[SOLVED] Poles, order of it and residue

**arbolis** · October 19th 2009, 02:09 PM

I must prove that the singularities of $f$ are poles. I also must determine their order and the corresponding residue.

$f(z)=\frac{z}{\cos z}$ .

My attempt :
I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
So I wrote $\cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$ .
So if I can find the order of the zeros of $\frac{1}{f(z)}$ I'm done.
I've found out that $\frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$
The first term of the series really bothers me. If it wasn't there then $\frac{1}{f(z)}$ would be equal to $z\cdot h(z)$ with $h(0) \neq$ $0$ and so $z=0$ would be a $0$ of order $1$ of $\frac{1}{f}$ , thus a pole of order $1$ of $f$ . But of course I should take in count the $\frac{1}{z}$ term and look for $z=k \frac{\pi}{2}$ , $k \in \mathbb{Z}$ , but I'm stuck.

**mr fantastic** · October 19th 2009, 03:51 PM

Originally Posted by arbolis

I must prove that the singularities of $f$ are poles. I also must determine their order and the corresponding residue.

$f(z)=\frac{z}{\cos z}$ .

My attempt :
I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
So I wrote $\cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$ .
So if I can find the order of the zeros of $\frac{1}{f(z)}$ I'm done.
I've found out that $\frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$
The first term of the series really bothers me. If it wasn't there then $\frac{1}{f(z)}$ would be equal to $z\cdot h(z)$ with $h(0) \neq$ $0$ and so $z=0$ would be a $0$ of order $1$ of $\frac{1}{f}$ , thus a pole of order $1$ of $f$ . But of course I should take in count the $\frac{1}{z}$ term and look for $z=k \frac{\pi}{2}$ , $k \in \mathbb{Z}$ , but I'm stuck.

f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

The singularities are $z = z_m = \frac{(2m+1)\pi}{2}$ where m is an integer.

So you need to show that $\lim_{z \rightarrow z_m} (z - z_m)^n f(z)$ is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

eg. see - Wolfram|Alpha

**arbolis** · October 25th 2009, 04:09 PM

Originally Posted by mr fantastic

f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

The singularities are $z = z_m = \frac{(2m+1)\pi}{2}$ where m is an integer.

So you need to show that $\lim_{z \rightarrow z_m} (z - z_m)^n f(z)$ is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

eg. see - Wolfram|Alpha

I wasn't able to solve the problem using your tip.
Here's what I've done :
Considering the singularity at $z=\frac{\pi}{2}$ , the series expansion of $\cos z$ around this singularity (singularity for $f$ ) is $1+ \left ( z- \frac{\pi}{2} \right )^2 h(z)$ where $h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-...$ so $h\left ( \frac{\pi}{2} \right ) \neq 0$ .
--------------------------------------------------------------------
I've a side-note question : From this can I conclude that the order of the pole of $r(z)=\frac{1}{\cos z}$ when $z=\frac{\pi}{2}$ is 2?
--------------------------------------------------------------------
Now I wrote that the pole of $f$ occurs when $\frac{1}{f}$ has a zero. That is, when $\frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0$ . I go nowhere.

No let me try again you way : $\lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]$ .
For $n=1$ , the limit is worth $0$ . Does that mean the order of the pole of $f$ at $z=\frac{\pi}{2}$ is $1$ , and that I've successfully solved the exercise?

**mr fantastic** · October 26th 2009, 03:39 AM

Originally Posted by arbolis

I wasn't able to solve the problem using your tip.
Here's what I've done :
Considering the singularity at $z=\frac{\pi}{2}$ , the series expansion of $\cos z$ around this singularity (singularity for $f$ ) is $1+ \left ( z- \frac{\pi}{2} \right )^2 h(z)$ where $h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-...$ so $h\left ( \frac{\pi}{2} \right ) \neq 0$ .
--------------------------------------------------------------------
I've a side-note question : From this can I conclude that the order of the pole of $r(z)=\frac{1}{\cos z}$ when $z=\frac{\pi}{2}$ is 2?
--------------------------------------------------------------------
Now I wrote that the pole of $f$ occurs when $\frac{1}{f}$ has a zero. That is, when $\frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0$ . I go nowhere.

No let me try again you way : $\lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]$ .
For $n=1$ , the limit is worth $0$ . Does that mean the order of the pole of $f$ at $z=\frac{\pi}{2}$ is $1$ , and that I've successfully solved the exercise?

The link I posted shows that the limit exists and is finite. Therefore there is a simple pole. You should consider the general limit $\lim_{z \rightarrow z_m} (z - z_m) f(z)$ to show that f(z) has a simple pole at $z = z_m$ .

**arbolis** · October 26th 2009, 05:57 AM

Ah I see, I should find that the limit is worth $-\frac{\pi}{2}$ instead of $0$ .
Thanks once again.

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