JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · 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 \overset{z}{ˉ}^{3} + \overset{z}{ˉ} z^{3} = 350

Answer: (A)

Solution

The equation can be rewritten as:

z \overset{z}{ˉ} (\overset{z}{ˉ}^{2} + z^{2}) = 350

Let

z = x + i y

. Then

z \overset{z}{ˉ} = x^{2} + y^{2}

and

\overset{z}{ˉ}^{2} + z^{2} = 2 (x^{2} - y^{2})

.

So:

(x^{2} + y^{2}) \cdot 2 (x^{2} - y^{2}) = 350

(x^{2} + y^{2}) (x^{2} - y^{2}) = 175 = 5 \times 5 \times 7

For integer solutions, we need:

x^{2} + y^{2} = 25, x^{2} - y^{2} = 7

Solving:

2 x^{2} = 32 \Rightarrow x^{2} = 16 \Rightarrow x = \pm 4

2 y^{2} = 18 \Rightarrow y^{2} = 9 \Rightarrow y = \pm 3

The four roots are

(\pm 4, \pm 3)

, forming a rectangle with dimensions

8 \times 6

.
Area

= 8 \times 6 = 48

square units.

JEE Advanced · 2009 · 12 Apr 2009 · Paper 1 · Single correct option