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.

If $z=x+yi$, you have, equivalent to $|z^2-1|<1$,
$$
(x^2-y^2-1)^2+4x^2y^2<1
$$
or
$$
(x^2-y^2)^2+4x^2y^2-2(x^2-y^2)<0
$$
that can also be written as
$$
(x^2+y^2)^2-2(x^2-y^2)<0
$$
which is the set of interior points to a Bernoulli lemniscate:

