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Using Classification of Poles to Calculate Residues

shen07

Member
Aug 14, 2013
54
Hi guys i wanted to clear out some confusion, Suppose $\text{z=a}$ is a simple pole, my Professor classify it as follows:

Simple Pole at z=a

Type 1: $$f(z)=(z-a)^{-1}g(z),g\in H(D(a;r)):$$

$$Res\{f(z);a\}=g(a)$$​

Type 2: $$f(z)=\frac{h(z)}{k(z)},h(a)\neq 0,k(a)=0,k^{'}(a)\neq 0:$$
$$Res\{f(z);a\}=\frac{h(a)}{k^{'}(a)}$$​

Now my question is how do i know when should i use the Type 1 or Type 2??

Please Help
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,901
Hi guys i wanted to clear out some confusion, Suppose $\text{z=a}$ is a simple pole, my Professor classify it as follows:

Simple Pole at z=a

Type 1: $$f(z)=(z-a)^{-1}g(z),g\in H(D(a;r)):$$

$$Res\{f(z);a\}=g(a)$$​

Type 2: $$f(z)=\frac{h(z)}{k(z)},h(a)\neq 0,k(a)=0,k^{'}(a)\neq 0:$$
$$Res\{f(z);a\}=\frac{h(a)}{k^{'}(a)}$$​

Now my question is how do i know when should i use the Type 1 or Type 2??

Please Help
Suppose you pick $k(z)=z-a$ in Type 2.
Which residue do you get?
Recognize it?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667

shen07

Member
Aug 14, 2013
54
May I ask what do you mean by that ?

Hi Zaid, its a notation: g is holomorphic(H) in the Disc center a,radius r

- - - Updated - - -

Suppose you pick $k(z)=z-a$ in Type 2.
Which residue do you get?
Recognize it?
What you mean to say is that, both are the same?? bt in what case should we use each type?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,901
What you mean to say is that, both are the same?? bt in what case should we use each type?
Type 2 is a more general formulation of the same thing.
Use whatever is easiest for the problem at hand.

Although I do not think that $h(a) \ne 0$ is a necessary precondition to use type 2.
It should suffice that h(a) exists.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Type 1 is a special case of Type 2 by letting \(\displaystyle k(z) = z-a \) . Of course we must have that \(\displaystyle h(z) \) is holomorphic in a neighborhood of \(\displaystyle a\).

Assume that \(\displaystyle z=a\) is a simple pole of order $1$ then we can have

\(\displaystyle f(z) = \frac{g(z)}{z-a}\)

Then the residue at \(\displaystyle z=a\) is \(\displaystyle f(z) = \frac{g(a)}{(z-a)'}=g(a)\)

If \(\displaystyle f(z) = \frac{h(z)}{k(z)}\) where \(\displaystyle h(z)\) has a simple zero of order \(\displaystyle 1\) at \(\displaystyle z=a\) then

\(\displaystyle \text{Res}(f(z);a) = \lim_{z \to a}(z-a) \frac{h(z)}{k(z)-k(a)}=\lim_{z \to a} \frac{h(z)}{\frac{k(z)-k(a)}{z-a}}=\frac{h(a)}{k'(a)}\)
 

shen07

Member
Aug 14, 2013
54
Suppose i have like a function

$$f(z)=\frac{(z-i)^2}{(z^3+1)}$$

then using Type 2 is much easier here than using Type 1, $$Res\{f(z);-1\}=\frac{h(-1)}{k^{'}(-1)},\text{ where }h(z)=(z-i)^{2}\text{ & }k(z)=z^{3}+1$$

Using type 1 would complicate things,Right??
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You could use Type 1 by factorization

\(\displaystyle z^3+1= (z+1)(z^2-z+1)\)

It won't be complicated.
 

shen07

Member
Aug 14, 2013
54
What about the Residues of Multiple Poles:

Multiple pole at z=a order m

Type 1:
$$f(z)=(z-a)^{m}g(z),g \in H(D(a;r)):$$
$$\text{Res}\{f(z);a\}=\frac{g^{m-1}(a)}{(m-1)!}$$

Type 2:
$$\text{Res}\{f(z);a\}=\text{Coefficient }C_{-1}\text{ of }\frac{1}{z-a}$$
in the Laurent expansion of f(z) about z=a

When to use each type??
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
For example :

\(\displaystyle f(z) = \frac{\cos(z)}{z^3}\) then you have two choices and that depends on the easieness of expanding the function .

Since we know that \(\displaystyle \cos(z) = 1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots \)

Then the Laurent expansion of the function around \(\displaystyle z=0\) will be

\(\displaystyle f(z) = \frac{ 1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots }{z^3}=\frac{1}{z^3}-\frac{1}{2! \, z}+\frac{1}{4!}\, z+\cdots \)

Clearly the residue will be \(\displaystyle \frac{-1}{2!}\)

To use Type 1 we use the formula which is easy since \(\displaystyle \cos^{(2)}(z) = -\cos(z) \) so the residue will be \(\displaystyle -\frac{1}{2!} \cos(0) = \frac{-1}{2!}\)