Formes exponentielles (2) - Corrigé

Énoncé

Écrire les nombres complexes suivants sous forme exponentielle.

1.  $z_1=(-\sqrt{3}+i)(2-2i)$

2.  $z_2={\displaystyle \frac{3i-\sqrt{3}}{1+i}}$

3.  $z_3={\left({\displaystyle \frac{i}{2}}\right)}^{15}$

4.  $z_4=(1-\sqrt{3}i{)}^{2024}$

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

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

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

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

3. On a : $\begin{array}{rl}{z}_3={\left(\frac{i}{2}\right)}^{15}& =\frac{i^{15}}{2^{15}}=\frac{{\text{e}}^{\frac{15i\pi }{2}}}{32768}=\frac{1}{32768}{\text{e}}^{i\pi }.\end{array}$

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