# Branch points of $\sqrt{z^2}$

branch-pointscomplex-analysis

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.

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.