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 $
$\Rightarrow \left(\bar{z} z\right)z^{2}+\left(z \bar{z}\right)\bar{z}^{2}=350$
$\Rightarrow \left|z\right|^{2} \left(x+iy\right)^{2}+\left|z\right|^{2} \left(x+iy\right)^{2}=350$
$\Rightarrow \left(x^{2}+y^{2}\right)$
$\left[x^{2}-y^{2}+2ixy+x^{2}-y^{2}-2ixy\right]=350$
$\Rightarrow 2\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)=350$
$\Rightarrow x^{4}-y^{4}=175$
$\Rightarrow x=\pm4$,
$y=\pm3$
$\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 $=8\times 6=48$