Bonjour !
a)
[tex]\dfrac{(1+2i)(-2+5i)}{(7+3i)(-12+i)}\\ \\On \ d \acute{e}veloppe\ le \ num\acute{e}rateur \ \\\\=\dfrac{-2+5i-4i-10}{(7+3i)(-12+i)} \\\\=\dfrac{-12+i}{(7+3i)(-12+i)} \\\\=\dfrac{1}{7+3i} \\\\=\dfrac{1(7-3i)}{(7+3i)(7-3i)} \\\\=\dfrac{7-3i}{7^2+3^2} \\\\\boxed{=\dfrac{7-3i}{58}} = > forme \ alg\acute{e}brique[/tex]
b)
[tex]\dfrac{1}{(1+2i)(3-i)} \\\\On \ d\acute{e}veloppe \ le \ d\acute{e}nominateur.\\\\=\dfrac{1}{3-i+6i+2}\\\\ =\dfrac{1}{5+5i} \\\\=\dfrac{1(5-5i)}{(5+5i)(5-5i)} \\\\=\dfrac{5-5i}{5^2+5^2} \\\\=\dfrac{5-5i}{50}\\\boxed{=\dfrac{1-i}{10} } = > forme \ alg\acute{e}brique[/tex]
c)
[tex]e^{i\pi}=-1 \ \ \ (identit\acute{e} \ d'Euler)[/tex]
Autre méthode :
[tex]e^{i\pi}\\=1(cos(\pi)+i\ sin(\pi))\\=\underbrace{cos(\pi)}_{=-1}+i\ \underbrace{sin(\pi)}_{=0}\\=-1+0i\\\boxed{=-1}[/tex]
Bonne journée
