Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# If z = x + iy is 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 (1) 48 (2) 32 (3) 40 (4) 80

Given

$$\begin{array}{l}z\bar{z}^{3}+\bar{z}z^{3}=350\end{array}$$

$$\begin{array}{l}z\bar{z}(\bar{z}^{2}+z^{2})=350\end{array}$$

Take z = x+iy

(x+iy)(x-iy)(x-iy)2+(x+iy)2 = 350

(x2+y2)(2x2-2y2) = 350

(x2+y2)(x2-y2) = 175 = (25)*7

x2+y2 = 25 (i)

x2-y2 = 7 ..(ii)

Adding (i) and (ii), we get

2x2 = 32

x = ±4

And y = ±3

Hence the vertices of the triangle are (4,3), (4, -3), (-4, -3) and (-4,3).

Area of rectangle = length * breadth

= 8*6

= 48

Hence option (1) is the answer.