2 ways to find a Laurent series? A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.




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(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $\frac1{z+1} = \frac1{2+z-1}$?

I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $\emptyset$!)
For (Q1)


*

*Take out $z-1$
$$\frac1{z+1} = \frac1{2+z-1} = \frac1{(z-1)(\frac{2}{z-1}+1)} \to \ \text{a Laurent series for} \ |z-1| > 2$$

*Take out $2$
$$\frac1{z+1} = \frac1{2+z-1} = \frac1{(2)(1+\frac{z-1}{2})} \to \ \text{a Laurent series for} \ |z-1| < 2$$
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(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?

For (Q2)
$$\frac{z-2}{z+1} = 1 + \frac{-3}{z+1} \ \text{only}?$$


(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?

Rewrite $\frac1{(z^2-4)(z-2)} = \frac1{(z-2)^2(z+2)}$. Rewrite $\frac1{z+2} = \frac1{z-2+4}$:


*

*Take out $z-2$
$$\frac1{z+2} = \frac1{z-2+4} = \frac1{(z-2)(\frac{4}{z-2}+1)} \to \ \text{a Laurent series for} \ |z-2| > 4$$


--> This is not the book's answer, and it doesn't seem to have a $c_{-1}$. It looks like the integral will be 0.

(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] \subsetneq \{|z-2| > 4\}$, so it doesn't apply or something.



*Take out $4$


$$\frac1{z+2} = \frac1{z-2+4} = \frac1{(4)(1+\frac{z-2}{4})} \to \ \text{a Laurent series for} \ |z-2| < 4$$
--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $\frac{- \pi i}{8}$

(Q3.2) Book says that for $\frac1{(4)(1+\frac{z-2}{4})}$, the region of convergence is $\color{red}{0 <} |z-2| < 4$. Why $0 <$?

 A: Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.

Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
  \begin{align*}
D_1:&\quad 0<  |z-1|<2\\
D_2:&\quad |z-1|>2
\end{align*}
  
  
*
  
*The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc.  It admits for the fraction with pole  at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
  
*The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.

We have in $D_1$ the representation
\begin{align*}
\frac{1}{(z-1)(z+1)}&=\frac{1}{z-1}\cdot\frac{1}{2+(z-1)}\\
&=\frac{1}{z-1}\cdot\frac{1}{2\left(1+\frac{z-1}{2}\right)}\\
&=\frac{1}{2}\cdot\frac{1}{z-1}\sum_{j=0}^\infty(-1)^j\left(\frac{z-1}{2}\right)^j\\
&=-\sum_{j=-1}^\infty\left(-\frac{1}{2}\right)^j(z-1)^j
\end{align*}
and in $D_2$ we have the representation
\begin{align*}
\frac{1}{(z-1)(z+1)}&=\frac{1}{z-1}\cdot\frac{1}{2+(z-1)}\\
&=\frac{1}{(z-1)^2}\cdot\frac{1}{1+\frac{2}{z-1}}\\
&=\frac{1}{(z-1)^2}\sum_{j=0}^\infty(-1)^j\left(\frac{2}{z-1}\right)^j\\
&=\frac{1}{4}\sum_{j=-\infty}^{-2}\left(-\frac{1}{2}\right)^j(z-1)^j
\end{align*}

Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
  \begin{align*}
\frac{1}{(z^2-4)(z-2)}&=\frac{1}{(z-2)^2}\cdot\frac{1}{z+2}\\
&=\frac{1}{(z-2)^3}\cdot\frac{1}{1+\frac{4}{z-2}}\\
&=\frac{1}{(z-2)^3}\sum_{j=0}^\infty(-4)^j\frac{1}{(z-2)^j}\tag{1}\\
&=\cdots
\end{align*}
Ad Q.3.2: The function under consideration
  \begin{align*}
\frac{1}{(z^2-4)(z-2)}
\end{align*}
  has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.

Hint: This answer with some more detailed considerations might be helpful.
