# Thread: cosecant series for complex analysis

1. ## cosecant series for complex analysis

Hey guys. My textbook has the following exercise:

Originally Posted by Priestley, Intro to Complex Analysis 2nd ed
Define

$\displaystyle f(z)=\frac{1}{z}-2z\sum_{n=1}^\infty \frac{(-1)^n}{n^2\pi^2-z^2}$

(i) Prove that $f$ is holomorphic in $G:=\mathbb{C}\setminus\{n\pi:n\in\mathbb{Z}\}$.

(ii) Prove that $f$ has a simple pole at each point $k\pi$, $k\in\mathbb{Z}$.

(iii) Deduce that $f(z)=\text{cosec }z$ for all $z\in\mathbb{C}$.
I've done parts (i) and (ii), but I'm stuck on part (iii). It looks like it should be easy. Usually when a textbook has as the last part of an exercise the phrase "deduce that..." it's something simple. But I just don't get this one.

Any help would be much appreciated!

EDIT: thanks for the latex fix! Now if I can only get this exercise...

2. Originally Posted by hatsoff
P.S. what's the story with the latex errors?
Using [tex] instead of [tex] is a temporary solution.

-Dan

3. Originally Posted by hatsoff

I've done parts (i) and (ii), but I'm stuck on part (iii). It looks like it should be easy. Usually when a textbook has as the last part of an exercise the phrase "deduce that..." it's something simple. But I just don't get this one.

Use the Mitagg-Lefler's expansion theorem.

4. You have to start from the 'funny identity'...

(1)

... and then consider the 'infinite products'...

(2)

(3)

Now if You compute using (2) and (3) ...

(4)

... and derive it You arrive at the expression of Your textbook. The details are left to You...

Kind regards

$\chi$ $\sigma$

5. Thanks guys, but I think I'm supposed to use at least one of the following two facts in my proof:

(i) $f$ is holomorphic in everywhere except for $k\pi$, $k\in\mathbb{Z}$

(ii) Each $k\pi$ is a simple pole.

I was thinking that maybe I could show $f(z)\sin z$ is bounded on $\mathbb{C}$. Then I could observe that $f(z)\sin z$ is holomorphic everywhere, and conclude by Liouville that it is a constant function. Finally, I could plug in zero to show that the constant is 1. But unfortunately I don't know how to show that it's bounded (without first showing that $f(z)=1/\sin z$ anyway).

I appreciate the suggestions though.

EDIT: I asked my prof, and he says I can use the fact that the identity is valid in $\mathbb{R}$, and of course from this the proof is obvious.

Thanks again guys!