Let $z\in\Bbb C\setminus\{0,1\}$, and let $\sqrt z$ be a square root of $z$. Then the points $1$, $\sqrt z$, $-\sqrt z$, and $z$ are either concyclic and colinear. This follows from the fact that four points of the complex plane are concyclic or collinear if and only if their cross-ratio is a real number. Now, assume that $z\notin\Bbb R$. Then it is easy to see that those four points cannot be collinear; therefore, there is some circle $c$ to which all of them belong.
My question is: is it possible to construct $c$ from $z$ alone (that is, without finding first one of its square roots) using compass and ruler alone? I am quite sure that I have already seen this done. And it was done in order to construct $\pm\sqrt z$: these are the points at which $c$ intersects the straight line which bissects the angle $\angle 1\hat0z$.
The center of the circle is the intersection of the perpendicular bisectors of the two segments joining $1$ with $z$ and $-\sqrt z$ with $\sqrt z$, respectively. For the latter one needs to cut the argument of $z$ in half, but the points $\pm \sqrt z$ themselves are not needed. All this can be done with compass and ruler.
Here is a simple attempt to visualize the construction with Geogebra: