√22 cos(74π) +i sin(74π

5.11 1) |z₁|=|1 +√ 3i|=

q

1²+ (√

3)² =√

1 + 3 =√ 4 = 2 z₁ = 1 +√

3i= 2 ¹₂ +i

√3 2

= 2 cos(^π₃) +i sin(^π₃)

|z₂|=|¹2 − ¹2i|=|¹2(1−i)|=|¹2| |1−i|= ¹₂p

1² + (−1)² = ¹₂√

2 = ^√₂²

z₂ = ¹₂ − ¹₂i= ^√₂² 1

√2 2 2

+i

−^√122 2

= ^√₂^{2 √1}₂ +i(−^√1₂)

=

√2 2

√2

2 +i(−^√₂²)

= ^√₂² cos(⁷₄^π) +i sin(⁷₄^π)

2) (a) Forme algébrique z₁z₂ = (1 +√

3i) (¹₂ − ¹₂i) = ¹₂ − ¹₂i+^√₂³i− ^√₂³i² = ^√3+1₂ + ^√3−1₂ i (b) Forme trigonométrique

z₁z₂ = 2 cos(^π₃) +i sin(^π₃)

· ^√₂² cos(⁷₄^π) +i sin(⁷₄^π)

= 2· ^√₂² cos(^π₃ + ⁷₄^π) +i sin(^π₃ +⁷₄^π)

=√

2 cos(²⁵₁₂^π) +i sin(²⁵₁₂^π)

√ =

2 cos(₁₂^π + 2π) +i sin(₁₂^π + 2π)

=√

2 cos(₁₂^π) +i sin(₁₂^π)

3) L’égalité ^√3+1₂ + ^√3−1₂ i=√

2 cos(₁₂^π) +isin(₁₂^π)

donne (a) ^√3+1₂ =√

2 cos(₁₂^π) cos(₁₂^π) = ^√3+1

2√

2 = ^{(√3+1)·√2}

2√ 2·√

2 = ^√6+₄^√2 (b) ^√3−1₂ =√

2 sin(₁₂^π) sin(₁₂^π) = ^√₂^3√−₂¹ = ⁽

√3−1)·√ 2 2√

2·√

2 = ^√6−₄^√2

Algèbre : nombres complexes — forme trigonométrique Corrigé 5.11