Let \(z=x+y i\), where, \(x, y\) are integers and \(i=\sqrt{-1}\) the area of the rectangle whose vertices…
Let \(z=x+y i\), where, \(x, y\) are integers and \(i=\sqrt{-1}\) the area of the rectangle whose vertices are the roots of the equation \(\bar{z} z^3+z(\bar{z})^3=700\) is
32
40
48
80
Solution
It is given, for a complex number \(z=x+i y\),
where \(x, y\) are integers such that
\(\begin{aligned}
& \bar{z} z^3+z(\bar{Z})^3=350 \\
& \Rightarrow \bar{Z} \bar{Z}\left(z^2+(\overline{(})^2\right)=350 \\
& \Rightarrow\left(x^2+y^2\right) 2\left(x^2-y^2\right)=350 \quad {\left[\because z \bar{z}=|z|^2\right]} \\
& \Rightarrow \quad\left(x^2+y^2\right)\left(x^2-y^2\right)=175 \\
& \therefore \quad x^2+y^2=25 \text { and } \\
& x^2-y^2=7 \quad {[\because x, y \in \text { Integer}]} \\
\end{aligned}\)
So, \(x^2=16 \text { and } y^2=9\)
Therefore vertices of the rectangle are \((\pm 4, \pm 3)\), so area of the rectangle is \(4 \times 4 \times 3=48\).
Hence, option (c) is correct.