Multiplicity of a pole

For [tex]f(z)=\frac{1}{(z-z_0)^n}[/tex], the pole at [tex]z=z_0[/tex] has multiplicity n

 

i.e. the order of the pole of f at z, equals the multiplicity of the zero of 1/f, at z.

 

May this be extended beyond polynomials?

 

it makes sense for the quotient of two holomorphic power series, yielding what is called a laurent series, i.e., a series with at most a finite number of terms of negative powers.

on any open set, the fraction field of the holomorphic functions form what is called the field of meromorphic functions on that set. those have at worst poles as non holomorphic points.

e^(1/z) has a worse than pole point at z = 0. isolated non holomorphic points are called (isolated) singularities. the simplest actual singularities of functions defined by power series, possibly infinite in both directions, i.e. summed over all integer powers of z, are the poles.

 

I.e. the multiplicity is the power of the term with the largest negative power in the laurent series of the function?

Does this also mean that an isolated/(essential?) singularity is a pole with infinite multiplicity?

 

More generally, a function f(z) has a pole at z= a of multiplicity n if and only if
[itex]\lim_{z\rightarrow a} (z-a)^nf(z)[/itex] exists but [itex]\lim_{z\rightarrow a} (z-a)^{n-1}f(z)[/itex] does not.

 

That definition put everything in place, thanks!