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Adtht12
@Adtht12
May 2019
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Bonjour, pouvez vous m’aidez pour mon dm de mathématiques de terminal S pour l’exercice 2 question b) et c) svp
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scoladan
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Bonjour,
2b)i) en posant a = A + iB :
(z - 3)(z + a)
= z² + (a - 3)z - 3a
= z² + (A + iB - 3)z - 3(A + iB)
⇒ par analogie des termes :
A - 3 + iB = -(1 + 3i)
et -3A - 3iB = -6 + 9i
⇒ A = 2 et B = -3
⇒ a = 2 - 3i
ii) en posant b = A' + B'i :
(z - 4i)((z + b)
= z² + (b - 4i)z - 4bi
= z² + (A' + B'i - 4i)z - 4(A' + B'i)i
⇒ A' + (B' - 4)i = -1 - 3i
et 4B' - 4A'i = 4 + 4i
⇒ A' = -1 et B' = 1
⇒ b = -1 + i
c) L'équation devient :
(z - 3)(z + a)(z - 4i)(z + b) = 0
⇒ solutions : 3, -a, 4i, et -b
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Verified answer
Bonjour,2b)i) en posant a = A + iB :
(z - 3)(z + a)
= z² + (a - 3)z - 3a
= z² + (A + iB - 3)z - 3(A + iB)
⇒ par analogie des termes :
A - 3 + iB = -(1 + 3i)
et -3A - 3iB = -6 + 9i
⇒ A = 2 et B = -3
⇒ a = 2 - 3i
ii) en posant b = A' + B'i :
(z - 4i)((z + b)
= z² + (b - 4i)z - 4bi
= z² + (A' + B'i - 4i)z - 4(A' + B'i)i
⇒ A' + (B' - 4)i = -1 - 3i
et 4B' - 4A'i = 4 + 4i
⇒ A' = -1 et B' = 1
⇒ b = -1 + i
c) L'équation devient :
(z - 3)(z + a)(z - 4i)(z + b) = 0
⇒ solutions : 3, -a, 4i, et -b