[（2+2i)/(√3-i)]^7=[(1+i)(√3+i)/2]^7
=(√2)^7*[(√2/2+√2/2i)(√3/2+1/2i)]^7
=(√2)^7*[(cosπ/4+isinπ/4)(cosπ/6+isinπ/6)]^7
=(√2)^7*(cos5π/12+isin5π/12)^7
=(√2)^7*(cos35π/12+isin35π/12)
=(√2)^7*(-cosπ/12+isinπ/12)
同理[(2-2i)/(1+√3i)^7=(√2)^7*(cosπ/12-isinπ/12)
[（2+2i)/(√3-i)]^7-[(2-2i)/(1+√3i)^7
=(√2)^7*（-2cosπ/12+2isinπ/12）
cosπ/12,sinπ/12 自己用半角公式算.OK.