bonjour j'aurais vraiment besoin d'aide sur ce petit exercice sur les nombre complexe s'il vous plait car je n'y comprend rien et en plus je suis nul l'exo est

Bonjour

1) [tex]Z=\dfrac{1+i\sqrt{3}}{\sqrt{2}-i\sqrt{2}}\\\\Z=\dfrac{(1+i\sqrt{3})(\sqrt{2}+i\sqrt{2})}{(\sqrt{2}-i\sqrt{2})(\sqrt{2}+i\sqrt{2})}\\\\Z=\dfrac{\sqrt{2}+i\sqrt{2}+i\sqrt{6}-\sqrt{6}}{2+2}\\\\Z=\dfrac{(\sqrt{2}-\sqrt{6})+i(\sqrt{2}+\sqrt{6})}{4}\\\\\boxed{Z=\dfrac{\sqrt{2}-\sqrt{6}}{4}+i\dfrac{\sqrt{2}+\sqrt{6}}{4}}[/tex]

2) [tex]z_1=1+i\sqrt{3}\\\\z_1=2(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2})\\\\z_1=2[\cos(\dfrac{\pi}{3})+i\sin(\dfrac{\pi}{3})]\\\\\boxed{z_1=2e^{i(\dfrac{\pi}{3})}}[/tex]


[tex]z_2=\sqrt{2}-i\sqrt{2}\\\\z_2=2(\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})\\\\z_2=2[\cos(\dfrac{\pi}{4})-i\sin(\dfrac{\pi}{4})]\\\\\boxed{z_2=2e^{-i(\dfrac{\pi}{4})}}[/tex]

[tex]Z=\dfrac{z_1}{z_2}\\\\Z=\dfrac{e^{i(\dfrac{\pi}{3})}}{e^{-i(\dfrac{\pi}{4})}}\\\\Z=e^{i(\dfrac{\pi}{3})+i(\dfrac{\pi}{4})}\\\\Z=e^{i(\dfrac{4\pi}{12})+i(\dfrac{3\pi}{12})}\\\\\boxed{Z=e^{i(\dfrac{7\pi}{12})}}[/tex]

3) [tex]Z=e^{i(\dfrac{7\pi}{12})}=\cos(\dfrac{7\pi}{12})+i\sin(\dfrac{7\pi}{12})\\\\Z=\dfrac{\sqrt{2}-\sqrt{6}}{4}+i\dfrac{\sqrt{2}+\sqrt{6}}{4}[/tex]

En comparant les deux écritures de Z, nous en déduisons que :

[tex]\cos(\dfrac{7\pi}{12})=\dfrac{\sqrt{2}-\sqrt{6}}{4}\\\\\sin(\dfrac{7\pi}{12})=\dfrac{\sqrt{2}+\sqrt{6}}{4}[/tex]

4) Constructions en pièces jointe.

5) [tex]Z^{2018}=(e^{i(\dfrac{7\pi}{12})})^{2018}\\\\Z^{2018}=e^{i(\dfrac{7\pi}{12}\times2018)}\\\\Z^{2018}=e^{i(\dfrac{1426\pi}{12})}\\\\Z^{2018}=e^{i(\dfrac{713\pi}{6})}\\\\Z^{2018}=e^{i(\dfrac{708\pi}{6}+\dfrac{7\pi}{6})}[/tex]

[tex]Z^{2018}=e^{i(118\pi+\dfrac{7\pi}{6})}\\\\Z^{2018}=e^{i(59\times2\pi+\dfrac{7\pi}{6})}\\\\Z^{2018}=e^{i(59\times2\pi)}\times e^{i(\dfrac{7\pi}{6})}\\\\Z^{2018}=(e^{i(2\pi)})^{59}\times e^{i(\dfrac{7\pi}{6})}\\\\Z^{2018}=1^{59}\times e^{i(\dfrac{7\pi}{6})}[/tex]

[tex]\\\\Z^{2018}=1\times e^{i(\dfrac{7\pi}{6})}\\\\Z^{2018}=e^{i(\dfrac{7\pi}{6})}\\\\Z=\cos(\dfrac{7\pi}{6})+i\sin(\dfrac{7\pi}{6})\\\\Z=-\dfrac{\sqrt{3}}{2}+i\times(-\dfrac{1}{2})\\\Z=-\dfrac{\sqrt{3}}{2}-\dfrac{i}{2}[/tex]