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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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Knowledge Check

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

    A
    `{z : |z| =1}`
    B
    `{ z:z=bar(z) }`
    C
    `{ z : z ne 1}`
    D
    `{ z : |z| = 1, z ne1}`

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

    A
    `|z|=1,z ne 2`
    B
    `|z|=1 and z ne 1`
    C
    `z= barz `
    D
    None of these

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