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.
 A: If you compute the square of the modulus, you get 
$$r^4\cos(2\theta)^2 -2r^2\cos(2\theta) + 1  +r^4\sin(2\theta)^2  < 1$$
Cancellations ...
$$ 2r^2\cos(2\theta) >r^4$$
$$ r^2<2\cos(2\theta)$$
Now find the values of $r,\theta$ where this is true...
A: Hint
Taking the square, we get after simplifying
$r^2<2\cos(2\theta)$
The set is inside the curve defined by its polar equation
$r=\sqrt{2\cos(2\theta)}$
A: If you use $re^{i\theta}$ as the polar representation than you get $$|r^2e^{2i\theta}-1|<1\Rightarrow
|(re^{i\theta}+1)(re^{i\theta}-1)|<1$$ which might be enlightening....
A: 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?
A: 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:

