Understanding solution set for $|z^2 - 1| < 1$ for $z \in \mathbb{C}$

Question

I've been having some trouble seeing where the solution for the above problem comes from. The given solution is for $z = r(\cos\theta + i\sin\theta)$ that $r^2< 2\cos2\theta$ and $r \neq 0$.

I'm not sure how to make any progress from the polar representation below, any hints or new directions would be great:

$|z^2 -1| = |r^2(\cos2\theta + i\sin2\theta)-1| < 1$

Answer 1

Note $|z^2 - 1| = |z-1||z+1|$. So your solution set is the interior of a Cassini oval with foci at $1$ and $-1$.

Answer 2

$$|z^2-1| < 1 \iff |z^2-1|^2 <1.$$ Now, $z^2-1 = r^2 (\cos(2\theta)+i\sin(2\theta))-1$, and $|z^2-1|^2 = (z^2-1)(\overline{z^2-1})$. Therefore, $$|z^2-1|^2 = (r^2 (\cos(2\theta)+i\sin(2\theta))-1)(r^2 (\cos(2\theta)-i\sin(2\theta))-1)=r^4-2r^2\cos(2\theta)+1.$$ Consequently, $$|z^2-1|^2 <1 \iff r^4 -2 r^2 \cos(2\theta) <0 \iff r^2 < 2\cos(2\theta) \text{ and } r\neq0.$$