Question

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 $z{\overline{z}}^3+\overline{z}{z}^3=350$, is

• A. 48
• B. 32
• C. 40
• D. 80

Answer 1

The correct option is A

48

Since, $z{\overline{z}}^3+\overline{z}{z}^3=350z\overline{z}(z^2+{\overline{z}}^2)=350$
$\Rightarrow 2(x^2+y^2)(x^2-y^2)=350\Rightarrow (x^2+y^2)(x^2-y^2)=175$
Since, $x,yϵI$, 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)
$x^2=16,y^2=9\Rightarrow x=\pm 4,y=\pm 3$
$\therefore$ Area of rectangle $=8\times 6=48$.