复数ω=(2i+z)/(1+z),z=x+iy
问题描述:
复数ω=(2i+z)/(1+z),z=x+iy
(1) 求证,ω = [ (x²+x+y²+2y) + (2x+y+2) i ] / [ (x+1)² + y² ]
(2) 已知 arg(z)=arg(ω)=π/4,求|z| (z的模)
答
复数ω=(2i+z)/(1+z),z=x+iy
(1) 求证,ω = [ (x²+x+y²+2y) + (2x+y+2) i ] / [ (x+1)² + y² ]
(2) 已知 arg(z)=arg(ω)=π/4,求|z| (z的模)