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

If `w=alpha+ibeta`, where `beta ne 0` and `z ne 1`, satisfies the condition that `((w-barwz)/(1-z))` is purely real, then the set of values of z is

A

`{z:abs(z)=1}`

B

`{z:z=bar(z)}`

C

`{z:z ne 1}`

D

`{z:abs(z)=1,z ne 1}`

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The correct Answer is:

D

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