Question

(2+2i/根号3-i)^7-(2-2i/1+根号3i)^7等于多少

Answer 1

(2+2i)/(3^.5-i) = (2+2i)(3^.5+i)/ [ 3+1] =0.5(1+i)(3^.5+i)

(2-2i)/(1+3^.5i) = (2-2i)(1-3^.5i)/[1+3] = 0.5(1-i)(1-3^.5i)=-0.5(1+i)(3^.5+i)

因此,原式 = 2*[0.5*(1+i)(3^.5+i)]^7 =1/64*(1+i)^7*(3^.5+i)^7

(1+i)^2=(1+i)*(1+i)=2i,因此 (1+i)7=(2i)^3*(1+i)=-8i*(1+i)=8-8i

(3^.5+i)^2=(2+2*3^.5*i),(3^.5+i)^4 = (4-12+8*3^.5*i)=-8+8*3^.5*i

(3^.5+i)^6=(-8+8*3^.5*i)(2+2*3^.5i)=-16-48=-64

(3^.5+i)^7=-64*3^.5-64i

原式 = 1/64*(8-8i)(-64*3^.5-64i)=-8*(1-i)(3^.5+i)=-8[3^.5+1+(-3^.5+1)i]

原式 = -8*3^.5-8 + (8*3^.5-8)i