InterviewSolution
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InterviewSolution
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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 ifA. `-1 lt lambda lt 1`B. `lambda gt 1`C. `lambda lt 1`D. none of these |
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Answer» Correct Answer - A `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)` |
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