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Suppose I have a function of the form $$f(x)=\frac{1}{(x-a)(x-b)^2(x-c)^3}$$

Clearly, I have a simple pole at $a$, and poles of order 2,3 at $b,c$, respectively. By definition, the residue at $x=\alpha$ is the coeffecient of the term $(x-\alpha)^{-1}$ in the Laurent expansion of $f(x)$ centered at $x=\alpha$. From my experience, this residue at $x=\alpha$ is equivalent to the coefficient of the term $(x-\alpha)^{-1}$ in the partial fraction decomposition of $f(x)$. However, the partial fraction of $f(x)$ will not be a Laurent series since it has no center (I.e., there are denominators of the form $x-\beta$ with $\beta \not = \alpha$). Is there a rule that says that the coefficient of $(x-\alpha)^{-1}$ in the partial fraction decompositon is the residue at $x=\alpha$, even if the partial fraction decompoition is not a Laurent expansion?

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3 Answers 3

up vote 3 down vote accepted

I'm sure you'll find the following theorem very useful.

Theorem: If $f(z)$ has a pole of order $m$ at $z_0$, then $$R [ f,z_0 ] = \lim_{z \to z_0} \frac{1}{( {m - 1} )!} \frac{d^{m - 1}}{dz^{m - 1}} ( ( z - z_0 )^m f( z )).$$

Proof:

The Laurent series for $f(z)$ around $z_0$ is given by $$f( z ) = \frac{a_{ - m}}{( z - z_0 )^m} + \cdots + \frac{a_{ - 2}}{( z - z_0 )^2} + \frac{a_{ - 1}}{( z - z_0 )^1} + a_0 + a_1 ( z - z_0 ) + \cdots $$

Multiplying the series by $( z - z_0 )^m$, we get $$\begin{array}{c}{\left( {z - {z_0}} \right)^m}f\left( z \right) = {a_{ - m}} + {a_{ - m + 1}}\left( {z - {z_0}} \right) + \ldots + {a_{ - 2}}{\left( {z - {z_0}} \right)^{m - 2}}\\ + {a_{ - 1}}{\left( {z - {z_0}} \right)^{m - 1}} + {a_0}{\left( {z - {z_0}} \right)^m} + a_1 ( z - z_0)^{m + 1} + \ldots \end{array}$$

Differentiating $m-1$ times, we obtain $$\frac{d^{m - 1}}{dz^{m - 1}} ( ( z - z_0 )^m f ( z ) ) = ( m - 1 )! a_{ - 1} + m! a_0 ( z - z_0 ) + \frac{( m + 1 )!}{2}{a_1}{( z - z_0 )^2} + \cdots $$

Hence, it follows that $$\lim_{z \to z_0} \frac{d^{m - 1}}{dz^{m - 1}} ( ( z - z_0 )^m f( z ) ) = ( m - 1 )! a_{ - 1}$$ and the theorem is proved.

From the previous theorem, you can conclude that if $f(z)$ has a simple pole at $z_0$, then $$R\left[ f,z_0 \right] = \lim_{z \to z_0} ( z - z_0 )f( z ).$$

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1  
Please. Look at my edits to your answer. Your style of writing TeX code is bizarre. It contains zillions of purposeless complications that can only make editing harder. I haven't cleaned up all of it; some of it is still there. \mathop{\lim}\limits_{z \to {z_{0}} could have been just \lim_{z \to z_0}, and lots of it is even much worse than that. –  Michael Hardy Feb 3 at 22:50
    
\frac{{{a_{ - m}}}}{{{{\left( {z - {z_0}} \right)}^m}}} could be coded as \frac{a_{ - m}}{( z - z_0 )^m}. –  Michael Hardy Feb 3 at 22:53
    
I'm really sorry. As I don't master LaTeX yet, sometimes I resort to MathType (dessci.com/EN/products/mathtype). To be honest, perhaps I'm just lazy :/ –  el.Salvador Feb 3 at 22:56
    
Maybe I'll complain to the authors of that software. It should be possible for them to do far better than that. To write \mathop{\lim}\limits_{} is to neglect really basic stuff. –  Michael Hardy Feb 3 at 23:06
    
It can be my fault... I cut/past/delete/undo/redo lots of stuff, so the code just gets messy and messier. –  el.Salvador Feb 3 at 23:10

Yes, there is such a rule. If you have a partial fraction decomposition

$$f(z) = h_\alpha(z) + h_\beta(z) + h_\gamma(z) + g(z),$$

where $h_w$ is the principal part of $f$ in $w$, and $g$ is the remaining holomorphic part of $f$, then you can develop $h_\beta,\, h_\gamma$, and $g$ in Taylor series about $\alpha$ to obtain the Laurent series of $f$ in the annulus $0 < \lvert z-\alpha < d$, where $d$ is the distance to the nearest other singularity ot the boundary of the domain of $f$, whatever is smaller.

The point is that everything but the fractions $\frac{c_k}{(z-\alpha)^k}$ in the partial fraction decomposition of $f$ is holomorphic in a neighbourhood of $\alpha$, so can be expanded into a power series around $\alpha$ and therefore doesn't influence the coefficient of $(z-\alpha)^{-1}$.

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Theorem: If $f(x) = g(x)h(x)$ and $g$ is holomorphic near $a$ and $h$ has a simple pole at $a$, then the residue of $f$ at $a$ is $g(a)$ times the residue of $h$ at $a$.

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@DanielFischer : When I wrote "Space-aliens were behind the JFK assassination", it was a typo. I really meant "All recursively enumerable sets are Diophantine." And in this answer, I've added the word "simple". –  Michael Hardy Feb 3 at 22:27
    
That is an impressive typo. Your cat must be proud. –  Daniel Fischer Feb 3 at 22:30

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