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If w=alpha+ibeta, where beta!=0 and z!=1...

If `w=alpha+ibeta,` where `beta!=0` and `z!=1` , satisfies the condition that `((w- bar w z)/(1-z))` is a purely real, then the set of values of `z` is `|z|=1,z!=2` (b) `|z|=1a n dz!=1` (c)`z=bar z ` (d) None of these

A

|z| = 1 and z = 2

B

|z| = 1 and ` z ne 1`

C

`z = bar(z)`

D

None of these

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The correct Answer is:
B
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