Let w=α+i β, β ≠ 0 and z ≠ 1. If (w- barw z/1-z) is purely real, then the set of value of z is

Question

Let w=α+i β, β ≠ 0 and z ≠ 1. If (w- barw z/1-z) is purely real, then the set of value of z is (A)  z:|z|=1  (B)  z: barz=z  (C)  z:|z| ≠ 1  (D)  z:|z|=1, z ≠ 1

Tardigrade Admin · Accepted Answer

As (w- barw z/1-z) is purely real, (w- barw z/1-z)=( barw-w barz/1- barz) ⇒(1- barz)(w- barw z)=(1-z)( barw-w barz) ⇒(w- barw)(1-z barz)=0 text As w ≠ barw, text we get z barz=1 Thus, set of values of z is z:|z|=1, z ≠ 1 .

Q. Let $w = α + i β, β \neq = 0$ and $z \neq = 1$ . If $\frac{w - w ˉ z}{1 - z}$ is purely real, then the set of value of $z$ is

${z : ∣ z ∣ = 1}$

${z : \overset{z}{ˉ} = z}$

${z : ∣ z ∣ \neq = 1}$

${z : ∣ z ∣ = 1, z \neq = 1}$

Q. Let w=α+iβ,β=0 and z=1. If 1−zw−wˉz​ is purely real, then the set of value of z is

{z:∣z∣=1}

{z:zˉ=z}

{z:∣z∣=1}

{z:∣z∣=1,z=1}

As 1−zw−wˉz​ is purely real, 1−zw−wˉz​=1−zˉwˉ−wzˉ​ ⇒(1−zˉ)(w−wˉz)=(1−z)(wˉ−wzˉ) ⇒(w−wˉ)(1−zzˉ)=0 As w=wˉ, we get zzˉ=1 Thus, set of values of z is {z:∣z∣=1,z=1}.

Q. Let $w = α + i β, β \neq = 0$ and $z \neq = 1$ . If $\frac{w - w ˉ z}{1 - z}$ is purely real, then the set of value of $z$ is

${z : ∣ z ∣ = 1}$

${z : \overset{z}{ˉ} = z}$

${z : ∣ z ∣ \neq = 1}$

${z : ∣ z ∣ = 1, z \neq = 1}$