已知|z|=1,k是任意的复数,求证:|(z-k)/(k*z'-1)|=1(z'表示z的共轭复数)

Question

已知|z|=1,k是任意的复数,求证:|(z-k)/(k*z'-1)|=1(z'表示z的共轭复数)

Answer 1

设z=x+yi z'=x-yi |z|=1 x^2+y^2=1
z-k=(x-k)+yi
k*z'-1=(kx-1)-kyi
|z-k|=√[(x-k)^2+y^2]=√[x^2+y^2-2kx+k^2]=√(k^2-2kx+1)
|k*z'-1|=√[(kx-1)^2^+k^2y^2]=√[k^2x^2+k^2y^2-2kx+1]=√(k^2-2kx+1)
|(z-k)/(k*z'-1)|
=|z-k|/||k*z'-1
=1