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Let z(1) and z(2) be complex numbers suc...

Let `z_(1)` and `z_(2)` be complex numbers such that `z_(1)^(2)-4z_(2)=16+20i` and the roots `alpha` and `beta` of `x^(2)+z_(1)x+z_(2)+m=0` for some complex number `m` satisfies `|alpha-beta|=2sqrt(7)`.
The locus of the complex number `m` is a curve

A

A. straight line

B

B. circle

C

C. ellipse

D

D. hyperbola

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` We have `alpha+beta=-z_(1)` and `alphabeta=z_(2)+m`
`(alpha-beta)^(2)=(alpha+beta)^(2)-4alphabeta`
`=z_(1)(^(2)-4z_(1)-4m`
`=16+20i-4m`
Since`|alpha-beta|=2sqrt(7)`, we have `|4+5i-m|=7`……`(i)`
Therfore, `m` lies on a circle having centre at `(4,5)` and radius `=7`.

`|m|_(max)=OB=7+sqrt(41)`
Maximum argument of `m` is `pi` which occurs at point `A`,
Let `m=x+iy`
For point `A`, `m=x`
So from `(i)`, `|4+5i-x|=7`
or `(x-4)^(2)=24`
`impliesx-4=+-2sqrt(6)`
`impliesx=4+-2sqrt(6)`
For point `A`, `x=2sqrt(6)-4`
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