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alcath08
@alcath08
January 2021
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bonsoir,niveau terminale s
quelqu'un peut m'aider svp merci
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taalbabachir
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A) z² = - 1 on sait que i² = - 1
z² = i² ⇔ z² - i² = 0 ⇔ (z - i)(z + i) = 0 ⇒ z = i ou z = - i
b) z² - 4 z + 13 = 0
Δ = 16 - 4*13 = 16 - 52 = - 36 on pose i² = - 1 ; √(6i)² = 6i
z1 = (4 + 6i)/2 = 2 + 3i
z2 = (4 - 6i)/2 = 2 - 3i
c) z - 3i = - 2iz + 4 ⇔ z + 2iz = 4 + 3i ⇔ z(1 + 2i) = 4 + 3i ⇒ z = (4 + 3i)/(1+2i)
⇒ z = (4 + 3i)/(1+2i) ⇔ (4 + 3i)(1 - 2i)/(1 + 2i)(1 - 2i) = (4 - 8i + 3i - 6i²)/(1-4i²
⇒ z = (10 - 5i)/5 = 2 - i
vous faite la suite
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Thanks 1
alcath08
merci beaucoup
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Verified answer
A) z² = - 1 on sait que i² = - 1z² = i² ⇔ z² - i² = 0 ⇔ (z - i)(z + i) = 0 ⇒ z = i ou z = - i
b) z² - 4 z + 13 = 0
Δ = 16 - 4*13 = 16 - 52 = - 36 on pose i² = - 1 ; √(6i)² = 6i
z1 = (4 + 6i)/2 = 2 + 3i
z2 = (4 - 6i)/2 = 2 - 3i
c) z - 3i = - 2iz + 4 ⇔ z + 2iz = 4 + 3i ⇔ z(1 + 2i) = 4 + 3i ⇒ z = (4 + 3i)/(1+2i)
⇒ z = (4 + 3i)/(1+2i) ⇔ (4 + 3i)(1 - 2i)/(1 + 2i)(1 - 2i) = (4 - 8i + 3i - 6i²)/(1-4i²
⇒ z = (10 - 5i)/5 = 2 - i
vous faite la suite