If w=α+iβ,whereβ≠0andz≠1, satisfies the condition that (w−¯wz1−z) is purely real, then the set of values of z is
A
|z|=1,z≠2
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B
|z|=1,andz≠1
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C
z=¯z
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D
None of these
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Solution
The correct option is B|z|=1,andz≠1 Letz1=w−¯wz1−zbepurelyreal⇒z1=¯z1∴w−¯wz1−z=¯w−w¯z1−¯z⇒w−w¯z−¯wz+¯wz.¯z=¯w−z¯w−w¯z+wz.¯z⇒(w−¯w)+(¯w−w)|z|2=0⇒(w−¯w)(1−|z|2)=0⇒|z|2=1[asw−¯w≠0,sinceβ≠0]⇒|z|=1andz≠1