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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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You made the same mistake here as the one I pointed out in your previous question: $|z^2-1|^2=\lambda^2$ (not $|z^2-1|$). –  joriki Sep 28 '12 at 9:56
    
Ups. You right. Thankyou @joriki. But is the rest correct? –  Mykolas Sep 28 '12 at 9:58
    
The last equation looks OK, but in the middle one it should be $-2x^2y^2$. –  joriki Sep 28 '12 at 10:02
    
OK. That's just typeing mistake. Is there anyway to see this geometricly directly without getting all this equations? –  Mykolas Sep 28 '12 at 10:04
    
I don't know of one. –  joriki Sep 28 '12 at 10:09

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