设ω=2+mi,则
2+mi=[(a+i)/(1+i)]^2
=[(a-ai+i+1)/2]^2
=[(a+1)^2+2(a+1)(1-a)i-(1-a)^2]/4
=[4a+2(1-a^2)i]/4
=a+[(1-a^2)/2]i.
∴a=2,虚部
m=(1-a^2)/2=-3/2.
设ω=2+mi,则
2+mi=[(a+i)/(1+i)]^2
=[(a-ai+i+1)/2]^2
=[(a+1)^2+2(a+1)(1-a)i-(1-a)^2]/4
=[4a+2(1-a^2)i]/4
=a+[(1-a^2)/2]i.
∴a=2,虚部
m=(1-a^2)/2=-3/2.