[JEE Advanced 2009] Let $z=x+iy$ with $x,y$ integers. The area of the rectangle whose vertices are the roots of $z\bar z^3+\bar z z^3=350$ is
Let $z=x+iy$ with $x,y$ integers. The area of the rectangle whose vertices are the roots of $z\bar z^3+\bar z z^3=350$ is
(a) $48$
(b) $32$
(c) $40$
(d) $80$
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 1mo ago
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Correct answer: (a) $48$
$z\bar z^3+\bar z z^3=2|z|^2(x^2-y^2)=350\Rightarrow(x^2+y^2)(x^2-y^2)=175$; integer solution $x^2=16,y^2=9$. Vertices $(\pm4,\pm3)$ give area $8\times6=48$.
JEE Advanced 2009 · Complex Numbers — verified solution by the Vidaara Team.
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