若(x^2+2xy+y^2+x^2y^2)/(x+y+xyi)=(27-8i)/(3+2i)(x,y是实数),则x+yi

先将(27-8i)/(3+2i)化简,分子分母同时乘以3-2i得(27-8i)(3-2i)/(3^2+2^2)
整理后得5-6i
(x^2+2xy+y^2+x^2y^2)/(x+y+xyi)
=((x+y)^2+x^2y^2)/(x+y+xyi)
(分子分母同时乘以(x+y-xyi))并整理得
(x+y-xyi)
=(27-8i)/(3+2i)=5-6i
于是x+y=5,xy=6
x+yi的穆9就是x^2+y^2=(x+y)^2-2xy=5^2-2*6=13

(x^2+2xy+y^2+x^2y^2)(3+2i)=(x+y+xyi)*(27-8i)
(x^2+2xy+y^2+x^2y^2)*3=(x+y)*27+8xy
(x^2+2xy+y^2+x^2y^2)*2=(x+y)*(-8)+27xy
3(x+y)^2+3(xy)^2=27(x+y)+8(xy)
2(x+y)^2+2(xy)^2=-8(x+y)+27(xy)