# Is $\sqrt{z}$ an analytic function?

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?

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

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