InterviewSolution
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InterviewSolution
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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 value of `|a-b|` isA. `5sqrt(5)`B. `sqrt(130)`C. 12D. `sqrt(175)` |
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Answer» Correct Answer - B 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. |
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