If w=α+iβ,where β=0and z=1, satisfies the condition that (1−zw−wz)is a purely real, then the set of values of zis ∣z∣=1,z=2 (b) ∣z∣=1andz=1z=z (d) None of these

Name: Video solution 1: If w=α+iβ,where β=0and z=1, satisfies the condition that (1−zw−wz)is a purely real, then the set of values of zis ∣z∣=1,z=2 (b) ∣z∣=1andz=1z=z (d) None of these
Uploaded: 2021-01-01T18:41:02.000Z
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Description: Video solution 1: If w=α+iβ,where β=0and z=1, satisfies the condition that (1−zw−wz)is a purely real, then the set of values of zis ∣z∣=1,z=2 (b) ∣z∣=1andz=1z=z (d) None of these

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If {w}=+{i},where {0}and {z}{1}, satisfies the condition that

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If $w = α + i β,$ where $β \neq = 0$ and $z \neq = 1$ , satisfies the condition that $(\frac{w - w z}{1 - z})$ is a purely real, then the set of values of $z$ is $∣ z ∣ = 1, z \neq = 2$ (b) $∣ z ∣ = 1 a n d z \neq = 1$ $z = z$ (d) None of these

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Levi Discussed

w = α + i β,

where

β \neq = 0

and

z \neq = 1

, satisfies the condition that

(\frac{w - w z}{1 - z})

is a purely real, then the set of values of

z

∣ z ∣ = 1, z \neq = 2

(b)

∣ z ∣ = 1 a n d z \neq = 1

z = z

(d) None of these

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Question Text	If $w = α + i β,$ where $β \neq = 0$ and $z \neq = 1$ , satisfies the condition that $(\frac{w - w z}{1 - z})$ is a purely real, then the set of values of $z$ is $∣ z ∣ = 1, z \neq = 2$ (b) $∣ z ∣ = 1 a n d z \neq = 1$ $z = z$ (d) None of these
Answer Type	Video solution: 1
Upvotes	150

If w=α+iβ,where β=0and z=1, satisfies the condition that (1−zw−wz​)is a purely real, then the set of values of zis ∣z∣=1,z=2 (b) ∣z∣=1andz=1z=z (d) None of these

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If $w = α + i β,$ where $β \neq = 0$ and $z \neq = 1$ , satisfies the condition that $(\frac{w - w z}{1 - z})$ is a purely real, then the set of values of $z$ is $∣ z ∣ = 1, z \neq = 2$ (b) $∣ z ∣ = 1 a n d z \neq = 1$ $z = z$ (d) None of these