TS – Complexes 3 – Ex 1

Exercice 1

Quelle est la forme trigonométrique de : $z_1 = -1 + \ic \sqrt{3}$ et $z_2 = 3 – 3\ic$?

Correction

$|z_1| = \sqrt{1 + 3} = 2$

donc $z_1 = 2\left(\dfrac{-1}{2} + \dfrac{\sqrt{3}}{2}\ic\right) = 2\left(\cos \dfrac{2\pi}{3} + \ic \sin \dfrac{2\pi}{3}\right)$

$\quad$

$|z_2| = \sqrt{9 + 9} = 3\sqrt{2}$

donc

$\begin{align} z_2 &= 3\sqrt{2}\left(\dfrac{1}{\sqrt{2} }- \dfrac{\ic}{\sqrt{2}}\right) \\\\
&= 3\sqrt{2} \left(\dfrac{\sqrt{2}}{2} – \dfrac{\sqrt{2}}{2} \ic\right) \\\\
& = 3\sqrt{2} \left( \cos \dfrac{-\pi}{4} + \ic \sin \dfrac{-\pi}{4}\right)
\end{align}$