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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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Verified Answer
The correct answer is:
$z \bar{W}=1$
$\begin{array}{ll} & \left(|z|^2+1\right) w=\left(|w|^2+1\right) z \\ \therefore & \text { Let } z=k w, k \neq 1, k \neq 0 \\ \therefore & k^2|w|^2 w-k|w|^2 w=k w-w \\ \therefore & k|w|^2 w(k-1)=(k-1) w \\ \Rightarrow & k=\frac{1}{|w|^2} \\ \therefore & z=\frac{w}{|w|^2} \\ & \quad z \bar{w}=\frac{w}{|w|^2} \cdot \bar{w}=\frac{|w|^2}{|w|^2}=1\end{array}$
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