Regarding evaluation of residuals for functions with simple poles.

Let's say $m$ represents the order of the pole, then in order to find the residual at each pole/the pole (if only one pole) we have the following equation:

$$ \text{where } m=1,2,\ldots\qquad\underset{z=z_0}{\operatorname{Res}} f(z) = \frac 1 {(m-1)!} \lim_{z\to z_0} \left( \frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)] \right) $$

If the order of the pole $m$ is equal to $1$ then the above equation reduces to:

$$ \text{simple pole }(m=1) \qquad \underset{z=z_0}{\operatorname{Res}} f(z) = \lim_{z\to x_0} (z-z_0) f(z). $$

Now I've seen the following equation for finding the residue at a simple pole for a function $f(z)$ given by:

$$ f(z) = \frac{p(z)}{q(z)} \qquad \underset{z=z_0}{\operatorname{Res}} f(z) = \frac{p(z_0)}{q'(z_0)}. $$

I'm eager to get this clarified once and for all, because it seems a bit confusing, seeing different ways to solve the below in different places, just wondering if it doesn't matter or if it actually does make a difference.

Now my question is if the function you are taking into account has one simple pole, would you use the same equation as in the case of a function with several simple poles?

In other words, is there any difference for finding residue(s) when you have one simple pole (if only one pole) versus a function with several simple poles, (and a third scenario would be a function with several poles, only one of which is simple)?

Or will any of the equations above return the same result in any of the mentioned cases?


1 Answer 1


You should be aware that a simple pole is a point in which the Laurent series of $f$ is $$ f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\ldots $$ The residual is $a_{-1}$, the coefficient of $1/(z-z_0)$. Now, we have $$ (z-z_0)f(z)=a_{-1}+a_0(z-z_0)+a_1(z-z_0)^3+a_2(z-z_0)^3+\ldots $$ that gives $a_{-1}$ when you do the limit $z\to z_0$.

When $f$ is a rational function, ratio of two polynomials $p$ and $q$, with a single pole, $q$ should be of the form $$ q(z)=(z-z_0)Q_1(z) $$ with $Q_1(z_0)\neq0$, and $$ q'(z)=Q_1(z)+(z-z_0)Q_1'(z) $$ so that the limit for $z\to z_0$ is $q'(z_0)=Q_1(z_0)$. At this point $$ (z-z_0)f(z)=(z-z_0)\frac{p(z)}{(z-z_0)Q_1(z)}=\frac{p(z)}{Q_1(z)} $$ and for $z\to z_0$ we have $$ \frac{p(z_0)}{Q_1(z_0)}=\frac{p(z_0)}{q'(z_0)} $$ I add that you could use the same formula for each single pole, also if the function has more than one single pole.


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