Modules, arguments et opérations - Corrigé

Énoncé

Déterminer le module et un argument des nombres complexes suivants.

1. \(z_1=\dfrac{\sqrt{6}-i\sqrt{2}}{2(1+i)}\)

2. \(z_2= \left( \dfrac{\sqrt{2}-\sqrt{2}i}{1-i\sqrt{3}} \right)^3\)

Solution

1. On a : \(\left\vert z_1 \right\vert= \left\vert \dfrac{\sqrt{6}-i\sqrt{2}}{2(1+i)} \right\vert= \dfrac{\left\vert \sqrt{6}-i\sqrt{2} \right\vert}{\left\vert 2(1+i) \right\vert}= \dfrac{\left\vert \sqrt{6}-i\sqrt{2} \right\vert}{\left\vert 2 \right\vert \left\vert 1+i \right\vert}\)
et  \(\arg(z_1)\equiv \arg\left(\dfrac{\sqrt{6}-i\sqrt{2}}{2(1+i)}\right)\equiv \arg(\sqrt{6}-i\sqrt{2})-\arg(2(1+i)) \ [2\pi]\equiv \arg(\sqrt{6}-i\sqrt{2})- \arg(2)- \arg(1+i) \ [2\pi]\)

• \(\left\vert 2 \right\vert = 2 \,\text {et} \arg(2) \equiv 0 \ [2\pi] \,\text {car}\, 2 \in \mathbb{R}.\)
  
• \(\left\vert 1+i \right\vert = \sqrt{2} \,\text {et} \arg(1+i) \equiv \dfrac{\pi}{4} \ [2\pi]\)
  
•   \(\left\vert \sqrt{6}-i\sqrt{2} \right\vert= \sqrt{(\sqrt{6})^2-(\sqrt{2})^2}= \sqrt{6+2}= \sqrt{8}= 2\sqrt{2}\) .
  

Soit \(\theta\) un argument de \(\sqrt{6}-i\sqrt{2}\) . On a alors :
\(\left\lbrace \begin{array}{l}\cos\theta=\dfrac{\sqrt{6}}{2 \sqrt{2}}=\dfrac{\sqrt{3}}{2}=\cos \dfrac{-\pi}{6}\\\sin\theta_=\dfrac{-\sqrt{2}}{2 \sqrt{2}}= -\dfrac{1}{2}=\sin \dfrac{-\pi}{6}\end{array} \right.\)
donc \(\theta \equiv -\dfrac{\pi}{6} \ [2\pi]\) .

On en déduit que :
\(\left\vert z_1 \right\vert=\dfrac{2 \sqrt{2}}{2\sqrt{2}}=1\)   et   \(\arg(z_1)\equiv \theta - \dfrac{\pi}{4}\equiv - \dfrac{2\pi}{12}-\dfrac{3\pi}{12}\equiv -\dfrac{5\pi}{12} \ [2\pi] .\)

2.  On a :   \(\left\vert z_2 \right\vert= \left\vert \left( \dfrac{\sqrt{2}-\sqrt{2}i}{1-i\sqrt{3}} \right)^3 \right\vert= \left\vert \dfrac{\sqrt{2}-\sqrt{2}i}{1-i\sqrt{3}} \right\vert^3= \dfrac{\left\vert \sqrt{2}-\sqrt{2}i \right\vert^3}{\left\vert 1-i\sqrt{3} \right\vert^3}= \dfrac{\left\vert \sqrt{2}(1-i) \right\vert^3}{\left\vert 1-i\sqrt{3} \right\vert^3}= \dfrac{\left\vert \sqrt{2} \right\vert^3 \left\vert 1-i \right\vert^3}{\left\vert 1-i\sqrt{3} \right\vert^3}\)

et   \(\arg(z_2)\equiv \arg \left( \dfrac{\sqrt{2}-\sqrt{2}i}{1-i\sqrt{3}} \right)^3\equiv 3 \left( \arg(\sqrt{2}- \sqrt{2}i)-\arg(1-i\sqrt{3}) \right) \ [2\pi]\)

et \(\arg(\sqrt{2}- \sqrt{2}i)= \arg(\sqrt{2} (1-i) )= \arg(\sqrt{2}) + \arg(1-i )= \arg(1-i )\)

Donc   \(\arg(z_2)\equiv 3 \left( \arg(1-i ) -\arg(1-i\sqrt{3}) \right) \ [2\pi]\)

Or,

• \(\left\vert \sqrt{2} \right\vert^3 = 2 \sqrt{2}\)
  
• \(\left\vert 1-i \right\vert = \sqrt{2} \,\text {et} \arg(1-i) \equiv \dfrac{-\pi}{4} \ [2\pi]\)
  
• \(\left\vert 1-i \sqrt{3} \right\vert = 2 \,\text {et} \arg(1-i \sqrt{3}) \equiv \dfrac{-\pi}{3} \ [2\pi]\)
  

On en déduit que :
\(\left\vert z_2 \right\vert=\dfrac{2 \sqrt{2} \times \left( \sqrt{2} \right)^3}{2^3}= \dfrac{8}{8}= 1\)
et  \(\arg(z_2)\equiv 3 \left( - \dfrac{\pi}{4} - \frac{\pi}{3} \right)\equiv 3 \left( - \dfrac{7\pi}{12} \right)\equiv -\dfrac{7\pi}{4}\equiv \dfrac{\pi}{4} \ [2\pi] .\)