Terminale
C & E & D & TI
Mathématiques
Correction exercice
Bonjour ! Groupe telegram de camerecole, soumettrez-y toutes vos préoccupations. forum telegram

### 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
$${\left( {{Z_3}} \right)^{24}} =$$ $$\cos \left( {\frac{{24\pi }}{{12}}} \right)$$ $$+ i\sin \left( {\frac{{24\pi }}{{12}}} \right)$$ $$= \cos \left( {2\pi } \right)$$ $$+ i\sin \left( {2\pi } \right)$$ $$= 1$$