Bonsoir excusez moi j’aurais besoin d’aide je ne réussi pas du tout cet exercice merci d’avance

[tex]Bonsoir; \\\\\\ z_1=2-2i =2 (1-i)=2 \sqrt{2} (\dfrac{ \sqrt{2} }{2} -i\dfrac{ \sqrt{2} }{2}) \\\\\\ = 2 \sqrt{2} (cos( \dfrac{\pi}{4} )-i\ sin( \dfrac{\pi}{4} )) = 2 \sqrt{2} e^{-i\dfrac{\pi}{4}} .[/tex]

[tex]z_2=1+ i\sqrt{3} =2(\dfrac{ 1 }{2}+i\dfrac{ \sqrt{3} }{2})=2(cos(\dfrac{\pi}{3})+isin(\dfrac{\pi}{3})) = 2e^{i\dfrac{\pi}{3}} .[/tex]

[tex]\textit{On a : } 1+i = \sqrt{2} (\dfrac{ \sqrt{2} }{2}+i\dfrac{ \sqrt{2} }{2}) = \sqrt{2} (cos(\dfrac{\pi}{4})+isin(\dfrac{\pi}{4})) = \sqrt{2} e^{i\dfrac{\pi}{4}} ; \\\\\\ \textit{donc on a : } z_3 = (1+i)^2 = 2e^{i\dfrac{\pi}{2}} .[/tex]

[tex]\textit{a) } z_1z_2 = 2 \sqrt{2} e^{-i\dfrac{\pi}{4}} \times 2e^{i\dfrac{\pi}{3}} = 4 \sqrt{2} e^{i(-\dfrac{\pi}{4}+\dfrac{\pi}{3})} = 4 \sqrt{2} e^{i\dfrac{\pi}{12}} .[/tex]

[tex]\textit{b) On a : } \overline {z_2} = 2e^{-i\dfrac{\pi}{3}} ;\\\\\\ \textit{donc : } z_1\overline {z_2} = 2 \sqrt{2} e^{-i\dfrac{\pi}{4}} \times 2e^{-i\dfrac{\pi}{3}} = 4 \sqrt{2} e^{-i(\dfrac{\pi}{4}+\dfrac{\pi}{3})} = 4 \sqrt{2} e^{-i\dfrac{7\pi}{12}} [/tex]

[tex]\textit{c) }\dfrac{z_1}{z_3} = \dfrac{2 \sqrt{2} e^{-i\dfrac{\pi}{4}}}{2e^{i\dfrac{\pi}{2}}} = \sqrt{2} e^{-i\dfrac{\pi}{4}} \times e^{-i\dfrac{\pi}{2}} \\\\\\ = \sqrt{2} e^{-i(\dfrac{\pi}{4}+\dfrac{\pi}{2})} = \sqrt{2} e^{-i\dfrac{3\pi}{4} } = \sqrt{2} e^{-i(\dfrac{3\pi}{4}-2\pi )} \\\\\\= \sqrt{2} e^{-i(\dfrac{3\pi}{4}-\dfrac{8\pi}{4})} = \sqrt{2} e^{i\dfrac{5\pi}{4}} .[/tex]

[tex]\textit{d) On a : } z_3^2=(2e^{i\dfrac{\pi}{2}})^2 = 4e^{i\pi} = -4 ;\\\\\\ \textit{donc : } z_2z_3^2 = - 4z_2 . \\\\\\ \textit{Conclusion : a est vraie ; b est fausse ; c est vraie et d est vraie .}[/tex]