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Vous êtes ici : AccueilCLASSESCorrection exercices sur les nombres complexes : Module et argument d'un nombre complexe
Terminale
C & E & D & TI
Mathématiques
Correction exercice
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Correction exercice I

Déterminons le module des nombres complexes suivants :
a) \(Z = - i \Rightarrow \) \(\left| Z \right| = \) \(\sqrt {{0^2} + {{\left( { - 1} \right)}^2}} \) \( = 1\)
b) \(Z = \sqrt 3 \Rightarrow \) \(\left| Z \right| = \) \(\sqrt {{{\left( {\sqrt 3 } \right)}^2} + {0^2}} \) \( = \sqrt 3 \)
c) \(Z = \frac{{1 - i}}{2}\) \( \Rightarrow \left| Z \right| = \) \(\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {{\left( { - \frac{1}{2}} \right)}^2}} \) \( = \frac{{\sqrt 2 }}{2}\)
d) \(Z = 1 - \) \(i\sqrt 3 \Rightarrow \) \(\left| Z \right| = \) \(\sqrt {{1^2} + {{\left( { - \sqrt 3 } \right)}^2}} \) \( = 2\)
e) \(Z = 1 - i\) \( \Rightarrow \left| Z \right| = \) \(\sqrt {{1^2} + {1^2}} \)

Correction exercice II

Déterminons le module des nombres complexes suivants
a) \({Z_1} = 1 - i\sqrt 3 \) \( \Rightarrow \left| {{Z_1}} \right| = \) \(\sqrt {{1^2} + {{\left( { - \sqrt 3 } \right)}^2}} \) \( = 2\)
b) \({Z_2} = 2 + \) \(2i \Rightarrow \) \(\left| {{Z_2}} \right| = \) \(\sqrt {{2^2} + {2^2}} \) \( = 2\sqrt 2 \)
c) \({Z_3} = - 1 - \) \(i\sqrt 3 \Rightarrow \) \(\left| {{Z_3}} \right| = \) \(\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - \sqrt 3 } \right)}^2}} \) \( = 2\)
d) \({Z_4} = \frac{{{Z_1}}}{{{Z_2}}}\) \( \Rightarrow \left| {{Z_4}} \right| = \) \(\left| {\frac{{{Z_1}}}{{{Z_2}}}} \right| = \) \(\frac{{\left| {{Z_1}} \right|}}{{\left| {{Z_2}} \right|}} = \) \(\frac{{\sqrt 2 }}{2}\)
e) \({Z_5} = {\left( {{Z_2}} \right)^2}\) \( \times {Z_3} \Rightarrow \) \(\left| {{Z_5}} \right| = \) \(\left| {{{\left( {{Z_2}} \right)}^2} \times {Z_3}} \right| = \) \(\left| {{{\left( {{Z_2}} \right)}^2}} \right| \times \) \(\left| {{Z_3}} \right| = \) \({\left| {{Z_2}} \right|^2} \times \left| {{Z_3}} \right|\) \( = 16\)
f) \({Z_6} = {\left( {\frac{{{Z_3}}}{{{Z_1}}}} \right)^2}\) \( \Rightarrow \left| {{Z_6}} \right| = \) \(\left| {{{\left( {\frac{{{Z_3}}}{{{Z_1}}}} \right)}^2}} \right| = \) \(\frac{{{{\left| {{Z_3}} \right|}^2}}}{{{{\left| {Z1} \right|}^2}}} = 1\)

Correction exercice III

Déterminons le module et un argument des nombres complexes suivants
a) \({Z_1} = - 1 - \) \(i\sqrt 3 \Rightarrow \) \(\left| {{Z_1}} \right| = \) \(\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - \sqrt 3 } \right)}^2}} \) \( = 2\)
\(\arg ({Z_1})\) est tel que : \(\left\{ \begin{array}{l}\cos \theta = \frac{{ - 1}}{2}\\\sin \theta = \frac{{ - \sqrt 3 }}{2}\end{array} \right.\) \( \Rightarrow \theta = \pi \) \( + \frac{\pi }{3} = \frac{{4\pi }}{3}\)
b) \({Z_2} = - \frac{1}{2} + \) \(i\frac{{\sqrt 3 }}{2} \Rightarrow \) \(\left| {{Z_2}} \right| = \) \(\sqrt {{{\left( { - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}} \) \( = 1\)
\(\arg ({Z_2})\) est tel que : \(\left\{ \begin{array}{l}\cos \theta = - \frac{1}{2}\\\sin \theta = \frac{{\sqrt 3 }}{2}\end{array} \right.\) \( \Rightarrow \theta = \pi \) \( - \frac{\pi }{3} = \) \(\frac{{2\pi }}{3}\)
c) \({Z_3} = 1 - i\) \( \Rightarrow \left| {{Z_3}} \right| = \) \(\sqrt {{1^2} + {1^2}} = \) \(\sqrt 2 \)
\(\arg ({Z_3})\) est tel que : \(\left\{ \begin{array}{l}\cos \theta = \frac{{\sqrt 2 }}{2}\\\sin \theta = \frac{{\sqrt 2 }}{2}\end{array} \right.\) \( \Rightarrow \theta = \frac{\pi }{4}\)
d) \({Z_4} = 1 - \) \(i\sqrt 3 \Rightarrow \) \(\left| {{Z_4}} \right| = 2\)
\(\arg ({Z_4})\) est tel que : \(\left\{ \begin{array}{l}\cos \theta = \frac{1}{2}\\\sin \theta = \frac{{ - \sqrt 3 }}{2}\end{array} \right.\) \( \Rightarrow \theta = \) \( - \frac{\pi }{3}\)

Correction exercice IV

Déterminons le module et un argument des nombres complexes suivants :
a) \({Z_1} = 1 - \) \(i\sqrt 3 \Rightarrow \) \(\left\{ \begin{array}{l}\left| {{Z_1}} \right| = 2\\\theta = \frac{{ - \pi }}{3}\end{array} \right.\)
b) \({Z_2} = 2 + \) \(2i \Rightarrow \) \(\left\{ \begin{array}{l}\left| {{Z_2}} \right| = 2\sqrt 2 \\\theta = \frac{\pi }{4}\end{array} \right.\)
c) \({Z_3} = - 1\) \( - i\sqrt 3 \Rightarrow \) \(\left\{ \begin{array}{l}\left| {{Z_3}} \right| = 2\\\theta = \frac{{4\pi }}{3}\end{array} \right.\)
d) \({Z_4} = \frac{{{Z_1}}}{{{Z_2}}}\) \( \Rightarrow \) \(\left\{ \begin{array}{l}\left| {{Z_4}} \right| = \frac{{\sqrt 2 }}{2}\\\theta = - \frac{{7\pi }}{{12}}\end{array} \right.\)
\(\arg \left( {{Z_4}} \right) = \) \(\arg \left( {\frac{{{Z_1}}}{{{Z_2}}}} \right)\) \( = \arg \left( {{Z_1}} \right)\) \( - \arg \left( {{Z_2}} \right) = \) \( - \frac{\pi }{3} - \frac{\pi }{4}\) \( = - \frac{{7\pi }}{{12}}\)
e) \({Z_5} = {\left( {{Z_2}} \right)^2}\) \( \times {Z_3} \Rightarrow \left| {{Z_5}} \right|\) \( = {\left| {{Z_2}} \right|^2} \times \) \(\left| {{Z_3}} \right| = 16\)
\(\arg \left( {{Z_5}} \right) = \) \(\arg \left( {{{\left( {{Z_2}} \right)}^2} \times {Z_3}} \right)\) \( = \arg \left( {{{\left( {{Z_2}} \right)}^2}} \right)\) \( + \arg \left( {{Z_3}} \right)\) \( = 2\arg \left( {{Z_2}} \right)\) \( + \arg \left( {{Z_3}} \right)\) \( = \frac{{11\pi }}{6}\)
\(\left\{ \begin{array}{l}\left| {{Z_5}} \right| = 16\\\arg \left( {{Z_5}} \right) = \frac{{11\pi }}{6}
\end{array} \right.\)
f) \({Z_6} = {\left( {\frac{{{Z_3}}}{{{Z_1}}}} \right)^2}\) \( \Rightarrow \) \(\left\{ \begin{array}{l}\left| {{Z_6}} \right| = 1\\\arg \left( {{Z_6}} \right) = \frac{{10\pi }}{3}\end{array} \right.\)