若复数z=(1+(根号3)i)/2,且(x-z)^4=a(0)x^4+a(1)x^3+a(2)x^2+a(3)x+a(4),则a2=

x/(24-y)=20/(24-8)
xy=(4/5)(30x-x²)
xy=-(4/5)(x-15)²+180
x=15,y=12
xy=180

z=(1+(根号3)i)/2=cos60+i*sin60,
根据二项式定理,可得
T(r+1)=C(n,r)*a^(n-r)*b^r,
a(2)*x^2,则有4-r=2,r=2.
即有,a2=z^2*C(4,2)=(cos120+i*sin120)*6=-3+3√3*i,
则a2=-3+3√3*i,