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Fiche mémoire sur les nombres complexes et la trigonométrie
En raison de limitations techniques, la typographie souhaitable du titre, «
Fiche : Nombres complexes et trigonométrieNombres complexes/Fiche/Nombres complexes et trigonométrie », n'a pu être restituée correctement ci-dessus.
∀
z
∈
C
,
R
e
(
z
)
=
z
+
z
¯
2
et
I
m
(
z
)
=
z
−
z
¯
2
i
{\displaystyle \forall z\in \mathbb {C} ,Re(z)={\frac {z+{\bar {z}}}{2}}{\text{ et }}Im(z)={\frac {z-{\bar {z}}}{2i}}}
Inégalité triangulaire :
|
z
+
z
′
|
⩽
|
z
|
+
|
z
′
|
{\displaystyle |z+z'|\leqslant |z|+|z'|}
Pour tous réels
a
,
b
{\displaystyle a,b}
a
2
+
b
2
=
1
{\displaystyle a^{2}+b^{2}=1}
θ
{\displaystyle \theta }
a
=
cos
θ
{\displaystyle a=\cos \theta }
b
=
sin
θ
{\displaystyle b=\sin \theta }
cos
′
(
x
)
=
−
sin
(
x
)
{\displaystyle \textstyle \cos '(x)=-\sin(x)}
sin
′
(
x
)
=
cos
(
x
)
{\displaystyle \textstyle \sin '(x)=\cos(x)}
tan
′
(
x
)
=
1
cos
2
(
x
)
=
1
+
tan
2
(
x
)
{\displaystyle \textstyle \tan '(x)={\frac {1}{\cos ^{2}(x)}}=1+\tan ^{2}(x)}
cos
(
a
+
b
)
=
cos
(
a
)
cos
(
b
)
−
sin
(
a
)
sin
(
b
)
{\displaystyle \textstyle \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)}
cos
(
a
−
b
)
=
cos
(
a
)
cos
(
b
)
+
sin
(
a
)
sin
(
b
)
{\displaystyle \textstyle \cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)}
sin
(
a
+
b
)
=
sin
(
a
)
cos
(
b
)
+
cos
(
a
)
sin
(
b
)
{\displaystyle \textstyle \sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)}
sin
(
a
−
b
)
=
sin
(
a
)
cos
(
b
)
−
cos
(
a
)
sin
(
b
)
{\displaystyle \textstyle \sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)}
tan
(
a
+
b
)
=
tan
(
a
)
+
tan
(
b
)
1
−
tan
(
a
)
tan
(
b
)
{\displaystyle \textstyle \tan(a+b)={\frac {\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}}}
tan
(
a
−
b
)
=
tan
(
a
)
−
tan
(
b
)
1
+
tan
(
a
)
tan
(
b
)
{\displaystyle \textstyle \tan(a-b)={\frac {\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}}}
cos
(
a
)
cos
(
b
)
=
1
2
(
cos
(
a
−
b
)
+
cos
(
a
+
b
)
)
{\displaystyle \textstyle \cos(a)\cos(b)={\frac {1}{2}}(\cos(a-b)+\cos(a+b))}
sin
(
a
)
sin
(
b
)
=
1
2
(
cos
(
a
−
b
)
−
cos
(
a
+
b
)
)
{\displaystyle \textstyle \sin(a)\sin(b)={\frac {1}{2}}(\cos(a-b)-\cos(a+b))}
sin
(
a
)
cos
(
b
)
=
1
2
(
sin
(
a
+
b
)
+
sin
(
a
−
b
)
)
{\displaystyle \textstyle \sin(a)\cos(b)={\frac {1}{2}}(\sin(a+b)+\sin(a-b))}
cos
(
a
)
sin
(
b
)
=
1
2
(
sin
(
a
+
b
)
−
sin
(
a
−
b
)
)
{\displaystyle \textstyle \cos(a)\sin(b)={\frac {1}{2}}(\sin(a+b)-\sin(a-b))}
sin
(
p
)
+
sin
(
q
)
=
2
sin
(
p
+
q
2
)
cos
(
p
−
q
2
)
{\displaystyle \textstyle \sin(p)+\sin(q)=2\sin({\frac {p+q}{2}})\cos({\frac {p-q}{2}})}
sin
(
p
)
−
sin
(
q
)
=
2
cos
(
p
+
q
2
)
sin
(
p
−
q
2
)
{\displaystyle \textstyle \sin(p)-\sin(q)=2\cos({\frac {p+q}{2}})\sin({\frac {p-q}{2}})}
cos
(
p
)
+
cos
(
q
)
=
2
c
o
q
(
p
+
q
2
)
cos
(
p
−
q
2
)
{\displaystyle \textstyle \cos(p)+\cos(q)=2coq({\frac {p+q}{2}})\cos({\frac {p-q}{2}})}
cos
(
p
)
−
cos
(
q
)
=
−
2
sin
(
p
+
q
2
)
sin
(
p
−
q
2
)
{\displaystyle \textstyle \cos(p)-\cos(q)=-2\sin({\frac {p+q}{2}})\sin({\frac {p-q}{2}})}
cos
(
2
x
)
=
cos
2
(
x
)
−
sin
2
(
x
)
=
2
cos
2
(
x
)
−
1
=
1
−
2
sin
2
(
x
)
{\displaystyle \textstyle \cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)=2\cos ^{2}(x)-1=1-2\sin ^{2}(x)}
sin
(
2
x
)
=
2
sin
(
x
)
cos
(
x
)
{\displaystyle \textstyle \sin(2x)=2\sin(x)\cos(x)}
tan
(
2
x
)
=
2
tan
(
x
)
1
−
tan
2
(
x
)
{\displaystyle \textstyle \tan(2x)={\frac {2\tan(x)}{1-\tan ^{2}(x)}}}
cos
2
(
x
)
=
1
+
cos
(
2
x
)
2
{\displaystyle \textstyle \cos ^{2}(x)={\frac {1+\cos(2x)}{2}}}
sin
2
(
x
)
=
1
−
cos
(
2
x
)
2
{\displaystyle \textstyle \sin ^{2}(x)={\frac {1-\cos(2x)}{2}}}
On pose
t
=
tan
(
x
2
)
{\displaystyle \scriptstyle t=\tan({\frac {x}{2}})}
sin
(
x
)
=
2
t
1
+
t
2
{\displaystyle \textstyle \sin(x)={\frac {2t}{1+t^{2}}}}
cos
(
x
)
=
1
−
t
2
1
+
t
2
{\displaystyle \textstyle \cos(x)={\frac {1-t^{2}}{1+t^{2}}}}
tan
(
x
)
=
2
t
1
−
t
2
{\displaystyle \textstyle \tan(x)={\frac {2t}{1-t^{2}}}}
cos
(
x
)
=
R
e
(
e
i
θ
)
=
e
i
θ
+
e
−
i
θ
2
{\displaystyle \textstyle \cos(x)=Re(e^{i\theta })={\frac {e^{i\theta }+e^{-i\theta }}{2}}}
sin
(
x
)
=
I
m
(
e
i
θ
)
=
e
i
θ
−
e
−
i
θ
2
i
{\displaystyle \textstyle \sin(x)=Im(e^{i\theta })={\frac {e^{i\theta }-e^{-i\theta }}{2i}}}
Remarque : On utilisera ces deux formules pour linéariser des expressions de la forme
cos
p
(
θ
)
+
sin
n
(
θ
)
{\displaystyle \cos ^{p}(\theta )+\sin ^{n}(\theta )}
Soient
θ
∈
R
{\displaystyle \scriptstyle \theta \in \mathbb {R} }
n
∈
N
{\displaystyle \scriptstyle n\in \mathbb {N} }
cos
(
n
θ
)
+
i
sin
(
n
θ
)
=
e
i
n
θ
=
(
e
i
θ
)
n
=
(
cos
θ
+
i
sin
θ
)
n
{\displaystyle \cos(n\theta )+i\sin(n\theta )=e^{in\theta }=(e^{i\theta })^{n}=(\cos \theta +i\sin \theta )^{n}}
Rappel :
(
n
k
)
=
n
!
k
!
(
n
−
k
)
!
{\displaystyle \textstyle {\binom {n}{k}}={\frac {n!}{k!(n-k)!}}}
Soient
a
,
b
{\displaystyle a,b}
n
{\displaystyle n}
(
a
+
b
)
n
=
∑
k
=
0
n
(
n
k
)
a
k
b
n
−
k
=
∑
k
=
0
n
(
n
k
)
a
n
−
k
b
k
{\displaystyle (a+b)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}a^{k}b^{n-k}=\sum _{k=0}^{n}{\binom {n}{k}}a^{n-k}b^{k}}
Soient
θ
∈
R
{\displaystyle \scriptstyle \theta \in \mathbb {R} }
n
∈
N
{\displaystyle \scriptstyle n\in \mathbb {N} }
cos
(
n
θ
)
=
∑
k pair
k
=
0
n
n
(
n
k
)
cos
n
−
k
(
θ
)
i
k
sin
k
(
θ
)
{\displaystyle \cos(n\theta )=\sum _{\overset {k=0}{\text{k pair}}}^{\overset {\color {White}n}{n}}{\binom {n}{k}}\cos ^{n-k}(\theta )i^{k}\sin ^{k}(\theta )}
sin
(
n
θ
)
=
1
2
∑
k impair
k
=
0
n
n
(
n
k
)
cos
n
−
k
(
θ
)
i
k
sin
k
(
θ
)
{\displaystyle \sin(n\theta )={\frac {1}{2}}\sum _{\overset {k=0}{\text{k impair}}}^{\overset {\color {White}n}{n}}{\binom {n}{k}}\cos ^{n-k}(\theta )i^{k}\sin ^{k}(\theta )}
Soit
z
∈
C
∗
{\displaystyle \scriptstyle z\in \mathbb {C} ^{*}}
z
n
{\displaystyle \scriptstyle z^{n}}
n
{\displaystyle n}
n
{\displaystyle n}
ièmes 2 à 2 distinctes.
Les racines carrées des nombres complexes de
R
⋅
e
i
θ
{\displaystyle \scriptstyle R\cdot e^{i\theta }}
R
⋅
e
i
θ
2
{\displaystyle \scriptstyle {\sqrt {R}}\cdot e^{i{\frac {\theta }{2}}}}
−
R
⋅
e
i
θ
2
{\displaystyle \scriptstyle -{\sqrt {R}}\cdot e^{i{\frac {\theta }{2}}}}
Les racines de l'unité sont :
ω
k
=
e
2
i
k
π
n
,
k
∈
{
0
,
1
,
…
,
n
−
1
}
{\displaystyle \omega _{k}=e^{\frac {2ik\pi }{n}},k\in \{0,1,\ldots ,n-1\}}
Les racines cubiques de l'unité sont
1
{\displaystyle 1}
j
{\displaystyle j}
j
2
{\displaystyle j^{2}}
j
=
e
2
π
i
3
=
−
1
2
+
i
3
2
{\displaystyle j=e^{\frac {2\pi i}{3}}={\frac {-1}{2}}+i{\frac {\sqrt {3}}{2}}}
De plus, on a :
j
¯
=
j
2
{\displaystyle \textstyle {\bar {j}}=j^{2}}
j
2
+
j
+
1
=
0
{\displaystyle \textstyle j^{2}+j+1=0}
Pour obtenir les racines carrées de
Z
=
X
+
i
Y
∈
C
∗
{\displaystyle \scriptstyle Z=X+iY\in \mathbb {C} ^{*}}
z
2
=
Z
{\displaystyle z^{2}=Z}
z
2
=
Z
⇔
{
(
x
+
i
y
)
2
=
X
+
i
Y
|
z
|
2
=
|
Z
|
{\displaystyle z^{2}=Z\Leftrightarrow {\begin{cases}(x+iy)^{2}=X+iY\\|z|^{2}=|Z|\end{cases}}}
Soient
a
,
b
,
c
{\displaystyle a,b,c}
a
≠
0
{\displaystyle \scriptstyle a\neq 0}
a
z
2
+
b
z
+
c
=
0
{\displaystyle az^{2}+bz+c=0}
Δ
=
b
2
−
4
a
c
{\displaystyle \Delta =b^{2}-4ac}
Si
Δ
≠
0
{\displaystyle \scriptstyle \Delta \neq 0}
−
b
+
δ
2
a
{\displaystyle \scriptstyle {\frac {-b+\delta }{2a}}}
−
b
−
δ
2
a
{\displaystyle \scriptstyle {\frac {-b-\delta }{2a}}}
(où
δ
{\displaystyle \delta }
Δ
{\displaystyle \Delta }
Si
Δ
=
0
{\displaystyle \scriptstyle \Delta =0}
−
b
2
a
{\displaystyle \scriptstyle {\frac {-b}{2a}}}
Remarque : si
z
1
{\displaystyle z_{1}}
z
2
{\displaystyle z_{2}}
a
z
2
+
b
z
+
c
=
0
{\displaystyle \scriptstyle az^{2}+bz+c=0}
z
1
+
z
2
=
−
b
a
{\displaystyle \scriptstyle z_{1}+z_{2}={\frac {-b}{a}}}
z
1
z
2
=
c
a
{\displaystyle \scriptstyle z_{1}z_{2}={\frac {c}{a}}}
Si
z
1
{\displaystyle z_{1}}
z
2
{\displaystyle z_{2}}
S
{\displaystyle S}
P
{\displaystyle P}
z
1
{\displaystyle z_{1}}
z
2
{\displaystyle z_{2}}
z
2
−
S
z
+
P
=
0
{\displaystyle z^{2}-Sz+P=0}
Soient
A
(
a
)
,
B
(
b
)
,
M
(
z
)
{\displaystyle \scriptstyle A(a),\ B(b),\ M(z)}
P
{\displaystyle P}
M
≠
A
{\displaystyle \scriptstyle M\neq A}
M
≠
B
{\displaystyle \scriptstyle M\neq B}
M
A
M
B
=
|
z
−
a
z
−
b
|
{\displaystyle \scriptstyle {\frac {MA}{MB}}\;=\;|{\frac {z-a}{z-b}}|}
(
M
B
→
,
M
A
→
)
=
arg
(
z
−
a
z
−
b
)
{\displaystyle \scriptstyle ({\overrightarrow {MB}},{\overrightarrow {MA}})\;=\;\arg({\frac {z-a}{z-b}})}
Les points
A
,
B
,
M
{\displaystyle A,B,M}
z
−
a
z
−
b
∈
R
{\displaystyle \scriptstyle {\frac {z-a}{z-b}}\in \mathbb {R} }
L'application
C
→
C
z
→
z
+
b
{\displaystyle {\begin{array}{llll}_{\mathbb {C} \rightarrow \mathbb {C} }\\^{z\rightarrow z+b}\end{array}}}
translation de vecteur
R
e
(
b
)
i
→
+
I
m
(
b
)
j
→
{\displaystyle \scriptstyle Re(b){\vec {i}}+Im(b){\vec {j}}}
L'application
C
→
C
z
→
z
¯
{\displaystyle {\begin{array}{llll}_{\mathbb {C} \rightarrow \mathbb {C} }\\^{z\rightarrow {\bar {z}}}\end{array}}}
symétrie d'axe (Ox) .
L'application
C
→
C
z
→
a
z
{\displaystyle {\begin{array}{llll}_{\mathbb {C} \rightarrow \mathbb {C} }\\^{z\rightarrow az}\end{array}}}
M
→
M
′
{\displaystyle \scriptstyle M\rightarrow M'}
{
O
M
′
=
|
a
|
⋅
O
M
(
i
→
,
O
M
′
→
)
=
(
i
→
,
O
M
→
)
+
arg
(
a
)
[
2
π
]
{\displaystyle \textstyle {\begin{cases}OM'=|a|\cdot OM\\({\vec {i}},{\overrightarrow {OM'}})=({\vec {i}},{\overrightarrow {OM}})+\arg(a)~[2\pi ]\end{cases}}}
L'application
C
→
C
z
→
a
z
+
b
{\displaystyle {\begin{array}{llll}_{\mathbb {C} \rightarrow \mathbb {C} }\\^{z\rightarrow az+b}\end{array}}}
M
→
M
′
{\displaystyle \scriptstyle M\rightarrow M'}
Ω
M
′
→
=
|
a
|
⋅
e
i
arg
(
a
)
Ω
M
→
{\displaystyle \textstyle {\overrightarrow {\Omega M'}}=|a|\cdot e^{i\arg(a)}{\overrightarrow {\Omega M}}}
(où
Ω
{\displaystyle \Omega }
unique point fixe d'affixe
ω
=
b
1
−
a
{\displaystyle \scriptstyle \omega ={\frac {b}{1-a}}}
similitude directe de centre
M
{\displaystyle M}
arg
(
a
)
{\displaystyle \arg(a)}
|
a
|
{\displaystyle |a|}
Le disque ouvert de centre
a
{\displaystyle a}
r
{\displaystyle r}
D
(
a
,
r
)
=
{
z
∈
C
/
|
z
−
a
|
<
r
}
{\displaystyle \scriptstyle D(a,r)\;=\;\{z\in \mathbb {C} \;/\;|z-a|<r\}}
Le disque fermé de centre
a
{\displaystyle a}
r
{\displaystyle r}
D
′
(
a
,
r
)
=
{
z
∈
C
/
|
z
−
a
|
⩽
r
}
{\displaystyle \scriptstyle D'(a,r)\;=\;\{z\in \mathbb {C} \;/\;|z-a|\leqslant r\}}
Soient
u
→
(
a
)
{\displaystyle \scriptstyle {\vec {u}}(a)}
v
→
(
b
)
{\displaystyle \scriptstyle {\vec {v}}(b)}
P
{\displaystyle P}
u
→
⋅
v
→
=
R
e
(
a
b
¯
)
{\displaystyle \scriptstyle {\vec {u}}\cdot {\vec {v}}\;=\;Re(a{\bar {b}})}
Soient
A
(
a
)
,
B
(
b
)
,
M
(
z
)
{\displaystyle \scriptstyle A(a),\ B(b),\ M(z)}
P
{\displaystyle P}
M
A
→
{\displaystyle \scriptstyle {\overrightarrow {MA}}}
M
B
→
{\displaystyle \scriptstyle {\overrightarrow {MB}}}
R
e
(
(
z
−
a
)
(
z
¯
−
b
¯
)
)
=
0
{\displaystyle \scriptstyle Re((z-a)({\bar {z}}-{\bar {b}}))=0}
L'ensemble des racines
n
{\displaystyle n}
ièmes de l'unité forme un polygone régulier à
n
{\displaystyle n}
(racines cubiques : triangle ; racines 4e : carré ; ...) 
![{\displaystyle \textstyle {\begin{cases}OM'=|a|\cdot OM\\({\vec {i}},{\overrightarrow {OM'}})=({\vec {i}},{\overrightarrow {OM}})+\arg(a)~[2\pi ]\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0099258c1fb9e6cce9714c285a22d3a7a5c631e7)
