Finding the residue of a singularity

[annawells]

Homework Statement

The following function has a singularity at z=0
(e^z)/(1 - (e^z))
decide if its removable/a pole/essential, and determine the residue

The Attempt at a Solution

I played with the function and saw it can be re-written as: -1 /(z + z^2/2! + z^3/3! +...)

In this case, the function still does not "behave" at z=0, so do i need to find a different expansion? I am all out of ideas!

(note: i know that as well as z=0, we also have singularities at z=2ki'pi' for k in Z, but the question just wants the nature of the singularity at z=0)

Any help would be greatly appreciated, thanks :-)

 

[lanedance]

an essential singularity gives infinite negative power terms in the laurent series, so you could look at that...

though looking at the form you have found, which appears to be singular at z = 0, you could try multplying by z, then see if you still have a singular function...

 

[lanedance]

this has a handy way to look at it as well...
http://en.wikipedia.org/wiki/Essential_singularity

 

[annawells]

Thanks for the idea.
I am not quite sure how to pursue it. To show it is an essential singularity, don't i need to "produce" the laurent series, or at least show that it has infinite negative terms? I don't know how to break up the fraction to do that!? maybe this is more of a problem with my general algebra!
What do you mean when you say multiply by z? doesn't that just change the thing altogether?
On the wikipedia page you linked to, i tried to do it the lmiit way. but, in the form I have it in, limf(z) doesn't exist, but the lim(1/f(z)) does exists =0, so that applies to case 3 which says that f is a pole of f ? but then, isn't it a pole of infinite order...so it "becomes" an essential singularity?
thanks
Anna

 

[lanedance]

welcome to pf by the way... you can write equations pretty easy use tex tags, click on the equation below to see how its written

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

so, why do you think it is a pole of infinite order?

what I was saying, was have a look at how the function $g(z) = z f(z)$ behaves as $z \to 0$? what does that tell you about the nature of the function f?

 

[annawells]

Thanks, I was wondering how people made the functions look nice on here! i have often found answers here, but never posted before; thanks for the welcome :-)

well, when z --> 0, we have the zf(z)--> -1

i am not sure what that says about the nature of a function? if 0 were a simple pole, then that would tell us that it had a residue of -1 at z=0?

i thought it was a pole of infinite order because it had infinitely increasing powers of z in the denominator? but,i guess that they all correspond to the same singularity, does that mean its a simple pole?

Anna

 

[lanedance]

Been a while since I've done some complex analysis, but as you can write $f(z) = \frac{g(z)}{z}$, where g(z) is a well behaved function, then it shows z=0 is pole of order 1 for f(z).

In your expansion, close to zero, $|z^2| < |z|$, so its really only the first term that counts.

In an essential singularity $f(z)z^n$ is still singular, for any n

 

[annawells]

okay, thanks a lot! i didnt know that thing about g(z)/z , thanks. but then, what is the residue at the singularity? its the coefficient of the "1/z" term, and here that looks to be g(z) though its not exactly a coefficient,its a series!

 

[lanedance]

not quite, you would have to expand it to be in laurent series form, though it falls out when you multple the function by z
$$Res(f,0) = \lim_{z \to 0} zf(z) = g(0)$$

 

[annawells]

thanks a million!

 

[lanedance]

no worries ;)