Question

设m为实数,且复数（1+i）/（2+mi） +1/2的实部和虚部相等,求m

Answer 1

(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