Lycée Gustave Eiffel Nombres Complexes Exercice 3. Équations

(3−2 )(2+3 )
2 +
3−
√2 +
2√2 +
1+2
√3+2
(2 −)3
(1 + 2 )4
7
X
k=0
(2 )k
52 2097 2001 314
zC
2z+ (2 + )¯z= 2 −
(3 + 2 )z+ (3 −)¯z= 3 + 4
(1 + )z+ (1 −)¯z= 3
(2 + 3 )z+ (2 −3 )¯z= 3
(1 + 2 )z+ (2 −1)¯z= 2
(1 + )z¯z+ 2¯z= 0
(1 + )z¯z+ (4 −2)z−4 + 4 = 0
z∈R
(1 + )z+ (1 −)¯z
−z+ 2 z+ ¯z+ 2 ¯z
3z2+z2−¯z2+ 3¯z2
5z2+ 5z2+ 10¯zz −5 ¯z2+ 5¯z2
z3+z2¯z+z+ ¯z+z¯z2+ ¯z3
2z3¯z3+z4¯z2+z3¯z+z2¯z4+ 2z2¯z2+z¯z3+z¯z
z3+z2¯z+z+z−¯z3+z¯z2−¯z+ ¯z
z
2z−1
z2z= ¯z z + ¯z= 2z¯z
z
2z−1
z2
(u, v)∈C2
|u+v|2+|u−v|2= 2 |u|2+|v|2
a b 
a
|a|2−b |b|2
=|a−b|
|a||b|
1
a1z6= 1 1 + a
1−a∈R
z∈R1 + z
1−z1
a b 1ab 6=−1
a+b
1 + ab ∈R
a b a¯ b6= 1 z=a−b
1−a¯ b
|z|= 1 ⇐⇒ |a|= 1 |b|= 1
z∈C
z+1
z∈R⇐⇒ (z∈R∗|z|= 1)
(|z|= 1 z6= 1) ⇐⇒ ∃x∈Rz=x+
x−1
z∈R⇐⇒ z2− |z|2= 0
1
a b c 1
|ab +ac +bc|=|a+b+c|

n
n∈Nz1, . . . , znn1
z= n
X
k=1
zk! n
X
k=1
1
zk!0≤z≤n2
t∈RAcos(ωt +ϕ)
A∈R+ω∈R+
cos(t) + √3 sin(t) 3 cos(2t)−3 sin(2t)
sin(2a) cos(2a)
sin(3a) cos(2a)
cos(5a)
sin(4a)
x∈R
sin(2x) + sin(3x) + sin(4x)=0
cos(2x) + cos(4x) + cos(6x)=0
sin(3x) + 4 sin2(x)−sin(x)−2=0
2 cos3(x)−3 cos(2x) + cos(3x)=0
x∈R
sin2(x) cos3(x)
sin3(x)
sin5(x) +
cos4(x) sin(x)
sin2n(x)n∈N
sin2n+1(x)n∈N
n
X
k=1
cos(kx)
n
X
k=1
sin(kx)
n
X
k=1
kcos(kx)
n
X
k=1
cosk(x) cos(kx)
n
X
k=1
cosk(x) sin(kx)
n
X
k=0 n
kcos(kx)
n
X
k=0 n
ksin(kx)
n
n
X
k=0 n
kcos(a+kx)=0
n−1
X
k=0
cos(kx)
cosk(x)= 0
n∈N
1
−1
2
−2
−3+3
(1 −√3)11
3√3−
√3+2 !32
(1 + )25
1−
1−4
√3−
−1!n
1 + eθ
(1 + eθ)n
1 + √2 +
2 + √3−!n
(1 + tan(α))2
1 + tan2(α)
1−sin(θ) + cos(θ)
1−sin(θ)−cos(θ)
eθ−eθ0
(1 + )n−(1 −)n
q2 + √2+ q2−√2
θ
cos( π
12) sin( π 12) tan( π 12)
cos( π
12) sin( π 12) tan( π 12)
z= 1 + z0=√3 +
z z0z
z0
cos( π
12) sin( π 12)
z∈C
z= 2 π
3
2z= 3 π
8
2z+z+ 1 = 0
3z−22z+ 2 z= 0
−3+4
1+4√3−7−24 2π
6
72 π
8
−2+2√3

C
z2+ (2 −3 )z−1 + 5 = 0
z2+ (1 + 8 )z−17 + = 0
z2+ (4 −3)z+−5=0
z4−(7 + 7 )z2+ 25 = 0
z−2
z−12
−(3 + 2 ) z−2
z−1+ 3 −1=0
z+ 1
z−22
−(1 + 2 ) z+ 1
z−2+−1=0
z3+ (−3 + 2 )z2+ (−1−5 )z+ 6 + 2 = 0
z3+ (1 + )z2+ (4 −)z+ (12 −6 ) = 0
z3−(4 + 9 )z2+ (18 + 31 )z−(18 + 39 ) = 0
z4+ 2(1 + )z3+ 3 z2+ (−1 + )z−2(1 + 3 )
2z4−6z3+ 9z2−6z+ 2 = 0 1 +
n∈N
(z+ 1)n= (z−1)nzn= (1 + z)n27(z−1)6−(z+ 1)6= 0
α−π
2< α < π 2
1 + z
1−z4
=1 + tan(α)
1−tan(α)
z+ 1
z−1n
+z−1
z+ 1n
= 1
3
P P (X) = X3+ 3X−10
z∈CP
u v P (u+v) = 0
uv =−1z=u+v
u v P (u+v) = 0 uv =−1
u3+v3
u3v3
(u, v)
CP(z)=0 z∈C
A B C −3−6−5 3 −4
ABC
A B C a b c
g G ABC
h H ABC a b
c
G H a +b+c= 0
G H
1
a+1 b+1 c
b
a+c a
bc
a2
b c a
ABC
ABCD A0B0C0D0
[AB] [BC] [CD] [DA]A0B0C0D0
z
1z1 + z2
z z2z3z
A B C
a b c ABC
a2+b2+c2−(ab +bc +ca) = 0
a b θ
zarg z−a
z−b=θ[π]

F M z C\ {−1−3}
M0
z0=z+7
2−3
z+ 1 + 3
F
M
|z0|= 1
z0∈R
IC\ {0}C\ {0}I(z) = 1
zD
CCA a r
06∈ D
b D
|z|=|z−b|
I(D)
0∈D I(D)
06∈ C I(C)
0∈ C I(C)
a r
z0= (1 −)z+ 2
z0= (√3 + )z+ 3 + 3 (1 −√3)
z0=π
4z+
z0=1 + i
1 + √3z+ 1 −√3 + (3 −√3)
2
π
3Ω 1 + 2
θ1∈Rθ2∈RΩ1ω1Ω2
ω2r1Ω1θ1r2θ2
r=r1◦r2
r
r
T
z0= (1 + )z−z∈CM M0M0=T(M)
T
A T
AMM0
D0D y =x T
B A
z0B z0
0B0=T(B)z0z0 0= 1
A0A O
A A0B B0
A B 1O
M z A B M1M2
BM M1AM M2
(−−−→ M1B, −−−→ M1M)=(−−−→ M2M, −−−→ M2A) = π
2
z1z2M1M2
M1M2M T1
T2M
M OM1M2
OM1=OM2M(∆)
OM1=M1M2M(Γ)
M OM1M2