Let w (Im, w neq 0) be a complex number. Then | vedclass

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(D) Consider the equation $w - \overline{w}z = k(1 - z)$, where $k \in \mathbb{R}$.Since $Im\, w 
eq 0$, $w 
eq \overline{w}$, so $z 
eq 1$.Thus, $

Vedclass · Accepted Answer

$\{z : |z| = 1, z 
eq 1\}$

Vedclass · Answer

(D) Consider the equation $w - \overline{w}z = k(1 - z)$, where $k \in \mathbb{R}$.Since $Im\, w 
eq 0$, $w 
eq \overline{w}$, so $z 
eq 1$.Thus, $k = \frac{w - \overline{w}z}{1 - z}$.Since $k$ is real, we have $\frac{w - \overline{w}z}{1 - z} = \overline{\left( \frac{w - \overline{w}z}{1 - z} ight)} = \frac{\overline{w} - w\overline{z}}{1 - \overline{z}}$.Cross-multiplying gives $(w - \overline{w}z)(1 - \overline{z}) = (\overline{w} - w\overline{z})(1 - z)$.Expanding both sides: $w - w\overline{z} - \overline{w}z + \overline{w}z\overline{z} = \overline{w} - \overline{w}z - w\overline{z} + w\overline{z}z$.Simplifying, we get $w + \overline{w}|z|^2 = \overline{w} + w|z|^2$.Rearranging terms: $(w - \overline{w})|z|^2 = w - \overline{w}$.Since $Im\, w 
eq 0$, $w - \overline{w} 
eq 0$, so $|z|^2 = 1$, which implies $|z| = 1$.Also, from the original equation, $z 
eq 1$ must hold.Therefore, the set is $\{z : |z| = 1, z 
eq 1\}$.

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

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