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 $D(0,2)\setminus \{0\}$

I can't say if this is "easier" or not, but here is my approach.
Let $w = z^2$. Then $\{w: |w-1|<1\}$ is a circle of radius $1$ centered at $(1,0)$. The mapping $w: z \mapsto z^2$ transforms the original set into this circle.
In polar form, $z = r_ze^{i\theta_z}$, and $w = r_we^{i\theta_w} = r_z^2e^{i2\theta_z}$.
You can find the polar form of the circle
$$ (x_w-1)^2 + y_w^2 = 1 \implies r_w^2 - 2r_w\cos(\theta_w) = 0 \implies r_w = 2\cos(\theta_w) $$
Converting back to $z$-space gives the polar form
$$ r_z^2 = 2\cos(2\theta_z) \implies r_z = \sqrt{2\cos(2\theta_z)} $$
Note that $w^{1/2}$ has two roots, so the left and right halves of your $z$-plot both map to the same $w$-circle. The right half maps to $-\pi < \arg(w) < \pi$ and the left half maps to $\pi < \arg(w) < 3\pi$
