# General method of determining order of poles

1. Mar 27, 2006

### sachi

In Boas on p.595 there's an FCV proof for finding the order of a pole.
It says to write f(z) as g(z)/[(z-zo)^n] and then write g(z) as a0 + a1(z-z0) .... etc. and that we can deduce that the Laurent series for f(z) begins with (z-z0)^(-n) unless a0 = 0 i.e g(z0) = 0. Therefore the order of the pole is n. However, how can we be sure that g(z) does not contain terms of the form (z-z0)^(-n) ? Is this just by assumption?

2. Mar 27, 2006

### HallsofIvy

Staff Emeritus
Definition of order of a pole: z0 is a pole of f(z) of order n if and only if (z-z0)nf(z) is analytic at z0 but (z-z0)n-1f(z) is not.
The fact that (z-z0)nf(z) is analytic means that is equal to its power series in some neighborhood of z0:
$$f(z)(z-z_0)^n= a_0+ a_1(z-z_0)+ a_2(z-z_0)^2+ ...$$
and so
$$f(z)= a_0(z-z_0)^{-n}+ a_1(z-z_0)^{1-n}+ a_2(z-z_0){2-n}+...$$

From the definition of order of a pole and the fact that an analytic function is equal to its Taylor series, it follows that the Laurent series has no term with exponent less than -n.