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 $z {¯ z}^{3} + ¯ z z^{3} = 350$ , is

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Solution

The correct option is A 48
Since, $z {¯ z}^{3} + ¯ z z^{3} = 350$
$\Rightarrow 2 (x^{2} + y^{2}) (x^{2} - y^{2}) = 350 \Rightarrow (x^{2} + y^{2}) (x^{2} - y^{2}) = 175$
Since, $x, y ϵ I$ , the only possible case which gives integral solution, is
$x^{2} + y^{2} = 25$ ...(1)
$x^{2} - y^{2} = 7$ ...(2)
From Eqs. (1) and (2)
$x^{2} = 16, y^{2} = 9 \Rightarrow x = \pm 4, y = \pm 3$
$∴$ Area of rectangle $= 8 \times 6 = 48$ .

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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¯z3+¯zz3=350, is

The correct option is A 48Since, z¯z3+¯zz3=350 ⇒2(x2+y2)(x2−y2)=350⇒(x2+y2)(x2−y2)=175 Since, x,yϵI, the only possible case which gives integral solution, is x2+y2=25 ...(1) x2−y2=7 ...(2) From Eqs. (1) and (2) x2=16,y2=9⇒x=±4,y=±3 ∴ Area of rectangle =8×6=48.

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 $z {¯ z}^{3} + ¯ z z^{3} = 350$ , is