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For homework, I've been asked to obtain the Laurent series expansions for the following function: $$f(z) = \frac{1}{z^2(1-z)}$$

The question says to use Laurent's Theorem (not geometric series), which has been given as
$$\tag{1}f(z) = \sum\limits_{n= -\infty}^\infty A_nz^n $$
where
$$\tag{2}A_n = \frac{1}{2\pi i} \oint\limits_\gamma \frac{f(z)}{z^{n+1}} dz$$


My initial approach to the problem was to use Cauchy’s residue theorem to evaluate the integral in $(2)$, this theorem states:
$$\oint\limits_\gamma f(z) dz = 2 \pi i \sum\limits_{a_k \in A} \mathrm{Res}_{z = a_k} f(z) $$

So in combination with $(2)$ I have,
$$A_n = \sum\limits_{a_k \in A} \mathrm{Res}_{z = a_k} \frac{f(z)}{z^{n+1}}$$
to evaluate the residue I'm using the fact that
$$\mathrm{Res}_{z = a} f(z) = \frac{1}{(m-1)!} \lim_{z \to a} \left(\frac{\partial}{\partial z}\right)^{m-1} ((z-a)^m f(z))$$
where $a$ is the pole and $m$ its corresponding order.


$f(z)$ has poles at $z = 0$ order $2$ and $z=1$ order one (simple pole),

so for the first region, $|z| < 1$, only the pole at $z = 0$ needs to be considered in the sum.

My problem is that when solving to get a function for $A_n$ without the Residual, the limit comes to $\dfrac{n+1}{0} = \infty$

Can anyone tell me if my method is wrong? or if I've failed to differentiate properly

(I apologise for how the question is formatted, I'm new to this)


Here is my working to get $A_n$
(INCORRECT, corrected below)

$$ A_n = \lim_{z \to 0} \frac{\partial}{\partial z} z^2 \frac{1}{z^{n+3}(1-z)} $$ therefore $$ A_n = \lim_{z \to 0} \frac{\partial}{\partial z} \frac{1}{z^{n+1}(1-z)} $$ by quotient rule $$ A_n = \lim_{z \to 0} \frac{(z^{n+1}(1-z)(0) - (1)((n+1)z^n - (n+2)z^{n+1}))}{(z^{n+1}(1-z))^2} $$ .. $$ A_n = \lim_{z \to 0} \frac{z^n((n+1) - (n+2))z)}{z^{2n+2}(1-z)^2} $$ canceling $$ A_n = \lim_{z \to 0} \frac{(n+1) - (n+2)z}{z^{n+2}(1-z)^2} $$ applying limit $$ A_n = \lim_{z \to 0} \frac{n+1}{(0)(1)^2} = \infty $$


Correct Answer
For $|z| < 1$, a = 0, m = n+3 $$ A_n = \frac{1}{((n+3) - 1)!} \lim_{z \to 0} (\frac{\partial}{\partial z})^{n+3 - 1} z^{n+3} \frac{1}{z^{n+3}(1-z)} $$ .. $$ A_n = \frac{1}{(n+2)!} \lim_{z \to 0} (\frac{\partial}{\partial z})^{n+2} \frac{1}{(1-z)} $$ .. $$ A_n = \frac{1}{(n+2)!} \lim_{z \to 0} \frac{(n+2)!}{(1-z)^{n+1}} $$ apply limit $$ A_n = \frac{(n+2)!}{(n+2)!} = 1$$ Which is as expected due to the geometric series solution, $n \ge -2$ is a condition for the factorial function
Therefore subbing into $$\tag{1}f(z) = \sum\limits_{n= -\infty}^\infty A_nz^n $$ gives $$f(z) = \sum\limits_{n= -2}^\infty A_nz^n, |z| < 1 $$ THANK YOU

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    $\begingroup$ Your LaTeX was fine, see my edit for little ameliorations. (\operatorname{} is often used instead of \mathrm{}). $\endgroup$ Aug 21, 2013 at 14:33
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    $\begingroup$ You should be differentiating $\frac{1}{1-z} = z^{n+2} \cdot \frac{f(z)}{z^n}$ and you shouldn't get $\infty$, but $1$ each time. $\endgroup$
    – Cocopuffs
    Aug 21, 2013 at 14:56

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The problem is that although $f$ has a double pole at $z=0$, $f(z)/z^{n+1}$ has a pole or order $n+3$, and it's the residue of this that is relevant, not the residue of $f$ itself.

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  • $\begingroup$ Thank you, I'll correct it and see what happens $\endgroup$
    – Morgan
    Aug 21, 2013 at 15:25

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