Singularities

[Bella123456789]

How would you find and classify the singularities of the function f(z) = cot(z)/z^4 ?

Is it correct to say as f(z)=cos(z)/z^4 sin(z) that there is a singularity if z^4 = 0, which is a pole of order 4?

 

[galactus]

\(\displaystyle cot(z)=\frac{cos(z)}{sin(z)}\)

Thus, It has a singularity wherever sin(z)=0.

But \(\displaystyle \frac{cot(z)}{z^{4}}=\frac{1}{z^{4}tan(z)}\)


See?.

 

[Bella123456789]

Yes, I see. So there are 2 singularities? Is it correct to say sin (z) is an essential singulairty? and that z^4 is a pole of order 4 then?

As lim (z tends to 0) z^4 g(z) = z^4/tan(z) which does not = 0

And then showing the laurent expanision for sin(z) we see that since f(z)=sin(z) g(z) and that g(z) is holomorphic at z=0 (when g(z) = cos(z)/z^4) that the laurent expansion will have infintely many negative powers of z

 

[RobertPaulson]

I'd like to know if the above response is correct, if possible

 

[Deleted member 4993]

I'd like to know if the above response is correct, if possible

Which above???

With answer/s - you should be able to now check your textbook to find out the correctness of the answer/s.

 

[RobertPaulson]

I'd like to know if the above response is correct, if possible

Which above???

With answer/s - you should be able to now check your textbook to find out the correctness of the answer/s.

The textbook i have is very vague on singularities, and the notes i have are ok but very short with limited examples (none on trig); so i'm not quite sure where i would go from the last bit of help. Any help would be greatly appreciated

 

[galactus]

\(\displaystyle f(z) = cot(z) = cos(z)= sin(z)\) has poles at the zeros of sin(z), namely at \(\displaystyle z=n{\pi} \;\ ,n = 0, \;\ \pm 1. \;\ \pm 2,\)

For instance, \(\displaystyle \frac{cot(z)}{z^{2}}=\frac{cos(z)}{z^{2}sin(z)}\) has simple poles at \(\displaystyle \pm 1, \;\ \pm 2, \;\ \pm 3, \;\ .................\)

and a triple pole at 0. \(\displaystyle \frac{cos(z)}{z^{4}}\) has a pole of order 5 at 0.

Note, \(\displaystyle \frac{cot(z)}{z^{4}}=\frac{1}{z^{5}}-\frac{1}{3z^{3}}-\frac{1}{45z}-\frac{2z}{945}-\frac{z^{3}}{4725}-............\)

The highest power term of z in the denominators is \(\displaystyle \frac{1}{z^{5}}\)

The residue at each non-zero integer n is \(\displaystyle \frac{1}{n^{2}}\), and at 0 the residue is \(\displaystyle \frac{-1}{3}\)

Therefore, \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{6}\)

Ever seen that before?. that is Euler's famous infinite sum, the Basel problem.

In the case of \(\displaystyle \frac{cos(z)}{z^{4}}\), the residue at 0 is \(\displaystyle \frac{-1}{45}\)

We can use this to show that \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{{\pi}^{4}}{90}\)

 

[galactus]

As a tidbit to add, the Riemann Zeta function, along with cot(z) can be used to find \(\displaystyle {\zeta}(k)=\sum_{n=1}^{\infty}\frac{1}{n^{k}}\)

It is rather long and complicated to post, but a very cool and interesting thing.

You may know that \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{{\pi}^{2}}{6}\).

But, using the proper rectangular region (call the region \(\displaystyle {\gamma}\)), residues, and \(\displaystyle \int_{\gamma}\frac{{\pi}cot({\pi}z)}{z^{k}}dz\)

we can work it down to \(\displaystyle 2{\zeta}(k)=\frac{-(2{\pi}i)^{k}B_{k}}{k!}\), where \(\displaystyle B_{k}\) are the Bernouilli numbers.

The Bernouilli numbers are all 0 at odd k. In the case of z^4, we get \(\displaystyle {\zeta}(4)=\sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac{{\pi}^{4}}{90}\)

Which is the sum of the reciprocals of the fourth powers.

 

[RobertPaulson]

Thanks so much this has been a big help;

Does that mean the poles are the only singularities in \(\displaystyle \frac{cos(z)}{z^{4}}\) ? And that \(\displaystyle sin(z)\) isn't an essential singularity?

 

[galactus]

In order to be an essential singularity, the laurent series must have infinite negative values.

For instance, \(\displaystyle cos(\frac{1}{z})=1-\frac{z^{-2}}{2!}+\frac{z^{-4}}{4!}+............\)

 

[RobertPaulson]

aaah, thanks again; all the definitions i have for singularities are all maths based and sometimes i can't get my head round them.

 

[RobertPaulson]

Sorry to keep bugging you but, how did you get that the residue of the non-zero integers for the simple poles is 1/n^2 ?

 

[galactus]

For \(\displaystyle \frac{cos(z)}{z^{4}}\), it is \(\displaystyle \frac{1}{n^{4}}\)

 

[RobertPaulson]

For \(\displaystyle \frac{cos(z)}{z^{4}}\), it is \(\displaystyle \frac{1}{n^{4}}\)

Is there a simple way of getting this? As i've been going through what you said and my notes and i can't arrive at this number

 

[Deleted member 4993]

For \(\displaystyle \frac{cos(z)}{z^{4}}\), it is \(\displaystyle \frac{1}{n^{4}}\)

Is there a simple way of getting this? As i've been going through what you said and my notes and i can't arrive at this number

Galactus has explained in detail - in the following response...

 

[galactus]

The series for \(\displaystyle \frac{cot(z)}{z^{4}}\) is \(\displaystyle z^{-5}-\frac{1}{3}z^{-3}-\frac{1}{45}z^{-1}+...............\)

Looking at the coefficient of \(\displaystyle z^{-1}\), we can see the residue is -1/45.

Now, for the residues at k we can use a slightly different approach.

Suppose the power series for f(k+z) was \(\displaystyle a_{-1}z^{-1}+a_{0}+a_{1}z+a_{2}z^{2}+..............\)

\(\displaystyle zf(k+z)=a_{-1}+a_{0}z+a_{1}z^{2}+a_{2}z^{3}+..............\)

We have:

\(\displaystyle a_{-1}=\lim_{z\to 0}zf(k+z)=\lim_{z\to 0}z\cdot \frac{cot(k+z)}{(k+z)^{4}}=\frac{z}{sin(k+z)}\cdot \frac{cos(k+z)}{(k+z)^{4}}\)

Applying L'Hopital to the first factor and find that

\(\displaystyle a_{-1}=\lim_{z\to 0}\frac{1}{(k+z)^{4}}=\frac{1}{k^{4}}\)

This calculation appears knowing in advance that the power series for f(k+z) has only one term with a negative power of z.

Why were there no terms involving \(\displaystyle z_{-2}, \;\ z_{-3},\)?. Since zf(k+z) has a limit at 0, its power series can not have any terms with

negative powers of z. Therefore, every term with the series for f(k+z) must have an exponent of at least -1.

 

[RobertPaulson]

The series for \(\displaystyle \frac{cot(z)}{z^{4}}\) is \(\displaystyle z^{-5}-\frac{1}{3}z^{-3}-\frac{1}{45}z^{-1}+...............\)

Looking at the coefficient of \(\displaystyle z^{-1}\), we can see the residue is -1/45.

Now, for the residues at k we can use a slightly different approach.

Suppose the power series for f(k+z) was \(\displaystyle a_{-1}z^{-1}+a_{0}+a_{1}z+a_{2}z^{2}+..............\)

\(\displaystyle zf(k+z)=a_{-1}+a_{0}z+a_{1}z^{2}+a_{2}z^{3}+..............\)

We have:

\(\displaystyle a_{-1}=\lim_{z\to 0}zf(k+z)=\lim_{z\to 0}z\cdot \frac{cot(k+z)}{(k+z)^{4}}=\frac{z}{sin(k+z)}\cdot \frac{cos(k+z)}{(k+z)^{4}}\)

Applying L'Hopital to the first factor and find that

\(\displaystyle a_{-1}=\lim_{z\to 0}\frac{1}{(k+z)^{4}}=\frac{1}{k^{4}}\)

This calculation appears knowing in advance that the power series for f(k+z) has only one term with a negative power of z.

Why were there no terms involving \(\displaystyle z_{-2}, \;\ z_{-3},\)?. Since zf(k+z) has a limit at 0, its power series can not have any terms with

negative powers of z. Therefore, every term with the series for f(k+z) must have an exponent of at least -1.

Thanks again so much, you've been extremely helpful and very patient; sorry you had to put it in such laymens terms, i guess i should pay attention more in lectures.

 

[galactus]

May I recommend a very nice book on Complex Analysis. It is for beginners and is about as nice and comprehensive as can be found.

"Complex Variables Demystified" by David McMahon. ISBN #: 978-0-07-154920-2

Go to Amazon and order it. It is inexpensive. Only around $15-$20.