1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.