Q48 / 120 · JEE Advanced 2006 · Mathematics

If $w=\alpha+i \beta$, 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 $z$ is

  • A
    $|z|=1$ and $z \neq 2$
  • Correct answer
    B
    $|z|=1$ and $z \neq 1$
  • C
    $z=\bar{Z}$
  • D
    None of these

Solution

Let $z_1=\frac{w-\bar{w}_z}{1-z}$, be purely real. $$ \begin{array}{rlrl} \Rightarrow & z_1 & =\bar{z}_1 \\ & \therefore & \frac{w-\bar{w} z}{1-z} & =\frac{\bar{w}-w \bar{z}}{1-\bar{z}} \\ \Rightarrow & & w-w \bar{z}-\bar{w} z+\bar{w} z \cdot \bar{z} & =\bar{w}-z \bar{w}-w \bar{z}+w z \cdot \bar{z} \\ \Rightarrow & & (w-\bar{w})+(\bar{w}-w)|z|^2 & =0 \\ \Rightarrow & & (w-\bar{w})\left(1-|z|^2\right) & =0 \\ \Rightarrow & & |z|^2 & =1 \\ \Rightarrow & & |z| & =1 \text { and } z \neq 1 . \end{array} $$ $$ \text { [as, } w-\bar{w} \neq 0 \text {, since } \beta \neq 0 \text { ] } $$

Hence, the correct answer is option: 'B'.

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