If w=α+i β , where β ≠ 0 and z ≠ 1 satisfies the condition that ( w- w z 1-z ) is purely real, then the set of values of z is

Correct answer: B. |z|=1 and z ≠ 1 Let z_1= w- w _z 1-z , be purely real. array rlrl & z_1 & = z _1 & & w- w z 1-z & = w -w z 1- z & & w-w z - w z+ w z · z & = w -z w -w z +w z · z & & (w- w )+( w -w)|z|^2 & =0 & & (w- w )(1-|z|^2) & =0 & & |z|^2 & =1 & & |z| & =1 and z ≠ 1 . array [as, w- w ≠ 0 , since β ≠ 0 ]

JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Complex Number

JEE Advanced 2006 · Mathematics · Complex Number

If w=α+i β , where β ≠ 0 and z ≠ 1 satisfies the condition that (...

w=\alpha+i \beta

, where

\beta \neq 0

and

z \neq 1

satisfies the condition that

\left(\frac{w-\bar{w} z}{1-z}\right)

is purely real, then the set of values of

z

Options

A $|z|=1$ and $z \neq 2$
B $|z|=1$ and $z \neq 1$
C $z=\bar{Z}$
DNone of these

Correct answer

B. $|z|=1$ and $z \neq 1$

Step-by-step solution

Let

z_1=\frac{w-\bar{w}_z}{1-z}

, be purely real.

\begin{array}{rlrl} \Rightarrow & z_1 & =\bar{z}_1 \\ & \therefore & \frac{w-\bar{w} z}{1-z} & =\frac{\bar{w}-w \bar{z}}{1-\bar{z}} \\ \Rightarrow & & w-w \bar{z}-\bar{w} z+\bar{w} z \cdot \bar{z} & =\bar{w}-z \bar{w}-w \bar{z}+w z \cdot \bar{z} \\ \Rightarrow & & (w-\bar{w})+(\bar{w}-w)|z|^2 & =0 \\ \Rightarrow & & (w-\bar{w})\left(1-|z|^2\right) & =0 \\ \Rightarrow & & |z|^2 & =1 \\ \Rightarrow & & |z| & =1 \text{ and } z \neq 1 . \end{array}

\text{ [as, }

w-\bar{w} \neq 0

\text{, since }

\beta \neq 0

\text{ ] }

Practice Complex Number on Quantrex Academy →

More from Complex Number