Arguments - Corrigé

Énoncé

Donner un argument des nombres complexes suivants.

1.  \(z_1=-\sqrt{3}+i\)

2.  \(z_2=-3i\)

3.  \(z_3=\sqrt{3}+3i\)

4.  \(z_4=5i\)

5.  \(z_5=-2+2i\)

6.  \(z_6=-1\)

Solution

1.  \(\lvert z_1 \lvert= \sqrt{(-\sqrt{3})^2+1^2}= \sqrt{3+1}= \sqrt{4}= 2\)
Soit  \(\theta_1\)  un argument de  \(z_1\) .
\(\left\{ \begin{array}{l}\cos\theta_1=\dfrac{-\sqrt{3}}{2}=\cos\dfrac{5\pi}{6}\\\sin\theta_1=\dfrac{1}{2}=\sin\dfrac{5\pi}{6}\end{array} \right.\\\)
Ainsi :  \(\theta_1 \equiv \dfrac{5\pi}{6} \ [2\pi]\) .

2.  \(\lvert z_2 \lvert= \sqrt{0^2+(-3)^2}= \sqrt{9}= 3\)
Soit  \(\theta_2\)  un argument de  \(z_2\) .
\(\left\{ \begin{array}{l}\cos\theta_2=\dfrac{0}{3}=0=\cos\dfrac{-\pi}{2}\\\sin\theta_2=\dfrac{-3}{3}=-1=\sin\dfrac{-\pi}{2}\end{array} \right.\)
Ainsi :  \(\theta_2 \equiv \dfrac{-\pi}{2} \ [2\pi]\)

3.  \(\lvert z_3 \lvert= \sqrt{(\sqrt{3})^2+3^2}= \sqrt{3+9}= \sqrt{12}= 2\sqrt{3}\)
Soit  \(\theta_3\)  un argument de  \(z_3\) .
\(\left\{ \begin{array}{l}\cos\theta_3=\dfrac{\sqrt{3}}{2\sqrt{3}}=\dfrac{1}{2}=\cos\dfrac{\pi}{3}\\\sin\theta_3=\dfrac{3}{2\sqrt{3}}=\dfrac{\sqrt{3}}{2}=\sin\dfrac{\pi}{3}\end{array} \right.\)
Ainsi  \(\theta_3 \equiv \dfrac{\pi}{3} \ [2\pi]\) .

4.  \(\lvert z_4 \lvert= \sqrt{0^2+5^2}= \sqrt{25}= 5\)
Soit  \(\theta_4\)  un argument de  \(z_4\) .
\(\left\{ \begin{array}{l}\cos\theta_4=\dfrac{0}{5}=0=\cos\dfrac{\pi}{2}\\\sin\theta_4=\dfrac{5}{5}=1=\sin\dfrac{\pi}{2}\end{array} \right.\)
Ainsi  \(\theta_4 \equiv \dfrac{\pi}{2} \ [2\pi]\) .

​​​​​​​​​​​​​​5.  \(\lvert z_5 \lvert= \sqrt{(-2)^2+2^2}= \sqrt{4+4}= \sqrt{8}= 2\sqrt{2}\)
Soit  \(\theta_5\)  un argument de  \(z_5\) .
\(\left \{ \begin{array}{l}\cos\theta_5=\dfrac{-2}{2\sqrt{2}}=\dfrac{-1}{\sqrt{2}}=\cos\dfrac{3\pi}{4}\\\sin\theta_5=\dfrac{2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\sin\dfrac{3\pi}{4}\end{array} \right.\)
Ainsi  \(\theta_5 \equiv \dfrac{3\pi}{4} \ [2\pi]\) .

​​​​​​​​​​​​​​6.  \(\lvert z_6 \lvert= \sqrt{(-1)^2+0^2}= \sqrt{1}= 1\)
Soit  \(\theta_6\)  un argument de  \(z_6\) .
\(\left\{ \begin{array}{l}\cos\theta_6=\dfrac{-1}{1}=-1=\cos(\pi)\\\sin\theta_6=\dfrac{0}{1}=0=\sin(\pi)\end{array} \right.\)
Ainsi  \(\theta_6 \equiv \pi \ [2\pi]\) .