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)]

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)]