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Let $z=x+i y$ be a complex number, where $x$ and $y$ are integers. Then, the area of the rectangle whose vertices are the roots of the equation $z \bar{z}^3+\bar{z} z^3=350$ is
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The correct answer is:
48
48
Since, $z \bar{z}\left(z^2+\bar{z}^2\right)=250$
$$
\begin{aligned}
& \Rightarrow \quad 2\left(x^2+y^2\right)\left(x^2-y^2\right)=350 \\
& \Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=175
\end{aligned}
$$
Since, $x, y \in I$, the only possible case which gives integral solution, is and $\quad \begin{aligned} x^2+y^2 & =25 \\ x^2-y^2 & =7\end{aligned}$
From Eqs. (i) and (ii), we get $x^2=16 ; y^2=9 \Rightarrow x=\pm 4 ; y=\pm 3$
$\therefore$ Area of rectangle $=8 \times 6=48$
$$
\begin{aligned}
& \Rightarrow \quad 2\left(x^2+y^2\right)\left(x^2-y^2\right)=350 \\
& \Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=175
\end{aligned}
$$
Since, $x, y \in I$, the only possible case which gives integral solution, is and $\quad \begin{aligned} x^2+y^2 & =25 \\ x^2-y^2 & =7\end{aligned}$
From Eqs. (i) and (ii), we get $x^2=16 ; y^2=9 \Rightarrow x=\pm 4 ; y=\pm 3$
$\therefore$ Area of rectangle $=8 \times 6=48$
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