Finding and classifying singularities

Hi. I have to find and classify the isolated singularities for:

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

So, I've found that $\displaystyle e^z=-1\rightarrow{z=i(2k+1)\pi},k\in{Z}$

Now, I think I should make the Laurent series for f(z), but I don't know how to handle this function to get it's Laurent series representation.

I also know, of course, that: $\displaystyle e^z=\sum_0^{\infty}\frac{z^n}{n!}$

Re: Finding and classifying singularities

Quote:

Originally Posted by

**Ulysses** Hi. I have to find and classify the singularities for:

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

So, I've found that $\displaystyle e^z=-1\rightarrow{z=i(2k+1)\pi},k\in{Z}$

Now, I think I should make the Laurent series for f(z), but I don't know how to handle this function to get it's Laurent series representation.

I also know, of course, that: $\displaystyle e^z=\sum_0^{\infty}\frac{z^n}{n!}$

Do you know how to prove that a singularity is a pole **without** finding the Laurent series? (btw the poles are simple).

Re: Finding and classifying singularities

I was looking at my notebook. Should I use a thing called "principle of isolated zeroes" or something like that?

I was trying to determine the order of the pole

EDIT: Alright, I did it this way:

I took $\displaystyle \displaystyle\frac{1}{f(z)}=\displaystyle\frac{1+e ^z}{e^z}=g(z)\Rightarrow{g'(z)=1-\displaystyle\frac{1+e^z}{e^z}}\Rightarrow{g'(z_0) }=1\neq{0}$

Then it's a pole of order 1.

Re: Finding and classifying singularities

Quote:

Originally Posted by

**Ulysses** I was looking at my notebook. Should I use a thing called "principle of isolated zeroes" or something like that?

I was trying to determine the order of the pole

EDIT: Alright, I did it this way:

I took $\displaystyle \displaystyle\frac{1}{f(z)}=\displaystyle\frac{1+e ^z}{e^z}=g(z)\Rightarrow{g'(z)=1-\displaystyle\frac{1+e^z}{e^z}}\Rightarrow{g'(z_0) }=1\neq{0}$

Then it's a pole of order 1.

$\displaystyle \lim_{z \to i(2k+1)\pi} \frac{e^z(z - i(2k+1)\pi}{1 + e^z} = .... $

Therefore ....