# Plot of a domain in the complex plane

I am trying to plot the following domain in the complex plane:

$\lbrace x\in\mathbb{C}|\: |x^{2}-1|<r\rbrace$

for some $r>1$.

I know that in general to take a square root of a complex number $z$ one would have to go to polar coordinates and the get $\sqrt{z}=r^{1/2}\exp (i\theta/2)$.

How do I proceed in taking a "square root of a disk"? in the sense that I can also rewrite it as $\lbrace \sqrt{w}|\: |w-1|<r\rbrace$. What kind of geometrical object is it? Still a disk?

(it's not a homework question, I'm just curious)

• I do not understand your question. What is this square root of a disk in relation to the set you have given? – BoZenKhaa Apr 14 '14 at 8:49
• well, my question is, geometrically what is the set of all $x\in\mathbb{C}$ satisfying $|x^{2}-1|<r$? – GregVoit Apr 14 '14 at 8:54
• @BoZenKhaa I agree it was a bit vague. Just edited the question – GregVoit Apr 14 '14 at 8:56

Let be $x=u+iv:$ $$\sqrt{(u^2-v^2-1)^2+(2uv)^2}=|u^2-v^2-1+2uvi|=|x^2-1|<r.$$
• Thanks, but from what I understand this is just a different algebraic way of writing my inequality using $x=u+iv$. But what kind of object is the inequality? Is it a disk? Half a disk? @Martín-Blas Pérez Pinilla – GregVoit Apr 14 '14 at 10:08
• The border is $(u^2-v^2-1)^2+(2uv)^2=r^2$. Expand and find $v$ as function of $u$ (you can, is a biquadratic equation). – Martín-Blas Pérez Pinilla Apr 14 '14 at 10:17
• Just one more question @Martín-Blas Pérez Pinilla. Now the two curves $v(u)$ I got are $v^{2}=-1-u^{2}+\sqrt{4u^{2}+r^{2}}$ (the other two roots don't exist). I then plotted it in Mathematica for $r=2.5$ and $u$ between -2 and 2. Got something like "oval shape". However, by the structure of the equation, that's clearly not an ellipse, right? So what would be the correct name for the shape? – GregVoit Apr 16 '14 at 9:36