JEE Advanced · Mathematics · 3. Complex Numbers
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
- A 48
- B 32
- C 40
- D 80
Answer & Solution
Correct Answer
(A) 48
Step-by-step Solution
Detailed explanation
Since, \(z \bar{z}\left(z^2+\bar{z}^2\right)=250\)
\(\begin{aligned}& \Rightarrow 2\left(x^2+y^2\right)\left(x^2-y^2\right)=350 \\& \Rightarrow \left(x^2+y^2\right)\left(x^2-y^2\right)=175
\end{aligned}\)
Since, \(x, y \in I\), the only possible case which gives integral solution, is and
\( \begin{aligned} x^2+y^2 & =25 \\ x^2-y^2 & =7\end{aligned}\)
From Eqs. (i) and (ii), we get \(x^2=16 ; y^2=9 \Rightarrow x=\pm 4 ; y=\pm 3\)
\(\therefore\) Area of rectangle \(=8 \times 6=48\)
\(\begin{aligned}& \Rightarrow 2\left(x^2+y^2\right)\left(x^2-y^2\right)=350 \\& \Rightarrow \left(x^2+y^2\right)\left(x^2-y^2\right)=175
\end{aligned}\)
Since, \(x, y \in I\), the only possible case which gives integral solution, is and
\( \begin{aligned} x^2+y^2 & =25 \\ x^2-y^2 & =7\end{aligned}\)
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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