Singularities, Residues, and Computation: Analyzing $f(z)$ without Prefix

In summary, the function $f(z)=\frac{z+5}{e^\frac{1}{z}-3}$ has singularities at $0$, $\infty$, and the points $z=(log(3)+2k\pi i)^{-1}$ where $k$ is an integer. The singularity at $0$ is essential, the singularity at $\infty$ is a simple pole, and the singularities at the points $z=(log(3)+2k\pi i)^{-1}$ are simple poles. To compute the residues, one can use the McLaurin expansion of $g(s)=\frac{1}{e^{s}-3}$ to find the coefficient of the
  • #1
pantboio
45
0
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?
 
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  • #2
pantboio said:
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

$\chi$ $\sigma$
 
  • #3
chisigma said:
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 ?
 
  • #4
pantboio said:
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 $\chi$ $\sigma$
 
  • #5
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.
 

FAQ: Singularities, Residues, and Computation: Analyzing $f(z)$ without Prefix

What is a singularity in mathematics?

A singularity in mathematics is a point in a function where it is undefined or infinite. It is often represented by a vertical asymptote on a graph.

How are singularities related to complex numbers?

Singularities are a concept in complex analysis, which deals with functions of complex numbers. In complex analysis, singularities are points where a function is not holomorphic, meaning it does not have a derivative at that point.

What is a residue in complex analysis?

In complex analysis, a residue is the value of a function at a singularity. It is calculated using the Cauchy residue theorem, which states that the value of a function at a singularity can be found by evaluating a contour integral around the singularity.

How are singularities and residues used in physics?

In physics, singularities and residues are often used in the study of black holes and other extreme phenomena. Singularities are often associated with the center of a black hole, where the curvature of space-time is infinite. Residues are used in the calculation of quantum scattering amplitudes in particle physics.

Can singularities and residues be visualized?

While singularities and residues themselves cannot be visualized, their effects can be seen on graphs of functions. Singularities often manifest as vertical asymptotes, while residues can be seen as spikes or dips in the graph near a singularity.

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