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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- barw 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|=1,z ne 2`

B

`|z|=1 and z ne 1`

C

`z= barz `

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
Let `z_1=(w-barw z )/(1-z)` be purely real `rArr z_1=barz_1`
`therefore " "(w-barwz)/(1-z)=(barw-w bar z)`
`rArr w-barwz-barwz+barwz.z=barw-zbarw-wbarz+wz.barz`
`rArr (w-barw)+(barw-w)|z|^2=0`
`rArr (w-barw)(1-|z|^2)=0`
`rArr |z|^2=1[as w-barw ne 0,since betane 0 ]`
`rArr |z|=1 and z ne 1`
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