The roots of the equation `z^(4) + az^(3) + (12 + 9i)z^(2) + bz = 0` (where a and b are complex numbers) are the vertices of a square. Then The area of the square is

Options

A. 25 sq.units
B. 20 sq.units
C. 5 sq.unit
D. 4 sq .units

Answer

C

Explanation

Given equation is `z(z^(3) + az^(3) +(12 + 9i)z + b) =0`
So , either z =0
or `z^(3) + az^(2) + (12 + 9i) z + b = 0`
So, one of the vertice of the square is origin .
Therefore, other three verticles are `z_(1),iz_(1)` and `(z_(1) + iz_(1))` which are the roots equation (1).
`therefore z_(1)(iz_(1)) + (iz_(1))(z_(1) + iz_(1)) + z_(1)(z_(1) + iz_(1)) = 12 +9i`
`rArr_(1)^(2)(i + i+ i^(2)+1+i)= 12+9i`
`rArr 3iz_(1)^(2) = 12+9i`
`rArr z_(1)^(2) = 3 - 4i`
` rArr z_(1) = sqrt(3-4 i) = pm (2-i)`
`therefore z_(1) = 2 - i or -2 +i or -2 + i ` (both value will give the same result )
Also, `-a= z_(1) + iz_(1) + z_(1)(1+i) = 2z_(1)(1+i)`
`= 2(2-i) (1+i) = 6 + 2i`
` therefore a= - 6- 2i`
Futher, `(z_(1))(iz_(1))(1+i) z_(1) = b`
`therefore - b = z_(1)^(3)(-1+i)`
or `b = (2-i)^(3) (1-i) = -9-13i`
` b = -9-13i`
`|z - b|=|3+11i|= sqrt(130)`
`|z_(1)| = sqrt(5)`
Therefore, area of square is 5sq. units.