$|z^{2}-1|=\lambda$
My aprouch: taking $z=x+iy$ $$z^{2}=(x^{2}-y^{2})+i(2xy)$$ Then : $$|z^{2}-1|^{2}=x^{4}-2x^{2}y^{2}+2y^{2}-2x^{2}+y^{4}+1+4x^{2}y^{2}=\lambda^{2}$$ Now taking $x^{2}=\alpha,y^{2}=\beta$: $$\alpha^{2}+2\alpha\beta+\beta^{2}-2\alpha+2\beta+(1-\lambda^{2})=0$$ This is equation of a parabola since the discriminant is equal to $0$ Would this be correct? And is there, maybe more "geometrical" way to see this?