En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Calcul de modules et d'argumentsApplications techniques des nombres complexes/Exercices/Calcul de modules et d'arguments », n'a pu être restituée correctement ci-dessus.
Calcul du module
modifier
Déterminer le module des nombres complexes suivants (on demande des valeurs exactes).
Contrôler sur une figure les résultats obtenus.
a)
z
=
1
+
i
{\displaystyle z=1+i}
Solution
z
=
1
+
i
{\displaystyle z=1+i}
|
z
|
=
1
2
+
1
2
=
2
{\displaystyle |z|={\sqrt {1^{2}+1^{2}}}={\sqrt {2}}}
b)
z
=
−
5
i
{\displaystyle z=-5i}
Solution
z
=
−
5
i
{\displaystyle z=-5i}
|
z
|
=
0
2
+
(
−
5
)
2
=
5
{\displaystyle |z|={\sqrt {0^{2}+(-5)^{2}}}=5}
c)
z
=
−
1
−
i
{\displaystyle z=-1-i}
Solution
z
=
−
1
−
i
{\displaystyle z=-1-i}
|
z
|
=
(
−
1
)
2
+
(
−
1
)
2
=
2
{\displaystyle |z|={\sqrt {(-1)^{2}+(-1)^{2}}}={\sqrt {2}}}
d)
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
Solution
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
|
z
|
=
1
2
+
(
−
3
)
2
=
4
=
2
{\displaystyle |z|={\sqrt {1^{2}+(-{\sqrt {3}})^{2}}}={\sqrt {4}}=2}
e)
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
Solution
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
|
z
|
=
(
−
3
)
2
+
1
2
=
4
=
2
{\displaystyle |z|={\sqrt {(-{\sqrt {3}})^{2}+1^{2}}}={\sqrt {4}}=2}
f)
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
Solution
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
|
z
|
=
(
cos
(
π
4
)
)
2
+
(
−
sin
(
π
4
)
)
2
=
1
=
1
{\displaystyle |z|={\sqrt {\left(\cos \left({\frac {\pi }{4}}\right)\right)^{2}+\left(-\sin \left({\frac {\pi }{4}}\right)\right)^{2}}}={\sqrt {1}}=1}
Calcul d'un argument
modifier
Déterminer un argument des nombres complexes suivants (on demande des valeurs exactes, en utilisant les valeurs de l'exercice précédent pour le module.
Contrôler sur la figure précédente les résultats obtenus.
a)
z
=
1
+
i
{\displaystyle z=1+i}
Solution
z
=
1
+
i
{\displaystyle z=1+i}
cos
(
θ
)
=
1
|
z
|
=
1
2
{\displaystyle \cos(\theta )={\frac {1}{|z|}}={\frac {1}{\sqrt {2}}}}
Donc
θ
=
π
4
~ou~
θ
=
−
π
4
{\displaystyle \theta ={\frac {\pi }{4}}{\textrm {~ou~}}\theta =-{\frac {\pi }{4}}}
Comme b > 0 , on a
θ
=
π
4
{\displaystyle \theta ={\frac {\pi }{4}}}
b)
z
=
−
5
i
{\displaystyle z=-5i}
Solution
z
=
−
5
i
{\displaystyle z=-5i}
a = 0 et b < 0 donc
θ
=
−
π
2
{\displaystyle \theta =-{\frac {\pi }{2}}}
c)
z
=
−
1
−
i
{\displaystyle z=-1-i}
Solution
z
=
−
1
−
i
{\displaystyle z=-1-i}
cos
(
θ
)
=
−
1
|
z
|
=
−
1
2
{\displaystyle \cos(\theta )={\frac {-1}{|z|}}=-{\frac {1}{\sqrt {2}}}}
Donc
θ
=
3
π
4
~ou~
θ
=
−
3
π
4
{\displaystyle \theta ={\frac {3\pi }{4}}{\textrm {~ou~}}\theta =-{\frac {3\pi }{4}}}
Comme b < 0 , on a
θ
=
−
3
π
4
{\displaystyle \theta =-{\frac {3\pi }{4}}}
d)
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
Solution
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
cos
(
θ
)
=
1
|
z
|
=
1
2
{\displaystyle \cos(\theta )={\frac {1}{|z|}}={\frac {1}{2}}}
Donc
θ
=
π
3
~ou~
θ
=
−
π
3
{\displaystyle \theta ={\frac {\pi }{3}}{\textrm {~ou~}}\theta =-{\frac {\pi }{3}}}
Comme b < 0 , on a
θ
=
−
π
3
{\displaystyle \theta =-{\frac {\pi }{3}}}
e)
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
Solution
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
cos
(
θ
)
=
−
3
|
z
|
=
−
3
2
{\displaystyle \cos(\theta )={\frac {-{\sqrt {3}}}{|z|}}={\frac {-{\sqrt {3}}}{2}}}
Donc
θ
=
5
π
6
~ou~
θ
=
−
5
π
6
{\displaystyle \theta ={\frac {5\pi }{6}}{\textrm {~ou~}}\theta =-{\frac {5\pi }{6}}}
Comme b > 0 , on a
θ
=
5
π
6
{\displaystyle \theta ={\frac {5\pi }{6}}}
f)
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
Solution
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
cos
(
θ
)
=
cos
(
π
4
)
|
z
|
=
2
2
{\displaystyle \cos(\theta )={\frac {\cos \left({\frac {\pi }{4}}\right)}{|z|}}={\frac {\sqrt {2}}{2}}}
Donc
θ
=
π
4
~ou~
θ
=
−
π
4
{\displaystyle \theta ={\frac {\pi }{4}}{\textrm {~ou~}}\theta =-{\frac {\pi }{4}}}
Comme b < 0 , on a
θ
=
−
π
4
{\displaystyle \theta =-{\frac {\pi }{4}}}
Forme trigonométrique
modifier
En utilisant les résultats des deux précédents exercices, mettre les nombres complexes suivants sous forme trigonométrique
a)
z
=
1
+
i
{\displaystyle z=1+i}
Solution
z
=
1
+
i
{\displaystyle z=1+i}
z
=
2
(
cos
(
π
4
)
+
i
sin
(
π
4
)
)
{\displaystyle z={\sqrt {2}}\left(\cos \left({\frac {\pi }{4}}\right)+i\sin \left({\frac {\pi }{4}}\right)\right)}
b)
z
=
−
5
i
{\displaystyle z=-5i}
Solution
z
=
−
5
i
{\displaystyle z=-5i}
z
=
5
i
sin
(
−
π
2
)
{\displaystyle z=5i\sin \left(-{\frac {\pi }{2}}\right)}
c)
z
=
−
1
−
i
{\displaystyle z=-1-i}
Solution
z
=
−
1
−
i
{\displaystyle z=-1-i}
z
=
2
(
cos
(
−
3
π
4
)
+
i
sin
(
−
3
π
4
)
)
{\displaystyle z={\sqrt {2}}\left(\cos \left(-{\frac {3\pi }{4}}\right)+i\sin \left(-{\frac {3\pi }{4}}\right)\right)}
d)
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
Solution
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
z
=
2
(
cos
(
−
π
3
)
+
i
sin
(
−
π
3
)
)
{\displaystyle z=2\left(\cos \left(-{\frac {\pi }{3}}\right)+i\sin \left(-{\frac {\pi }{3}}\right)\right)}
e)
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
Solution
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
z
=
2
(
cos
(
5
π
6
)
+
i
sin
(
5
π
6
)
)
{\displaystyle z=2\left(\cos \left({\frac {5\pi }{6}}\right)+i\sin \left({\frac {5\pi }{6}}\right)\right)}
f)
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
Solution
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
z
=
cos
(
−
π
4
)
+
i
.
sin
(
−
π
4
)
{\displaystyle z=\cos \left(-{\frac {\pi }{4}}\right)+i.\sin \left(-{\frac {\pi }{4}}\right)}
Calcul avec Arctan
modifier
En utilisant la fonction
A
r
c
t
a
n
{\displaystyle Arctan}
θ
{\displaystyle \theta }
θ
+
π
{\displaystyle \theta +\pi }
a)
z
=
1
+
i
{\displaystyle z=1+i}
b)
z
=
−
5
i
{\displaystyle z=-5i}
Solution
z
=
−
5
i
{\displaystyle z=-5i}
a = 0 et b < 0 donc
θ
=
−
π
2
{\displaystyle \theta =-{\frac {\pi }{2}}}
c)
z
=
−
1
−
i
{\displaystyle z=-1-i}
d)
z
=
1
−
i
3
{\displaystyle z=1-i{\sqrt {3}}}
e)
z
=
−
3
+
i
{\displaystyle z=-{\sqrt {3}}+i}
f)
z
=
cos
(
π
4
)
−
i
.
sin
(
π
4
)
{\displaystyle z=\cos \left({\frac {\pi }{4}}\right)-i.\sin \left({\frac {\pi }{4}}\right)}
Passage de la forme trigonométrique à la forme algébrique
modifier
Soit
z
{\displaystyle z}
r
{\displaystyle r}
θ
{\displaystyle \theta }
Écrire
z
{\displaystyle z}
a
+
b
i
{\displaystyle a+bi}
r
1
=
3
{\displaystyle r_{1}=3}
θ
1
=
π
6
{\displaystyle \theta _{1}={\frac {\pi }{6}}}
Solution
z
1
=
r
1
(
c
o
s
θ
1
+
i
s
i
n
θ
1
)
=
3
×
(
3
2
+
1
2
i
)
=
3
3
2
+
3
2
i
{\displaystyle {\begin{aligned}z_{1}&=r_{1}(cos\theta _{1}+isin\theta _{1})\\&=3\times ({\frac {\sqrt {3}}{2}}+{\frac {1}{2}}i)\\&={\frac {3{\sqrt {3}}}{2}}+{\frac {3}{2}}i\end{aligned}}}
r
2
=
2
{\displaystyle r_{2}=2}
θ
2
=
3
π
4
{\displaystyle \theta _{2}={\frac {3\pi }{4}}}
Solution
z
2
=
r
2
(
c
o
s
θ
2
+
i
s
i
n
θ
2
)
=
2
×
(
−
2
2
+
i
2
2
)
=
−
2
+
i
2
{\displaystyle {\begin{aligned}z_{2}&=r_{2}(cos\theta _{2}+isin\theta _{2})\\&=2\times (-{\frac {\sqrt {2}}{2}}+i{\frac {\sqrt {2}}{2}})\\&=-{\sqrt {2}}+i{\sqrt {2}}\end{aligned}}}
r
3
=
2
{\displaystyle r_{3}=2}
θ
3
=
−
π
3
{\displaystyle \theta _{3}={\frac {-\pi }{3}}}
Solution
z
3
=
r
3
(
c
o
s
θ
3
+
i
s
i
n
θ
3
)
=
2
×
(
1
2
−
i
3
2
)
=
1
−
i
3
{\displaystyle {\begin{aligned}z_{3}&=r_{3}(cos\theta _{3}+isin\theta _{3})\\&=2\times ({\frac {1}{2}}-i{\frac {\sqrt {3}}{2}})\\&=1-i{\sqrt {3}}\end{aligned}}}
r
4
=
2
,
5
{\displaystyle r_{4}=2,5}
θ
4
=
−
5
π
6
{\displaystyle \theta _{4}={\frac {-5\pi }{6}}}
Solution
z
4
=
r
4
(
c
o
s
θ
4
+
i
s
i
n
θ
4
)
=
5
2
×
(
−
3
2
−
i
1
2
)
=
5
3
4
−
5
4
i
{\displaystyle {\begin{aligned}z_{4}&=r_{4}(cos\theta _{4}+isin\theta _{4})\\&={\frac {5}{2}}\times (-{\frac {\sqrt {3}}{2}}-i{\frac {1}{2}})\\&={\frac {5{\sqrt {3}}}{4}}-{\frac {5}{4}}i\end{aligned}}}
On donne :
z
A
=
−
4
−
4
i
{\displaystyle z_{A}=-4-4i}
z
B
=
−
3
+
2
i
{\displaystyle z_{B}=-3+2i}
a) Placer les points A et B dans un repère orthonormé d'origine O et d'unité 2 cm
b) Calculer
|
z
A
|
{\displaystyle \left|z_{A}\right|}
|
z
B
|
{\displaystyle \left|z_{B}\right|}
Solution
|
z
A
|
=
(
−
4
)
2
+
(
−
4
)
2
=
16
+
16
=
32
=
4
2
{\displaystyle {\begin{aligned}|z_{A}|&={\sqrt {(-4)^{2}+(-4)^{2}}}\\&={\sqrt {16+16}}\\&={\sqrt {32}}\\&=4{\sqrt {2}}\end{aligned}}}
|
z
B
|
=
(
−
3
)
2
+
2
2
=
9
+
4
=
13
{\displaystyle {\begin{aligned}|z_{B}|&={\sqrt {(-3)^{2}+2^{2}}}\\&={\sqrt {9+4}}\\&={\sqrt {13}}\end{aligned}}}
c) Calculer
|
z
A
−
z
B
|
{\displaystyle \left|z_{A}-z_{B}\right|}
Solution
|
z
A
−
z
B
|
=
(
−
4
−
(
−
3
)
)
2
+
(
−
4
−
2
)
2
=
1
+
36
=
37
{\displaystyle {\begin{aligned}|z_{A}-z_{B}|&={\sqrt {(-4-(-3))^{2}+(-4-2)^{2}}}\\&={\sqrt {1+36}}\\&={\sqrt {37}}\end{aligned}}}
d) Le triangle OAB est-il rectangle ? Justifier.
On donne
z
A
=
−
1
−
i
3
{\displaystyle z_{A}=-1-i{\sqrt {3}}}
1) On pose
z
B
=
2
i
×
z
A
{\displaystyle z_{B}=2i\times z_{A}}
z
B
=
2
3
−
2
i
{\displaystyle z_{B}=2{\sqrt {3}}-2i}
Solution
z
B
=
2
i
×
z
A
=
−
2
i
+
2
3
{\displaystyle {\begin{aligned}z_{B}&=2i\times z_{A}\\&=-2i+2{\sqrt {3}}\end{aligned}}}
2) a) Calculer un argument de chacun des nombres complexes
z
A
{\displaystyle z_{A}}
z
B
{\displaystyle z_{B}}
Solution
z
A
=
−
1
−
i
3
{\displaystyle z_{A}=-1-i{\sqrt {3}}}
cos
(
θ
)
=
−
1
|
z
A
|
=
−
1
2
{\displaystyle \cos(\theta )={\frac {-1}{|z_{A}|}}={\frac {-1}{2}}}
Donc
θ
=
2
π
3
~ou~
θ
=
−
2
π
3
{\displaystyle \theta ={\frac {2\pi }{3}}{\textrm {~ou~}}\theta =-{\frac {2\pi }{3}}}
Comme b < 0 , on a
a
r
g
(
z
A
)
=
−
2
π
3
{\displaystyle arg(z_{A})=-{\frac {2\pi }{3}}}
z
B
=
2
3
−
2
i
{\displaystyle z_{B}=2{\sqrt {3}}-2i}
cos
(
θ
)
=
2
3
|
z
B
|
=
3
2
{\displaystyle \cos(\theta )={\frac {2{\sqrt {3}}}{|z_{B}|}}={\frac {\sqrt {3}}{2}}}
Donc
θ
=
π
6
~ou~
θ
=
−
π
6
{\displaystyle \theta ={\frac {\pi }{6}}{\textrm {~ou~}}\theta =-{\frac {\pi }{6}}}
Comme b < 0 , on a
a
r
g
(
z
B
)
=
−
π
6
{\displaystyle arg(z_{B})=-{\frac {\pi }{6}}}
b) Placer dans un repère orthonormé d'origine O et d'unité 2 cm A et B .
3) Démontrer que le triangle OAB est rectangle en O .
Solution
On calcule l'angle
(
O
B
→
,
O
A
→
)
{\displaystyle ({\overrightarrow {OB}},{\overrightarrow {OA}})}
(
O
B
→
,
O
A
→
)
=
a
r
g
(
z
A
)
−
a
r
g
(
z
B
)
=
−
π
2
{\displaystyle ({\overrightarrow {OB}},{\overrightarrow {OA}})=arg(z_{A})-arg(z_{B})=-{\frac {\pi }{2}}}
Donc le triangle OAB est rectangle en O.
