如果是 x/(1+i)=3/(2-i)+y/(1-i),可化为
x(1-i)/2=3(2+i)/5 +y(1+i)/2
x/2 - i/2 = 6/5+y/2 +(3/5+y/2)i
实数和虚数各相等,有
x/2=6/5+y/2
-1/2=3/5+y/2
解方程 y=-2.2,x=0.2
如果是 x/(1+i)=3/(2-i)+y/(1-i),可化为
x(1-i)/2=3(2+i)/5 +y(1+i)/2
x/2 - i/2 = 6/5+y/2 +(3/5+y/2)i
实数和虚数各相等,有
x/2=6/5+y/2
-1/2=3/5+y/2
解方程 y=-2.2,x=0.2