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I want to find the Laurent expansion of $$f(z)=\frac{z-1}{z^2}, \qquad |z-1|>1.$$ Since the annulus is described for $z-1$, I substitute $w=z-1$ to use a geometric series in terms of $w$: $$f(w)=\frac{w}{w^2+2w+1}=\frac{w}{(w+1)^2}$$ Usually I would take partial fractions if there are two linear factors in the denominator to split them up, but because these are the same linear factor the partial fractions are $\frac{1}{w+1}-\frac{1}{(w+1)^2}$ which does not help.

Instead, I seek to turn this into the form for a geometric series. I pull out all but one linear term and find $f(w)=\frac{w}{w+1}\cdot\frac{1}{w+1}$. Since $|w|>1$, I must pull out $\frac{1}{w}$: $$\frac{1}{w+1}\cdot\frac{1}{1+\frac{1}{w}}=\frac{1}{w+1}\cdot\sum\limits_{j=0}^\infty\left(\frac{-1}{w}\right)^j=\sum\limits_{j=0}^\infty\frac{(-1)^j}{w^j(w+1)}$$ Putting this back in terms of $z$: $$f(z)=\sum\limits_{j=0}^\infty\frac{(-1)^j}{(z-1)^jz}$$

Is this the correct Laurent expansion? Usually each value of $j$ in the sum would give a $\frac{1}{z^n}$ term and a $z^n$ term, ensuring the sum continues both to $-\infty$ and $\infty$, but this one does not.

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You have not correctly represented $f$. To expand $f$, first write

$$f(z) = \frac{z - 1}{[(z - 1) + 1]^2} = \frac{\frac{1}{z-1}}{\left(1 + \frac{1}{z-1}\right)^2}.$$

Note

$$\frac{x}{(1 - x)^2} = \sum_{n = 1}^\infty nx^n,\quad |x| < 1.\tag{*}$$

Since $|z - 1| > 1$, $|1/(z - 1)| < 1$. So we may set $x = -1/(z - 1)$ in $(*)$ to obtain

$$f(z) = -\sum_{n = 1}^\infty n\left(-\frac{1}{z-1}\right)^n = \sum_{n = 1}^\infty (-1)^{n-1}n(z - 1)^{-n}.$$

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Hint: Considering $f(z) = \frac{z - 1}{z^2} $.

$$\frac{1}{1 + \frac{1}{z-1}} = \sum_{n=0}^{\infty}(-1)^n \frac{1}{(z-1)^n}, \,\,\,\,|1/(z-1)| < 1 $$

and $$\frac{d}{dz} \Bigg(\frac{1}{1 +\frac{1}{z-1}}\Bigg) = \frac{1}{z^2}$$

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  • $\begingroup$ @AccioHogwarts Feel free to ask if you have any questions $\endgroup$ – Aaron Maroja Apr 5 '15 at 20:32

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