TD 4

2015 −16
3x2−6x+ 2 = 0, x2+ (1 + i)x+i= 0, x2+ (3 + 4i)x−1 + 5i= 0
32
242,31
p(X) = adXd+· · · +a0∈Z[X]d
α
β∈Qp(x)α a0β ad
d= 3
ad= 1 −1p(X)
a0
A1=
5−3 2
6−4 4
4−4 5
A2=
1 1 −1
111
−1 1 1
A3=
4 2 −4
1 4 −3
1 1 0
P−1AP P
2×2
3×3
2×2
3×3
A
s s A
unvn
n p = 0,03
q= 0,95
unvn

2
A=a b
c d 
ρeiθ ρe−iθ
ACP
P−1AP =ρcos θ ρ sin θ
−ρsin θ ρ cos θ