Soit \(\alpha\in\left]\frac{\pi}{2},\frac{\pi}{2}\right[\). \(\frac{1+i\tan\alpha}{1-i\tan\alpha}=\frac{\cos\alpha+i\sin\alpha}{\cos\alpha-i\sin\alpha}=e^{2i\alpha}\). Donc,

\[\left(\frac{1+iz}{1-iz}\right)^3=\frac{1+i\tan\alpha}{1-i\tan\alpha}\Leftrightarrow\exists k\in\{-1,0,1\}/\;\frac{1+iz}{1-iz}=e^{i(\frac{2\alpha}{3}+\frac{2k\pi}{3})}=\omega_k\Leftrightarrow\exists k\in\{-1,0,1\}/\;i(\omega_k+1)z=\omega_k-1.\] Maintenant, pour \(k\in\{-1,0,1\}\),

\[\omega_k=-1\Leftrightarrow\frac{2\alpha}{3}+\frac{2k\pi}{3}\in\pi+2\pi\Zz\Leftrightarrow\alpha\in-k\pi+\frac{3\pi}{2}+3\pi\Zz,\] ce qui est exclu pour \(\alpha\in]-\frac{\pi}{2},\frac{\pi}{2}[\). Donc,

\[\begin{align*} \left(\frac{1+iz}{1-iz}\right)^3=\frac{1+i\tan\alpha}{1-i\tan\alpha}&\Leftrightarrow\exists k\in\{-1,0,1\}/\;z=\frac{\omega_k-1}{i(\omega_k+1)}\Leftrightarrow\exists k\in\{-1,0,1\}/\;z=\frac{e^{i(\frac{\alpha}{3}+\frac{k\pi}{3})}}{e^{i(\frac{\alpha}{3}+\frac{k\pi}{3})}} \frac{e^{i(\frac{\alpha}{3}+\frac{k\pi}{3})}-e^{-i(\frac{\alpha}{3}+\frac{k\pi}{3})}}{i(e^{i(\frac{\alpha}{3} +\frac{k\pi}{3})}+e^{-i(\frac{\alpha}{3}+\frac{k\pi}{3})})}\\ &\Leftrightarrow\exists k\in\{-1,0,1\}/\;z=\frac{2i\sin(\frac{\alpha}{3}+\frac{k\pi}{3})}{i(2\cos(\frac{\alpha}{3}+\frac{k\pi}{3}))}\Leftrightarrow \exists k\in\{-1,0,1\}/\;z=\tan(\frac{\alpha}{3}+\frac{k\pi}{3}) \end{align*}\]

exo7 5122