Question

求15题解答

Answer 1

15.
设x=a+bi，(b≠0)
(x+1)/(x-1)
=(a+bi+1)/(a+bi-1)
=(a+bi+1)(a-1-bi)/[(a-1)²+b²]
=[(a²+b²-1)-2bi]/[(a-1)²+b²]
(x+1)/(x-1)是纯虚数，则a²+b²=1
令a=cosθ，b=sinθ
z=x+1+2i
=a+bi+1+2i
=(a+1)+(b+2)i
=(cosθ+1)+(sinθ+2)i
令x=cosθ+1，y=sinθ+2
cosθ=x-1，sinθ=y-2
sin²θ+cos²θ=1
(x-1)²+(y-2)²=1
复数z对应复平面上点的集合是以(1，2)为圆心，1为半径的圆。