若w=-2/(1+根号3i),则1+w+w^2等于

w=-2/(1+根号3i)
=-2(1-根号3i)/[(1+根号3i)(1-根号3i)]
=-2(1-根号3i)/[1²-(根号3i)²]
=-2(1-根号3i)/[1-(-3)]
=-2(1-根号3i)/[1+3]
=-2(1-根号3i)/4
=-(1-根号3i)/2
=(根号3i-1)/2
w^2=(根号3i-1)²/4
=[(根号3i)²-2根号3i+1]/4
=[(-3)-2根号3i+1]/4
=[-2-2根号3i]/4
=-(1+根号3i)/2
1+w+w^2
=1+(根号3i-1)/2+[-(1+根号3i)/2]
=1+(根号3i-1-1-根号3i)/2
=1+(-1-1)/2
=1-1
=0

还没学，对不起

w=-2/(1+根号3i)=(根号3i-1)/2
w^2=-(1+根号3i)/2
1+w+w^2等于=0