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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 $$\sum_{n=0}^{\infty} \left((-1)^n\frac{2(n+1)}{4^{n+3}}+\frac{5}{2^{n+1}}\right)(z-2)^n.$$

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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 $\lvert z-2\rvert<4$.

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  • $\begingroup$ Yes, I have noted this. But does this converge in the region where $z=1$ (which is what the question asks)? $\endgroup$
    – user557493
    Sep 27, 2018 at 8:53
  • $\begingroup$ @Bell Since $\lvert1-2\rvert=1<4$, yes. $\endgroup$ Sep 27, 2018 at 8:55
  • $\begingroup$ 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? $\endgroup$
    – user557493
    Sep 27, 2018 at 9:05
  • $\begingroup$ @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. $\endgroup$ Sep 27, 2018 at 9:35
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    $\begingroup$ @Bell Actually, $\lvert1.5-2\rvert=0.5$, but yes. $\endgroup$ Sep 27, 2018 at 11:05

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