Presque tous les exercices qui suivent sont Hors
Programme. L’exercice 12 est particulièrement ardu...
Avant de les attaquer, étudier le paragraphe IV. du
document “structures algébriques”
1, i, 3 + 4i, 8
6i,
3
10i
4i
24
7 + 24i
1+i
p
2
cos(⇡/8)
sin(⇡/8)
cos(⇡/12)
sin(⇡/12)
az 2 + bz + c = 0
a b c
C
z2 + z + 1 = 0 ;
z2
(5 14i)z
z2
(1 + 2i)z + i
2(5i + 12) = 0 ; z 2
4
(3 + 4i)z
2
z + 10z + 169 = 0 ;
x4
1=0 ;
z2
p
3z
1 + 5i = 0 ; 4z 2
4
i=0 ;
2z + 1 = 0 ;
2
z + 2z + 4 = 0.
30x2 + 289 = 0.
z 2 C \ {2i}
f (z) =
2z i
.
z 2i
z 2 = i, z 2 C.
f (z) = z, z 2 C \ {2i}.
j=e
j
2⇡
3
.
j2
1 + j + j2 = 0
z3
p
1+i
p 3
3+i
z2 + z + 1 = 0
8i
1 + i 7 + 24i i 5 + 12i
z2 + z
z2
(5
2=0
14i)z
2(5i + 12) = 0
z 2 + 4z + 5 = 0
z2
(3 + 4i)z
z4
i)z 2
(1
1 + 5i = 0
i=0
z 4 + 4z 3 + 6z 2 + 4z
15 = 0
C
z2
z3
(11
5i)z + 24
+ 3z
27i = 0
2i = 0
C
z2
(E)
(1 + a) (1 + i) z + 1 + a2 i = 0,
a
a2R
Z1
M
z2
z1
a)2
z1
[Z1 , Z2 ]
R
a
↵2R
z 2n
(E)
2i(1
Z2
M
1 = 0.
z2
C
2 cos(↵)z n + 1 = 0,
z 2 2 cos(↵)z +
n
P↵ (z) = z 2n
2 cos(↵)z n + 1.
P↵
⇣
P↵ (z) = z 2
2 cos
1
⇣↵⌘
n
+1
⌘✓
z2
2 cos
✓ ◆
✓
cos ✓ = 2 sin
,
2
2
✓
↵ 2⇡
+
n
n
◆
◆
✓
+ 1 . . . z2
✓ 2 R.
P↵ (1)
✓
◆
⇣↵⌘
⇣↵
sin2
⇡⌘
↵
(n 1)⇡
2
2
sin
sin
+
. . . sin
+
=
2n
2n n
2n
n
4n
2
↵
2
1
.
2 cos
✓
↵ 2(n 1)⇡
+
n
n
↵
]0, ⇡[
n
2
✓
◆
✓
◆
⇣↵
⇡⌘
↵
2⇡
↵
(n 1)⇡
Hn (↵) = sin
+
sin
+
. . . sin
+
.
2n 2n
2n
n
2n
n
↵
2n
1
Hn (↵) =
sin(↵/2)
.
sin(↵/2n)
Hn (↵)
↵
0
n
sin
⇣⇡ ⌘
n
sin
✓
2⇡
n
◆
. . . sin
✓
(n
1)⇡
n
2
◆
=
z 2 C
z, p, q
n
.
2n 1
p, q
z
C
14i)z 2
z4
(5
2(5i + 12) = 0
C z 4 + 6z 3 + 9z 2 + 100 = 0
im)z
m2C
(1 + im) = 0
|u|
↵,
u, v 2 C
✓
|u + v|2 + |u
2 2
|v|
=
2
2
2C
C
|u + v|2 + |u
◆2
+z+1=0
+ 2z + 1 = 0
z2
2z cos ✓ + 1 = 0 ✓
z2
(6 + i)z + (11 + 13i) = 0
(2 +
4|uv|2 .
=0
u, v 2
v|2 = 2|u|2 + 2|v|2
C
2z 2
v|2
z 2 + ↵z +
↵, 2 C m = ↵+
µ
2
|↵| + | | = |m + µ| + |m µ|
z2
z2
↵
2z 2
(7 + 3i)z + (2 + 4i) = 0
z = e2i⇡/5
a = z + z4
cos
sin
⇡
5
2⇡
5
sin
⌦
(Ox)
I
2⇡
5
b = z2 + z3
a
b
4⇡
cos 5 sin 4⇡
5
1
2
cos
⇡
5
M
i
OI +OJ = OI.OJ =
J
1
O
1
1
[AC]
AF
AC
(ABCDE)
F
G
FG
AF
C
0
z 4 (5 14i)z 2 2(5i+12) =
z = a + ib
a + ib
! =↵+i
|!|2 = |z|
1+i
p
2
az 4 + bz 2 + c = 0
(↵ + i )2 =
⇡
= ei 4
Z = z2
z = a + ib
! 2 C
a, b 2 R
! =↵+i
!2 = z
! 2 = z , (↵ + i )2 = a + ib
, ↵2
,
(
2
+ 2i↵ = a + ib
2 =a
↵2
2↵ = b
|!|2 = |z|
8
p
2
2 =
>
a 2 + b2
<↵ +
2 =a
, ↵2
>
:
2↵ = b
p
8
2 = a+ a2 +b2
>
↵
<
2
p
2 = a+ a2 +b2
,
2
>
:
2↵ = b
q p
8
a+ a2 +b2
>
>
<↵ = ±q
2
p
a+ a2 +b2
,
=±
>
2
>
:
↵
! =↵+i
b
(↵, )
z
z =
8
6i
! 2 = z , (↵ + i )2 = 8
6i
2
, ↵2
+ 2i↵ = 8 6i
(
2 =8
↵2
,
2↵ = 6
8
p
2+ 2 =
>
↵
82 + ( 6)2 = 10
<
2 =8
, ↵2
>
:
2↵ = 6
8
2
>
<2↵ = 18
2 =1
,
>
:
2↵ = 6
8
p
>
<↵ = ± 9 = ±3
,
= ±1
>
:
↵
8
>
↵=3
= 1
<
,
>
:
↵= 3
= +1
z=8
6i
!1 = 3
i
1
+1
1
p
2
i
(1
+
i)
2
3 + 4i
2+i
7 + 24i
4 + 3i
2
!, !
z=
i
2+i
5
z
!2 =
p
2
2 (1
+ i)
4
3i
2
i
i
5+i
1+i
p
2
!=
!=
sp
1
2
z
2+1
p +i
2 2
sp
2 1
p ,
2 2
q
q
p
1
2+ 2+i
2
2
⇡
z = ei 4
p
2.
!1 =
3+i
⇡
ei 8
⇣
⇡
ei 8
!
⇡
ei 8
⌘2
=e
2i⇡
8
⇡
1
cos =
8
2
⇡
ei 8 = cos ⇡8 + i sin ⇡8
z
i ⇡8
e =!
cos ⇡8 > 0
!
⇡
= ei 4 .
q
p
2+ 2
⇡
1
sin =
8
2
P (z) = az 2 + bz + c
<0
z
q
2
= b2
P
p
2.
4ac
0
P (z) = 0
P (z) = az 2 + bz + c = az 2 + bz + c = P (z) = 0.
z
z 6= z
z
z
P
z
<0
P
a 6= 0
=
2
2
a, b, c 2 C
az 2 + bz + c = 0
4ac
=
z1 =
z2
b2
2
p
3z
b+
2a
i=0
b
2a
z2 =
.
= 3 + 4i
=2+i
z1 =
p
3+2+i
2
z2 =
z2 + z + 1 = 0
z 2 p(1 + 2i)z + i 1 = 0
z2
3z i = 0
z 2 (5 14i)z 2(5i + 12) = 0
z 2 (3 + 4i)z 1 + 5i = 0
4z 2 2z + 1 = 0
z 4 + 10z 2 + 169 = 0
2 + 3i
p
3
2
2
i
.
p
p
1 + i 3) 12 ( 1 i 3)
p 1 +1i i p
1
3+i) 2 ( 2
3 i)
2 (2
5 12i 2i
p 2 +1 3i 1 +
pi
1
(1
+
i
3)
(1
i
3)
4
4
2 + 3i 2 3i 2 3i
1
2(
p
2
2 (1 + i
z 4 + 2z 2 + 4 = 0
p
p
p
p
2
2
(
1
+
i
3)
i 3)
2
2 ( 1
= 2i
z1 = 5 2i
z1 = i
z i
z2 = 3
3)
2
2 (1
p
i 3)
i
1+i
p
1
z2 = 6
3i
z2 = i
)
z1 =
p
2i
|z| = 1
2i z3 =
z = 1 ± 2i z =
z3 =
z4 = 1
i
4 ± 2i
m = 2i
1
4
= t↵2 , t
↵=0
|↵|2 +| |2 +2 |↵ | = |m
|{z} |
|µ|2
z 2 +z+1 = 0 , z =
j2
0
= 12
2=
z2
2|m|2 +2|µ|2
p
3
1
2 +i 2 = j
1 = i2
z1 = 12 ( 1 + i)
✓2R
µ|2 + |m + µ|2 +2 |m2 µ2 |
{z
}
| {z }
|↵
z=
z2 = 12 ( 1
1
2
|2 /4
p
i 23 =
i)
z
2z cos ✓ + 1 = (z
= (z
cos ✓)2 + 1
cos ✓
cos2 ✓ = (z
i sin ✓)(z
cos ✓)2 + sin2 ✓ = (z
cos ✓ + i sin ✓) = (z
ei✓ )(z
0
= cos2 ✓ 1 = sin2 ✓
✓2
/ ⇡Z
2
(E)
z
(6 + i)z + (11 + 3i) = 0
= (6 + i)2 4(11 + 13i) = 9 40i
40 = 2 ⇥ 20 = 2 ⇥ (4 ⇥ 5)
2
2
4
5 = 16 25 = 9
=
(4 5i)2
(E)
C
5i
z1 = 6+i+4
= 5 2i z2 = 6+i 24+5i = 1 + 3i
2
z1 = ei✓
z2 = e
i✓
2z 2 (7 + 3i)z + (2 + 4i) = 0
=
8(2 + 4i) = 24 + 10i
10 = 2 ⇥ 5 = 2 ⇥ (5 ⇥ 1)
2
2
5
1 = 24
= (5 + i)2
C
z1 =
7+3i+5+i
7+3i 5 i
1
= 3 + i z2 =
= 2 (1 + i)
4
4
(7 + 3i)2
cos ✓)2
e
i✓
)
(i sin ✓)2
2⇡
5
a = 2 cos
z4
z3 + z4 = 0
4⇡
5
b = 2 cos
1
C
a + b = z + z2 + z3 + z4 =
1 z z2 z3
1 + z + z2 +
1
ab = (z+z 4 )(z 2 +z 3 ) = z 3 +z 4 +z 6 +z 7 = z+z 2 +z 3 +z 4 =
a
b
p
1+ 5
2
cos 2⇡
5
a>0
sin 2⇡
5 >0
sin
✓
2⇡
5
◆
=+
p
1
10
4
sin
⇡
5
=
cos
1
2⇡
5
=
p
5 1
4
p
p
2 5
= sin ⇡
5
2 p
1+ 5
4
v
u
u
2⇡
cos2 ( ) = t1
5
r
cos
1
X2 + X
p
p
5
⇡
5
=
= sin
4⇡
5
cos
⇡
5
sin
2⇡
5
4⇡
5
=
1+
4
p !2
5
1
4
=
5 cos
cos ⇡
p
= 14 10
xI = x⌦ + R = 1+2 5
xJ = x⌦
4⇡
xI = 2 cos 2⇡
x
=
2
cos
J
5
5
[O, I] [O, J]
1 = 0
⇤ ⇡⇥
2 0, 2
5
4
1
4
=
q
p
10 + 2 5.
p
p
10 + 2 5
4⇡
5 =
⇡
=
5 p
p
R = ⌦O2 + OM 2 =
p
1
2⇡
5p
z 5 = 1).
1(
p
1
5
sin
4⇡
5
=
4
cos
4⇡
5
p
1+ 5
4
2 5
p
5
.
2
p
1
R =
5
2
O
1
=
M
J
I
O
⌦
x=
AF
AC
x=
x2
AF
HK
FG
AC
=
=
=
AC
HC
FC
AC
3x + 1 = 0
AG
AC AF
=1
AC =
AC
p
p
3+ 5
1+ 5
2
=
2
2
x=
2AF
1 2x
=
.
AF
1 x
x<1 x=
p
1+ 5
2
FG
AF
=
3
p
2
AC 2AF
AF
5
=
1
x
2=
3
2p
2=
5
B
C
G
F
A
K
H
D
E
C
A
•
C
•
[A, B]
=
x
a
=
a x
x
B
•
BC
AC
a = AB
x 2 x
a +a
1=0
x
a
>0
=
AC
AB
x = AC
x
a
=
p
1+ 5
2
p
1+ 5
2
a
x
1, 618...
= 0, 618...
=
p
1+ 5
2
=
Z 2 (5 14i)Z 2(5i+12) = 0
= (5
Z2 =
z
14i)2 + 8(5i + 12) =
5 14i 5+10i
2
75
100i = 25( 3
Z1 =
=
2i
, z2 = 5
,z=3
4i) = (5(1
5 14i+5 10i
2
12i = (3
2i
z=
=5
2i)2
3 + 2i
2i))2
12i
z2 =
2i = (1
z=1
i
i)2
z=
1 + i.