已知复数z满足lzl=1,z(z^6+1)=1,求z

Question

已知复数z满足lzl=1,z(z^6+1)=1,求z

Answer 1

|z|=1,得 |z^6|=1
∵z(z^6+1)=1
∴|z(z^6+1)| = 1,|z| * |z^6+1| =1
∴ |z^6+1|=1
假设 z^6=a+bi
有 a^2+b^2=1 (a+1)^2+b^2 = 1
两者相减,得 (a+1)^2-a^2=0
2a+1=0,得 a=-0.5 ,b=±√0.75
因此 z^6 = -0.5±√0.75i = cos(2/3*π)±sin(2/3*π)i
z = cos[(2nπ+2/3*π)/6]±sin[(2nπ+2/3*π)/6]i,n=0,1,2,3,4,5
但上述z解没有一个满足 z(z^6+1)=1,因此无解.