Nombres complexes

Sachant que

\(\underline{z}_{1} = \sqrt{2} e^{\left(j \frac{5 \pi}{4}\right)} = \sqrt{2} \left(\cos \frac{5 \pi}{4} +j \sin \frac{5 \pi}{4}\right) = -1-j\)

\(\underline{z}_{2} = \sqrt{3} + j= 2 \left(\cos \frac{\pi}{6} +j \sin \frac{\pi}{6}\right) = 2 e^{\left(j\frac{\pi}{6}\right)}\)

nous calculons :

a. \(\underline{Z}_{1} = \underline{z}_{1} + \underline{z}_{2} = -1+\sqrt{3}\) ( 2 points )

formes algébrique et trigonométrique ( 2 points ) car \(\underline{Z}_{1}\in \mathbb{R}\)

b. \(\underline{Z}_{2} = \underline{z}_{1} \cdot \underline{z}_{2} = \left(-1-j\right)\left(\sqrt{3} + j\right) = 1 - \sqrt{3} - j \left(1+\sqrt{3}\right)\)

forme algébrique ( 2 points )

\(\underline{Z}_{2} = \underline{z}_{1} \cdot \underline{z}_{2} = \sqrt{2} e^{\left(j \frac{5 \pi}{4}\right)} \cdot 2 e^{\left(j\frac{\pi}{6}\right)} = 2 \sqrt{2} e^{j\left(\frac{5 \pi}{4} + \frac{\pi}{6}\right)} = 2 \sqrt{2} e^{\left(j \frac{17 \pi}{12}\right)}\)

forme trigonométrique ( 2 points )

c. \(\underline{Z}_{3} = \frac{\underline{z}_{1}}{\underline{z}_{2}} = \frac{-1-j}{\sqrt{3} + j} = - \frac{1}{4} \left(3 + 1 + j \left(\sqrt{3} - 1 \right)\right)\)

forme algébrique ( 2 points )

\(\underline{Z}_{3} = \frac{\underline{z}_{1}}{\underline{z}_{2}} = \frac{\sqrt{2} e^{\left(j \frac{5 \pi}{4}\right)}}{2 e^{\left(j\frac{\pi}{6}\right)}} = \frac{\sqrt{2}}{2} e^{j\left(\frac{5\pi}{4} - \frac{\pi}{6}\right)} = \frac{\sqrt{2}}{2} e^{\left(j \frac{13 \pi}{12}\right)}\)

forme trigonométrique ( 2 points )