TITLE: Computing the square root of a circle
QUESTION [6 upvotes]: I've noticed that if I consider the equation
$$
z^2=c
$$
where $c \in \mathbb{C}$ then i'm able to find $z$ by
$$
z= \pm \sqrt{\left|c\right|}e^{i\frac{\text{arg}\left(c\right)}{2}}
$$
Hence using Geogebra, I've found that if I consider $f : z \mapsto \sqrt{z}$ the image of a disk $D$ of radius $R$ is

composed of one structure if $0 \in D$ like here

composed of two structures ( that look like disks symmetric to $0$ ) if $0 \notin D$ like here


Can someone explain to me this phenomenon and how I can methematically explain this ? Further more, can we prove that those structures are disks if $0 \notin D$ and are a connected structure if $0 \in D$ ?
EDIT
I've now shown ( thanks to Quinn ) that the cartesian equation for the curve if we start from a circle of center $(a,b)$ and radius $R$ is
$$
\left(u^2-v^2-a\right)^2+\left(2uv-b\right)^2=R^2
$$
Is there a wa with this equation to show that it is split into two parts if and only if $0 \in D$ ?

REPLY [2 votes]: WLOG, the center of the disk is on the imaginary axis (otherwise, you can rotate both axis).
If you consider the real axis, of equation $\Re(z)=0$, the square root of this locus is made of the two lines $\Re(z)=0$ and $\Im(z)=0$, i.e. the two coordinates axis.
So if the disk doesn't overlap the origin, it doesn't cross the real axis and the square root is made of two disjoint components in opposite quadrants.
I suspect that the transformed circles are Cassini ovals (and a lemniscate when the circle is through the origin).

If you consider the circle of equation
$$z=e^{i4\theta}+1$$ which goes through the origin, the square root is
$$\sqrt z=\sqrt{\cos4\theta+1+i\sin4\theta}=\sqrt{2\cos2\theta}\sqrt{\cos2\theta+i\sin2\theta}=\sqrt{2\cos2\theta}\,(\cos\theta+i\sin\theta)$$
and this is indeed a lemniscate of polar equation
$$r^2=2\cos2\theta.$$
It is the limit between the connected and disconnected images, and has a double point.

(For the center on the imaginary axis, rotate by $45°$.)