TITLE: Is $\sqrt{z}$ an analytic function?
QUESTION [6 upvotes]: 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?

REPLY [7 votes]: Hint: If $z\neq 0$ and $r=|z|$ and $\arg(z)=\theta$, then $z=r(\cos(\theta)+i\sin(\theta))$. Hence $\sqrt{r}e^{\frac{i\theta}{2}}$ 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.