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.
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}
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}}}
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)}
En utilisant la fonction
A
r
c
t
a
n
{\displaystyle Arctan}
de la calculatrice, donner un argument des nombres complexes suivants. Penser à décider entre
θ
{\displaystyle \theta }
et
θ
+
π
{\displaystyle \theta +\pi }
grâce au signe de la partie réelle.
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)}
Soit
z
{\displaystyle z}
un nombre complexe de module
r
{\displaystyle r}
et d'argument
θ
{\displaystyle \theta }
.
Écrire
z
{\displaystyle z}
sous forme algébrique
a
+
b
i
{\displaystyle a+bi}
dans les cas suivants.
r
1
=
3
{\displaystyle r_{1}=3}
et
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}
et
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}
et
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}
et
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}
et
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|}
et
|
z
B
|
{\displaystyle \left|z_{B}\right|}
.
Que représentent ces quantités géométriquement ?
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}}}
Ces résultats représentent les distances qui séparent le point d'origine aux points A et B.
c) Calculer
|
z
A
−
z
B
|
{\displaystyle \left|z_{A}-z_{B}\right|}
Interpréter géométriquement ce résultat.
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}}}
Ce résultat représente la distance entre les points A et B.
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}}
, démontrer que
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}}
et
z
B
{\displaystyle z_{B}}
(on demande des valeurs exactes).
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 les points 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.