What is the Topological Interpretation of Orders of Poles in Rational Functions?

[dyn]

Hi.
If I look at the function $(z^2+z-2)/(z-1)^2$ it appears to have a double pole at z=1 but if I factorise the numerator I get $z^2+z-2 = (z+2)(z-1)$ and it is a simple pole.
Is it wrong to say it is a double pole ?
If I overestimate the order of the pole in this case as 2 and calculate the residue using limits and differentials I still get the correct answer. Is this always true ?
In this case the numerator was easy to factorise. If it was a complicated function involving higher powers that couldn't be factorised is it possible to say for certain what the order of the pole is just by looking at the denominator ?
Thanks

 

[fresh_42]

This is the same as in real analysis. $\dfrac{z^2+z-2}{(z-1)^2}$ has a pole of order two, $\dfrac{z+2}{z-1}$ one of order one.

Here is how the various singularities are classified:
https://en.wikipedia.org/wiki/Singularity_(mathematics)

 

[FactChecker]

I would say that $\frac {z^2+z-2}{(z-1)^2}$ has a pole of order one. In defining the order of a pole of $f(z)$, it is usual to say that $(z-a)^nf(z)$ is holomorphic and nonzero for a pole of order n.
(seehttps://en.wikipedia.org/wiki/Zeros_and_poles#Definitions )

 

[dyn]

Hence my confusion !

 

[fresh_42]

I would say that $\frac {z^2+z-2}{(z-1)^2}$ has a pole of order one. In defining the order of a pole of $f(z)$, it is usual to say that $(z-a)^nf(z)$ is holomorphic and nonzero for a pole of order n.
(seehttps://en.wikipedia.org/wiki/Zeros_and_poles#Definitions )

But we have $(z-1)^2f(z)=(z-1)^2\dfrac{[(z-1)(z+2)]}{(z-1)^2}$. The numerator is not defined at $z=1$ hence it cannot be equal to $z+2$.

It is a rather academic question and of not much use: What if we combine a removable singularity with a non removable? A bit like discussing why units are not prime.

 

[jasonRF]

Interesting … I learned that the order of a pole of a function that is analytic in a closed region except at an isolated point, is defined as the largest negative exponent in the Laurent expansion about that point. My old copy of "A Course of Modern Analysis" by Whittaker and Watson also has this definition. Since the Laurent expansion of the function about $z=1$ is $1 + 3/(z-1)$, the pole at $z=1$ is of order 1. Practically, I use the approach given by FactChecker.

fresh_42

Jason

 

[fresh_42]

Yes, it's probably reasonable to cancel the quotient if there is still the same pole available, so that the same undefined points are still there.

 

[HallsofIvy]

For a simpler example, f(z)= \frac{z- 1}{z- 1} has a "removable discontinuity" at z= 1, not a pole.

 

[dyn]

Hi.
If I look at the function $(z^2+z-2)/(z-1)^2$ it appears to have a double pole at z=1 but if I factorise the numerator I get $z^2+z-2 = (z+2)(z-1)$ and it is a simple pole.
Is it wrong to say it is a double pole ?

So it seems the function has a simple pole and it is wrong to say it is a double pole. But if I overestimate the pole as a double pole and calculate the residue using limits and derivatives I get the correct answer. Does this method of calculating residues always work if the order of poles is overestimated eg. by not spotting a factorisation ?
Thanks

 

[FactChecker]

Yes, you do not have to factor it. The residue at $z_0$ only depends on the coefficient of the $\frac 1{z-z_0}$ term in the Laurent series. So the effect of any numerator/denominator zeros that cancel each other will disappear.

 

[dyn]

Thank you

 

[mathwonk]

the order of the pole is a measure of the rate of growth of the function as you approach the given point. In particular it depends only on the values of the function away from the given point, not on the specific representation of the function. Since the two representations (z^2+z-2)/(z-1)^2 and (z+2)/(z-1) define the same functional values away from z=1, the functions they represent are the same and have the same order of pole at z=1. Similarly any calculations that depend only on the values away from z=1 (such as taking limits) will give the same result.

A simpler example that illustrates the same point is to ask yourself whether the function (z-1)/(z-1) has a simple pole at z=1. Since this is (or extends continuously to) the constant function 1, it has no poles at all.

 

[FactChecker]

the order of the pole is a measure of the rate of growth of the function as you approach the given point. In particular it depends only on the values of the function away from the given point, not on the specific representation of the function.

And more generally, how the function behaves in the complex plane around the pole. If it is a pole at $z_0$ of order $n$, then the function behaves like $\frac {b_n} {(z-z_0)^n}$ near $z_0$.

 

[mathwonk]

yes, for example it wraps a small circle around z0, n times around the point at infinity.

 

[mathwonk]

Indeed there is a topological interpretation of orders of poles. A rational function defines a continuous map from the riemann sphere to itself, and hence has a certain degree. That degree is the number of preimages of any given point, and in particular is the sum of the orders of all the poles, i.e. the number of preimages of infinity, properly counted.