Nombres complexes

Cas de la multiplication :

Module : \(\left\arrowvert \underline{Z} \right\arrowvert = \left\arrowvert \underline{z_{1}} \right\arrowvert \cdot \left\arrowvert \underline{z_{2}} \right\arrowvert = 4\sqrt{3}\) , argument : \(\arg (\underline{Z}) = \arg (\underline{z_{1}}) + \arg(\underline{z_{2}}) = \pi\)

\(\underline{Z} = 4 \sqrt{3}\left( \cos \pi + j \sin \pi\right)\)

Cas de la division :

Module : \(\left\arrowvert \underline{Z'} \right\arrowvert = \frac{\left\arrowvert \underline{z_{1}} \right\arrowvert}{\left\arrowvert \underline{z_{2}} \right\arrowvert} = \frac{\sqrt{3}}{3}\), argument : \(\arg (\underline{Z'}) = \arg (\underline{z_{1}}) - \arg(\underline{z_{2}}) = \frac{\pi}{3}\)

\(\underline{Z'} = \frac{\sqrt{3}}{3}\left( \cos \frac{\pi}{3} + j \sin \frac{\pi}{3}\right)\)

Transformons les formes algébriques de \(\underline{z_{1}}\) et \(\underline{z_{2}}\) en formes trigonométriques :

Module de \(\underline{z_{1}}\): \(\left\arrowvert \underline{z_{1}} \right\arrowvert = \sqrt{(-1)^{2} + (\sqrt{3})^{2}} = 2\) , argument de \(\underline{z_{1}}\): \(\arg(\underline{z_{1}}) = \alpha_{1}\) est tel que \(\cos \alpha_{1} = \frac{-1}{2} \quad \sin \alpha_{1} = \frac{\sqrt{3}}{2}\)

Soit \(\alpha_{1} = \frac{2 \pi}{3}\) et \(\underline{z_{1}} = 2\left(\cos \frac{2 \pi}{3} + j \sin \frac{2 \pi}{3}\right)\)

Module de \(\underline{z_{2}}\): \(\left\arrowvert \underline{z_{2}} \right\arrowvert = \sqrt{3}\sqrt{1^{2} + (\sqrt{3})^{2}} = 2\sqrt{3}\) , argument de \(\underline{z_{2}}\): \(\arg(\underline{z_{2}}) = \alpha_{2}\) est tel que \(\cos \alpha_{2} = \frac{1}{2} \quad \sin \alpha_{2} = \frac{\sqrt{3}}{2}\)

Soit \(\alpha_{2} = \frac{\pi}{3}\) et \(\underline{z_{2}} = 2\sqrt{3}\left(\cos \frac{\pi}{3} + j \sin \frac{\pi}{3}\right)\)

Cas de la multiplication : \(\underline{Z} = \underline{z_{1}} \cdot \underline{z_{2}}\)

Module : \(\left\arrowvert\underline{Z}\right\arrowvert =\left\arrowvert \underline{z_{1}} \right\arrowvert \cdot \left\arrowvert\underline{z_{2}} \right\arrowvert = 2 \times 2 \sqrt{3} = 4 \sqrt{3}\), argument : \(\arg( \underline{Z}) = \arg(\underline{z_{1}}) + \arg(\underline{z_{2}}) = \frac{2\pi}{3} + \frac{\pi}{3} = \pi\)

d'où la forme trigonométrique de \(\underline{Z}\) : \(\underline{Z} = 4 \sqrt{3} \left(\cos \pi + j \sin \pi \right)\)

( La forme algébrique sera :\(\underline{Z} = - 4 \sqrt{3}\))

Cas de la division : \(\underline{Z}' = \frac{\underline{z_{1}}}{\underline{z_{2}}}\)

Module : \(\arrowvert\underline{Z}'\arrowvert = \frac{\arrowvert\underline{z_{1}}\arrowvert}{\arrowvert\underline{z_{2}}\arrowvert} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\), argument : \(\arg \left(\underline{Z}'\right) = \arg\left(\underline{z_{1}}\right)- \arg\left(\underline{z_{2}}\right) = \frac{2 \pi}{3} - \frac{\pi}{3} = \frac{\pi}{3}\)

d'où la forme trigonométrique de \(\underline{Z'}\): \(\underline{Z'} = \frac{\sqrt{3}}{3} (\cos \frac{\pi}{3} + j \sin \frac{\pi}{3})\)

( La forme algébrique sera : \(\underline{Z'} = \frac{\sqrt{3}}{6} + j \frac{1}{2}\))

Les nombres complexes \(\underline{z_{1}}\) et \(\underline{z_{2}}\) étant conjugués nous avons les relations entre :

Les modules : \(\arrowvert\underline{Z}\arrowvert = \frac{\arrowvert\underline{z_{1}}\arrowvert}{\arrowvert\underline{z_{2}}\arrowvert} = 1\) car \(\arrowvert\underline{z_{1}}\arrowvert = \arrowvert\underline{z_{2}}\arrowvert\)

Les arguments : \(\arg(\underline{Z}) = \arg(\underline{z_{1}}) - \arg(\underline{z_{2}}) = \frac{5 \pi}{4} \quad[2 \pi]\)

Car \(\arg(\underline{z_{1}}) = - \arg(\underline{z_{2}}) = - \frac{3 \pi}{8}\)

Calcul du module de \(\underline{z_{1}}\):

\(\arrowvert \underline{z_{1}}\arrowvert = \sqrt{1^{2} + \tan^{2} \frac{5 \pi}{8}} = \frac{1}{\sqrt{\cos^{2} \frac{5 \pi}{8}}} = \frac{1}{\left\arrowvert \cos \frac{5 \pi}{8}\right\arrowvert} = - \frac{1}{\cos \frac{5 \pi}{8}}\)

( \(\cos \frac{5 \pi}{8} <0\) car \(\frac{\pi}{2} < \frac{5 \pi}{8} < \pi\) )

\(\arrowvert\underline{z_{1}}\arrowvert = \arrowvert\underline{z_{2}}\arrowvert\) car \(\underline{z_{1}}\) et \(\underline{z_{2}}\) sont conjugués d'où \(\arrowvert\underline{Z}\arrowvert = \frac{\arrowvert\underline{z_{1}}\arrowvert}{\arrowvert\underline{z_{2}}\arrowvert} = 1\)

Calcul de l'argument de \(\underline{z_{1}}\):

\(\underline{z_{1}} = 1 + j \tan \frac{5 \pi}{8} = - \frac{1}{\cos \frac{5 \pi}{8}}(- \cos \frac{5 \pi}{8} - j \sin \frac{5 \pi}{8} ) = \frac{-1}{\cos \frac{5 \pi}{8}}(+\cos \frac{13 \pi}{8} + j \sin \frac{13 \pi}{8})\)

d'où \(\arg(\underline{z_{1}}) = \frac{13 \pi}{8} \quad [2 \pi]\)

\(\arg (\underline{z_{1}}) = - \arg(\underline{z_{2}})\) car \(\underline{z_{1}}\) et \(\underline{z_{2}}\) sont conjugués et

\(\arg\left(\underline{Z}\right) = \arg (\underline{z_{1}}) - \arg(\underline{z_{2}}) = 2 \arg (\underline{z_{1}}) = \frac{13 \pi}{4} \textrm{ ou }\frac{5 \pi}{4} \quad [2 \pi]\)

(car \(\frac{13 \pi}{4} = \frac{8 \pi + 5 \pi}{4} = 2 \pi + \frac{5 \pi}{4}\))

\(\arg\left(\underline{Z}\right) = \frac{5 \pi}{4} \quad [2 \pi]\)

Pour \(\alpha \in \left[0, \frac{\pi}{2} \right]\) : \(\left\arrowvert \underline{Z} \right\arrowvert = 2 \cos \alpha \textrm{ et }\arg \left(\underline{Z}\right) = \alpha + \frac{\pi}{2} ~\left[2 \pi \right]\)

Pour \(\alpha \in \left]\frac{\pi}{2} , \pi\right]\) : \(\left\arrowvert \underline{Z} \right\arrowvert =- 2 \cos \alpha \textrm{ et }\arg \left(\underline{Z}\right) = \alpha + \frac{3 \pi}{2} ~\left[2 \pi \right]\)

Transformons l'expression de \(\underline{Z}\) pour obtenir une forme trigonométrique.

Application de la formule de duplication : \(\sin 2 \alpha = 2 \sin \alpha \cos \alpha\)

D'où \(\underline{Z} = - \sin 2 \alpha + j 2 \cos^{2} \alpha = -2 \sin \alpha \cos \alpha + j 2 \cos^{2} \alpha = 2 \cos \alpha \left[\cos\left(\alpha + \frac{\pi}{2} \right) + j \sin \left( \alpha + \frac{\pi}{2} \right) \right]\)

car \(-\sin \alpha = \cos \left( \alpha + \frac{\pi}{2} \right) \textrm{ et }\cos \alpha = \sin \left(\alpha + \frac{\pi}{2}\right)\)

Le facteur \(2\cos \alpha\) représentera un module si \(\cos \alpha\) est positif ou nul.

1er cas : \(\alpha \in \left[0, \frac{\pi}{2} \right] \Rightarrow \cos \alpha \ge 0\) et \(\left\arrowvert \underline{Z} \right\arrowvert = 2 \cos \alpha \textrm{ et }\arg \left(\underline{Z}\right) = \alpha + \frac{\pi}{2} ~\left[2 \pi \right]\)

2ème cas : \(\alpha \in \left]\frac{\pi}{2},\pi \right] \Rightarrow \cos \alpha < 0\) et \(\left\arrowvert \underline{Z} \right\arrowvert = -2 \cos \alpha \textrm{ et }\arg \left(\underline{Z}\right) = \alpha + \frac{3\pi}{2} ~\left[2 \pi \right]\)

car \(\underline{Z} = -2 \cos \alpha (-\cos(\alpha + \frac{\pi}{2}) - j \sin(\alpha + \frac{\pi}{2}))\)

\(=-2 \cos \alpha \left[\cos \left(\alpha + \frac{\pi}{2} + \pi \right) + j \sin \left(\alpha + \frac{\pi}{2} + \pi \right) \right]\)