Plot $|z^2-1|=1$

Question

$$|z^2-1|=1$$

$|x^2-y^2-1+2xyi|=1$

$\sqrt{(x^2-y^2-1)^2+(2yx)^2}=1$

${(x^2-y^2-1)^2+4x^2y^2}=1$

$x^4+y^4+2x^2y^2+2x^2-2y^2=1$

$(x^2+y^2)^2=2(x^2-y^2)+1$

I am trying to bring it to the form of $(x^2+y^2)^2=2a(x^2-y^2)$ but cant get rid of the $1$

Answer 1

Hint

$$(x^2-y^2-1)^2+4x^2y^2=1\to (x^2-y^2)^2-2(x^2-y^2)+1+4x^2y^2=1$$

$$(x^2+y^2)^2-2(x^2-y^2)=0\to (x^2)^2+x^2(2y^2-2)+y^4+2y^2=0$$