(1+i）/（2+mi） +1/2
=(1+i)(2-mi)/(2+mi)(2-mi)+1/2
=(2+m+2i-mi)/(2+m^2)+1/2
=(2+m)/(2+m^2)+(2-m)i/(2+m^2)=1/2
所以(2+m)/(2+m^2)+1/2=(2-m)/(2+m^2)
2(2+m)+(2+m^2)=2(2-m)
m^2+4m+2=0
m=-2±√2