I am trying to find the Laurent series for the function $$f(z)=\frac{2}{(z+1)^2}-\frac{5}{z-5},$$ in powers of $(z-1)$ that converges when $z=4$.
The series will converge when $z=4$ is in the region $2<|z-1|<4$. Hence, \begin{align} f(z)&=-2\frac{d}{dz}\left(\frac{1}{z+1}\right)+5\left(\frac{1}{5-z}\right) \\ &=-2\frac{d}{dz}\left(\frac{1}{(z-1)}\frac{1}{1+\frac{2}{(z-1)}}\right)+\frac{5}{4}\left(\frac{1}{1-\frac{(z-1)}{4}}\right) \\ &=-2\frac{d}{dz}\left(\frac{1}{(z-1)}\sum_{n=0}^{\infty}\frac{(-1)^n2^n}{(z-1)^n}\right)+\frac{5}{4}\sum_{n=0}^{\infty}\frac{(z-1)^n}{4^n} \\ &=\sum_{n=0}^{\infty}\frac{(-1)^n2^{n+1}(n+1)(z-1)^n}{(z-1)^{2(n+1)}}+5\sum_{n=0}^{\infty}\frac{(z-1)^n}{4^{n+1}}. \end{align} Is this solution correct?