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corrigé epreuve 2008 : Equation dans C écriture complexe similitude

Détails: Catégorie : algèbre

1) (E) : z $^3$ + (-6 - 4i)z $^2$ + (12 + 21i)z + 9 - 45i = 0

a) Déterminer la solution imaginaire pure z $_0$ de (E)

on a z $_0$ = ai

$z_0$ est solution de (E) $\Longrightarrow$ -a $^3$ i + (-6-4i)(-a $^2$ ) + (12 + 21i)ia + 9 - 45i = 0

$\Longrightarrow$ -ia $^3$ + 6a $^2$ + 4ia $^2$ + 12ia - 21a + 9 - 45i = 0

$i(-a^3$ + 4a $^2$ + 12a - 45) + 6a $^2$ - 21a + 9 = 0

$\Longrightarrow \left\{ \begin{array}{l} -a^3 + 4a^2 + 12 - 45 = 0 \\ 6a^2 - 21a + 9 = 0 \end{array} \right. \Longrightarrow \left\{ \begin{array}{l} -a^3 + 4a^2 + 12a - 45 = 0 \\ 2a^2 - 7a + 3 = 0 \end{array} \right.$

2a $^2$ - 7a + 3 = 0 $\Delta$ = b $^2$ - 4ac = 49 - 24 = 25

$a_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$ ,

$a_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{7+5}{4} = \frac{12}{4} = 3$

Calculons $-a_1^3 + 4a_1^2 + 12a_1 - 45 = -\frac{1}{8} + \frac{4}{4} + \frac{12}{2} - 45 = -\frac{1}{8} + 1 + 6 - 45 = -\frac{1}{8} - 38 \neq 0$

$-a_2^3 + 4a_2^2 + 12a_2 - 45 = -27 + 36 + 36 - 45 = 72 - 72 = 0$

donc z $_0$ = 3i

b) Achever la résolution de (E)

(E) $\Longleftrightarrow$ (z - 3i)(z $^2$ + az + b) = 0

z $^3$ + az $^2$ + bz - 3iz $^2$ - 3iaz - 3ib = 0

z $^3$ + (a - 3i)z $^2$ + (b - 3ia)z - 3ib = 0

$\Longrightarrow \left\{ \begin{array}{l} a - 3i = -6 - 4i \\ b - 3ia = 12 + 21i -3ib = 9 - 45i \end{array} \right. \Longrightarrow \left\{ \begin{array}{l} a = -6 - i \\ b = 9 - \frac{45i}{-3i} b - 3ia = 12 + 21i \end{array} \right.$

$\Longrightarrow \left\{ \begin{array}{l} a = -6-i \\ b= +15 + 3i\\ 15 + 3i - 3i(-6-i) = 12 + 21i \end{array} \right.$ $\Longrightarrow \left\{ \begin{array}{l}a = -6-i \\b = 15 + 3i \\15 + 3i + 18i - 3 = 12 + 21i \end{array} \right.$