Exercices : Nombres Réels et Complexes

√2
f:Q→Q
∀x, y ∈Q, f(x+y) = f(x) + f(y)
f C C
f(0)
∀x∈Q, f(−x) = −f(x)
∀n∈N,∀x∈Q, f(nx) = nf(x)
n∈Z
a=f(1) ∀x∈Q, f(x) = ax
2
3+41 81r5 31/3
+2
3−41 81r5 31/3
err∈Q∗
a, b ∈N∗
Pn(x) = 1
n!xn(bx −a)n
x= 0
x=a/b
r=a/b n ∈N∗
In=Zr
0
Pn(t)etdt
In→0
er=p/q p, q ∈N∗qIn∈Z
√2 + 3 √3
f:R→R
∀(x, y)∈R2,f(x+y) = f(x) + f(y);
∀(x, y)∈R2,f(xy) = f(x)f(y);
∃x∈R,f(x)6= 0
f(0) f(1) f(−1)
f(x)x∈Zx∈Q
∀x≥0, f(x)≥0f
f= IdR
n∈N∗x1, . . . , xn∈R
n
X
k=1
xk=
n
X
k=1
x2
k=n
k∈ {1, . . . , n}xk= 1
x, y ∈[0 ; 1]
x2+y2−xy ≤1

∀u, v ≥0,1 + √uv ≤√1 + u√1 + v
(α, β)∈(R∗
+)2M∈R
∀x, y > 0, xαyβ≤M(x+y)
(x1, . . . , xn) (y1, . . . , yn)
1
n
n
X
k=1
xk! 1
n
n
X
k=1
yk!1
n
n
X
k=1
xkyk
∀x, y ∈[0 ; 1],minxy, (1 −x)(1 −y)≤1
4
∀x, y ∈R,bxc+bx+yc+byc≤b2xc+b2yc
∀x∈R,∀n∈N∗,
n−1
X
k=0x+k
n=bnxc
a≤b∈R
Card([a;b]∩Z) = bbc+b1−ac
n∈N∗
(an, bn)∈N∗2
(2 + √3)n=an+bn√3 3b2
n=a2 n−1
(2 + √3)n
θ∈Rn∈N
Cn=
n
X
k=0 n
kcos(kθ)Sn=
n
X
k=0 n
ksin(kθ)
BC
z∈B1−z+z2∈B1 + z+z2∈B
B
a, b, z
b
az−a z−b2
∈R∗
+
M z I i
M0iz
M0
z7→ 1
1−z

M z
z+z=|z|
a, b, c
t, u, v
tt =a2, uu =b2vv =c2
z=q2 + √2 + iq2−√2
z∈C∗z0∈C
|z+z0|=|z|+|z0| ⇐⇒ ∃λ∈R+, z0=λ.z
∀z, z0∈C,|z|+|z0|≤|z+z0|+|z−z0|
a∈C|a|<1
z
z−a
1−az ≤1
∀z1, z2∈C,|z1+z2|2+|z1−z2|2= 2 |z1|2+ 2 |z2|2
z1, z2∈C|z1| ≤ 1|z2| ≤ 1ε= 1
−1
|z1+εz2| ≤ √2
f:C→C
f(z) = z+|z|
2
f
|z+ 1|=|z|+ 1 z∈C
n∈N∗Unn
X
z∈Un|z−1|
n≥3ω1, . . . , ωnn ωn= 1
p∈Z
Sp=
n
X
i=1
ωp
i
T=
n−1
X
i=1
1
1−ωi

ω n
S=
n−1
X
k=0
(k+ 1)ωk
(1 −ω)S S
j(j+ 1) j
j2+1
j+1
j−1
n∈N∗
(z+ 1)n= (z−1)n
ω= ei2π
7
A=ω+ω2+ω4B=ω3+ω5+ω6
n∈Nn≥2ω= exp(2iπ/n)
z∈C, z 6= 1
n−1
Y
k=1
(z−ωk) =
n−1
X
`=0
z`
z= 1
n−1
Y
k=1
sin kπ
n=n 2n−1
sinπ
5=s5−√5 8
a∈Rx3+ 2x2+ 2ax −a2= 0 x= 1
C
z2−2iz−1 + 2i = 0
z4−(5 −14i)z2−2(12 + 5i) = 0
5−12i
z3−(1 + 2i)z2+ 3(1 + i)z−10(1 + i) = 0
C
x+y= 1 + i
xy = 2 −i
Z∈C∗ez=Z z ∈C
eiθ−1
eiθ+1 θ∈]−π;π[
ei.θ + ei.θ0θ, θ0∈R

√2∈Q
√2 = p/q p, q ∈N∗p q
2q2=p2
p p2p= 2k
k∈Nq2= 2k2
q p q
f(x+y) = f(x) + f(y)f C
C=C+C C = 0
x=y= 0 f(x+y) = f(x) + f(y)f(0) = 0
y=−x f(x+y) = f(x) + f(y) 0 = f(−x) + f(x)
f(−x) = −f(x)
∀n∈N,∀x∈Q, f(nx) = nf(x)
n∈Z−, n =−p p ∈N
f(nx) = f(−px) = −f(px) = −pf(x) = nf(x)
x=p/q p ∈Zq∈N∗
f(x) = f(p×1
q) = pf(1 q)
a=f(1) = f(q×1
q) = qf(1 q)
f(1
q) = a q
f(x) = ap
q=ax
x
1/3
x3
x
2
3+41 243√151/32 3−41 243√151/3
1
81486 + 123√151/3486 −123√151/3
(a−b)(a+b) = a2−b2
2
3+41 243√151/32 3−41 243√151/3
=7
27
x
x3=4
3+7 9x
(3x−4)(3x2+ 4x+ 3) = 0
3x2+ 4x+ 3 >0
x= 4/3