Formes exponentielles (1) - Corrigé

Énoncé

Écrire les nombres complexes suivants sous forme exponentielle.

1.  \(z_1=5i\)

2.  \(z_2=1-i\)

3.  \(z_3=-\sqrt{3}+i\)

4.  \(z_4=-3\)

5.  \(z_5=-\dfrac{1}{2}i\)

6.  \(z_6=4+4i\)

7.  \(z_7=\sqrt{2}+\sqrt{6}i\)

Correction

1.  On a : \(\left\vert z_1 \right\vert = \left\vert 5i \right\vert = 5\)   et donc   \(z_1 = 5(0+1i)= 5\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)= 5\text e^{\frac{i\pi}{2}}.\)

2.  On a : \(\left\vert z_2 \right\vert = \left\vert 1-i \right\vert= \sqrt{1^2+(-1)^2}= \sqrt{1+1}= \sqrt{2}\)  et donc
\(\begin{align*}z_2& = \sqrt{2}\left(\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)= \sqrt{2}\left(\cos\frac{-\pi}{4}+i\sin\frac{-\pi}{4}\right)= \sqrt{2}\text e^{\frac{-i\pi}{4}}.\end{align*}\)
3.  On a : \(\left\vert z_3 \right\vert= \left\vert -\sqrt{3} +i \right\vert= \sqrt{(-\sqrt{3})^2+1^2}= \sqrt{3+1}= \sqrt{4}= 2\)  et donc
\(\begin{align*}z_3& = 2\left(-\frac{\sqrt{3}}{2} +\frac{1}{2}i\right)= 2\left(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)= 2\text e^{\frac{5i\pi}{6}}.\end{align*}\)

4.  On a : \(\left\vert z_4 \right\vert= \left\vert -3 \right\vert= 3\)  et donc 
\(\begin{align*}z_4& = 3(-3+0i)= 3\left(\cos(\pi)+i\sin(\pi)\right)= 3\text e^{i\pi}.\end{align*}\)

5.  On a : \(\left\vert z_5 \right\vert= \left\vert -\dfrac{1}{2}i \right\vert= \dfrac{1}{2}\)   et donc   \(\begin{align*}z_5& = \frac{1}{2}(0-1i)= \frac{1}{2}\left(\cos\frac{-\pi}{2}+i\sin\frac{-\pi}{2}\right)= \frac{1}{2} \text e^{-\frac{i\pi}{2}}.\end{align*}\)

6.  On a :  \(z_6=4+4i\)  et donc  \(\left\vert z_6 \right\vert= \left\vert 4+4i \right\vert= \sqrt{4^2+4^2}= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}\)
\(\begin{align*}z_6& = 4\sqrt{2}\left(\frac{2}{4\sqrt{2}}+i\frac{2}{4\sqrt{2}}\right)= 4\sqrt{2}\left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)= 4\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)= 4\sqrt{2}\text e^{\frac{i\pi}{4}}.\end{align*}\)

7.  On a : \(\left\vert z_7 \right\vert= \left\vert \sqrt{2}+\sqrt{6}i \right\vert= \sqrt{\left( \sqrt{2}\right)^2+\left(\sqrt{6}\right)^2}= \sqrt{2+6}= \sqrt{8}= 2 \sqrt{2}\)  et donc
\(\begin{align*}z_7& = 2 \sqrt{2} \left(\frac{\sqrt{2}}{2 \sqrt{2}}+i\frac{\sqrt{6}}{2 \sqrt{2}}\right)= 2 \sqrt{2} \left(\frac{1}{2}+ i\frac{\sqrt{3}}{2}\right)= 2 \sqrt{2} \left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)= 2\sqrt{2} \text e^{\frac{i\pi}{3}}.\end{align*}\)