计算：
(1)[(1+i)7/(1-i)]+[(1一i)7/(1+i)]-[(3—4i)(2+2i)3/(4+3i)]；
(2)[一(√3/2)一(1/2)i]12+[(2+2i)/(1-√3i)]8．
(1)原式=[(1+i)2]3•[(1+i)/(1-i)]+[(1-i)2]3•[(1-i)/(1+i)]-[8(3-4i)(1+i)2(1+i)]/[(3-4i)i] =(2i)3•i+(-2i)3•(-i)-[8•2i(1+i)/i] =8+8-16-16i=-16i (2)[-(√3/2)-(1/2)i]12+[(2+2i1)/(1-√3i)]8 =i12•[-(1/2)+(√3/2)i]12+[(1+i)/(1/2)-(√3/2)i]8 ={[-(1/2)+(√3/2)i]3}4+[(1+i)2]4[(1/2)-(√3/2)i]/[(1/2)-{(√3/2)i]3}3 =1-(2i)4[(1/2)-(√3/4)i] =1-8+8√3i=-7+8√3i
