If `z_1a n dz_2` both satisfy `z+ z =2|z-1|` and `a r g(z_1-z_2)=pi/4` , then find `I m(z_1+z_2)` .

Options

A. 0
B. 1
C. 2
D. none of these

Answer

C

Explanation

Let `z_(1)=a_(1)+ib_(1)` and `z_(2)=a_(2)+ib_(2)` be the roots of `z+barz=2|z-1|`. Then,
`z_(1)+barz_(1)=2|z_(1)|` and `z_(2)+barz_(2)=2|z_(2)-1|`
`rArr 2a_(1)=2sqrt((a-(1)-1)^(2)+b_(1)^(2))` and `2a_(2)=sqrt(a_(2)-1)^(2)+b_(2)^(2)`
`rArr 2a_(1)=1+b_(1)^(2)` and `2a_(2_=1+b_(2)^(2)`
`rArr 2(a_(1)-a_(2))=b_(1)^(2)-b_(2)^(2)`
`rArr 2(a_(1)-a_(2))=(b_(1)+b_(2))(b_(1)-b_(2))`..................(i)
We have, `"arg"(z_(1)-z_(2))=pi/4`
`rArr tan^(-1)(b_(1)-b_(2))/(a_(1)-a_(2))=pi/4`
`rArr (b_(1)-b_(2))/(a_(1)-a_(2)) =1 rArr b_(1)-b_(2)=a_(1)-a_(2)`.............(ii)
From (i) and (ii), we obtain `b_(1)+b_(2)=2`
Hence, `"Im"(z_(1)+z_(2))=2`.