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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- 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` `z= z ` (d) None of these

A

`{z::|z|=1}`

B

`{z::z=barz}`

C

`{z::zzin1}`

D

`{z::|z|=1,z in1}`

Text Solution

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The correct Answer is:
D
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