# Residue of a complex function with essential singularity

by benygh2002
Tags: complex analysis, residue theorem
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 An essential singularity does not have a residue. However, your Laurent series should not have an infinite number of negative powers and so this is NOT an essential singularity. $$e^{Az^n+ Bz^{-n}}= e^{Az^n}e^{Bz^{-n}}$$ is analytic and so has a Taylor's series- with no negative power terms. Dividing by z gives a series have "$z^{-1}$" as its only negative power. The residue is the coefficient of $z^{-1}$.
 P: 3 Thanks for your reply. 1) You mentioned that "an essential singularity does not have a residue". I'm a bit confused about this. For example: $f\left(z\right)=e^{z+\frac{1}{z}}$ has an essential singularity at z=0. However. after writing Laurent expansion of f(z) the residue at z=0 can be obtained as a serie: $Res\left(f,0\right)=1+\frac{1}{2!}+\frac{1}{2!3!}+\frac{1}{3!4!}+...$ 2) Laurent expnasion of my desired function f(z) is: $f\left(z\right)=\frac{e^{Az^n+Bz^{-n}}}{z}$ $=\frac{\sum{(Az^n+Bz^{-n})}^k}{z k!}$ (k=0 to infinity) in which the coefficients of 1/z terms give the residue as a serie (but the serie does not converge). 3) I would be thankful if you could tell me what I should do to find the residue of the following function which is similar to the previous function f(z): $g\left(z\right)=\frac{e^{Az^n+Bz^{-n}}}{z-a}$ where a is a constant. What should I do to find the poles? At what point I should write the Taylor expansion? Thanks for your help.