Nombres complexes : point de vue algébrique

Equations du 1er degré dans C\mathbb{C} - Exercice 3

7 min
20
Question 1

Résoudre, dans C2\mathbb{C}^2, le système {3z1+z2=2i2z1+4z2=35i\left\{ \begin{array}{ccc}3z_1+z_2 & = & -2-i \\ 2z_1+4z_2 & = & 3-5i \end{array}\right. .

Correction
{3z1+z2=2i2z1+4z2=35i\left\{ \begin{array}{ccc}3z_1+z_2 & = & -2-i \\ 2z_1+4z_2 & = & 3-5i \end{array}\right.
{z2=2i3z12z1+4z2=35i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ 2z_1+4z_2 & = & 3-5i \end{array}\right.
{z2=2i3z12z1+4(2i3z1)=35i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ 2z_1+4\left(-2-i-3z_1\right) & = & 3-5i \end{array}\right.
{z2=2i3z12z184i12z1=35i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ 2z_1-8-4i-12z_1 & = & 3-5i \end{array}\right.
{z2=2i3z110z184i=35i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ -10z_1-8-4i & = & 3-5i \end{array}\right.
{z2=2i3z110z1=35i+8+4i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ -10z_1 & = & 3-5i+8+4i \end{array}\right.
{z2=2i3z110z1=11i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ -10z_1 & = & 11-i \end{array}\right.
{z2=2i3z1z1=11i10\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ z_1 & = & \frac{11-i}{-10} \end{array}\right.
{z2=2i3z1z1=11i10\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ z_1 & = & \frac{11-i}{-10} \end{array}\right.
{z2=2i3z1z1=1110+i10\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ z_1 & = & \frac{11}{-10}+\frac{-i}{-10} \end{array}\right.
{z2=2i3z1z1=1110+110i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3z_1 \\ z_1 & = & -\frac{11}{10}+\frac{1}{10}i \end{array}\right.
{z2=2i3×(1110+110i)z1=1110+110i\left\{ \begin{array}{ccc}z_2 & = & -2-i-3\times \left(-\frac{11}{10}+\frac{1}{10}i\right) \\ z_1 & = & -\frac{11}{10}+\frac{1}{10}i \end{array}\right.
{z2=2i+3310310iz1=1110+110i\left\{ \begin{array}{ccc}z_2 & = & -2-i+\frac{33}{10}-\frac{3}{10}i \\z_1 & = & -\frac{11}{10}+\frac{1}{10}i \end{array}\right.
{z2=20101010i+3310310iz1=1110+110i\left\{ \begin{array}{ccc}z_2 & = & -\frac{20}{10}-\frac{10}{10}i+\frac{33}{10}-\frac{3}{10}i \\ z_1 & = & -\frac{11}{10}+\frac{1}{10}i \end{array}\right.
{z2=13101310iz1=1110+110i\left\{ \begin{array}{ccc}z_2 & = & \frac{13}{10}-\frac{13}{10}i \\ z_1 & = & -\frac{11}{10}+\frac{1}{10}i \end{array}\right.
Le couple solution du système {3z1+z2=2i2z1+4z2=35i\left\{ \begin{array}{ccc}3z_1+z_2 & = & -2-i \\ 2z_1+4z_2 & = & 3-5i \end{array}\right. est alors
S={(1110+110i;13101310i)}S=\left\{\left(-\frac{11}{10}+\frac{1}{10}i;\frac{13}{10}-\frac{13}{10}i\right)\right\}