If `k>0`, `|z|=\w\=k`, and `alpha=(z-bar w)/(k^2+zbar(w))`, then `Re(alpha)` (A) 0 (B) `k/2` (C) `k` (D) None of these

Options

A. 0
B. `k//2`
C. k
D. none of these

Answer

A

Explanation

`alpha=(z-barw)/(k^(2)+zbarw)rArr baralpha=(barz-w)/(k^(2)+barzw)`
But `zbarz =wbarw =k^(2)`. Hence,
`rArr " "baralpha=((k^(2))/(z)-(k^(2))/(barw))/(k^(2)+(k^(2))/(z)(k^(2))/(barw))=(barw-z)/(zbarw+k^(2))=-alpha`
`rArralpha+baralpha=0`
`rArr Re(alpha)=0`