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

Question

Dr. Abubacker · Accepted Answer

(a)
Since, $z\overline{z}\left( z^{2} + {\overline{z}}^{2} ight) = 350$
$\Rightarrow \ \ \ \ \ 2\left( x^{2} + y^{2} ight)\left( x^{2} - y^{2} ight) = 350$
$\Rightarrow \ \ \ \ \ \left( x^{2} + y^{2} ight)\left( x^{2} - y^{2} ight) = 175$
Since, $x,\ y\  \in I$, the only possible case which gives integral solution, is
$x^{2} + y^{2} = 25$ …(i)
$x^{2} - y^{2} = 7$ …(ii)
From Eqs. (i) and (ii), we get
$x^{2} = 16,\ \ \ \ y^{2} = 9$
$\Rightarrow \ \ \ \ \ \ \ \ x = \pm 4,\ \ \ y = \pm 3$
$	herefore$ Area of rectangle$= 8 	imes 6 = 48$

Q. 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

Explanation

Reviewed by Dr. Abubacker

Want more practice?