When I was solving a problem in complex analysis, I encountered the function $\sqrt{z^2}$.

I am wondering if this function has a branch point.

**My attempt**

The definition of a branch point given by our instructor was as follows:

Definition: A point is a branch point if the multi-valued function $w

= f (z)$ is discontinuous upon traversing a small circuit around the point.

My guess is that there is no branch point because if we let $z=re^{i\theta+2n\pi}$, we have $z^2=r^2e^{2i\theta+4n\pi}$ so $\sqrt{z^2}=re^{i\theta+2n\pi}$. The final value does not change depending on $n$, so there is no discontinuity if we travel around $z=0$ (which I am guessing is the only candidate of the branch point). However, I am not sure if it is a good argument.

## Best Answer

There are two values of $\sqrt{z^2}$, namely $z$ and $-z$. If you start on one of them and travel continuously around $0$, you stay on that one and come back to your starting value. So there is no branch point.