If Re `(lambda Z + (2)/(Z))` is zero and Re(z) `!= 0` and |Z| = 2. Then `lambda` can be

Options

A. 2
B. `-2`
C. `(1)/(2)`
D. `-(1)/(2)`

Answer

D

Explanation

Let z = x + yi .
`= lambda (x+yi) + (2(x-yi))/(x^(2) + y^(2))`
` = x lambda + (2x)/(x^(2) + y^(2)) = 0 " "`[Re part =0]
` =lambda = - (2)/(|z|^(2)) rArr " " lambda = - (2)/(4) , " " lambda = - (1)/(2)`