Z=1/2+(√3/2)i, 那么Z+Z^2+.....+Z^11=? 怎样做哦?
1个回答
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z=1/2+根号3/2i
z^2=1/4-3/4+根号3/2i=-1/2+根号3/2i
z^3=z*z^2=-3/4-1/4=-1
z+z^2+z^3+...+z^11
=z+z^2+z^3+z^3(z+z^2+z^3)+z^6(z+z^2+z^3)+z^9*(z+z^2)
=根号3i-(根号3i-1)+(根号3i-1)-(根号3i)
=0
z^2=1/4-3/4+根号3/2i=-1/2+根号3/2i
z^3=z*z^2=-3/4-1/4=-1
z+z^2+z^3+...+z^11
=z+z^2+z^3+z^3(z+z^2+z^3)+z^6(z+z^2+z^3)+z^9*(z+z^2)
=根号3i-(根号3i-1)+(根号3i-1)-(根号3i)
=0
追问
答案是-1,阁下是不是算错了?
追答
算错了,应该是:
=()z+z^2+z^3)+z^3(z+z^2+z^3)+z^6(z+z^2+z^3)+z^9*(z+z^2)
=(根号3i-1)-(根号3i-1)+(根号3i-1)-(根号3i)
=-1
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