Question

Let $z=x+iy$ be a complex number where $x,y$ are integers then the area of the rectangle whose vertices are roots of the equation $z{\bar{z}}^3+\bar{z}{z}^3=1088$, is

• A. $15$
• B. $30$
• C. $60$
• D. None of these

Answer 1

Answer: (C)

Given, $z{\bar{z}}^3+\bar{z}{z}^3=1088$
$\Rightarrow z{\bar{z}}^2(z^2+{\bar{z}}^2)=1088$
$\Rightarrow z\bar{z}\{(z+\bar{z}{)}^2-2z\bar{z}\}=1088$
$\Rightarrow (x^2+y^2)[(2Re(z){)}^2-2(x^2+y^2)]=1088$
$\Rightarrow (x^2+y^2)(4x^2-2x^2-2y^2)=1088$
$(\because z+\bar{z}=2Rez=2x)$
$\Rightarrow (x^2+y^2)(x^2-y^2)=544=34\times 16$
or $(32\times 17$ which is impossible)
$\Rightarrow x^2+y^2=34$ and $x^2-y^2=16$
$\Rightarrow (x,y)=(\pm 5,\pm 3)=(5,-3),(-5,3),(-5,-3)$ & $(5,-3)$

$\therefore$ Required area (area of rectangle)
$=4(x\times y)=4\times 5\times 3=60$ units