Find coefficiant of Laurent series, without using residue

[ognik]

Hi - I admit to struggling a little with my 1st exposure to complex analysis and Laurent series in particular, so thought I'd try some exercises; always seem to help my understanding.

A function f(z) expanded in Laurent series exhibits a pole of order m at z=z₀. Show that the coefficient of $ (z-z_0)^{-1} , \: a_{-1} $ is given by $ {a}_{-1}=\frac{1}{(m-1)!} \frac{{d}^{m-1}}{{dz}^{m-1}}[(z-{z}_{o})^m f(z)]_{z={z}_{0}} $
with $ {a}_{-1} =[(z-{z}_{0)}f(z)]_{z={z}_{0}} $ when the pole is a simple pole (m=1)

Please check me below:
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I found the uniqueness of power series theorem, which states $ f(x)=\sum_{n=0}^{\infty} {a}_{n}{x}_{n} = \sum_{n=0}^{\infty}\frac{1}{n!}{f}^{n}(0){x}^{n} $

So, by analytic continuation (?), $ f(z)=\sum_{n=0}^{\infty} {a}_{n}({z-{z}_{0}})^{n} = \sum_{n=0}^{\infty}\frac{1}{n!}{f}^{n}({z}_{0})({z-{z}_{0}})^{n} $

So far it looks like a good approach, but Laurent series should sum from $ -\infty $, I am only summing from 0?

Also the question confuses me, they seem to have 2 different formula for $ a_{-1} $ - I decided to assume that the 1st formula is for $ a_{-m} $ - does that sound right?

So if we are looking at one term at a time, we can drop the summation (I still would like to know about not starting from -infinity), and with n=m-1 we would have the answer - but why would n = m-1?

 

[Fernando Revilla]

Perhaps the following solve your doubts.

1. Proposition. If $f$ is analytic on $r<\left|z-z_0\right|<R$ and $\displaystyle\sum_{-\infty}^{+\infty}a_n(z-z_0)^n$ is its Laurent series, then $z_0$ is a pole of order $m$ for $f$ iff $$\sum_{-\infty}^{+\infty}a_n(z-z_0)^n=\frac{a_{-m}}{(z-z_0)^m}+\cdots+\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots \;\;(a_{-m}\ne 0).$$

Now, supposing $z_0$ is a pole of order $m$ for $f,$

2. Definition. $\text{Res }\left[f,z_0\right]=\dfrac{1}{2\pi i}\displaystyle\int_{\gamma}f(z)\;dz$ where $\gamma$ is any circle $\left|z-z_0\right|=\rho$ with $r<\rho<R.$

3. Proposition. $\text{Res }\left[f,z_0\right]=a_{-1}.$

4. Proposition. $$\text{Res }\left[f,z_0\right]=\frac{1}{(m-1)!} \lim_{z\to z_0}\frac{{d}^{m-1}}{{dz}^{m-1}}\left((z-{z}_{o})^m f(z)\right).$$

 

[math771]

Hello! Your approach is on the right track, but there are a few things that need to be clarified.

First, the uniqueness of power series theorem only applies to functions that are analytic in a neighborhood of the point of expansion. In this case, we are dealing with Laurent series, which can have poles or other singularities. So the theorem does not directly apply here.

Second, the Laurent series expansion of $f(z)$ around $z_0$ is given by $f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$, where the coefficients $a_n$ are given by $a_n=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{(z-z_0)^{n+1}}dz$, where $C$ is a simple closed curve around $z_0$. So in this case, we are not just summing from $0$, but from $-\infty$ to $\infty$.

Now, for the formula for $a_{-1}$, you are correct that there are two different formulas given in the question. The first one is a general formula for the coefficient of $(z-z_0)^{-m}$, while the second one is a special case when $m=1$, which corresponds to a simple pole. So the first formula is more general and can be used to find the coefficients for any pole of order $m$, while the second formula is specific to simple poles.

To see why $n=m-1$, we can look at the definition of the coefficient $a_n$ given above. When $n=m-1$, we have $a_{m-1}=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{(z-z_0)^{m}}dz$. Now, using the Cauchy integral formula, we can write this as $a_{m-1}=\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\left[\frac{f(z)}{z-z_0}\right]_{z=z_0}$. And since we are only considering the coefficient of $(z-z_0)^{-1}$, the term inside the square brackets becomes $f(z_0)$, giving us the formula $a_{-1}=\frac