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 ? z ? 3 + z z 3 = 350 is [2009]

Question

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 [2009]

Smart Achievers · Accepted Answer

(1)

$Given: z = x + i y, where x and y are integers$

$Also, z {\bar{z}}^{3} + \bar{z} z^{3} = 350 \Rightarrow | z |^{2} ({\bar{z}}^{2} + z^{2}) = 350$

$\Rightarrow (x^{2} + y^{2}) (x^{2} - y^{2}) = 175$

$\Rightarrow (x^{2} + y^{2}) (x^{2} - y^{2}) = 25 \times 7 . . . (i)$

$or (x^{2} + y^{2}) (x^{2} - y^{2}) = 35 \times 5 . . . (i i)$

$∵$ $x, y are integers$

$∴$ $x^{2} + y^{2} = 25 and x^{2} - y^{2} = 7 [From eq (i)]$

$\Rightarrow x^{2} = 16, y^{2} = 9$

$\Rightarrow x = \pm 4, y = \pm 3$

$∴$ $Vertices of rectangle are$

$(4, 3), (4, - 3), (- 4, - 3), (- 4, 3)$

$∴$ $Area of rectangle = 8 \times 6 = 48 sq. units$

$Now from eq. (ii),$

$x^{2} + y^{2} = 35 and x^{2} - y^{2} = 5$

$\Rightarrow x^{2} = 20, which is not possible for any integral value of x .$

Solution

Q.
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 [2009]

1 48

2 32

3 40

4 80

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