Bonjour,
Soient z₁ = -2+3i et z₂ = 4-i
a) [tex]\frac{z_2}{z_1}=\frac{4-i}{-2+3i}=\frac{(4-i)(-2-3i)}{(-2+3i)(-2-3i)}=\frac{-8-12i+2i-3}{(-2)^2+3^2}=\frac{-11-10i}{13}=-\frac{11}{13}- \frac{10}{13}i [/tex]
b) [tex]\frac{z_1}{z_2}=\frac{-2+3i}{4-i}=\frac{(-2+3i)(4+i)}{(4-i)(4+i)}=\frac{-8-2i+12i-3}{4^2+(-1)^2}=\frac{-11+10i}{17}=-\frac{11}{17}+ \frac{10}{17}i [/tex]
c) [tex]\frac{3}{z_1+z_2}=\frac{3}{-2+3i+4-i}=\frac{3}{2+2i}=\frac{3(2-2i)}{(2+2i)(2-2i)}=\frac{6-6i}{2^2+2^2}=\frac{6-6i}{8}=\frac{3-3i}{4}[/tex][tex]=\frac{3}{4}- \frac{3}{4}i[/tex]
d) [tex]\frac{z_1+z_2}{z_1-z_2}=\frac{-2+3i+4-i}{-2+3i-4+i}=\frac{2+2i}{-6+4i}=\frac{(2+2i)(-6-4i)}{(-6+4i)(-6-4i)}=\frac{-12-8i-12i+8}{(-6)^2+4^2}[/tex][tex]=\frac{-4-20i}{52}=\frac{-1-5i}{13}=-\frac{1}{13}- \frac{5}{13}i[/tex]
