Let `a , b ,xa n dy` be real numbers such that `a-b=1a n dy!=0.` If the complex number `z=x+i y` satisfies `I m((a z+b)/(z+1))=y` , then which of the following is (are) possible value9s) of x?

Options

A. `1-sqrt(1+y^2)`
B. `-1sqrt(1-y^2)`
C. `1+sqrt(1+y^2)`
D. `-1+sqrt(1-y^2)`

Answer

B, D

Explanation

`(az+b)/(z+1)=(ax+b+aiy )/((x+1)+iy)=((ax+b+aiy)((x+1)-iy))/((x+1)^2+y^2)`
`therefore Im((az+b)/(z+1))=(-(ax+b)y+ay(x+1))/((x+1)^2+y^2)`
`rArr ((a-b)y)/((x+1)^2+y^2)=y`
`because " "a-b=1`
`therefore ( x+1)^2+y^2=1`
`therefore x=1pmsqrt(1-y^2)`