Let z be a complex number satisfying `z^(2) + 2zlambda + 1=0` , where `lambda ` is a parameter which can take any real value.
The roots of this equation lie on a certain circle if

Options

A. `-1 lt lambda lt 1`
B. `lambda gt 1`
C. `lambda lt 1`
D. none of these

Answer

A

Explanation

`z = - lambda pm sqrt(lambda^(2) -1)`
Case I:
When `-1 lt lambdalt 1`, we have
`lambda^(2) lt 1 rArr lambda^(2) - 1 lt 0`
`z = - lambda pm isqrt(1-lambda^(2))`
`rArr y^(2) = 1 - x^(2) or x^(2) y + y^(2) = 1`
Case II:
`lambda gt 1 rArr lambda^(2) - 1 gt0`
` z=- lambda pm sqrt(lambda^(2) -1)`
`or x = - lambda pm sqrt(lambda^(2) -1),y = 0`
Roots are`(-lambda + sqrt(lambda^(2) -1,0),(-lambda - sqrt(lambda^(2))-1,0)`.One root lies inside the units circle and the other root lies outside the unit circle.
Case III: When `lambda` is very large, then
`z = - lambda- sqrt(lambda^(2) -1) ~~ - 2lambda`
`z=- lambda+ sqrt(lamda^(2) -1) =((-lambda + sqrt(lambda^(2) -1))(-lambda -sqrt(lambda^(2)-1)))/((-lambda - sqrt(lambda^(2) -1)))`
`= (1)/(-lambda-sqrt(lambda^(2)-1)) =- (1)/(2lambda)`