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 $\bar{z}{z}^3+z{\bar{z}}^3=350$ is

• A. 48
• B. 32
• C. 40
• D. 80

Answer 1

Answer: (A)

Given, $z=x+iy$
Now, $\bar{z}{z}^3+z{\bar{z}}^3=350$
$\Rightarrow \left(\bar{z}z\right)z^2+\left(z\bar{z}\right){\bar{z}}^2=350$
$\Rightarrow {\left|z\right|}^2{\left(x+iy\right)}^2+{\left|z\right|}^2{\left(x+iy\right)}^2=350$
$\Rightarrow \left(x^2+y^2\right)$
$\left[x^2-y^2+2ixy+x^2-y^2-2ixy\right]=350$
$\Rightarrow 2\left(x^2+y^2\right)\left(x^2-y^2\right)=350$
$\Rightarrow x^4-y^4=175$
$\Rightarrow x=\pm 4$,
$y=\pm 3$
$\therefore$ Vertices are $\left(-4,-3\right),\left(-4,3\right),\left(4,-3\right)$ and $\left(4,3\right)$
$\therefore$ Length and breadth are $8$ and $6$
$\therefore$ Area of the rectangle $=8\times 6=48$