1.

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

Answer» Correct Answer - A
`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`


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