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