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Let $u,v$ be complex numbers such as $u,v\notin \mathbb{R} $, and : $$z= \frac{u-\overline{u}v}{1-v}$$

Prove that : $z\in\mathbb{R} \Longleftrightarrow |v|=1$.

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up vote 6 down vote accepted

$$|v|=1\Longleftrightarrow |v|^2=1\Longleftrightarrow v \bar{v}=1 \Longleftrightarrow \bar{v}=\frac{1}{v} \Longleftrightarrow \bar{z}=$$

$$\dfrac{\bar{u}-u \dfrac{1}{v}}{1-\dfrac{1}{v}} \Longleftrightarrow \bar{z}=\dfrac{\bar{u} v-u}{v-1} \Longleftrightarrow \bar{z}=z \Longleftrightarrow z \in \mathbb R$$

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@Gigili it looks more uglier now. –  Kaster Mar 11 '13 at 21:01
    
Nice, smooth and short. +1 –  DonAntonio Mar 11 '13 at 22:39
1  
@Kaster: I didn't ask for your opinion. –  Gigili Mar 12 '13 at 2:49
    
@Gigili well, you should have, or you correct posts just for yourself? –  Kaster Mar 12 '13 at 3:17
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@Kaster: For Michael specifically in this case, and everyone else except you. –  Gigili Mar 12 '13 at 13:18

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