Ok, so I have this function and I need to find the residues for it.

I developed the function into Laurent series and my question is, how can I find a(-1) which is the residues?

Thanks in advance.

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- Feb 10th 2009, 01:17 AMasi123residues 2
Ok, so I have this function and I need to find the residues for it.

I developed the function into Laurent series and my question is, how can I find a(-1) which is the residues?

Thanks in advance. - Feb 10th 2009, 02:44 AMOpalg
There is only one singularity, at z = 2. It is an essential singularity, but there are only even powers of z–2 in the Laurent series. So there is no term in $\displaystyle (z-2)^{-1}$ and therefore the residue is 0.

- Feb 10th 2009, 02:53 AMasi123
- Feb 10th 2009, 04:05 AMHallsofIvy
I, myself, would have said that if a point is an essential singularity the function does not HAVE a residue there but I note that Wikipedia asserts that the residue at an essential singularity, $\displaystyle z_0$ is still the coefficient of $\displaystyle (z- z_0)^{-1}$ in the Laurent series only noting that in this case, that is the only way to find the residue, the other formulas, such differentiating n times where n is the order of the pole, not applying. I guess I've been wrong all these years!

Therefore, as Opalg says, the Laurent series for your original function having only even powers, the coefficient of $\displaystyle (z-2)^{-1}$ is 0 and the residue is 0.

For your second function, the coefficient of $\displaystyle z^{-1}$ is 1, as you show, and so the residue is 1. - Feb 10th 2009, 08:19 AMOpalg
If you define the residue of f(z) at z=a to be the coefficient of $\displaystyle (z-a)^{-1}$ in the Laurent series about that point, then what I said before is correct. However, I see that Ahlfors defines it as "the unique complex number R which makes $\displaystyle f(z) - R/(z-a)$ the derivative of a single-valued function in an annulus $\displaystyle 0<|z-a|<\delta$." Ahlfors is evasive about whether such a number exists in the case of an essential singularity, commenting that in this case "there is no [...] procedure of any practical value" for computing residues, "and thus it is not surprising that the residue theorem is comparatively seldom used in the presence of essential singularities."