<s>[INST] Describe the region $\{|z^2 - 1|<1\}$ of the complex plane Since it's complex analysis, I assume there is an easy solution without much algebra.  But even in my algebra heavy solution there should be a more streamlined approach...  Either approach is appreciated as an answer, or verification that my approach is the standard one, thanks!

My Attempt:

We can equivalently look at the condition $|z^2 - 1|^2 - 1< 0$.  Note that $z^2 - 1 = (x^2 - y^2 - 1) +i(2xy)$, so 
\begin{align*}
|z^2 - 1|^2 - 1 & = (x^2 - y^2 - 1)^2 + 4x^2 y^2 - 1 \\
& = x^4 + y^4 -2x^2 - 2y^2 + 2x^2 y^2 \\
& = (x^2 + y^2)(x^2 + y^2 - 2) \\
& <0
\end{align*}
Case 1: $x^2 + y^2 < 0$ and $x^2 + y^2 - 2 >0$
This case is not possible
Case 2: $x^2 + y^2 > 0$ and $x^2 + y^2 - 2 <0$
Here, at most one of $x$ and $y$ can be zero, so $z\neq 0$.  Further, $x^2 + y^2 <2$ is the open disk radius $2$ centered at $0$
Therefore, The region is the punctured open disk $D(0,2)\setminus \{0\}$
 [/INST] Credit to @Reinhard Meier in the comments

There is an algebra heavy answer that you can find to give an exact, graphable answer: $$ x^4 - 2x^2 + y^4 + 2y^2 + 2x^2 y^2 < 0 $$
Although we will take another approach.  Rewrite the condition to be $$ |(z+1)(z-1)|<1 \implies |z+1|\cdot|z-1|<1  $$
We can see clearly 3 things:


*

*$z\neq 0$

*If $z\in\mathbb{R}$, then $0< z < \sqrt{2}$

*The region is has 2 ``centers'' $z=1$ and $z=-1$, and for any point in the region the product of distances to $1$ and $-1$ is never greater than $1$
From this we can describe the region as the interior of a "figure 8" that is is "centered" at the complex numbers $1$ and $-1$, and is not perfectly circular due to the third bullet point
Here is a picture of the region, obtained by drawing random points in the $[-2,2] \times [-2,2]$ square and painting them according to the desired
inequality.
R code:
n = 1e4
x = runif(n,-2,2)
y = runif(n,-2,2)
i = complex(real=0,imaginary=1)
z = x+i*y
ind = abs(z^2-1)<1
plot(x,y,col=ind+1,xlab='Re(z)',ylab='Im(z)')

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