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Q.
Let $w=\alpha+i \beta, \beta \neq 0$ and $z \neq 1$. If $\frac{w-\bar{w} z}{1-z}$ is purely real, then the set of value of $z$ is
Complex Numbers and Quadratic Equations
Solution:
As $\frac{w-\bar{w} z}{1-z}$ is purely real,
$\frac{w-\bar{w} z}{1-z}=\frac{\bar{w}-w \bar{z}}{1-\bar{z}} $
$\Rightarrow(1-\bar{z})(w-\bar{w} z)=(1-z)(\bar{w}-w \bar{z}) $
$\Rightarrow(w-\bar{w})(1-z \bar{z})=0 $
$\text { As } w \neq \bar{w}, \text { we get } z \bar{z}=1$
Thus, set of values of $z$ is $\{z:|z|=1, z \neq 1\}$.