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I have to calculate the Laurent series expansion of $$f(z) = \frac {2z−2}{(z+1)(z−2)}$$ in $1 < |z| < 2$ and $|z| > 3$.

For first annulus, I know I must manipulate the given expression to contain terms $1/z$ and $z/2$ so that some expansion is valid for $|\frac1{z}|<1$ and $|\frac z{2}|<1$, so I try doing that.

Decomposing into partial fractions,
$$f(z) = \frac 4{3} (\frac 1{z+1}) + \frac 2{3}(\frac 1{z-2})$$
$$= \frac4{3z} \frac1{1-\frac {(-1)}{z}} - \frac1{3} \frac 1{1 - \frac z{2}}$$

So can I now expand the $2$ terms by a G.P. for $|\frac1{z}|<1$ and $|\frac z{2}|<1$ to get the Laurent series?

For second case of $|\frac 3{z}|<1 $, how should I proceed?

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The function \begin{align*} f(z)&=\frac{2z-2}{(z+1)(z+2)}\\ &= \frac{4}{3}\left( \frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\ \end{align*} is to expand around the center $z=0$ in $1<|z|<2$ and $|z|>3$.

Since there are simple poles at $z=-1$ and $z=2$ we have to distinguish three regions of convergence \begin{align*} D_1:&\quad 0\leq |z|<1\\ D_2:&\quad 1<|z|<2\\ D_3:&\quad |z|>2 \end{align*}

  • The first region $D_1$ is a disc with center $z=0$, radius $1$ and the pole at $z=-1$ at the boundary of the disc. It admits for both fractions a representation as power series.

  • The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. 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 $z=2$ a power series.

  • The region $D_3$ contains all points outside the disc with center $z=0$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.

Since we are interested in an expansion for $1<|z|<2$ we consider the expansion in $D_2$.

Expansion in $D_2$:

\begin{align*} f(z)&=\frac{4}{3}\left(\frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\ &=\frac{4}{3z}\left(\frac{1}{1+\frac{1}{z}}\right)-\frac{1}{3}\left(\frac{1}{1-\frac{1}{2}z}\right)\\ &=\frac{4}{3}\sum_{n=0}^\infty\frac{(-1)^{n}}{z^{n+1}}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{2^n}z^n\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{2^n}z^n\\ \end{align*}

Since we are interested in an expansion for $|z|>3$ we consider the expansion in $D_3$.

Expansion in $D_3$:

\begin{align*} f(z)&=\frac{4}{3}\left(\frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{2}{3z}\left(\frac{1}{1-\frac{2}{z}}\right)\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=0}^\infty\frac{2^{n+1}}{z^{n+1}}\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=1}^\infty\frac{2^n}{z^{n}}\\ &=\frac{1}{3}\sum_{n=1}^\infty\left(2^n-4(-1)^{n}\right)\frac{1}{z^{n}}\\ \end{align*}

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  • $\begingroup$ Thank you, I was trying to figure out what that 3 had to do with it $\endgroup$ – john doe Nov 13 '17 at 22:21
  • $\begingroup$ @johndoe: You're welcome! $\endgroup$ – Markus Scheuer Nov 13 '17 at 22:30
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The trick when solving Laurent Series is to use the following geometric serie: $$\frac{1}{1-w} = \sum _0^{\infty} (w)^n \text{ for } | w | < 1$$ and let $w = z-z_0$, where $z_0$ is the point of expansion. But since you sometimes would like to have a series which is valid outside instead of inside the circle |z| = 1 we can insert $\frac{1}{w} = w$ in the geometric series above $$\frac{1}{1-\frac{1}{w}} = \sum _0^{\infty} (\frac{1}{w})^n \text{ for } |{w}| > 1$$ this is great since this series is valid outside of the circle |z| = 1. Also the regions for the series can be alternated by chaning the formula a bit, let $a$ and $b$ be constants, then:

1)$$\qquad \frac{1}{1-a \cdot w} = \sum _0^{\infty} (a \cdot w)^n \text{ for } |{w}| > \frac{1}{|a|}$$ 2)$$\qquad \frac{1}{1-\frac{b}{w}} = \sum _0^{\infty} (\frac{b}{w})^n \text{ for } |{w}| > |b| $$ hence, we are able to create series for whatever region you would like in the complex plan. This is the principal idea behind solving these kind of problems and everything is explained in this video, with some examples which might help you solve your examples: https://www.youtube.com/watch?v=RC15R-ktnUI

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  • $\begingroup$ But in the above question, I can only make terms $1/z$ and $2/z$ which can give $|z| >1$ and $|z|>2$, does that mean that now since the Laurent expansion will be valid for $|z| >2$ (intersection of the 2 conditions), then it is automatically valid for $|z| > 3$ also, and so I am done? $\endgroup$ – john doe Nov 13 '17 at 20:18
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    $\begingroup$ I guess you are refering to the second part of the problem, i.e. calculate the Laurent Series in the domain |z|>3, then you are completly correct. You start by using partial fraction and then you calculate the geometric series for both terms and then add them to get the end result and as you are saying it is only in the intersection that the sum of the two series are valid (|z|>2) and therfore the sum most also be valid for |z|>3, since this region is included in |z|>2. I hope this clarifies. $\endgroup$ – James Blond Nov 13 '17 at 20:51

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