1. \(z_A = \sqrt{2} \ \text{cis} \left( -\frac{\pi}{4}\right) = 1-\mathbf{i}\)
2. \(z_B = \text{cis} \left( \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}+\frac{1}{2}\mathbf{i}\)
3. \(z_C = 2 \ \text{cis} \left( \frac{2\pi}{3}\right) = -1+\sqrt{3}\mathbf{i}\)
4. \(z_D = \text{cis} \left( 0\right) = 1\)
5. \(z_E = \text{cis} \left( \frac{\pi}{2}\right) = \mathbf{i}\)
6. \(z_F = \text{cis} \left( \frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}-\frac{1}{2}\mathbf{i}\)
7. \(z_G = 2 \ \text{cis} \left( -\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\mathbf{i}\)
8. \(z_H = 2 \ \text{cis} \left( \pi\right) = -2\)

1) et 2) \begin{equation} \begin{array}{l|l|l|l} z_1 = 3 \ \text{cis} \left( 0\right) & z_2 = 4 \ \text{cis} \left( \pi\right) & z_3= \text{cis} \left( \frac{\pi}{2}\right) & z_4 = 3 \ \text{cis} \left( \frac{3\pi}{2}\right) \\ z_5 = 2\sqrt{2} \ \text{cis} \left( \frac{\pi}{4}\right) & z_6=2\sqrt{2} \ \text{cis} \left( \frac{7\pi}{4}\right) & z_7=2\sqrt{3} \ \text{cis} \left( \frac{2\pi}{3}\right) & \end{array} \end{equation}

Si \(z = r(\cos \phi + i \sin \phi)\) alors \(\overline{z} = r(\cos \phi - i \sin \phi) = r[\cos \phi + i(-\sin \phi)]\)

Or, \(\cos(-\phi) = \cos \phi\) et \(\sin(-\phi) = -\sin \phi\)

Donc \(\overline{z} = r[\cos(-\phi) + i \sin(-\phi)]\) et \(\arg \overline{z} = -\phi\) ou encore \(\bbox[lightblue,5px] {\arg(\overline{z}) = -\arg(z)}\)

Soit \(z=r \cdot \operatorname{cis}(\phi)\)
Sachant que \( \text{cis}(\theta_1) \cdot \text{cis}(\theta_2) = \text{cis}(\theta_1+\theta_2) \) et \(\operatorname{cis}(\pi) = \cos(\pi) + i \sin(\pi) = -1 \), on a : \[\begin{aligned} -z &= -1 \cdot r \cdot \operatorname{cis}(\phi) \\ &= \operatorname{cis}(\pi) \cdot r \cdot \operatorname{cis}(\phi) \\ &= r \cdot \operatorname{cis}(\pi) \cdot \operatorname{cis}(\phi) \\ &= r \cdot \operatorname{cis}(\pi+\phi) \\ \end{aligned}\] Par conséquent : \(\bbox[lightblue,5px] {\arg(−z) =} \pi+\phi =\bbox[lightblue,5px] { \pi+\arg(z)}\)

\[ \begin{aligned} \bbox[lightblue,5px] {\arg(-\overline{z})} &= \pi + \arg(\overline{z}) \quad \text{(car multiplier par } -1 \text{ revient à ajouter } \pi \text{ à l'argument)}\\ &= \bbox[lightblue,5px] {\pi - \arg(z)} \quad \text{(car prendre le conjugué revient à changer le signe de l'argument)} \end{aligned} \]

\(\arg(z_1)=\frac{\pi}{4}\) et \(\arg(z_2)=\frac{5\pi}{6}\)

note : \(z_2\) se trouve dans le deuxième quadrant → \(\arg(z_2)=\pi + \arctan\left( \frac{\sqrt{3}}{-3} \right)\)

\(\arg(z_1\cdot z_2) = \arg(z_1)+\arg(z_2) = \frac{\pi}{4}+\frac{5\pi}{6} = \frac{13\pi}{12}\)

\(\arg(−3 - i\sqrt{3}) = \arg(\overline{z_2}) = -\arg(z_2) = -\frac{5\pi}{6}\)

\(\arg\left(-\frac12 (1 + i)\right) =\arg\left(\left(-\frac12 + 0 \cdot i\right) (1 + i)\right) = \arg\left(-\frac12 + 0 \cdot i\right) +\arg (1 + i) = \pi + \arg (z_1) = \frac{5\pi}{4}\)

\(\arg(1 - i) = \arg(\overline{z_1}) = -\arg(z_1) = -\frac{\pi}{4}\)

\(\arg{\left(3 - i\sqrt{3}\right)} = \arg (-z_2) = \pi + \arg (z_2) = \pi + \frac{5\pi}{6} =\frac{11\pi}{6} \)

d'où \[\begin{aligned} \arg\left(\left(3 - i\sqrt{3}\right)^2\right) &= \arg\left(\left(3 - i\sqrt{3}\right)\cdot\left(3 - i\sqrt{3}\right)\right) \\ &= \arg{\left(3 - i\sqrt{3}\right)}+\arg{\left(3 - i\sqrt{3}\right)}\\ &= \frac{11\pi}{6}+\frac{11\pi}{6} = \frac{11\pi}{3} \end{aligned}\] de même, \[\begin{aligned} \arg\left(\left(1 - i\right)^3\right) &= \arg\left(\left(1 - i\right)\cdot\left(1 - i\right)\cdot\left(1 - i\right)\right) \\ &= -\frac{\pi}{4}-\frac{\pi}{4}-\frac{\pi}{4} = -\frac{3\pi}{4} \end{aligned}\] finalement, \[\begin{aligned} \arg\left(\dfrac{\left(3 - i\sqrt{3}\right)^2}{\left(1-i\right)^3}\right) &= \arg\left(\left(3 - i\sqrt{3}\right)^2\right) - \arg\left(\left(1 - i\right)^3\right) \\ &= \frac{11\pi}{3} - \left(-\frac{3\pi}{4}\right) = \frac{53\pi}{12} \end{aligned}\] cet argument peut être simplifié par \( \dfrac{5\pi}{12} \) car \[ \frac{53\pi}{12} = \frac{5\pi}{12} + 4\pi \]

Note : \(\arg\left(\dfrac{\left(3 - i\sqrt{3}\right)^2}{\left(1-i\right)^3}\right) = 2\arg\left(3 - i\sqrt{3}\right) - 3\arg\left(1-i\right)\) (→ voir Formule de Moivre)

\[\begin{aligned} z_{1}&=2\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\\ &= 2\ \text{cis} \left( \frac{\pi}{4}\right) \implies \text{ immédiat } : \bbox[lightblue,5px] {|z_1|=2 \ \text{ et } \ \arg(z_1) = \frac{\pi}{4}}\\ z_{2}&=-2\left(\cos \tfrac{\pi}{4}+i \sin \tfrac{\pi}{4}\right) \\ &=2\left(-\cos \tfrac{\pi}{4}-i \sin \tfrac{\pi}{4}\right) \quad \textbf{est dans le 3ème quadrant}\\ &=2\left(\cos \tfrac{5\pi}{4}+i \sin \tfrac{5\pi}{4}\right)\\ &= 2\ \text{cis} \left( \tfrac{5\pi}{4}\right) \implies \bbox[lightblue,5px] {|z_2|=2 \ \text{ et } \ \arg(z_2) = \tfrac{5\pi}{4}}\\ z_{3}&=2\left(-\cos \tfrac{\pi}{4}+i \sin \tfrac{\pi}{4}\right)\quad \textbf{est dans le 2ème quadrant}\\ &= 2\left(\cos \tfrac{3\pi}{4}+i \sin \tfrac{3\pi}{4}\right)\\ &= 2\ \text{cis} \left( \tfrac{3\pi}{4}\right) \implies \bbox[lightblue,5px] {|z_3|=2 \ \text{ et } \ \arg(z_3) = \tfrac{3\pi}{4}}\\ z_{4}&=2\left(\cos \tfrac{\pi}{4}-i \sin \tfrac{\pi}{4}\right)\quad \textbf{est dans le 4ème quadrant}\\ &= 2\ \text{cis} \left( \tfrac{7\pi}{4}\right) \implies \bbox[lightblue,5px] {|z_4|=2 \ \text{ et } \ \arg(z_4) = \tfrac{7\pi}{4}}\\ \end{aligned}\]