IIT-JEE 2009 · Paper 1 · Mathematics Q24

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MathematicsSingle correct+3 / -1Question 24 of 60

Let $z=x+iy$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation
$z\bar z^3+\bar zz^3=350$
is

Answer and worked solution

Official answer: A

Simplify. $z\bar z^3+\bar zz^3=|z|^2\left(z^2+\bar z^2\right)=(x^2+y^2)\cdot2(x^2-y^2)=350$ , so
$(x^2+y^2)(x^2-y^2)=175.$

Integers. $x^2+y^2>x^2-y^2>0$ and both have the same parity; $175=25\times7$ gives $x^2=16$ , $y^2=9$ (175 × 1 gives $x^2=88$ , not a square). The roots are $(\pm4,\pm3)$ .

Area. The rectangle has sides 8 and 6: area 48.

(A)

Topics:Algebra of complex numbers Number systems

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IIT-JEE 2009 · Paper 1, Mathematics question 24