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singularities and residues

pantboio

Member
Nov 20, 2012
45
Consider $$f(z)=\frac{z+5}{e^\frac{1}{z}-3}$$ Find and classify its singularities and compute residues.
I think singularities are: $0,\infty$ and zeroes of denominators. We have $e^{\frac{1}{z}}=3$ for $z=(log(3)+2k\pi i)^{-1}$. I think $0$ is essential, $\infty$ a simple pole and zeroes of denominator are simple poles.
How can i prove this facts, and how to compute residues?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Consider $$f(z)=\frac{z+5}{e^\frac{1}{z}-3}$$ Find and classify its singularities and compute residues.
I think singularities are: $0,\infty$ and zeroes of denominators. We have $e^{\frac{1}{z}}=3$ for $z=(log(3)+2k\pi i)^{-1}$. I think $0$ is essential, $\infty$ a simple pole and zeroes of denominator are simple poles.
How can i prove this facts, and how to compute residues?

Computing the residue in $z=0$ is highly simplified with the change of variable $s=\frac{1}{z}$, so that the function 'under investigation' is...

$\displaystyle f(s)= \frac{1 + 5\ s}{s\ (e^{s}-3)}$ (1)

It is clear that the residue of f(z) in z=0 is the coefficint of the first order of the Laurent expansion of f(s) in s=0, so that we first find the McLaurin expansion of $\displaystyle g(s)=\frac{1}{e^{s}-3}$. We obtain...

$\displaystyle g(s)= \frac{1}{e^{s}-3} \implies a_{0}= - \frac{1}{2}$

$\displaystyle g^{\ '}(s)= - \frac{e^{s}}{(e^{s}-3)^{2}} \implies a_{1}= - \frac{1}{4}$

$\displaystyle g^{\ ''}(s)= \frac{2\ e^{2\ s}}{(e^{s}-3)^{3}} + \frac{e^{s}}{(e^{s}-3)^{2}} \implies a_{2}=0$

... so that is...

$\displaystyle f(s) = \frac{1+5\ s}{s}\ \{-\frac{1}{2} - \frac{1}{4}\ s + 0\ s^{2} + \mathcal{O} (s^{3})\}$ (2)

... and from (1) we obtain easily that the residue of f(z) in z=0 is $r_{0}=- \frac{5}{4}$...

Kind regards

$\chi$ $\sigma$
 

pantboio

Member
Nov 20, 2012
45
Computing the residue in $z=0$ is highly simplified with the change of variable $s=\frac{1}{z}$, so that the function 'under investigation' is...

$\displaystyle f(s)= \frac{1 + 5\ s}{s\ (e^{s}-3)}$ (1)

It is clear that the residue of f(z) in z=0 is the coefficient of the first order of the Laurent expansion of f(s) in s=0, so that we first find the McLaurin expansion of $\displaystyle g(s)=\frac{1}{e^{s}-3}$. We obtain...

$\displaystyle g(s)= \frac{1}{e^{s}-3} \implies a_{0}= - \frac{1}{2}$

$\displaystyle g^{\ '}(s)= - \frac{e^{s}}{(e^{s}-3)^{2}} \implies a_{1}= - \frac{1}{4}$

$\displaystyle g^{\ ''}(s)= \frac{2\ e^{2\ s}}{(e^{s}-3)^{3}} + \frac{e^{s}}{(e^{s}-3)^{2}} \implies a_{2}=0$

... so that is...

$\displaystyle f(s) = \frac{1+5\ s}{s}\ \{-\frac{1}{2} - \frac{1}{4}\ s + 0\ s^{2} + \mathcal{O} (s^{3})\}$ (2)

... and from (1) we obtain easily that the residue of f(z) in z=0 is $r_{0}=- \frac{5}{4}$...

Kind regards

$\chi$ $\sigma$
thank you very much for the solution, but i have a question about it: we have the zeros of denominator which are of the form $z_k=(log3+2k\pi i)^{-1}$, and these points accumulate to zero when $k$ goes to $\infty$. Hence, i suppose, zero is not an isolated singularity for the function. So my question is: does it make sense to compute the residue in a point which is non an isolated singularity ?
 

chisigma

Well-known member
Feb 13, 2012
1,704
thank you very much for the solution, but i have a question about it: we have the zeros of denominator which are of the form $z_k=(log3+2k\pi i)^{-1}$, and these points accumulate to zero when $k$ goes to $\infty$. Hence, i suppose, zero is not an isolated singularity for the function. So my question is: does it make sense to compute the residue in a point which is non an isolated singularity ?
This new member [Italian nationality(Yes)...] is very able to propose 'very enbarassing questions'!... and that's good because nothing in Mathematics has to be consider as 'absolute true' and all must be subject to 'constructive critic'...


If $z_{0}$ isn't an insulated singularities of $f(z)$ my provisional opinion is that the Laurent expansion of $f(z)$ 'around' $z=z_{0}$ exists [and also its residue...] but it doen't converge for any value of z. The problem is not a 'butterfly' because we have to extablish if in computing an integral like $\displaystyle \int_{c} f(z)\ dz$, where $z_{0}$ is inside c, the residue in $z_{0}$ has to be considered or not (Malthe)...


Kind regards


$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
This is indeed an evil question. You can certainly have a residue at an essential singularity, provided that it is isolated. But as the OP points out, the essential singularity at 0 in this example is not isolated. I believe that most authors only define a residue at an isolated singularity. That is what Ahlfors does in his classic text Complex analysis, and it is also the context in which residues are defined in the Wikipedia page on residues. So, unless you are using a text that defines residues differently, I would say that in this example the residue at 0 is not defined.