What is a Pole of Order n in Complex Analysis?

[aaaa202]

If a complex function has the form:

f(z) = g(z)/(z-a)ⁿ then z=a is a pole of order n. I don't really understand all this fancy terminology. Isn't a pole just like when you for a real valued function g(x)/(x-a) don't want to divide by 0 and therefore the function is defined at x=a? If so what is then all this talk about a pole of order n, and how does poles at different orders distinguish from each other? Since you are classifying poles by order, my understanding of a pole as simply a point on which f is not defined is probably wrong or at least lacking something.

 

[Pzi]

If a complex function has the form:

f(z) = g(z)/(z-a)ⁿ then z=a is a pole of order n. I don't really understand all this fancy terminology. Isn't a pole just like when you for a real valued function g(x)/(x-a) don't want to divide by 0 and therefore the function is defined at x=a? If so what is then all this talk about a pole of order n, and how does poles at different orders distinguish from each other? Since you are classifying poles by order, my understanding of a pole as simply a point on which f is not defined is probably wrong or at least lacking something.

Are you familiar with zeros of higher order?
If z=a is a zero of n-th order for a function 1/f(z) , then z=a is a pole of n-th order for a function f(z). Those two are related like that.

 

[aaaa202]

Yes okay, but what is the idea of classifying the order of a zero? Surely (z-a)^n is zero for a=z no matter what n. I don't see how n can ever change the properties of the zero?

 

[Pzi]

Yes okay, but what is the idea of classifying the order of a zero? Surely (z-a)^n is zero for a=z no matter what n. I don't see how n can ever change the properties of the zero?

It tells how "powerful" the zero is.
Technically I'd say that for F(z)=(z-a)^n we have a zero of n-th order z=a because n-th derivative of F(z) at z=a is no longer equal to zero...
This works for all finite arguments.

How would you evaluate what is the order of zero z=0 for the function G(z)=sin(z) ?

You should not look for too much meaningfulness in the definition, but realizing such orders is quite an important thing in complex analysis.

 

[fortissimo]

Well, the order of a pole to a function is significant for its Laurent expansion (Taylor series with 1/(z-a)^n terms in it). This is important when you apply the residue theorem (a nice way to calculate integrals you never could do before).
And to elaborate a little on Pzi's post, you can calculate how "much zero" a function f is (or the order of the pole of 1/f) by computing its derivatives at a point. A double root means that you have f=f'=0, triple root f=f'=f''=0 and so on. Some functions have poles of infinite order, such a point to a function is called an essential singularity (like log z in the origin).

 

[jackmell]

log(z) has an algebraic branch point of infinite order at the origin. That's not an essential singularity. The function $e^{1/z}$ has an essential singularity at the origin.