Bonsoir,
1) [tex]z_2=e^{3i\frac{\pi}{4}}=cos(\dfrac{3\pi}{4})+isin(\dfrac{3\pi}{4})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}[/tex]
2) [tex]z_1\times z_2=(1-i\sqrt{3})(-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{6}i}{2}+\dfrac{\sqrt{6}}{2}\\\\=\dfrac{\sqrt{6}-\sqrt{2}}{2}+\dfrac{\sqrt{6}+\sqrt{2}}{2}i[/tex]
3) [tex]1-i\sqrt{3}=r\times cis(\theta)\ \ avec\ r = \sqrt{1^2+(\sqrt{3})^2}=\sqrt{1+3}=2\\\\\theta=tan^{-1}(-\dfrac{\sqrt{3}}{1})=-\dfrac{\pi}{3}\\\\\\1-i\sqrt{3}=2e^{-i\dfrac{\pi}{3}}[/tex]
4) [tex]z_1\times z_2=2e^{-i\dfrac{\pi}{3}}\times e^{i\dfrac{3\pi}{4}}=2e^{i(-\dfrac{\pi}{3}+\dfrac{3\pi}{4})}\\\\=2e^{i(-\dfrac{4\pi}{12}+\dfrac{9\pi}{12})}=2e^{i\dfrac{5\pi}{12}}[/tex]
5) [tex]z_1\times z_2= 2e^{i\dfrac{5\pi}{12}}=2(cos(\dfrac{5\pi}{12})+isin(\dfrac{5\pi}{12}))\\\\=2cos(\dfrac{5\pi}{12})+2isin(\dfrac{5\pi}{12})\\\\\\z_1\times z_2=\dfrac{\sqrt{6}-\sqrt{2}}{2}+\dfrac{\sqrt{6}+\sqrt{2}}{2}i[/tex]
En identifiant les parties réelles et imaginaires, nous en déduisons :
[tex]2cos(\dfrac{5\pi}{12})=\dfrac{\sqrt{6}-\sqrt{2}}{2}\\\\cos(\dfrac{5\pi}{12})=\dfrac{\sqrt{6}-\sqrt{2}}{4}\\\\\\2sin(\dfrac{5\pi}{12})=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\\\sin(\dfrac{5\pi}{12})=\dfrac{\sqrt{6}+\sqrt{2}}{4}[/tex]
