Odpowiedzi

2010-02-11T19:12:49+01:00
Z⁴ - iz² + 2 = 0
niech t=z²
t²-it+2 = 0
Δ=-1-8 = -9, √Δ=√-9 = 3i ∨ √Δ= -3i
t = (i-3i)/2 = -i ∨ t = (i+3i)/2 = 2i
z²=-i ∨ z²=2i
z=√-i = √1(cos((3/2 π+2kπ)/2) + i*sin((3/2 π+2kπ)/2)), k∈{0, 1}
z = cos(3π/4)+i*sin(3π/4) = -(√2/2)+i(√2/2) ∨
z = cos(7π/4)+i*sin(7π/4) = (√2/2)-i(√2/2)

z =√2i = √4(cos((π/2 +2kπ)/2) + i*sin((π/2 +2kπ)/2)), k∈{0, 1}
z = 2(cos(π/4)+i*sin(π/4)) = 2((√2/2) + i(√2/2)) = √2 + i√2 ∨
z = 2(cos(5π/4)+i*sin(5π/4)) = 2(-(√2/2) - i(√2/2)) = -√2 - i√2

odp: z∈{-(√2/2)+i(√2/2); (√2/2)-i(√2/2); √2 + i√2; -√2 - i√2}

2010-02-12T01:07:36+01:00
Z⁴ - iz² + 2 = 0
podstawmy:
z²=t
t²-jt+2=0
Δ=(-j)²-4*2=-1-8=-9=9j²
√Δ=j√9=3j
t₁=(j-3j)/2=-j
t₂=2j
2j=z²
2j=(a+bj)²
2j=a²+2abj-b²
2j=2abj
1=ab
a=1/b