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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|=1`

B

`z=bar(z)`

C

`z=-bar(z)`

D

`z=0`

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The correct Answer is:
A
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