Formes trigonométriques - Corrigé

Énoncé

Déterminer une forme trigonométrique des complexes suivants.

1.  `z_1=\sqrt{3}+i`  

2.  `z_2=4-4i`  

3.  `z_3=-6`  

4.  `z_4=3i`  

5.  `z_5=\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}`

6.  `z_6=\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}`

Solution

1.  On a : \(\left\vert z_1 \right\vert = \left\vert \sqrt{3}+i \right\vert = \sqrt{(\sqrt{3})^2+1^2} = \sqrt{3+1} = \sqrt{4} = 2\) .
On a alors : \(\begin{align*} z_1 =2 \left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right) =2 \left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right). \end{align*}\)

2.  On a :  \(\left\vert z_2 \right\vert = \left\vert 4-4i \right\vert = \sqrt{4^2+(-4)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}\) .
On a alors :
\(\begin{align*} z_2 =4\sqrt{2} \left(\frac{4}{4\sqrt{2}}+i\frac{-4}{4\sqrt{2}}\right) =4\sqrt{2} \left(\frac{1}{\sqrt{2}}+i\frac{-1}{\sqrt{2}}\right) =4\sqrt{2} \left(\cos\frac{-\pi}{4}+i\sin\frac{-\pi}{4}\right). \end{align*}\)

3.  On a :  \(\left\vert z_3 \right\vert = \left\vert -6 \right\vert = 6\) .
On a alors :
\(\begin{align*} z_3 =6 \left(\frac{-6}{6}+i\frac{0}{6} \right) =6 \left(-1+0i\right) =6 \left(\cos\pi+i\sin\pi\right). \end{align*}\)

4.  On a : \(\left\vert z_4 \right\vert = \left\vert 3i \right\vert = 3\) .
On a alors :
\(\begin{align*} z_4 =3 \left(\frac{0}{3}+i\frac{3}{3} \right) =3 \left(0+1i\right) =3 \left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right). \end{align*}\)

5.  On a : \(\begin{align*} z_5 =1\left(\cos\frac{\pi}{9}+i\sin\frac{\pi}{9}\right). \end{align*}\)

6.  On a : \(\begin{align*} z_6 =1\left(\cos\frac{\pi}{5}-i\sin\frac{\pi}{5}\right) =1\left(\cos\frac{-\pi}{5}+i\sin\frac{-\pi}{5}\right). \end{align*}\)