Let $z=x+i y$ be a complex number with $x, y \in Z$. Then, the area (in sq units) of the rectangle whose…
Let $z=x+i y$ be a complex number with $x, y \in Z$. Then, the area (in sq units) of the rectangle whose vertices are the roots of the equation $\bar{z} \cdot z^3+z \cdot \bar{z}^3=350$ is
$48$
$32$
$40$
$44$
Solution
$\bar{z} z^3+z \bar{z}^3=350$
$\Rightarrow z \bar{z}\left(z^2+\bar{z}^2\right)=350$
Let $\quad z=x+i y$
$\begin{array}{lc}\Rightarrow & \bar{z}=x-i y \\ \therefore & (x+i y)(x-i y)\left[(x+i y)^2+(x-i y)^2\right]=350\end{array}$
$\begin{aligned} & \Rightarrow \quad\left(x^2+y^2\right) \cdot 2\left(x^2-y^2\right)=350 \\ & \Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=175=25 \times 7\end{aligned}$
$\Rightarrow \quad x^2+y^2=25$...(i)
and $x^2-y^2=7$...(ii)
On adding Eqs. (i) and (ii), $2 x^2=32$
$\begin{array}{ll}\Rightarrow & x^2=16 \Rightarrow x= \pm 4 \\ \text { and } & y^2=9 \Rightarrow y= \pm 3\end{array}$
$\therefore$ Vertices of the rectangle are
$(4,3),(4,-3),(-4,-3)$ and $(-4,3)$.
$\begin{gathered}A B=6 \text { and } A D=8 \\ \therefore \text { Required area }=6 \times 8\end{gathered}$
$=48$ sq. units
