TS Correction Ex 2 DS3 Complexes

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Exercice 2 : (5 points)

Partie A :

1. z1 = a ¹ × z0 = (√³⁺¹

4 + i √^{3 1}

4 )(6 + 6i) = 6

√

3+6

4 + 6

√

3 6

4 i + 6

√

3+6

4 i – 6

√

3 6 4 = 3 + 3

√

^{3 i}

a² = ( √³⁺¹

4 + i √^{3 1}

4 )² = 1

16(3 + 1 + 2

√

3 – 3 – 1 + 2

√

3 + 2i(

√

3 – 1)(

√

3 + 1)) = 4

√

3

16 + 1

4 i =

√

3 4 + 1

4 i

2. | z1 | = | 3 + 3

√

^{3 i}^{| =}

√

³²⁺⁽³

^√

³⁾²⁼

^√

^{36 = 6}

et z1 = 3 + 3

√

3i = 6( 1 2 +

√

³

2 i) = 6 e

iπ 3. a² =

√

3

4 + 1 4 i = ¹

2 (

√

3 2 + ¹

2 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 e^{i( π}³^{+ π}⁶⁾ = 3 eⁱ

π 2

et z7 = a ⁶ × z1 = (a ²)³ × z1 = 1 8 eⁱ

π

2 × 6 eⁱ^π³ = 3 4 e⁵ⁱ

π 6 . 5.

Partie B :

Pour tout entier naturel, on pose | zn | = rn. 1. | a² | = ¹

2 donc | a | = ¹

√² ⁼

√

²

2

donc rn = | zn | = | a ⁿ × z0 | = | a ⁿ | × | z0 | = ( 1

√²⁾

n ×

√

^{72 =}⁽^√₂²⁾ⁿ × 6

√

^{2 = 12}⁽^√₂²⁾ⁿ⁺¹ ^.

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2. rn+1 = 12 (√²

2 )

n+2

=

√

²

2 × rn donc la suite ( rn) est géométrique de raison

√

²

2 et de premier terme | r0 = | z0 | = 6

√

2.

3. 0 <

√

²

2 < 1 donc la suite ( rn) tend vers 0.

4. Algorithme :