I am trying to find the Laurent series of, $$f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4},$$ in powers of $z-2$ that converges at $z=1$.
My attempt:
I think our radius for convergence is $|z-2|<2$. Now, \begin{align} f(z)&=\frac{2}{(z+2)^2}-\frac{5}{z-4} \\ &=-2\frac{d}{dz}\left(\frac{1}{z+2}\right)+\frac{5}{4-z} \\ &=-2\frac{d}{dz}\left(\frac{1}{4+(z-2)}\right)+\frac{5}{2-(z-2)} \\ &=-2\frac{d}{dz}\left(\frac{1}{4+(z-2)}\right)+\frac{5}{2}\sum_{n=0}^{\infty} \frac{(z-2)^n}{2^n} \end{align} I am stuck at this point. $-2\frac{d}{dz}\left(\frac{1}{4+(z-2)}\right)$ appears not to be convergent in the region $|z-2|<2$. It certainly converges in the region $2<|z-2|<4$, but this is not where $z=1$ is located. I am very confused.
further attempt
Using the information from a related post (Finding a series for $f(z)=\frac{2}{(z+2)^2}$), I believe the correct Laurent series is $$5\sum_{n=0}^{\infty}\frac{(z-2)^n}{2^{n+1}}-2\sum_{n=1}^{\infty} (-1)^n\frac{n(z-2)^{n-1}}{4^{n+1}}.$$ Is this correct? The answer that I have found in the book states the correct series is $$\sum_{n=0}^{\infty} \left((-1)^n\frac{2(n+1)}{4^{n+3}}+\frac{5}{2^{n+1}}\right)(z-2)^n.$$
Note that\begin{align}\frac1{4+(z-2)}&=\frac14\frac1{1+\frac{z-2}4}\\&=\frac14\left(1-\frac{z-2}4+\frac{(z-2)^2}{4^2}-\frac{(z-2)^3}{4^3}+\cdots\right)\\&=\frac14-\frac{z-2}{4^2}+\frac{(z-2)^2}{4^3}-\frac{(z-2)^3}{4^4}+\cdots\end{align}and that this series converges in the region $\lvert z-2\rvert<4$.