Réponse :
Bonjour
Exercice 8
1) j² = [tex](-\frac{1}{2}+i\frac{\sqrt{3} }{2} )^{2}[/tex] = [tex]\frac{1}{4}-i\frac{\sqrt{3} }{2}-\frac{3}{4}[/tex] = [tex]-\frac{1}{2}-i\frac{\sqrt{3} }{2}[/tex]
2) j² + j + 1 = [tex]-\frac{1}{2}-i\frac{\sqrt{3} }{2}-\frac{1}{2}+i\frac{\sqrt{3} }{2}+1[/tex] = -1 + 1 = 0
j³ = j²×j = [tex](-\frac{1}{2}-i\frac{\sqrt{3} }{2})(-\frac{1}{2}+i\frac{\sqrt{3} }{2})[/tex] = [tex](-\frac{1}{2}) ^{2}-(i\frac{\sqrt{3} }{2}) ^{2}[/tex] = [tex]\frac{1}{4}+\frac{3}{4}[/tex] = 1
[tex]\frac{1}{j}=\frac{1}{-\frac{1}{2}+i\frac{\sqrt{3} }{2} }[/tex] = [tex]\frac{-\frac{1}{2}-i\frac{\sqrt{3} }{2} }{(-\frac{1}{2}) ^{2}-(i\frac{\sqrt{3} }{2}) ^{2} }[/tex] = [tex]-\frac{1}{2}-i\frac{\sqrt{3} }{2}[/tex] = j² =j barre
Exercice 9
1) 4z + 6 - 2i = 5z - 4 + 2i
⇔ 4z - 5z = -4 + 2i - 6 + 2i
⇔ -z = -10 + 4i
⇔ z = 10 - 4i
2) (5 + 2i)z - 8 = 2 - i
⇔ (5 + 2i)z = 10 - i
⇔ z = (10 - i)/(5 + 2i)
⇔ z = [(10 - i)(5 - 2i)]/[(5 + 2i)(5 - 2i)]
⇔ z = (50 - 20i - 5i + 2i²)/(5² - 4i²)
⇔ z = (46 - 25i)/29
⇔ z = [tex]\frac{46}{29}-i\frac{25}{29}[/tex]
3) (9 + i)z + 6 - 2i = 5 - (2 + 2i)z + 4
⇔ (9 + i)z + (2 + 2i)z = 5 - 4 + 6 + 2i
⇔ (9 + i + 2 + 2i)z = 7 + 2i
⇔ (11 + 3i)z = 7 + 2i
⇔ z = (7 + 2i)/(11 + 3i)
⇔ z = [(7 + 2i)(11 - 3i)]/[(11 + 3i)(11 - 3i)]
⇔ z = (77 - 21i + 22i - 6i²)/(11² - 9i²)
⇔ z = (83 + i)/130
⇔ z = [tex]\frac{83}{130}+i\frac{1}{130}[/tex]
