# Laurent Series of $f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}$ that converges at $z=1$ in powers of $z-2$

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 $$\lvert z-2\rvert<4$$.
• Yes, I have noted this. But does this converge in the region where $z=1$ (which is what the question asks)?
• @Bell Since $\lvert1-2\rvert=1<4$, yes. Sep 27, 2018 at 8:55
• Okay. I was considering the regions $|z-2|<2, \ 2<|z-2|<4$ and $|z-2|>4$. Because $z=1$ lies within $|z-2|<2$, I was thinking that the series could only converge for $|z-2|<2$. This is incorrect?
• @Bell The domain of $f$ is $\mathbb{C}\setminus\{-2,4\}$. So, the Taylor series of $f$ centered at $2$ converges in the larges open disk centered at $2$ and contained in the domain, which is the disk centered at $2$ and with radius $2$. And $1$ belongs to that disk. Sep 27, 2018 at 9:35
• @Bell Actually, $\lvert1.5-2\rvert=0.5$, but yes. Sep 27, 2018 at 11:05