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Exercice6: Soit z = 5 +3i et z' = 2-i, deux nombres complexes. Calculer et mettre sous forme algébriques, en détaillant bien chaque calcul : 1. z-z' 2. 2z'-3z 3. z' .z 4. z3 ( au cube ) 5. z'/z​

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MOZI

Bonsoir,

1) z - z' = 5 + 3i - 2 + i = 5 - 2 + (3 + 1) i = 3 + 4i

2) 2z' - 3z = 2 (2 - i) - 3 (5 + 3i) = 4 - 2i - 15 - 9i = -11 - 11i

3) z' . z = (2 - i) (5 + 3i) = 10 + 6i - 5i - 3 i² = 10 + 3 + i = 13 + i

4) z³ = (5 + 3i)³ = (5 + 3i) (5 + 3i)² = (5 + 3i) (25 - 9 + 30i) = (5 + 3i) (16 + 30i)

z³ = 80 - 90 + 150i + 48i = -10 + 198i

5) z' / z = (2 - i) / (5 + 3i) = (2 - i) (5 - 3i) / [(5 + 3i) (5 - 3i)]

z' / z = (10 - 3 - 6i - 5i) / (25 - 9i²) = (7 - 11i) / (25 + 9) = 7/34 - 11/34 i