Bonjour,
Voici une piste:
$$z = 2 \frac {e^{i \theta}-1}{e^{i \theta}+1} = 2 \frac {e^{i \theta / 2}(e^{i \theta / 2}-e^{-i \theta / 2})}{e^{i \frac {\theta}{2}}(e^{i \theta}+e^{-i \frac {\theta}{2})}} = 2 \frac{2isin( \frac {\theta}{2})}{2cos( \frac {\theta}{2})} = 2tan( \frac {\theta}{2} ) e^{i {\frac {\pi}{2}}} = [2tan( \frac {\theta}{2}), \frac {\pi}{2}]$$