1.

Let z be a complex numbersatisfying z^(2) + 2zlambda + 1=0 , where lambdais a parameter whichcan take any realvalue. Forevery large value of lambda the roots are approximately.

Answer»

`-2lambda, 1//lambda`
`-lambda,-1//lambda`
`-2lambda, -(1)/(2lambda)`
none of these

Solution :`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 pmisqrt(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 pmsqrt(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 verylarge, 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)`


Discussion

No Comment Found

Related InterviewSolutions