Exercice 1. Exercice 2. Problème 1. Problème 2. Partie I.

y00 (x)+2y0(x)+2y(x) = 2 ch xcos x
ABC G
ABG BCG CAG
R
∀(x, y)∈R2:f(x+y) + f(x−y)=2f(x)f(y) (1)
f
f(0) = 1 f
∀(x, y)∈R2:f(x)f00 (y) = f00 (x)f(y)
∀x∈R:f(x) = 1
2eλx +e−λx
λ f00 (0)
λ
∀x∈R:f(x) = 1
2eλx +e−λx
P
(O, −→ i , −→ j)
u
u ju j2u
M3
M4
M2
OA
B
C
M
M1
M M1M2M3M4
a b s C C
∀z∈C:s(z) = az +b
s
∀z∈C:s◦s(z) = z
a b s
s
z s(z)

z s(z) = z
s(z) = eiαz α ∈R
s(z) = −b
bz+b b ∈C∗
PO A B C 0 1 j j2
S1S2S3S4
(OA) (OB) (OC) (BC)s1s2s3s4C C
M z M1M2M3M4
z1=s1(z), z2=s2(z), z3=s3(z), z4=s4(z)
z1z2z3z4z M1
M2M3
M2M3M4M
M2M3M4Ωω
M2M3M4
Ω (OM1)
ρ=|z|θ z
ω=−1+2ρcos θ
ρ+ 2 cos θe−iθ
R M2M3M4
−ρ+1+2ρcos θ
ρ+ 2 cos θ
R= 1 ρ= 1
(ρ+ cos θ)2= 1 −3 cos2θ
ΓM
M1M2M3M2M3M4
M4M
M1
A
B
C
M3
M2
M2M3M4