Forme exponentielle, module et arguments - Corrigé

Énoncé

Soit \(z_1=3\sqrt{3}-3i\) et \(z_2=1+i\) .

1. Déterminer la forme exponentielle de \(z_1\) et de \(z_2\) .

2. En déduire celles de :
    a. \(z_1z_2\)
    b. \(\dfrac{z_1}{z_2}\)
    c. \(\dfrac{z_2^5}{z_1^2}\)

Solution

1. On a : \(\left\vert z_1 \right\vert=\sqrt{(3\sqrt{3})^2+(-3)^2}=\sqrt{9 \times 3+9}=\sqrt{36}=6\) .
On a alors : \(\begin{align*}z_1=6\left(\frac{3\sqrt{3}}{6}-\frac{3}{6}i\right)=6\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)=6\left(\cos\frac{-\pi}{6}+i\sin\frac{-\pi}{6}\right)=6\text e^{-\frac{i\pi}{6}}.\end{align*}\)

On a : \(\left\vert z_2 \right\vert=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}\) .
On a alors :
\(\begin{align*}z_2=\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)=\sqrt{2}\text e^{\frac{i\pi}{4}}.\end{align*}\)

2. a. On a :  \(\begin{align*}z_1z_2& = 6\text e^{-\frac{i\pi}{6}} \times \sqrt{2}\text e^{\frac{i\pi}{4}}=6\sqrt{2}\text e^{-\frac{i\pi}{6}+\frac{i\pi}{4}}=6\sqrt{2}\text e^{\frac{i\pi}{12}}.\end{align*}\)

    b. On a :
\(\begin{align*}\frac{z_1}{z_2}& = \frac{6\text e^{-\frac{i\pi}{6}}}{\sqrt{2}\text e^{\frac{i\pi}{4}}}= \frac{6}{\sqrt{2}}\text e^{-\frac{i\pi}{6}-\frac{i\pi}{4}}= \frac{4\sqrt{2}}{2}\text e^{-\frac{5i\pi}{12}}= 2\sqrt{2}\text e ^{-\frac{5i\pi}{12}}.\end{align*}\)  

    c. On a :  \(\begin{align*}\dfrac{z_2^5}{z_1^2}& = \frac{\left(\sqrt{2}\text e^{\frac{i\pi}{4}}\right)^5}{\left(6\text e^{-\frac{i\pi}{6}}\right)^2}= \frac{(\sqrt{2})^5\text e^{\frac{5i\pi}{4}}}{6^2\text e^{-\frac{2i\pi}{6}}}= \frac{\sqrt{2}}{9}\text e^{\frac{5i\pi}{4}+\frac{i\pi}{3}}= \frac{\sqrt{2}}{9}\text e^{\frac{19i\pi}{12}}.\end{align*}\)