(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