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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 $\angle 1\hat0z$.

José Carlos Santos
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  • Would this work? The center of the circle is the intersection of the perpendicular bisectors of the segments from $1$ to $z$ and from $-\sqrt z$ to $\sqrt z$. For the latter you need to bisect the argument of $z$, but you don't need $\sqrt z$ itself. (My usage of English terms for these geometric constructions may be wrong.) – Martin R Sep 26 '22 at 10:13
  • Finding $\sqrt{|z|}$ with ruler and compass is not difficult - see for example https://math.stackexchange.com/questions/705/compass-and-straightedge-construction-of-the-square-root-of-a-given-line – Henry Sep 26 '22 at 10:54
  • @Henry I know that, but I wanted to find the square roots of $z$ through the method that I have outlined. – José Carlos Santos Sep 26 '22 at 12:37
  • @MartinR That's it! I feel so silly. Will you post that as an answer? – José Carlos Santos Sep 26 '22 at 12:38

1 Answers1

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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:

enter image description here

Martin R
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