计算4i/(1+i)^6-(2+i)/(1+2i)
问题描述：
计算4i/(1+i)^6-(2+i)/(1+2i)
答
(√3/2+1/2i)^6 =(√3/2+1/2i)^2 * (√3/2+1/2i)^2 *(√3/2+1/2i)^2 =[(√3/2+1/2i)^2]^3 =[3/4 +1/4 *i^2 +2*√3/4*i]^3 =(1/2+√3/2*i)^2 *(√3/2+1/2i) = ( 1/4 +3/4*i^2 + 2*√3/4*i)*(√3/2+1/2i) =(-1/2+√3/2*i)*(√3/2+1/2i) =(√3/2*i-1/2)*(√3/2+1/2i) =3/4i+√3/4*i^2-√3/4-1/4i =1/2i-√3/2 =cosπ+isinπ =-1
答
(1+i)^2=1+2i-1=2i
所以(1+i)^6=(2i)^3=8*i^3=-8i
所以原式=4i/(-8i)-(2+i)(1-2i)/(1+2i)(1-2i)
=-1/2-(2-4i+i+2)/(1+4)
=-1/2-(4-3i)/5
=-13/10+3i/5