Complex Numbers 2 Question 12
12.
If $w=\alpha+i \beta$, where $\beta \neq 0$ and $z \neq 1$, satisfies the condition that $(\frac{w-\bar{w} z}{1-z})$ is purely real, then the set of values of $z$ is
$(2006,3 M)$
(a) $|z|=1, z \neq 2$
(b) $|z|=1$ and $z \neq 1$
(c) $z=\bar{z}$
(d) None of these
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Answer:
Correct Answer: 12. (b)
Solution:
- Let $z _1=\frac{w-\bar{w} z}{1-z}$ be purely real $\Rightarrow \quad z _1=\bar{z} _1$
$\therefore \quad \frac{w-\bar{w} z}{1-z}=\frac{\bar{w}-w \bar{z}}{1-\bar{z}}$
$\Rightarrow \quad w-w \bar{z}-\bar{w} z+\bar{w} z \cdot \bar{z}=\bar{w}-z \bar{w}-w \bar{z}+w z \cdot \bar{z}$
$\Rightarrow \quad(w-\bar{w})+(\bar{w}-w)|z|^{2}=0$
$\Rightarrow \quad(w-\bar{w})\left(1-|z|^{2}\right)=0$
$\Rightarrow \quad|z|^{2}=1$
[as $w-\bar{w} \neq 0$, since $\left.\beta \neq 0\right]$
$\Rightarrow \quad|z|=1$ and $z \neq 1$
