Let w (Im, w neq 0) be a complex nu | vedclass

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Question

Consider the equation

$w-\bar{w} z=k(1-z), k \in R$

Clearly $z 
eq 1$ and $\frac{w-\bar{w} z}{1-z}$ is purely real

$	herefore \frac{\overli

Vedclass · Accepted Answer

$\left\{ {z:\left| z ight| = 1,z 
e 1} ight\}$

Vedclass · Answer

Consider the equation

$w-\bar{w} z=k(1-z), k \in R$

Clearly $z 
eq 1$ and $\frac{w-\bar{w} z}{1-z}$ is purely real

$	herefore \frac{\overline{w-\bar{w} z}}{1-z}=\frac{w-\bar{w} z}{1-z}$

$\Rightarrow \frac{\bar{w}-w \bar{z}}{1-\bar{z}}=\frac{w-\bar{w} z}{1-z}$

$\Rightarrow \bar{w}-\bar{w} z-w \bar{z}+w \overline{z z}$

$=w-w \bar{z}-\bar{w} z+\bar{w} z \bar{z}$

$\Rightarrow \bar{w}+w|z|^{2}=w+\bar{w}|z|^{2}$

$ \Rightarrow (w - \bar w)(|z{|^2}) = w - \bar w$

$\Rightarrow|z|^{2}=1$ $ \quad(\because \operatorname{Im} w 
eq 0)$

$\Rightarrow|z|=1$ and $z 
eq 1$

$	herefore$ The required set is $\{z:|z|=1, z 
eq 1\}$

Let $w$ $(Im\, w \neq 0)$ be a complex number. Then the set of all complex number $z$ satisfying the equation $w - \overline {w}z = k\left( {1 - z} \right)$ , for some real number $k$, is

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