复数z=reiθ证明Re[ln(z-1)]=1/2[\1ln(1+r^2-2rcosθ)]
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![复数z=reiθ证明Re[ln(z-1)]=1/2[\1ln(1+r^2-2rcosθ)]](/uploads/image/z/7264717-61-7.jpg?t=%E5%A4%8D%E6%95%B0z%3Drei%CE%B8%E8%AF%81%E6%98%8ERe%5Bln%28z-1%29%5D%3D1%2F2%5B%5C1ln%281%2Br%5E2-2rcos%CE%B8%29%5D)
复数z=reiθ证明Re[ln(z-1)]=1/2[\1ln(1+r^2-2rcosθ)]
复数z=reiθ证明Re[ln(z-1)]=1/2[\1ln(1+r^2-2rcosθ)]
复数z=reiθ证明Re[ln(z-1)]=1/2[\1ln(1+r^2-2rcosθ)]
设z-1=pe^iα,则由条件得
pcosα=rcosθ-1,
psinα=rsinθ,
两式平方相加得p^2=1+r^2-2rcosθ,
ln(z-1)=lnp+iα,∴实部就是lnp,即为
1/2[\1ln(1+r^2-2rcosθ)]