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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?

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    $\begingroup$ Analytic on $\mathbb C$, no. But it is analytic on disks that don't include $0$... $\endgroup$
    – 5xum
    Apr 18, 2016 at 8:34
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    $\begingroup$ When you ask whether it may be analytic, you should specify the domain of your function. $\endgroup$
    – Crostul
    Apr 18, 2016 at 8:42
  • $\begingroup$ note that it is analytic also on some weird Riemann surface en.wikipedia.org/wiki/… $\endgroup$
    – reuns
    Apr 18, 2016 at 9:02
  • $\begingroup$ hence when you apply the residue theorem to it, don't forget that you are in fact on that Riemann surface ! $\endgroup$
    – reuns
    Apr 18, 2016 at 9:07
  • $\begingroup$ 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 $\endgroup$
    – reuns
    Apr 18, 2016 at 9:12

1 Answer 1

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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.

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    $\begingroup$ or more simply by showing that it is discontinuous around $arg(z) = \pm\pi$ (or the boundary that you chose for your $arg(z)$) $\endgroup$
    – reuns
    Apr 18, 2016 at 9:04
  • $\begingroup$ That's probably easier :) $\endgroup$ Apr 18, 2016 at 9:06
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    $\begingroup$ but testing if $\int_C f(z) dz = 0$ on all possible (closed) contours $C$ is much more general $\endgroup$
    – reuns
    Apr 18, 2016 at 9:08

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