Question

Let $z=x+iy$ be a complex number where $x$ and $y$ are intergers. The area of the rectangle whose vertices are the roots of the equation $\overline{z}{z}^3+z{\overline{z}}^3=350$ is

• A. $48\text{sq. units}$
• B. $32\text{sq. units}$
• C. $40\text{sq. units}$
• D. $80\text{sq. units}$

Answer 1

The correct option is A $48\text{sq. units}$

Given equation is $\overline{z}{z}^3+z{\overline{z}}^3=350$
$\Rightarrow z\overline{z}({\overline{z}}^2+z^2)=350$
Putting $z=x+iy$, we get
$\Rightarrow (x+iy)(x-iy)[2x^2-2y^2]=350\Rightarrow (x^2+y^2)(x^2-y^2)=175$
By observation, we write
$\Rightarrow (x^2+y^2)(x^2-y^2)=25\times 7$
Or,
$\Rightarrow (x^2+y^2)(x^2-y^2)=175\times 1$
Or,
$\Rightarrow (x^2+y^2)(x^2-y^2)=35\times 5$

We only get integral values of $x$ and $y$, when
$x^2+y^2=25,x^2-y^2=7$
Therefore,
$x=\pm 4,y=\pm 3$


The length of rectangle $=8$
The width of rectangle $=6$

Hence, the area of rectangle $=6\times 8=48\text{sq. units}$