Réponse :
Explications étape par étape
Exercice 7
1)
[tex]z_{3} =\frac{z_{1} }{z_{2} } =\frac{1+\sqrt{3} i}{1+i} =\frac{(1+\sqrt{3} i)(1-i)}{(1+i)(1-i)}\\z_{3} = \frac{1-i+\sqrt{3} i+\sqrt{3} }{1+1} =\frac{1+\sqrt{3}}{2} + i \frac{\sqrt{3}-1}{2}[/tex]
2)
[tex]|z_{1} |=\sqrt{1+3}=2\\[/tex]
Soit [tex]\theta_{1}[/tex] un argument de Z1.
On a
[tex]tan(\theta_{1}) = \frac{Im(z_{1} )}{Re(z_{1}) } = \sqrt{3}\\ Or \ Re(z_{1})>0 => cos(\theta_{1})>0\\ Donc\ \theta_{1}=\frac{\pi }{3}[/tex]
[tex]z_{1} = 2(cos(\frac{\pi }{3})+i sin( \frac{\pi }{3}))[/tex]
[tex]|z_{2} |=\sqrt{1+1}=\sqrt{2} \\[/tex]
Soit [tex]\theta_{2}[/tex] un argument de Z1.
On a
[tex]tan(\theta_{2}) = \frac{Im(z_{2} )}{Re(z_{2}) } = 1\\ Or \ Re(z_{2})>0 => cos(\theta_{2})>0\\ Donc\ \theta_{2}=\frac{\pi }{4}[/tex]
[tex]z_{2} = \sqrt{2}(cos(\frac{\pi }{4})+i sin( \frac{\pi }{4}))[/tex]
[tex]|z_{3} |=\frac{|z_{1} |}{|z_{2} |} =\frac{2}{\sqrt{2} } =\sqrt{2}\\[/tex]
[tex]\theta_{3} = \theta _{1} -\theta _{2} = \frac{\pi }{3} -\frac{\pi }{4} =\frac{\pi }{12}[/tex]
[tex]z_{3} = \sqrt{2}(cos(\frac{\pi }{12})+i sin( \frac{\pi }{12}))[/tex]
3)
On a
[tex]z_{3} =\frac{1+\sqrt{3}}{2} + i \frac{\sqrt{3}-1}{2}\\z_{3} = \sqrt{2}(cos(\frac{\pi }{12})+i sin( \frac{\pi }{12}))[/tex]
Donc
[tex]\sqrt{2}*cos(\frac{\pi }{12}) =\frac{1+\sqrt{3}}{2}[/tex] donc [tex]cos(\frac{\pi }{12}) =\frac{\sqrt{2}+\sqrt{6}}{4}[/tex]
[tex]\sqrt{2}*sin(\frac{\pi }{12}) =\frac{\sqrt{3}-1}{2}[/tex] donc [tex]sin(\frac{\pi }{12}) =\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]
[tex]tan(\frac{\pi }{12} )=\frac{sin(\frac{\pi }{12} )}{cos(\frac{\pi }{12} )}=\frac{\frac{\sqrt{6}-\sqrt{2} }{4} }{\frac{\sqrt{6} +\sqrt{2} }{4} } = \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6} +\sqrt{2} } =\frac{(\sqrt{6}-\sqrt{2})^{2} }{6-2} = \frac{6-4\sqrt{3} +2}{4} =2-\sqrt{3}[/tex]
Exercice 8
[tex]c-a=1+2i-1-3i=-i\\b-a = \frac{2-\sqrt{3} }{2} +\frac{5}{2} i-1-3i=\frac{2}{2} - \frac{\sqrt{3} }{2} -1 +i (\frac{5}{2}-3)= - \frac{\sqrt{3} }{2}-\frac{1}{2} i\\[/tex]
[tex]|\frac{c-a}{b-a} | = 1\\arg(\frac{c-a}{b-a} ) =arg(c-a)-arg(b-a)\\arg(c-a)=arg(-i)=-\frac{\pi }{2} \\[/tex]
[tex]b-a = - \frac{\sqrt{3} }{2}-\frac{1}{2} i =-cos(\frac{\pi }{6}) -i sin(\frac{\pi }{6}) = cos(\frac{\pi }{6}-\pi ) +i sin(\frac{\pi }{6}-\pi ) \\b-a =cos(\frac{-5\pi }{6} ) +i sin(\frac{-5\pi }{6} )[/tex]
[tex]arg(\frac{c-a}{b-a} )=-\frac{\pi }{2} -(-\frac{5\pi }{6} )=\frac{\pi }{3}[/tex]
[tex]\frac{c-a}{b-a}= cos(\frac{\pi }{3})+i sin(\frac{\pi }{3})[/tex]
2
[tex]\overrightarrow{AB} (b-a)\\\overrightarrow{AC} (c-a)\\[/tex]
[tex]\widehat{\overrightarrow{AB},\overrightarrow{AC}}= \widehat{{(Ox)},\overrightarrow{AC}} - \widehat{{(Ox)},\overrightarrow{AB}} = arg(c-a)-arg(b-a)=\frac{\pi }{3}[/tex]
[tex]|\frac{c-a}{b-a} | = 1 \Rightarrow |c-a| = |b-a| \Rightarrow \left\lVert\overrightarrow{AC}\right\rVert = \left\lVert\overrightarrow{AB}\right\rVert\\ \Rightarrow AC=AB[/tex]
3
Le triangle est isocèle en A d'angle principale [tex]\frac{\pi }{3}[/tex]. C'est donc un triangle équilatéral
