$, but it can be expanded (I think) as a Taylor series that will converge, meaning is should in theory be called analytic. Is it common practice to call such a function analytic or not?

Quantum spaghettification
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    Analytic on $\mathbb C$, no. But it is analytic on disks that don't include $0$... – 5xum Apr 18 '16 at 08:34
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    When you ask whether it may be analytic, you should specify the domain of your function. – Crostul Apr 18 '16 at 08:42
  • note that it is analytic also on some weird Riemann surface https://en.wikipedia.org/wiki/Complex_logarithm#The_associated_Riemann_surface – reuns Apr 18 '16 at 09:02
  • hence when you apply the residue theorem to it, don't forget that you are in fact on that Riemann surface ! – reuns Apr 18 '16 at 09:07
  • and no its Taylor series won't converge everywhere, the radius of convergence of $(1+z)^{1/2} = \sum_{k=0}^\infty {1/2 \choose k} z^k$ is $1$, not $\infty$ as for an entire function – reuns Apr 18 '16 at 09:12
  • Is it common practice to call it analytic? A textbook or article would be more careful and specify the domain or surface. But in casual writing or discussion with people familiar with complex variables it might well be called analytic as a form of shorthand. – almagest Apr 18 '16 at 09:14
  • @almagest : I always searched for a word, I say "locally analytic" or "locally holomorphic" to say that the rest is obtained by analytic continuation, but I don't know how other people say – reuns Apr 18 '16 at 09:34

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Hint: If $z\neq 0$ and $r=|z|$ and $\arg(z)=\theta$, then $z=r(\cos(\theta)+i\sin(\theta))$. Hence $\sqrt{r}e^{\frac{1}{2}i\theta}$ is a square root of $z$. Using this branch of $\sqrt{z}$, you can show that $\sqrt{z}$ is not analytic by showing that $\int_C \sqrt{z}\mathrm{d}z\neq 0$ where $C$ is the unit circle.

If I remember correctly, this is an exercise from Foundations of Analysis.

TimeTravelPenguin
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