Exercice 6. Correction 7. 9

a+ib a, b ∈R
3+6i
3−4i;1 + i 2−i2
+3+6i
3−4i;2+5i 1−i+2−5i 1 + i.
z1
z2=z1 z2·¯z2 ¯z2=z1¯z2
|z2|2
a+ib
a, b ∈R
5+2i
1−2i; −1 2+i√3 2!3
;(1 + i)9
(1 −i)7.
a+ib
2π/3
3−π/8
cos(2θ) sin(2θ) cos θsin θ
z1=i, z2= 1 + i, z3=−2+2i, z4=e−iπ
3.
a+ib, a ∈Rb∈R.
−2
1−i√3
1
(1 + 2i)(3 −i)
1+2i
1−2i,2+5i 1−i+2−5i 1 + i.
z1=
3+3i z2=−1−√3i z3=−4
3i z4=−2z5=eiθ +e2iθ.
(1+i√3
2)2000
(3 + 2i)(1 −3i)
2π/3
3−5π/6
3+2i
1−3i
2π/3
3−5π/6
9−7i−6i−0,3+1,1i−√3
3−i 3

1 + i(1 + √2)
p10 + 2√5 + i(1 −√5)
tan ϕ−i
tan ϕ+iϕ
ρ=p4+2√2θ=3π
8ρ= 4 θ=−π 10 ρ= 1
θ= 2ϕ+π
1 + i; 1 + i√3 ; √3 + i;1 + i√3
√3−i.
(cos(π/7) + isin(π/7))(1−i√3
2)(1 + i) = √2(cos(5π/84) + isin(5π/84)),
(1−i)(cos(π/5)+isin(π/5))(√3−i) = 2√2(cos(13π/60)−isin(13π/60)),
√2(cos(π/12)+isin(π/12))
1+i=√3−i 2.
θ∈R
eiθ = cos θ+isin θ.
θ∈R
cos θ=eiθ +e−iθ
2sin θ=eiθ −e−iθ 2i.
a, b ∈R
2 cos acos b; 2 sin asin b; cos2a; sin2a.
eixeiy =ei(x+y)x, y ∈R
sin(x+y) cos(x+y) tan(x+y)
x y sin(2x)
cos(2x) tan(2x)x, y ∈R
cos xsin xtan x
2x6=π+ 2kπ , k ∈Z
θ∈R
(cos θ+isin θ)n= cos(nθ) + isin(nθ).
cos(3x) sin(3x)
sin xcos x

cos 5θcos 8θsin 6θsin 9θ
θ
sin3θsin4θcos5θcos6θ
θ
cos 5θsin 5θ
cos θsin θ
ei5θ= (eiθ)5