$z= \frac{u-\overline{u}v}{1-v}$ is real is equivalent to $|v|=1$.

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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$$|v|=1\Longleftrightarrow |v|^2=1\Longleftrightarrow v \bar{v}=1 \Longleftrightarrow \bar{v}=\frac{1}{v} \Longleftrightarrow \bar{z}=$$
$$\dfrac{\bar{u}-u \dfrac{1}{v}}{1-\dfrac{1}{v}} \Longleftrightarrow \bar{z}=\dfrac{\bar{u} v-u}{v-1} \Longleftrightarrow \bar{z}=z \Longleftrightarrow z \in \mathbb R$$