if y = 1 i, then y3 − 3y2
y^3-3y^2(1+i)^3-3(1+i)^2remember(1+i)^3=(1+i)(1+i)^2(1+i)(1+i)^2-3(1+i)^2factor out (1+i)^2(1+i)^2((1+i)-3)(1+i)^2(1+i-3)(1+i)^2(i-2)(1+i)(1+i)(i-2)(1+2i-1)(i-2)(2i)(i-2)-2-4i
y = 1i = iy³ - 3y² (i)³ - 3(i)² i² = -1, i³ = i²(i) = -ii³ - 3i² -i - 3(-1)-i + 3
