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