Is this set region? $$
|z^2 - 1| < 1
$$
Hint: use polar coordinates.
the answer is not a region.
I don't know how to start. Whenever I am trying to do, it failed.
*. z is complex number.
 A: Following the hint, we write $z=re^{i\theta}$ with $r\geq 0$ and $\theta\in[0,2\pi)$. Then
$$
z^2-1=r^2e^{2i\theta}-1=r^2\cos(2\theta)-1+ir^2\sin(2\theta).
$$
Then
$$
|z^2-1|^2=(r^2\cos(2\theta)-1)^2+(r^2\sin(2\theta))^2
$$
$$
=r^4-2r^2\cos(2\theta)+1.
$$
Since $|z^2-1|<1$ if and only if $|z^2-1|^2<1$, you are looking for all $r\geq 0$ and $\theta\in[0,2\pi)$ such that
$$
r^4-2r^2\cos(2\theta)=r^2(r^2-2\cos(2\theta))<0.
$$
This is equivalent to $r>0$ and $r^2-2\cos(2\theta)<0$, i.e
$$
0< r^2
<2\cos(2\theta).$$
Since $\cos(2\theta)$ must be positive, we need to restrict $\theta$ to $[0,\pi/4)\cup (3\pi/4,5\pi/4)\cup (7\pi/4,2\pi)$. Then
$$
0<r<\sqrt{2}\sqrt{\cos(2\theta)}.
$$
For $\theta\in (3\pi/4,5\pi/4)$, we get an open connected bounded set. For $\theta\in [0,\pi/4)\cup(7\pi/4,2\pi)$, we get another open connected bounded set. But the intersection is empty. So your set is open, bounded, and has two connected components.
A: Hint also in rectangular coordinates (good'ol analytic geometry). Put $\,z=x+yi\,$ :
$$|z^2-1|^2<1^2=1\iff (x^2-y^2 -1)^2+(2xy)^2<1\iff $$
$$(x^2+y^2)^2-2(x^2-y^2)<0$$
Having a nice geometric vision the above is the difference between a circle-like region and the intersection of two straight lines and, thus not connected.
But, of course, Julien's answer is neater.
