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 $48$

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

$z\overline{z}({\overline{z}}^2+z^2)=350$

Let $z=x+iy$

$\therefore (x+iy)(x-iy)\left(\right(x-iy{)}^2+(x+iy{)}^2)=350$

$(x^2+y^2)(2x^2-2y^2)=350$

$(x^2+y^2)(x^2-y^2)=175=25\ast 7$

$x^2+y^2=25$ ..... $\left(1\right)$

$x^2-y^2=7$ ....... $\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$, we get

$2x^2=32⟹x=\pm 4$

and $y=\pm 3$

Therefore the vertices of the rectangle are $(4,3),(4,-3),(-4,-3)$ and $(-4,3)$

Area of the rectangle $=length\times breadth=8\times 6=48$ sq.units