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)
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![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)](/uploads/image/z/10438250-50-0.jpg?t=%E5%A4%8D%E6%95%B0%CF%89%3D%282i%2Bz%29%2F%281%2Bz%29%2Cz%3Dx%2Biy%281%29+%E6%B1%82%E8%AF%81%2C%CF%89+%3D+%5B+%28x%26%23178%3B%2Bx%2By%26%23178%3B%2B2y%29+%2B+%282x%2By%2B2%29+i+%5D+%2F+%5B+%28x%2B1%29%26%23178%3B+%2B+y%26%23178%3B+%5D%282%29+%E5%B7%B2%E7%9F%A5+arg%28z%29%3Darg%28%CF%89%29%3D%CF%80%2F4%2C%E6%B1%82%7Cz%7C+%28z%E7%9A%84%E6%A8%A1%29)
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)
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)
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)
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