Question

证明：1/z²=∑(无穷,n=0)(n+1)(z+1)^n(|z+1|<1)

Answer 1

1/z²=1/(z+1-1)²
=1/[1-(z+1)]²
因1/(1-z)²=[1/(1-z)]'
=(1+z+z²+...)'
=1+2z+3z²+...
=∑(n=0→∞)(n+1)z^n，|z|<1
把z换成z+1即可

Answer 2

1/z=1/[i+(z-i)]=1/i×1/[1+(z-i)/i]=1/i×1/[1-(z-i)i]=-i×∑{n=0~∞}[(z-i)i]^n 1/z2=-(1/z)‘=-{-i×∑{n=0~∞}[(z-i)i]^n}’=i∑{n=1~∞}ni^n(z-i)^{n-1}=∑{n=1~∞}ni^{n+1}(z-i)^{n-1} 因此 1/z2(z-i)=1/(z-i)×1/z2=1/(z-i)×∑{n=1~∞}ni^{n+1}(z-i)^{n-1}=∑{n=1~∞}ni^{n+1}(z-i)^{n-2} 对z的要求是0