bonjour
rappel : i² = -1
l'équation z² = a <=> z = √a ou z = -√a
quand a < 0 les solutions sont complexes
1) z² = -3
<=> z = √-3 ou z = - √-3
or -3 = i² x 3
il y a deux solutions
z₁ = √(i² x 3) = i√3 et z₂ = -√(i² x 3) = -i√3
S = {-i√3 ; i√3}
2) z² - 3z + 3 = 0
Δ = b² − 4ac = (-3)² - 4*1*3 = 9 - 12 = -3 ; √Δ = √(-3) = i√3
z₁ = (3 + i√3) / 2 et z₂ = (3 - i√3)/2
S = { (3 + i√3) / 2 ; z₂ = (3 - i√3)/2}
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bonjour
rappel : i² = -1
l'équation z² = a <=> z = √a ou z = -√a
quand a < 0 les solutions sont complexes
1) z² = -3
<=> z = √-3 ou z = - √-3
or -3 = i² x 3
il y a deux solutions
z₁ = √(i² x 3) = i√3 et z₂ = -√(i² x 3) = -i√3
S = {-i√3 ; i√3}
2) z² - 3z + 3 = 0
Δ = b² − 4ac = (-3)² - 4*1*3 = 9 - 12 = -3 ; √Δ = √(-3) = i√3
z₁ = (3 + i√3) / 2 et z₂ = (3 - i√3)/2
S = { (3 + i√3) / 2 ; z₂ = (3 - i√3)/2}