I always get confused with these problems and factoring this gives me many problems $$f(z)=\frac{1}{(z-1)(z+i)^3}$$ find the Lauren series in $K_{1}(0,\sqrt{2})=\lbrace z \mid 0<\vert z-1\vert<\sqrt{2}\rbrace$
My exercise only need a factorization as follows with the next example: $g(z)=\frac{1}{(z)(z-i)^2}$ in $K_{i}(0,1)$ \begin{equation} g(z)=\frac{1}{(z)(z-i)^2}=\frac{1}{(z-i)^2}\frac{1}{z-i+i}=\frac{1}{(z-i)^2}\frac{-i}{1-i(z-i)}=\frac{1}{(z-i)^2}\sum_{k=0}^{\infty}i^{k-1}(z-i)^k=\sum_{k=-2}^{\infty}i^{k-1}(z-i)^k \end{equation}
Observe that: $$\frac{1}{(z-1)(z+i)^3} = \frac{1}{(z-1)((z-1)+(i+1))^3} = \frac{1}{(z-1)(i+1)^3\left(1+\frac{z-1}{i+1}\right)^3}.$$