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复数的三角形式运算:Z=[(-√2+√2)的5次方(√3-i)2次方]/(1/2-1/2i)8次方,急求解答,谢谢~~
人气:198 ℃ 时间:2019-11-06 11:53:55
解答
(-√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~~
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