Let z and w be two distinct non-zero complex | vedclass

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(C) Given the equation: $|z|^2 w - |w|^2 z = z - w$ Rearranging the terms: $|z|^2 w + w = |w|^2 z + z$ $(|z|^2 + 1) w = (|w|^2 + 1) z$ Since $z$ and $

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$z \bar{w} = 1$

Vedclass · Answer

(C) Given the equation: $|z|^2 w - |w|^2 z = z - w$ Rearranging the terms: $|z|^2 w + w = |w|^2 z + z$ $(|z|^2 + 1) w = (|w|^2 + 1) z$ Since $z$ and $w$ are non-zero,we can write: $\frac{z}{|z|^2 + 1} = \frac{w}{|w|^2 + 1}$ Let $\frac{z}{|z|^2 + 1} = \frac{w}{|w|^2 + 1} = k$ Then $z = k(|z|^2 + 1)$ and $w = k(|w|^2 + 1)$ Since $z$ and $w$ are complex numbers,$k$ must be a real number. If $k$ is real,then $\frac{z}{w} = \frac{|z|^2 + 1}{|w|^2 + 1}$,which is a real number. Thus,$z = cw$ for some real constant $c 
eq 1$. Substituting $z = cw$ into the original equation: $|cw|^2 w - |w|^2 (cw) = cw - w$ $c^2 |w|^2 w - c |w|^2 w = (c - 1) w$ Since $w 
eq 0$,we divide by $w$: $c^2 |w|^2 - c |w|^2 = c - 1$ $c |w|^2 (c - 1) = c - 1$ Since $z 
eq w$,$c 
eq 1$,so we divide by $(c - 1)$: $c |w|^2 = 1 \Rightarrow c = \frac{1}{|w|^2}$ Therefore,$z = \frac{w}{|w|^2} = \frac{w}{w \bar{w}} = \frac{1}{\bar{w}}$ This implies $z \bar{w} = 1$.

Let $z$ and $w$ be two distinct non-zero complex numbers. If $|z|^2 w - |w|^2 z = z - w$,then:

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