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$|z^{2}-1|=\lambda$

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?

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    Ok. Thankyou one more time @joriki2012-09-28

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