a complex number question

(8 + z^3) / (8 - z^3) = i
8 + z^3 = 8i - z^3i
z^3(1 + i) = 8i - 8
z^3 = 8(-1 + i)/(1 + i)
z^3 = 8(-1 + i)(1 - i)/[(1 + i)(1 - i)]
z^3 = 8(-1 + i + i + 1)/2 = 8i
z^3 = 8(cosπ/2 + i sinπ/2)
z = 2(cosπ/6 + i sinπ/6), 2(cos5π/6 + i sin5π/6), 2(cos-π/2 + i sin-π/2)

2009-11-07 23:23:12 補充：
(1 - i)^2 = 1 - 2i - 1 = 2i
(2i)^(1/3) has 3 possibilities for complex number, not just one!!!

Let x=(z/2)^3, then the equation becomes (x+1)/(1-x)=i
which has x=(i-1)/(1+i)=-(1-i)^2/2

So z=2x^1/3=-2^(2/3) (1-i)^(2/3)
Since (1-i)^2/3=√2^(2/3)[cos(-π/6)+isin(-π/6)]
z=-2^(2/3) (√2)^(2/3)[cos(-π/6)+isin(-π/6)]
=-2[cos(-π/6)+isin(-π/6)]
=2[cos(5π/6)+isin(5π/6)]