已知复数z = (1-i)^2+3(1+i) /2-i,若z^2+az+b=1-i,求实数a,b的值?/
问题描述：
已知复数z = (1-i)^2+3(1+i) /2-i,若z^2+az+b=1-i,求实数a,b的值?/
答
z=(1-i)²+3(1+i)/(2-i)=-2i+3(1+i)(2+i)/5=-2i+3(1+3i)/5=3/5-1/5 i若z²+az+b=1-i,则√[(3/5)²+(1/5)²]+3a/5+b-a/5 i=1-i即√10/5+3a/5+b-a/5 i=1-ia,b∈R,由复数相等定义得√10/5+3a/5+b=1 ①-...