问一道关于复数乘除运算的题x²-x+1y=f（x）=则f(1-i)为(x²+x+1我汗,怎么搞成了这种效果啊?我还是重发吧:x²-x+1y=f（x）=则f(1-i)为(x²+x+1

Question

问一道关于复数乘除运算的题
x²-x+1
y=f（x）= __________ 则f(1-i)为 (
x²+x+1
我汗,怎么搞成了这种效果啊?我还是重发吧:
x²-x+1
y=f（x）= __________ 则f(1-i)为 (
x²+x+1

Answer 1

由f（x）=（x²-x+1）/(x²+x+1)f(1-i)=[(1-i)²-(1-i)+1]/(1-i)²+(1-i)+1]=(1-2i-1-1+i+1)/(1-2i-1+1-i+1)=-i/(2-3i)=-i(2+3i)/(2-3i)(2+3i)=(3-2i)/7.