# Residues of an essential singularity and a simple pole

1. May 11, 2010

### daoshay

1. The problem statement, all variables and given/known data
Classify the isolated singularities and find the residues
$$\frac {\sin(\frac {1}{z})}{1-z}$$

2. Relevant equations
I know the Taylor series expansion for 1/(1-z) when |z|<1
and I think I know the Taylor series for sin(1/z). The reciprocal of each term of the Taylor series of sin(z), right?

3. The attempt at a solution
znot = 0 is an essential singularity and znot = 1 is a simple pole.
I've tried using the limit approach to find the singularity at znot = 1, but I keep getting -sin(1) as an answer. I am thinking I should change sin(1/z) into (e^iz - e^-iz) / 2i,but I'm not sure if that is the right direction. If someone could just nudge me in the right direction, I'd be pumped.

P.S. I gave up on latex for now, it was driving me nuts...I'm learning from looking at other people's code. It kept throwing in a sin function into an expression I never coded a sin in. I have some learning to do.

2. May 11, 2010

### vela

Staff Emeritus
No, that would be

$$\sum_{k=0}^\infty (-1)^k \frac{(2k+1)!}{z^{2k+1}}$$

which is not the Taylor series for sin(1/z). I'm guessing that's not what you meant though.
What's wrong with -sin(1)?
There's a problem with the forum not rendering equations on the page correctly. If you refresh the page, it usually fixes the problem.

3. May 11, 2010

### daoshay

Thanks, Vela. For now, I'm okay with the residue of 1 being -sin(1). I'm really uncertain about how to handle the 0 singularity. Do you suggest I expand the function into the Laurent series and find the -1 coefficient? If so, do I need to substitute

$$\frac{e^{i1/z}-e^{-i1/z}}{2i}$$

for

$$\sin\frac{1}{z}$$
?

Last edited: May 11, 2010
4. May 11, 2010

### jackmell

Whenever you have an essential singularity multiplied by an analytic function that's not a polynomial, the residue at zero is going to be an infinite sum:

$\left(a_0+a_1 z+a_2 z^2+\cdots\right)\left(\frac{b_1}{z}+\frac{b_2}{z^2}+\cdots\right)$

And you have:

$\sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n+1}(2n+1)!}\sum_{n=0}^{\infty} z^n$

Can you then form the Cauchy product of those sums and pick out all the terms with 1/z?

Last edited: May 11, 2010
5. May 11, 2010

### Count Iblis

A simpler way that requires almost no computation is to consider the contour integral along a circle with center the origin of radius smaller than one.

6. May 11, 2010

### daoshay

So, I've got:

$$\frac {1}{z}\left(1-\frac{1}{3!}+\frac{1}{5!}-\frac{1}{7!}+\cdots\right)$$

where the alternating series is my residue and that has a limit of... (I don't recognize this one except for the coefficients in sin(x))

7. May 11, 2010

### Count Iblis

You should recognize this as sin(1). That's the weakness of this method.

8. May 11, 2010

### daoshay

So, I've got -sin(1) and sin(1) as the residues? That feels awkward, only because none of the examples we did had residues like that. But I suppose those are numbers too.

9. May 11, 2010

### Count Iblis

You should do more practice problems! Let me suggest one. Do the same problem again, but now with sin(1/z) replaced by
f(1/z) where f(z) is any arbitrary analytic function.

10. May 12, 2010

### daoshay

Thanks count. Will I end up getting -f(1), f(1/1)? I haven't tried it, but based on this problem and the behavior of the essential singularity times the analytic function, that's what I'm guessing I'll get. I -will- try it later this week.

11. May 12, 2010

### HallsofIvy

Staff Emeritus
The Taylor's series for sin(z), about z= 0, is
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} z^{2n+1}$$

Therefore the Laurent's series for sin(1/z), about z= 0, is
$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{-(2n+1)$$

Az vela pointed out, that is not the same as "The reciprocal of each term of the Taylor series of sin(z)".

Since that has an infinite number of terms with negative exponent, z= 0 is an essential singularity.

12. May 12, 2010

### Count Iblis

You'll find that the residue at zero does not only involve f(1), so there was something special about the case f(z) = sin(z). You can also try to solve this more general problem: Find the residue at z = 0 of

$$\frac{f(\frac{1}{z})}{1-z^{n}}$$

where f(z) is an arbitrary analytic function and n an arbitrary integer.

13. May 12, 2010

### daoshay

Thanks, Halls. I realized as soon as Vela pointed that out that I shouldn't have reciprocated because of the factorials. Only the z value is reciprocated. I'd been staring at this stuff for a while and wasn't thinking clearly.

You guys have been a great help.

14. Jul 8, 2011

### benygh2002

Hello friends,
I'm a student of mechanical engineering and I have a problem with computing residues of a complex function. I ve read your useful comments. Now I ve got some ideas about essential singularity and series expansion in computing the residue. However, I still can't find the solution to my problem.
I arrived at a complex function in the process of finding a solution to a mechanical problem.
Then I have to obtain the residues to proceed to the next steps. The function has the following form:

f(z)=exp(A*Z^N+B*Z^-N)/z

where A, B and N are real constants (N>=3).

I want to compute the resiude at z=0. I wrote the Laurent serie of f but got an infinite sum. I do not even know if I am at the right direction.
Engineering students like me have always problems with math, let alone complex analysis.
Anyway, I would be really thankful if someone could give me hint on this and put me back in the right direction.