Module d'un nombre complexe
Définition :
Pour tout \(\underline{z} = a + j b \in \mathbb{C}\), \((a , b) \in R^{2}\), on définit le module \(\arrowvert\underline{z}\arrowvert\) de \(\underline{z}\) par :
\(\arrowvert \underline{z}\arrowvert = \sqrt{a^{2} + b^{2}}\) \(\qquad\) \(\left(\arrowvert \underline{z}\arrowvert \in R_{+} \right)\)
Propriété :
\(\forall \underline{z} \in \mathbb{C} \qquad \begin{array}{ll} \arrowvert \underline{z}\arrowvert = 0 \Leftrightarrow z = 0 \\ \arrowvert \underline{z} \arrowvert = \arrowvert \underline{z}^{\ast} \arrowvert = \arrowvert - \underline{z}\arrowvert = \arrowvert - \underline{z}^{\ast} \arrowvert \\ \arrowvert \underline{z} \arrowvert^{2} = \underline{z}~ \underline{z}^{*} \end{array}\)
\(\forall~ \underline{z} \in \mathbb{C}^{*} \qquad \arrowvert \underline{z} \arrowvert = 1 \Leftrightarrow \underline{z}^{\ast} = \frac{1}{\underline{z}}\)
\(\forall ~(\underline{z} , \underline{z}')\in \mathbb{C}^{2} \qquad \arrowvert \underline{z}~\underline{z}'\arrowvert =\arrowvert \underline{z} \arrowvert ~ \arrowvert \underline{z}' \arrowvert\)
Généralisation
\(\forall n \in N^{\ast}, \forall \left(\underline{z}_{1},\underline{z}_{2},...,\underline{z}_{n} \right) \in \mathbb{C}^{n} \Rightarrow \left \arrowvert \displaystyle{\prod^{n}_{i=1}\underline{z}_{i}} \right \arrowvert = \displaystyle{\prod^{n}_{i=1}} \left \arrowvert\underline{z}_{i}\right \arrowvert\)
\(\arrowvert \underline{z} + \underline{z}' \arrowvert \le \arrowvert \underline{z} \arrowvert + \arrowvert \underline{z}' \arrowvert\) inégalité triangulaire
\(\forall \underline{z} \in \mathbb{C}, \forall \underline{z}' \in \mathbb{C}^{\ast} \qquad \left\arrowvert\frac{\underline{z}}{\underline{z}'}\right\arrowvert = \frac{\arrowvert\underline{z}\arrowvert}{\arrowvert\underline{z}'\arrowvert}\)
Exemple :
\(\underline{Z}_{1} = 1-3j \qquad \underline{Z}_{2} = -2 + 5j\)
d'où \(\underline{Z}_{1} ~\underline{Z}_{2} = (1-3j)(-2+5j) = 13 + 11j\)
\(\frac{\underline{Z}_{1}}{\underline{Z}_{2}} = \frac{1-3j}{-2+5j} = \frac{(1-3j)~(-2-5j)}{(-2+5j)~(-2-5j)} = \frac{-17+j}{29}\)
\(\arrowvert\underline{Z}_{1}\arrowvert = \sqrt{1^{2}+(-3)^{2}}\quad \arrowvert\underline{Z}_{2}\arrowvert = \sqrt{(-2)^{2} + (5)^{2}} = \sqrt{29}\)
\(\arrowvert\underline{Z}_{1} ~\underline{Z}_{2} \arrowvert = \sqrt{13^{2} + 11^{2}} = \sqrt{290} = \sqrt{10}\sqrt{29} = \arrowvert\underline{Z}_{1} \arrowvert ~\arrowvert\underline{Z}_{2}\arrowvert\)
\(\left\arrowvert\frac{\underline{Z}_{1}}{\underline{Z}_{2}}\right\arrowvert = \frac{1}{29} \sqrt{(-17)^{2} + 1^{2}} = \frac{\sqrt{290}}{29} = \frac{\sqrt{10}\sqrt{29}}{\sqrt{29}\sqrt{29}} = \frac{\sqrt{10}}{\sqrt{29}} = \frac{\arrowvert\underline{Z}_{1}\arrowvert} {\arrowvert\underline{Z}_{2}\arrowvert}\)