Describe the set whose points satisfy the following relation: $|z^2 - 1| < 1$ There is a hint which states to use polar coordinates, but I feel like that complicates the problem more. As far as trying it myself, I get lost very early on. If we take $z = r(\cos{\theta} + i\sin{\theta})$, then we have
$|r^2(\cos{2\theta} + i\sin{2\theta}) - 1| < 1$
But I have no idea how to find the modulus of this point with that extra $-1$ in there.

Notice that $|a| < 1$ if and only if ${|a|}^2 < 1$. Moreover, ${|a+ib|}^2=a^2+b^2$. Hence, we're looking for $z$ such that:
$${\left|\big(r^2\cos(2\theta)-1\big) + i\sin(2\theta)\right|}^2={\big(r^2\cos(2\theta)-1\big)}^2+{\sin(2\theta)}^2\\
=r^4{\cos(2\theta)}^2-2r^2\cos(2\theta)+1+{\sin(2\theta)}^2<1\\
\Rightarrow r^4{\cos(2\theta)}^2-2r^2\cos(2\theta)+{\sin(2\theta)}^2<0\\
\Rightarrow r^4{\cos(2\theta)}^2-2r^2\cos(2\theta)+1<{\cos(2\theta)}^2\\
\Rightarrow {\left(r^2\cos(2\theta)-1\right)}^2<{\cos(2\theta)}^2$$
(We used that ${\sin(2\theta)}^2 = 1 - {\cos(2\theta)}^2$)
With $r^2=s$ and $\cos(2\theta)=t$, we may rewrite it as:
$$(st-1)^2<t^2$$
Do you think you can take it from here?
