Quick question on order of poles

**aman_cc** · October 4th 2009, 09:12 AM

For a complex function $f(z)$ which has an isolated singularity at $z=z_0$

1. How do I find out if $z=z_0$ is a pole or not?
2. If it is a pole what is the order of the pole

Consider for e.g. $f(z)=\frac{1}{zsin(z)}$ . Is $z=0$ pole? If yes what is the order of the pole?

Is there a way to find answer to the above question without really writing down the Laurent Series about $z=z_0$

A related question I would have is - do we have similar method to comment on essential and removable singularities?

Thanks

**Laurent** · October 4th 2009, 10:04 AM

Originally Posted by aman_cc

For a complex function $f(z)$ which has an isolated singularity at $z=z_0$

1. How do I find out if $z=z_0$ is a pole or not?
2. If it is a pole what is the order of the pole

Consider for e.g. $f(z)=\frac{1}{zsin(z)}$ . Is $z=0$ pole? If yes what is the order of the pole?

Is there a way to find answer to the above question without really writing down the Laurent Series about $z=z_0$

You don't need the whole Laurent Series, only the first non-zero term. If you know that $f$ is analytic in $U\setminus\{z_0\}$ and that $(z-z_0)^k f(z)\to c\neq 0$ when $z\to z_0$ , then $z_0$ is a pole of order $k$ for $f$ . And the residue is $c$ .

In your example, you may know that $\frac{\sin z}{z}\to 1$ when $z\to 0$ , hence $z^2\frac{1}{z\sin z}\to_{z\to 0} 1$ , hence 0 is a pole of order 2, with residue 1.

**aman_cc** · October 4th 2009, 10:12 AM

Originally Posted by Laurent

You don't need the whole Laurent Series, only the first non-zero term. If you know that $f$ is analytic in $U\setminus\{z_0\}$ and that $(z-z_0)^k f(z)\to c\neq 0$ when $z\to z_0$ , then $z_0$ is a pole of order $k$ for $f$ . And the residue is $c$ .

In your example, you may know that $\frac{\sin z}{z}\to 1$ when $z\to 0$ , hence $z^2\frac{1}{z\sin z}\to_{z\to 0} 1$ , hence 0 is a pole of order 2, with residue 1.

Let me just re-phrase (so that I get it)
1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $(z-z_0)^k f(z)$ when $z\to z_0$
Then 'k' is the order of the pole at $z_0$ . Have I got it correct?

Few questions (I tried to work out a logic for this)
1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
2. Why is the assumption $f$ is analytic in $U\setminus\{z_0\}$ important?

**aman_cc** · October 4th 2009, 10:41 AM

Thinking about this I got a related question (which might help me understand why the limit method to find the order works)

Is is correct to say

if $f(z) = \frac{a_1}{z}+\frac{a_2}{z^2}+\frac{a_3}{z^3}+...+ \frac{a_k}{z^k}$

and Limit $z \to 0:f(z)$ exists then
1. each of ${a_1},{a_2},...,{a_k} = 0$
2. Above is true in both cases:
a) $k$ is finite
b) $k$ in infinite

Originally Posted by aman_cc

Let me just re-phrase (so that I get it)
1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $(z-z_0)^k f(z)$ when $z\to z_0$
Then 'k' is the order of the pole at $z_0$ . Have I got it correct?

Few questions (I tried to work out a logic for this)
1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
2. Why is the assumption $f$ is analytic in $U\setminus\{z_0\}$ important?

**shawsend** · October 4th 2009, 10:58 AM

Originally Posted by Laurent

You don't need the whole Laurent Series, only the first non-zero term. If you know that $f$ is analytic in $U\setminus\{z_0\}$ and that $(z-z_0)^k f(z)\to c\neq 0$ when $z\to z_0$ , then $z_0$ is a pole of order $k$ for $f$ . And the residue is $c$ .

$c$ then is the coefficient on the $\frac{1}{(z-z_0)^k}$ term and the residue is:

$\lim_{z\to z_0} \frac{1}{(k-1)!} \frac{d^{k-1}}{dz} (z-z_0)^k f(z)$

**Laurent** · October 4th 2009, 12:43 PM

Originally Posted by aman_cc

Let me just re-phrase (so that I get it)
1. Keep putting higher k (starting with 1), till we get a finite not zero limit for $(z-z_0)^k f(z)$ when $z\to z_0$
Then 'k' is the order of the pole at $z_0$ . Have I got it correct?

You don't really have to check every $k$ one by one: if $k$ is more than the order, the limit is 0, and if it is less than the order, the limit doesn't exist. Therefore, if the limit is finite and non-zero, $k$ is the right order. In your example, I could have written " $\sin z\sim_{z\to 0} z$ , hence $\frac{1}{z\sin z}\sim\frac{1}{z^2}$ hence the order is 2. "

Few questions (I tried to work out a logic for this)
1. Isn't this based on the assumption that there is a 'pole' and only thing we need is order?
2. Why is the assumption $f$ is analytic in $U\setminus\{z_0\}$ important?

Your questions are related to each other. With your definition of a pole (using Laurent series), you have: "for all $z$ in a small disc centered at $z_0$ , $f(z)=\sum_{n=-k}^\infty a_n (z-z_0)^n$ and $a_{-k}\neq 0$ ". This definition concerns $f$ not only "at $z_0$ " but also on a neighbourhood of $f$ . And it implies that $f$ is analytic on a disc around $z_0$ .

The condition $(z-z_0)^k f(z)\to c\neq 0$ , on the other hand, does not say anything about the regularity of $f$ outside $z_0$ . If however you know (in addition) that $f$ is analytic on a subset (like a disc) around $z_0$ (but deprived of $z_0$ ), then you can see that the function $g:z\mapsto (z-z_0)^k f(z)$ is analytic on the same subset, and has a finite limit $c$ at $z_0$ , so that $z_0$ is a removable singularity of $g$ (this is a property you probably know). In other words, $g$ (with $g(z_0)=c$ ) is analytic in the previous subset plus the point $z_0$ . Hence $g(z)=\sum_{n=0}^\infty a_n (z-z_0)^n$ for some sequence $a_n$ with $a_0=g(z_0)=c\neq 0$ . And from the relation $g(z)=(z-z_0)^k f(z)$ , one gets $f(z)=\sum_{n=0}^\infty a_n (z-z_0)^{n-k}$ $=\sum_{n=-k}^\infty a_{n+k} (z-z_0)^n$ . Thus we find a Laurent series and again the initial definition of pole and order. This justifies my previous post.

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