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I want to find the Laurent series for $\frac{z}{(z-1)(z-2)}$ in the region $1 < |z| < 2$. This implies that $\frac{1}{|z|} < 1$, so noticing that $(z-1) = z(1 - \frac 1 z)$ I can rewrite the desired function as $$\frac{z}{z(1- \frac 1 z)(z-2)} = \frac{1}{z-2} \cdot \frac{1}{1 - \frac 1 z}.$$

Now using the definition of the geometric series I rewrite it as $$\sum_{k=0}^{\infty} \frac{1}{z^k(z-2)}$$ Is this the right Laurent series, and if not, where did I go wrong? Please note I am trying to understand where I made a mistake, not simply finding any solution.

I've read a similar question at Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ and one answer uses the fact that $\frac{|z|}2 < 1$, but I am not sure if my method is also valid.

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    $\begingroup$ This is not a Laurent series. $\endgroup$ Dec 12, 2022 at 0:34
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    $\begingroup$ Hint: First decompose the rational function using partial fractions. $\endgroup$ Dec 12, 2022 at 0:51
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    $\begingroup$ I don’t see a result yet. The $(z-2)$ factor in the denominator of each term in the series means that this expression is not a Laurent series. $\endgroup$ Dec 12, 2022 at 0:55
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    $\begingroup$ The series needs to look like $$\sum_{n \in \mathbb{Z}} a_n z^n$$ for some coefficients $\{\dots, a_{-2}, a_{-1}, a_0, a_1, a_2, \dots\}$. $\endgroup$ Dec 12, 2022 at 0:57
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    $\begingroup$ Sometimes this expression is “folded” to allow for it to be indexed just in one direction, with two terms per index: $$a_0 + \sum_{n=1}^{\infty} \bigl( a_n z^n + a_{-n} z^{-n} \bigr), $$ but this is just a difference of notation. $\endgroup$ Dec 12, 2022 at 1:00

3 Answers 3

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Since $\dfrac 1{z-2}=-\dfrac 12\dfrac 1{1-\frac z2}=-\dfrac 12\sum \left(\dfrac z2\right)^n$ converges for $\lvert \dfrac z2\rvert \lt1$, or $\lvert z\rvert \lt 2$, we can do this.

Because $$\dfrac z{(z-1)(z-2)}= z\cdot \left(\dfrac 1{z-2}-\dfrac 1{z-1}\right),$$ we get

$$ -\sum z^{-n}-\dfrac z2\cdot \sum\left(\dfrac z2\right)^n=-\sum_{n\ge0}z^{-n}-\left(\dfrac z2\right)^{n+1},$$ which is in powers of $z$ on the annulus $1\lt\lvert z\rvert \lt2$.

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For $1 < |z| < 2$, \begin{align*} \frac{1}{(z-1)(z-2)} &=-\frac{1}{z-1} + \frac{1}{z-2}\\ &=-\frac{1}{z}\left[\frac{1}{1-(1/z)}\right]-\frac{1}{2}\left[\frac{1}{1-(z/2)}\right]\\ &=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{z^n}-\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\\ \end{align*} \begin{align*} \frac{z}{(z-1)(z-2)} &=-\sum_{n=0}^{\infty} \frac{1}{z^n}-\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^{n+1}\\ &=-1-\sum_{n=1}^{\infty} \frac{1}{z^n}-\sum_{n=1}^{\infty} \left(\frac{z}{2}\right)^{n}\\ &=-\sum_{n=1}^{\infty} \frac{1}{z^n}-\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^{n}\\ \end{align*}

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Is this the right Laurent series, and if not, where did I go wrong?

The condition $1<|z|<2$ itself is sufficient to tell that each term of the Laurent series expansion must contain either negative or non-negative power of $z$ because the annular region $r<|z-z_0|<R$ between two concentric circles has center $z_0$ ; which is also the center of series. In your case, it is $0$. The correct expansion will be obtained by using partial fractions as displayed in other answers.

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