Exercice 2 : (5 points)
Partie A :
1. z1 = a 1 × z0 = (√3+1
4 + i √3 1
4 )(6 + 6i) = 6
√
3+64 + 6
√
3 64 i + 6
√
3+64 i – 6
√
3 6 4 = 3 + 3√
3 ia2 = ( √3+1
4 + i √3 1
4 )2 = 1
16(3 + 1 + 2
√
3 – 3 – 1 + 2√
3 + 2i(√
3 – 1)(√
3 + 1)) = 4√
316 + 1
4 i =
√
3 4 + 14 i
2. | z1 | = | 3 + 3
√
3 i | =√
32+(3√
3)2 =√
36 = 6et z1 = 3 + 3
√
3i = 6( 1 2 +√
32 i) = 6 e
iπ 3. a2 =
√
34 + 1 4 i = 1
2 (
√
3 2 + 12 i) = 1 2 ei
π 6 . 3. z3 = a 3 × z0 = a 2 × z1 et z7 = a 7 × z0 = a 6 × z1. 4. z3 = a 2 × z1 = 1
2 ei
π
6 × 6 ei
π
3 = 3 ei( π3+ π6) = 3 ei
π 2
et z7 = a 6 × z1 = (a 2)3 × z1 = 1 8 ei
π
2 × 6 eiπ3 = 3 4 e5i
π 6 . 5.
Partie B :
Pour tout entier naturel, on pose | zn | = rn. 1. | a2 | = 1
2 donc | a | = 1
√2 =
√
22
donc rn = | zn | = | a n × z0 | = | a n | × | z0 | = ( 1
√2)
n ×
√
72 = (√22)n × 6√
2 = 12 (√22)n+1 .2. rn+1 = 12 (√2
2 )
n+2
=
√
22 × rn donc la suite ( rn) est géométrique de raison
√
22 et de premier terme | r0 = | z0 | = 6
√
2.3. 0 <
√
22 < 1 donc la suite ( rn) tend vers 0.
4. Algorithme :
