Question

一道复数的数学竞赛小题.已知实数r＞1,若复平面上的点动点z满足,Z的绝对值为r,则动点

Answer 1

设w=x+yi,z=x'+y'i ,x,y,x',y'∈R

|z|^2=x'^2+y'^2=r^2>1

z+1/z=x'+y'i+1/(x'+y'i)

=x'+y'i+(x'-y'i)/r^2

=x'(1+1/r^2)+y'(1-1/r^2)i

∴x=x'(1+1/r^2)

y=y'(1-1/r^2)

∴x'=r^2*x/(r^2+1) ①

y'=r^2*y/(r^2-1) ②

①^2+②^2:

x'^2+y'^2=r^4/(r^2+1)^2*x^2+r^4/(r^2-1)^2*y^2=r^2

∴r^2/(r^2+1)^2*x^2+r^2/(r^2-1)^2*y^2=1

轨迹为椭圆