计算(2+2i/根号3-i)^7-(2-2i/1+根号3i)^7

(2+2i)/(√3-i) = (2+2i)(√3+i)/ (3+1) =0.5(1+i)(√3+i)
(2-2i)/(1+√3i) = (2-2i)(1-√3i)/(1+3) = 0.5(1-i)(1-√3i)=-0.5(1+i)(√3+i)
因此,
原式
=[0.5(1+i)(√3+i)]^7-[-0.5(1+i)(√3+i) ]^7
= 2*[0.5*(1+i)(√3+i)]^7
=1/64*(1+i)^7*(√3+i)^7
∵(1+i)^2=(1+i)(1+i)=2i
∴(1+i)^7=(2i)^3*(1+i)=-8i*(1+i)=8-8i
∵(√3+i)^2=2+2√3i,(√3+i)^4 = (4-12+8√3i)=-8+8√3i
(√3+i)^6=(-8+8√3i)(2+2√3i)=-16-48=-64
(√3+i)^7=-64√3-64i
∴原式
= 1/64*(8-8i)(-64√3-64i)
=-8(1-i)(√3+i)
=-8[√3+1+(-√3+1)i]
= -8√3-8 + (8√3-8)i