帮忙算一道复数减法题.(-1+根号3i)^3/(1+i)^6-(-2+2i)/(1+2i)=我算了好多遍都跟答案不一样.想让大家算一下来对对.

Question

帮忙算一道复数减法题.
(-1+根号3 i)^3/(1+i)^6 - (-2+2i)/(1+2i) =
我算了好多遍都跟答案不一样.想让大家算一下来对对.

Answer 1

(1+i)^2=1+2i-1=2i
所以(1+i)^6=(2i)^3=8*(-i)=-8i
-1+i√3=2(-1/2+i√3/2)
=2(cos2π/3+isin2π/3)
所以(-1+i√3)^3=2^3*[cos(3*2π/3)+isin(3*2π/3)]
=8*(cos2π+isin2π)
=8
所以(-1+根号3 i)^3/(1+i)^6=8/(-8i)=-1/i=-i/i^2=i
(-2+2i)/(1+2i)
=(-2+2i)(1-2i)/(1+2i)(1-2i)
=(-2+4i+2i+4)/(1+4)
=(2+6i)/5
所以原式=i+(2+6i)/5=(2+11i)/5