Lycée Gustave Eiffel Formulaire sur les nombres complexes

2=−1
1=−
4k= 1 4k+1 =
4k+2 =−14k+3 =−k∈Z
(a1+b1)+(a2+b2)=(a1+a2) + (b1+b2)
(a1+b1).(a2+b2)=(a1.a2−b1.b2) + (a1.b2+b1.a2)
(a+b).(a−b) = a2+b2
(a+b)−1=a
a2+b2−b a2+b2=1 a2+b2(a−b)
z2
1+z2 2= (z1+z2)(z1−z2)
z= (z) + (z)
z=z0⇐⇒ (z) = (z0) (z) = (z0)
(z+z0) = (z) + (z0) (λz) = λ(z)
(λz) = λ(z)λ∈R (z+z0) = (z) + (z0)
(z) = −(z) ( z) = (z)
¯
¯z=z z +z0= ¯z+¯ z0
z.z0= ¯z. ¯ z0¯ zn=zn
λz =λ¯z, λ ∈Rz=−¯z
(z) = 1
2(z+ ¯z) (z) = 1 2(z−¯z)
z∈R⇐⇒ (z)=0 ⇐⇒ z= ¯z
z∈R⇐⇒ (z)=0 ⇐⇒ z=−¯z
~
~
×
M(z)
|z|
×
M0(¯z)
O
(z)
(z)
−(z)
arg(z)
z=a+b(a, b)∈R2,|z| ≥ 0z= 0 ⇐⇒ |z|= 0
|z|=√a2+b2≥0|z|=OM |z|2=z.¯z
(|(z)|≤|z|
|(z)| ≤ |z||¯z|=|z|1
z=¯z |z|2
|z.z0|=|z|.|z0| |zn|=|z|n, n ∈Z|z+z0| ≤ |z|+|z0|
U={z∈C/|z|= 1}
∀z∈Uz−1= ¯z∈U
∀z∈C∗z∈U⇐⇒ z−1= ¯z
arg(zz0) = arg(z) + arg(z0)
arg(z−1) = arg(¯z) = −arg(z)
arg(zn) = narg(z)
x+y=x y z+z0=z0
(θ+θ0)=eθeθ0(θ)−1=eθ=e−θ
cos(θ) = eθ+e−θ
2sin(θ) = eθ−e−θ 2
nθ = ( iθ)ncos(nθ) + sin(nθ) = (cos(θ) + sin(θ))n
θ=θ0⇐⇒ θ=θ0[2π]
|z|=(z)z=¯z
az2+bz +c= 0 (E)
∆ = b2−4ac. δ ∈Cδ2= ∆
∆ = 0 (E){− b
2a}
∆6= 0 (E){−b+δ
2a,−b−δ 2a}
az2+bz +c=az−−b+δ
2az−−b−δ 2a
z1z2(E)
1) z1+z2=−b
a2) z1z2=c a

n
zn= 1 ⇐⇒ ∃k∈Zz=2kπ
n
⇐⇒ ∃k∈ {0, . . . , n −1}z=2kπ
n
zn
0=ω
zn=ω⇐⇒ ∃k∈Zz=z0
2kπ
n
⇐⇒ ∃k∈ {0, . . . , n −1}z=z0
2kπ
n
ω=rθ
zn=ω⇐⇒ ∃k∈Zz=n
√rθ+2kπ
n
⇐⇒ ∃k∈ {0, . . . , n −1}z=n
√rθ+2kπ
n
A1(a1), . . . , An(an)n
(αk)n
k=1 ∈Rn
n
X
k=1
αk6= 0
z0(A1, α1) (An, αn)
z0=
n
X
k=1
αkak
n
X
k=1
αk
A(a)B(b)C(c)
−→
v(z)−→ v0(z0)
k−→ vk=|z|
AB =|b−a|
(−→ v , −→ v0) = arg z0
z[2π]
(−−→ AB, −→ AC) = arg c−a
b−a[2π]
−→
v−→ v0z0
z∈R
−→
v−→ v0z0
z∈R
−→
v(z0) Ω(ω)θ∈Rλ∈R
−→
v z0=z+z0
Ωθ z0=θ(z−ω) + ω
Ωλ z0=λ(z−ω) + ω
(a, b)∈C∗×C
F
z0=az +b
a= 1 F b
a6= 1 FΩ(ω)θ λ
ω=aω +b θ = arg(a)[2π]λ=|a|
F θ
λΩ