Correction exercice I
Crierons sous forme trigonométrique les nombres complexes suivants :
a) \({Z_1} = 1 - i\sqrt 3 \) ;
\(\left| {{Z_1}} \right| = 2\)
\(\left\{ \begin{array}{l}\cos \theta = \frac{1}{2}\\\sin \theta = \frac{{ - \sqrt 3 }}{2}\end{array} \right.\) \( \Rightarrow \theta = \) \( - \frac{\pi }{3}\)
\({Z_1} = 2\) \(\left[ {\cos \left( { - \frac{\pi }{3}} \right) + i\sin \left( { - \frac{\pi }{3}} \right)} \right]\)
b) \({Z_2} = 2 + 2i\) ;
\(\left| {{Z_2}} \right| = 2\sqrt 2 \)
\(\left\{ \begin{array}{l}\cos \theta = \frac{2}{{2\sqrt 2 }}\\\sin \theta = \frac{2}{{2\sqrt 2 }}
\end{array} \right.\) \( \Rightarrow \theta = \frac{\pi }{4}\)
\({Z_2} = 2\) \(\left[ {\cos \left( {\frac{\pi }{4}} \right) + i\sin \left( {\frac{\pi }{4}} \right)} \right]\)
c) \({Z_3} = - 1 - i\sqrt 3 \) ;
\(\left| {{Z_3}} \right| = 2\)
\(\left\{ \begin{array}{l}\cos \theta = \frac{{ - 1}}{2}\\\sin \theta = \frac{{ - \sqrt 3 }}{2}
\end{array} \right.\) \( \Rightarrow \theta = \pi \) \( + \frac{\pi }{3} = \frac{{4\pi }}{3}\)
\({Z_2} = 2\) \(\left[ {\cos \left( {\frac{{4\pi }}{3}} \right) + i\sin \left( {\frac{{4\pi }}{3}} \right)} \right]\)
d) \({Z_4} = \frac{{{Z_1}}}{{{Z_2}}}\) ;
\(\left| {{Z_4}} \right| = \frac{{\sqrt 2 }}{2}\)
\({Z_4} = \frac{{\sqrt 2 }}{2}\)
\(\left[ {\cos \left( {\frac{{5\pi }}{{12}}} \right) + i\sin \left( {\frac{{5\pi }}{{12}}} \right)} \right]\)
e) \({Z_5} = {\left( {{Z_2}} \right)^2} \times {Z_3}\) ;
\(\left| {{Z_5}} \right| = 16\)
\(\arg \left( {{Z_5}} \right) = \frac{{11\pi }}{6}\)
\({Z_5} = 16\) \(\left[ {\cos \left( {\frac{{11\pi }}{6}} \right) + i\sin \left( {\frac{{11\pi }}{6}} \right)} \right]\)
f) \({Z_6} = {\left( {\frac{{{Z_3}}}{{{Z_1}}}} \right)^3}\).
\(\left| {{Z_6}} \right| = 1\)
\(\arg \left( {{Z_6}} \right) = \frac{{4\pi }}{3}\)
\({Z_6} = 1\) \(\left[ {\cos \left( {\frac{{4\pi }}{3}} \right) + i\sin \left( {\frac{{4\pi }}{3}} \right)} \right]\)
Correction exercice II
Donnons la forme polaire des nombres complexes suivants
a) \({Z_1} = - 1 - i\) ;
\(\left| {{Z_1}} \right| = \sqrt 2 \)
\(\theta = \frac{{5\pi }}{4}\)
\({Z_1} = \) \(\left[ {\sqrt 2 ;\frac{{5\pi }}{4}} \right]\)
b) \({Z_2} = - \frac{1}{2} + \frac{i}{2}\) ;
\(\left| {{Z_2}} \right| = \frac{{\sqrt 2 }}{2}\)
\(\theta = \frac{{3\pi }}{4}\)
\({Z_2} = \) \(\left[ {\frac{{\sqrt 2 }}{2};\frac{{3\pi }}{4}} \right]\)
c) \({Z_3} = 1 - i\sqrt 3 \) ;
\(\left| {{Z_3}} \right| = 2\)
\(\theta = - \frac{\pi }{3}\)
\({Z_3} = \) \(\left[ {2; - \frac{\pi }{3}} \right]\)
d) \({Z_1} \times {Z_2}\) ;
\(\left| {{Z_1} \times {Z_2}} \right| = \) \(\left| {{Z_1}} \right| \times \left| {{Z_2}} \right|\) \( = \sqrt 2 \times \frac{{\sqrt 2 }}{2}\) \( = 1\)
\(\arg \left( {{Z_1} \times {Z_2}} \right)\) \( = \arg \left( {{Z_1}} \right)\) \( + \arg \left( {{Z_2}} \right)\) \( = \frac{{5\pi }}{4} + \) \(\frac{{3\pi }}{4} = 2\pi \)
\({Z_1} \times {Z_2} = \) \(\left[ {1;2\pi } \right]\)
e) \(\frac{{{Z_3}}}{{{Z_2}}}\) ;
\(\left| {\frac{{{Z_3}}}{{{Z_2}}}} \right| = \) \(\frac{{\left| {{Z_3}} \right|}}{{\left| {{Z_2}} \right|}} = \) \(\frac{2}{{\frac{{\sqrt 2 }}{2}}} = \) \(2\sqrt 2 \)
\(\arg \left( {\frac{{{Z_3}}}{{{Z_2}}}} \right) = \) \(\arg \left( {{Z_1}} \right) - \) \(\arg \left( {{Z_2}} \right) = \) \( - \frac{{13\pi }}{{12}}\)
\(\frac{{{Z_3}}}{{{Z_2}}} = \) \(\left[ {2\sqrt 2 ; - \frac{{13\pi }}{{12}}} \right]\)
f) \(Z_3^4\).
\(\left| {Z_3^4} \right| = \) \({\left| {{Z_3}} \right|^4} = \) \({2^4}\)
\(\arg \left( {Z_3^4} \right) = \) \(4\arg \left( {{Z_3}} \right) = \) \( - \frac{{4\pi }}{3}\)
\(Z_3^4 = \) \(\left[ {{2^4}; - \frac{{4\pi }}{3}} \right]\)
Correction exercice III
Complétons le tableau suivant
| Formes algé briques de \(Z\) | \( - 5(1\) \( + i\sqrt 3 )\) | \(2( - 1 + \) \(i\sqrt 3 )\) | \(\sqrt 2 (\) \(1 - i)\) |
| Formes trigono métriques de \(Z\) | \(10[\) \(\cos \left( {\frac{{4\pi }}{3}} \right)\) \( + i\) \(\sin \left( {\frac{{4\pi }}{3}} \right)]\) | \(4[\) \(\cos \left( {\frac{{2\pi }}{3}} \right)\) \( + i\) \(\sin \left( {\frac{{2\pi }}{3}} \right)]\) | \(2(\) \(\cos \left( { - \frac{\pi }{4}} \right)\) \( + i\) \(\sin \left( { - \frac{\pi }{4}} \right))\) |
| Formes expo nentielles de \(Z\) | \(10{e^{i\frac{{4\pi }}{2}}}\) | \(4{e^{i\left( {\frac{{2\pi }}{3}} \right)}}\) | \(2{e^{\frac{{ - i\pi }}{4}}}\) |
Correction exercice IV
1) Ecrirons \({Z_1}\), \({Z_2}\) et \({Z_3}\) sous forme trigonométrique.
\({Z_1} = \sqrt 2 [\) \(\cos \left( { - \frac{\pi }{6}} \right)\) \( + i\) \(\sin \left( { - \frac{\pi }{6}} \right)]\)
\({Z_2} = \sqrt 2 [\) \(\cos \left( { - \frac{\pi }{4}} \right)\) \( + i\) \(\sin \left( { - \frac{\pi }{4}} \right)]\)
\({Z_3} = \) \(\cos \left( {\frac{\pi }{{12}}} \right) + \) \(i\sin \left( {\frac{\pi }{{12}}} \right)\)
2) Ecrirons \({Z_3}\) sous forme algébrique.
\({Z_3} = \) \(\frac{{\sqrt 6 + \sqrt 2 }}{4}\) \( + i\frac{{\sqrt 6 - \sqrt 2 }}{4}\)
3) Déduisons les valeurs exactes de \(\cos \frac{\pi }{{12}}\) et \(\sin \frac{\pi }{{12}}\)
Par identification des formes trigonométrique et algébrique, on a :
\(\cos\left( {\frac{\pi }{{12}}} \right) = \) \(\frac{{\sqrt 6 + \sqrt 2 }}{4}\)
\(\sin \left( {\frac{\pi }{{12}}} \right) = \) \(\frac{{\sqrt 6 - \sqrt 2 }}{4}\)
4) Calculons \({\left( {{Z_3}} \right)^{24}}\)
En utilisant la formule de Moivre
\({\left( {{Z_3}} \right)^{24}} = \) \(\cos \left( {\frac{{24\pi }}{{12}}} \right)\) \( + i\sin \left( {\frac{{24\pi }}{{12}}} \right)\) \( = \cos \left( {2\pi } \right)\) \( + i\sin \left( {2\pi } \right)\) \( = 1\)
