Calculs d'arguments - Corrigé

Énoncé

Déterminer un argument des complexes suivants.

1.  `z_1=-3-3i`

2.  `z_2=-\sqrt{6}+i\sqrt{2}`  

3.  `z_3=2-2i\sqrt{3}`

Solution

1.  On a : \(\left\vert z_1 \right\vert = \sqrt{(-3)^2+(-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}\) .
Soit  `\theta_1` un argument de `z_1` . On a alors :
\(\left\lbrace \begin{array}{l} \cos\theta_1 =\dfrac{-3}{3\sqrt{2}} =\dfrac{-1}{\sqrt{2}} =\cos\dfrac{-3\pi}{4} \\ \sin\theta_1 =\dfrac{-3}{3\sqrt{2}} =\dfrac{-1}{\sqrt{2}} =\sin\dfrac{-3\pi}{4} \end{array} \right.\)
donc `\theta_1 \equiv \frac{-3\pi}{4} \ [2\pi]` .

2.  On a : \(\left\vert z_2 \right\vert = \sqrt{(-\sqrt{6})^2+(\sqrt{2})^2} = \sqrt{6+2} = \sqrt{8} = 2\sqrt{2}\) .
Soit `\theta_2`  un argument de `z_2` . On a alors :
\(\left\lbrace \begin{array}{l} \cos\theta_2 =\dfrac{-\sqrt{6}}{2\sqrt{2}} =\dfrac{-\sqrt{3}}{2} =\cos\dfrac{5\pi}{6} \\ \sin\theta_2 =\dfrac{\sqrt{2}}{2\sqrt{2}} =\dfrac{1}{2} =\sin\dfrac{5\pi}{6} \end{array} \right.\)
donc `\theta_2 \equiv \frac{5\pi}{6} \ [2\pi]` .

3.  On a : \(\left\vert z_3 \right\vert = \sqrt{2^2+(-2\sqrt{3})^2} = \sqrt{4+12} = \sqrt{16} = 4\) .
Soit `\theta_3` un argument de `z_3` . On a alors :
\(\left\lbrace \begin{array}{l} \cos\theta_3 =\dfrac{2}{4} = \dfrac{1}{2} =\cos \left( -\dfrac{\pi}{3} \right) \\ \sin \theta_3 =\dfrac{-2\sqrt{3}}{4} = \dfrac{-\sqrt{3}}{2} =\sin \left( - \dfrac{\pi}{3} \right) \end{array} \right.\)
donc `\theta_3 \equiv - \frac{\pi}{3} \ [2\pi]` .