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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Mathematics
Chapter
Complex Number

JEE Advanced 2009 · Paper 1 · Mathematics · Complex Number

Let z=x+i y be a complex number, where x and y are integers. Then,...

Let z=x+iyz=x+i y be a complex number, where xx and yy are integers. Then, the area of the rectangle whose vertices are the roots of the equation zzˉ3+zˉz3=350z \bar{z}^3+\bar{z} z^3=350 is

Options

  1. A48
  2. B32
  3. C40
  4. D80

Correct answer

A. 48

Step-by-step solution

Since, zzˉ(z2+zˉ2)=250z \bar{z}\left(z^2+\bar{z}^2\right)=250⇒2(x2+y2)(x2−y2)=350⇒(x2+y2)(x2−y2)=175\begin{aligned}& \Rightarrow \quad 2 \left(x^2+y²\right)\left(x^2-y²\right)=350 \\ & \Rightarrow \quad \left(x^2+y²\right)\left(x^2-y²\right)=175 \end{aligned}Since,Since,x, y \in I,theonlypossiblecasewhichgivesintegralsolution,isand, the only possible case which gives integral solution, is and\quad x2+y2=25x2−y2=7\begin{aligned} x^2+y^2 & =25 \\ x^2-y^2 & =7\end{aligned} From Eqs. (i) and (ii), we getx^2=16 ; y^2=9 \Rightarrow x=\pm 4 ; y=\pm 3\therefore Area of rectangle =8 \times 6=48
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