(-√2+√2i)^5(√3-i)^2 2^8.√2(i-1)^5(2-2√3i)
z= --------------------------- = ------------------------------- (分母通分)
(1/2-1/2i)^8 (i-1)^8
2^8.√2(2-2√3i) 2^8√6i-2^8√2
= --------------------------(约分) = ---------------------------,(最后用除法公式)剩下的交给你啦
(i-1)^3 （i+1）你看我做的有点问题：[2(cos3TT/4+isin3TT/4)]^5×[2(cos(-TT/6)+isin(-TT/6]^2----------------------------------------------------------------[√2/2[cos(-TT/4)+isin(-TT/4)]^8结果算出来是2^11cos(5TT+12/5TT),这个cos(5TT+12/5TT），可以帮我解答一下吗？你这是。。。理科题？cos（π+α）= -cosαcos（2kπ+α）= cosα cos(5TT+12/5TT）=cos(37TT/5)=cos(3*2TT+TT+2TT/5)=-cos(2TT/5)= -cos(72)查表 -cos(72) = 0.96725058827388谢了啊～～