Formes exponentielles (2) - Corrigé

Énoncé

Écrire les nombres complexes suivants sous forme exponentielle.

1.  \(z_1=(-\sqrt{3}+i)(2-2i)\)

2.  \(z_2=\dfrac{3i-\sqrt{3}}{1+i}\)

3.  \(z_3=\left(\dfrac{i}{2}\right)^{15}\)

4.  \(z_4=(1-\sqrt{3}i)^{2024}\)

5.  \(z_5=\left(\dfrac{-\sqrt{3}+i}{1+i}\right)^{8}\)

Solution
1. \(\left\vert -\sqrt{3}+i \right\vert= \sqrt{(-\sqrt{3})^2+1^2}= \sqrt{3+1}= \sqrt{4}= 2\)
donc   \(-\sqrt{3}+i= 2 \left(\dfrac{-\sqrt{3}}{2}+\dfrac{1}{2}i\right)= 2 \left( \cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)= 2 \text e^{\frac{5 i \pi}{6}}\)
Et on a : \(\left\vert 2-2i \right\vert= \sqrt{2^2+(-2)^2}= \sqrt{4+4}= \sqrt{8}= 2\sqrt{2}\)
donc \(2-2i= 2\sqrt{2}\left(\dfrac{2}{2\sqrt{2}}+i\dfrac{-2}{2\sqrt{2}}\right)= 2\sqrt{2}\left( \dfrac{1}{\sqrt{2}}+i\dfrac{-1}{\sqrt{2}}\right)= 2\sqrt{2}\left( \cos\dfrac{-\pi}{4}+i\sin\dfrac{-\pi}{4}\right)= 2 \sqrt{2} \text e^{\frac{-i \pi}{4}}\)
Ainsi,
\((-\sqrt{3}+i) \times ( 2-2i)= 2 \text e^{\frac{5 i \pi}{6}} \times 2 \sqrt{2} \text e^{\frac{-i \pi}{4}}= 4 \sqrt{2} \text e^{i\left( \frac{5 i \pi}{6} - \frac{\pi}{4} \right)}= 4 \sqrt{2} \text e^{ \frac{14i\pi}{24} }= 4 \sqrt{2} \text e^{ \frac{7i\pi}{12} }\)

2. On a : \(\left\vert 3i - \sqrt{3} \right\vert= \left\vert -\sqrt{3}+3i \right\vert= \sqrt{(-\sqrt{3})^2+3^2}= \sqrt{3+9}= \sqrt{12}= 2\sqrt{3}\)
donc
\(3i - \sqrt{3} = -\sqrt{3}+3i= 2\sqrt{3} \left(\dfrac{-\sqrt{3}}{2\sqrt{3}}+\dfrac{3}{2\sqrt{3}}i\right)= 2\sqrt{3} \left(\dfrac{-1}{2}+\frac{\sqrt{3}}{2}i\right)= 2\sqrt{3} \left( \cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}\right)= 2\sqrt{3} \text e^{\frac{2i \pi}{3}}\) et  \(\left\vert 1+i \right\vert= \sqrt{1^2+1^2}= \sqrt{1+1}= \sqrt{2}\)
donc  \(1+i= \sqrt{2}\left(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}\right)= \sqrt{2}\left(\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}\right)= \sqrt{2}\left( \cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)= \sqrt{2} \text e^{\frac{i \pi}{4}}\) .

Ainsi  \(\dfrac{ 3i - \sqrt{3} }{ 1+i }= \dfrac{ 2\sqrt{3} \text e^{\frac{2 i \pi}{3}} }{ \sqrt{2} \text e^{\frac{i \pi}{4}} }= \sqrt{6} \text e^{ i \left( \frac{2 \pi}{3} - \frac{ \pi}{4} \right) }= \sqrt{6} \text e^{i \frac{5 \pi}{12} }\)

3. On a : \(\begin{align*}z_3=\left(\frac{i}{2}\right)^{15}& = \frac{i^{15}}{2^{15}}= \frac{\text e^{ \frac{ 15 i \pi}{2} }}{32\,768}= \frac{1}{32\,768}\text e^{i\pi}.\end{align*}\)

4. On a :  \(\left\vert 1-i\sqrt{3} \right\vert= \sqrt{1^2 + (-\sqrt{3})^2}= \sqrt{1+3}= \sqrt{4}= 2\)
soit  \(1-i\sqrt{3}= 2 \left(\dfrac{1}{2} -i \dfrac{\sqrt{3}}{2}\right)= 2\left( \cos\dfrac{-\pi}{3}+i \sin \dfrac{-\pi}{3}\right)= 2 \text e^{ \frac{-i\pi}{3}}\)
donc  \(z_4=(1-\sqrt{3}i)^{2024} = (2 \text e^{ \frac{-i\pi}{3}})^{2024}=2^{2024} \text e^{ \frac{-2024i\pi}{3}}\) .

5. On a  \(-\sqrt{3}+i = 2 \text e^{\frac{5 i \pi}{6}}\)    et   \(1+i = \sqrt{2} \text e^{\frac{i \pi}{4}}\) .
Donc 
\(z_5= \left(\dfrac{-\sqrt{3}+i}{1+i}\right)^{8}= \left(\dfrac{2 \text e^{\frac{5 i \pi}{6}}}{\sqrt{2} \text e^{\frac{i \pi}{4}}}\right)^{8}= \dfrac{ \left( 2 \text e^{\frac{5 i \pi}{6}} \right)^8 }{ \left( \sqrt{2} \text e^{\frac{i \pi}{4}} \right)^8 }= \dfrac{ 2^8 \text e^{\frac{40 i \pi}{6}} }{ (\sqrt{2})^8 \text e^{\frac{8i \pi}{4}} }= \dfrac{ 2^8 \text e^{\frac{4 i \pi}{6}} }{ 2^4 \text e^{2i \pi } }= 16 \text e^{\frac{2 i \pi}{3}}\)