Réponse :
1) Prenons soin de distinguer partie réelle et imaginaire de u.
[tex]u\times v = (\frac{\sqrt{2} }{2} -i\frac{\sqrt{2}}{{2}})(1-i\sqrt{3})\\u\times v = \frac{\sqrt{2} }{2} - i\frac{\sqrt{2} }{2} \sqrt{3} -i\frac{\sqrt{2}}{{2}} + i^2\frac{\sqrt{2} }{2} \sqrt{3}\\u\times v = \frac{\sqrt{2} }{2}- \frac{\sqrt{6}}{2} - i(\frac{\sqrt{2} }{2} + \frac{\sqrt{6}}{2})\\u\times v = \frac{\sqrt{2} -\sqrt{6}}{2}- i(\frac{\sqrt{2} +\sqrt{6}}{2})\\[/tex]
2)
[tex]\frac{u}{v} =\frac{\frac{\sqrt{2} -i\sqrt{2} }{2} }{1-i\sqrt{3} } \\\frac{u}{v} =\frac{{\sqrt{2} -i\sqrt{2} } }{2(1-i\sqrt{3}) } \times\frac{1+i\sqrt{3} }{1+i\sqrt{3} } \\\frac{u}{v} = \frac{\sqrt{2}+i\sqrt{2} \sqrt{3} - i\sqrt{2} -i^2 \sqrt{2} \sqrt{3} }{2(1^2+\sqrt{3} ^2)} \\\frac{u}{v} = \frac{\sqrt{2} +\sqrt{6}+i(\sqrt{6}-\sqrt{2}) }{8} \\\frac{u}{v} = \frac{\sqrt{6} +\sqrt{2}}{8} +i\frac{\sqrt{6}-\sqrt{2}}{8}[/tex]
3) [tex]|u|=|\frac{\sqrt{2}}{2} -i\frac{\sqrt{2}}{2}|=\sqrt{(\frac{\sqrt{2}}{2})^2+(-\frac{\sqrt{2}}{2})^2} = \sqrt{\frac{2}{4} +\frac{2}{4}} = 1[/tex]
4) Avec u = a + ib on a :
[tex]cos(\theta)=\frac{a}{|u|} \\cos(\theta)=\frac{ \frac{\sqrt{2} }{2} }{1} = \frac{\sqrt{2} }{2} \\\\sin(\theta) = \frac{b}{|u|} \\sin(\theta) = \frac{ -\frac{\sqrt{2} }{2} }{1} = -\frac{\sqrt{2} }{2} \\\\[/tex]
On en déduit que
[tex]\theta = -\frac{\pi}{4} \\arg(u)= -\frac{\pi}{4} [2 \pi][/tex]
5)
u = |u|×[cos(θ) + i×sin(θ)]
[tex]u=1\times(cos\frac{-\pi}{4} +i\times sin(\frac{-\pi}{4} ))\\u=cos\frac{-\pi}{4} +i\times sin(\frac{-\pi}{4} )[/tex]
6)
[tex]|v| = |1-i\sqrt{3} | = \sqrt{1^2+(-\sqrt{3})^2} =2[/tex]
7) Avec v = a + ib on a :
[tex]cos(\theta)=\frac{a}{|v|} \\cos(\theta)=\frac{ 1 }{2} \\\\sin(\theta) = \frac{b}{|v|} \\sin(\theta) = -\frac{ \sqrt{3} }{2} }\\[/tex]
On en déduit que
[tex]\theta = -\frac{\pi}{3} \\arg(v)= -\frac{\pi}{3} [2 \pi][/tex]
8)
v = |v|×[cos(θ) + i×sin(θ)]
[tex]v=2\times(cos(\frac{-\pi}{3}) +i\times sin(\frac{-\pi}{3} ))[/tex]
9)
|u×v|= |u| × |v|
|u×v|= 1 × 2 = 2
arg(u×v) = arg(u) + arg(v)
[tex]arg(u\times v) = -\frac{\pi}{4} + (-\frac{\pi}{3} )\\\\arg(u\times v) =-\frac{7\pi}{12} [2\pi]\\[/tex]
Une forme trigonométrique de u×v est
[tex]u\times v = 2[cos(-\frac{7\pi}{12}) +i\times sin(-\frac{7\pi}{12})][/tex]
10)
On a
[tex]u\times v = (\frac{\sqrt{2} }{2}- \frac{\sqrt{6}}{2}) - i(\frac{\sqrt{2} }{2} + \frac{\sqrt{6}}{2})\\[/tex]
et
[tex]u\times v = 2[cos(-\frac{7\pi}{12}) +i\times sin(-\frac{7\pi}{12})]\\u\times v = 2cos(-\frac{7\pi}{12}) +i\times 2sin(-\frac{7\pi}{12})][/tex]
En comparant les parties réelles et imaginaires entre elles, on en déduit :
[tex]\frac{\sqrt{2} }{2} -\frac{\sqrt{6} }{2} =2cos(-\frac{7\pi}{12})\\ cos(-\frac{7\pi}{12}) = \frac{\frac{\sqrt{2} }{2} -\frac{\sqrt{6} }{2} }{2} \\ cos(-\frac{7\pi}{12}) = \frac{\sqrt{2}-\sqrt{6} }{4}[/tex]
et
[tex]-(\frac{\sqrt{2} }{2} +\frac{\sqrt{6} }{2}) =2sin(-\frac{7\pi}{12})\\ sin(-\frac{7\pi}{12}) = \frac{-\frac{\sqrt{2} }{2} -\frac{\sqrt{6} }{2} }{2} \\ sin(-\frac{7\pi}{12}) = \frac{-\sqrt{2}-\sqrt{6} }{4}[/tex]
11)
[tex]|\frac{u}{v} |=\frac{|u|}{|v|}= \frac{1}{2}[/tex]
[tex]arg(\frac{u}{v} )=arg(u)-arg(v)\\arg(\frac{u}{v} )=-\frac{\pi}{4}-(-\frac{\pi}{3} )=\frac{\pi}{12}[/tex]
[tex]\frac{u}{v} = \frac{1}{2}[cos(\frac{\pi}{12})+i\times sin( \frac{\pi}{12})][/tex]
12)
On a
[tex]\frac{u}{v} = \frac{\sqrt{6} +\sqrt{2}}{8} +i\frac{\sqrt{6}-\sqrt{2}}{8}[/tex]
et
[tex]\frac{u}{v} = \frac{1}{2} [cos(\frac{\pi}{12})+i\times sin( \frac{\pi}{12})]\\\frac{u}{v} = \frac{1}{2}cos(\frac{\pi}{12})+i\times \frac{1}{2}sin( \frac{\pi}{12})[/tex]
En comparant les parties réelles et imaginaires entre elles, on en déduit :
[tex]\frac{1}{2}cos(\frac{\pi}{12} ) = \frac{\sqrt{6} +\sqrt{2} }{8} \\cos(\frac{\pi}{12} ) = \frac{\sqrt{6} +\sqrt{2} }{4}[/tex]
et
[tex]\frac{1}{2}sin(\frac{\pi}{12} ) = \frac{\sqrt{6} -\sqrt{2} }{8} \\sin(\frac{\pi}{12} ) = \frac{\sqrt{6} -\sqrt{2} }{4}[/tex]
