将函数f(z)=1/(1+z^2),0<|z-i|<2及|z-i|>2展开为罗朗级数

1) 0<|z-i|<2时，即|-(z-i)/2i|<1时，
f(z)=[(z-i)(z+i)]^(-1)
=-{4*[(z-i+2i)/2i][(z-i)/2i]}^(-1)
=(1/4)*{[-(z-i)/2i]^(-1)}*1/{1-[-(z-i)/2i]}
=(1/4)*{[-(z-i)/2i]^(-1)}*∑[-(z-i)/2i]^n
=∑[(z-i)^(n-1)]/[(-2i)^(n+1)]

2) |z-i|>2时 即|-2i/(z-i)|<1时，
f(z)=[(z-i)(z-i+2i)]^(-1)
=1{[-2i/(z-i)]^(-2)}*1/{1-[-2i/(z-i)]}
=(-1/4)*{[-2i/(z-i)]^2}*∑[-2i/(z-i)]^n
=(-1/4)* ∑[(-2i)^(n+2)]/[(z-i）^(n+2)]
= ∑ [(-2i)^n]*[(z-i)^(-n-2)]

注：n=0到∞