Computing the square root of a circle

Question

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 $0 \in D$ ?

Answer 1

The complex numbers can be parameterized by z = a+bi, but they can also be parametrized with z = $re^{i\theta}$. This does have the issue that (r,$\theta$) ->z is not one-to-one; ($0,\theta$) goes to zero for all values of $\theta$, and go to zero, and ($r,\theta$), ($r,\theta+ 2\pi$) go to the same value. Suppose we define $f_p(r,\theta) = (\sqrt x, \theta /2)$. Then finding the image of a region under f can be done by finding the corresponding region in the polar coordinate system, apply $f_p$, and then seeing what region that corresponds to in rectangular coordinates.

Another way of thinking about is is imagining a helical sheet; imagine taking a the polar graph, and wrapping it around on top of itself so that each curve with constant r being a helix, the set of point ($r,\theta + n\pi$) all lying on top of each other. If we want to convert from this coordinate system to the rectangular one, we simply project onto one layer, taking every point that appears on at least one of the layers.

Taking the square root halves the value of $\theta$. If a point lies on an even layer (that is, $\theta = 2n(2 \pi)+\theta'$), then we end up at half the angle at a layer half as high (that is, now we have $n(2 \pi)+\theta'$/2). If it's on an odd layer, then we still take half as many layers, but now we have the extra half-layer that shows up as a rotation of $\pi$, so $(2n+1)(2 \pi)+\theta'$ turns into $n(2 \pi)+\theta'/2 + \pi$. Originally, we had a copy of our region in each layer, each copy corresponding to the same region in the rectangular coordinates. Now half of the copies correspond to one region, while the other half correspond to another.

Another way of thinking of it is that when you take the square root, you halve the angle to the x-axis. The complication is that both the angle clockwise, and the angle counter-clockwise, is halved. In your example, the center of your disk is 90 degrees counter-clockwise from the x-axis, so halving that ends up with it being 45 degrees counter-clockwise from the x-axis. But it's also 270 degrees clockwise from the x-axis, so you end up with a region 135 degrees clockwise from the x-axis.

As for whether the image consists of disks, consider the points A = 11+0i, B = 10+1i, C = 9+0i, D = 10-1i. These four point all lie on the circle centered at 10+0i having radius 1. If you take the principle square root, you end up with

$\sqrt A = 3.316624790+0i$
$ \sqrt B = 3.166218219+0.1579171i$
$ \sqrt C = 3+0i$
$ \sqrt D = 3.166218219+0.1579171i$.

Let's take
$H = \sqrt A - \sqrt C = 0.3166247903554$
$V = \sqrt B - \sqrt D = 0+0.3158342005574769i$.

H is the horizontal diameter of the image, while V is the vertical diameter. If this is a circle, these should have the same magnitude. But H is slightly larger than V; the circle gets compressed slightly more in the direction perpendicular to the displacement from the origin. For circles far from the origin, this difference is small, and the image appears to be made up of circles. But as you get closer to the origin, the distortion gets larger and larger. By the time you're including the origin, the distortion is quite substantial. When you include the origin in the original region, the image consists of two overlapping regions, but those two regions no longer look anything like circles, and so the whole region doesn't look like two overlapping circles.

Answer 2

In the first diagram, you have the graph of $$|z-2i|=3$$ together with the transformation $$w^2=z\implies(u+iv)^2=x+iy$$ from which we obtain$$x=u^2-v^2$$ and $$y=2uv$$

So the image of the circle, which in Cartesian form is $$x^2+(y-2)^2=9$$ is then $$(u^2-v^2)^2+(2uv-2)^2=9$$

This is the curve plotted in the thicker line on the same diagram.

Answer 3

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).

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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 $45°$.)