If $w=\alpha+i\beta$, where $\beta \neq 0$ and $z \neq 1$, satisfies the condition that $\left(\frac{w-\overline{w}z}{1-z}\right)$ is purely real, then the set of values of z is

JEE Advanced · 2006 · 09 Apr 2006 · Complex Numbers

w = α + i β

, where

β \neq = 0

and

z \neq = 1

, satisfies the condition that

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

is purely real, then the set of values of z is

Answer: (D)

For

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

to be purely real, it must equal its conjugate:

\frac{w - w z}{1 - z} = \frac{w - w z}{1 - z}

Cross-multiplying and simplifying:

(w - \overline{w} z) (1 - \overline{z}) = (\overline{w} - w \overline{z}) (1 - z)

Expanding and collecting terms leads to:

(z \overline{z} - 1) (\overline{w} - w) = 0

Since

β \neq = 0

, we have

\overline{w} - w = - 2 i β \neq = 0

.

Therefore,

z \overline{z} = 1

, which means

∣ z ∣^{2} = 1

, so

∣ z ∣ = 1

.

Combined with the given condition

z \neq = 1

, the answer is

{z : ∣ z ∣ = 1, z \neq = 1}

JEE Advanced · 2006 · 09 Apr 2006 · Single correct option