MHB Using Classification of Poles to Calculate Residues

[shen07]

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

 

[I like Serena]

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?

 

[alyafey22]

$$\cdots , g\in H(D(a;r)):$$

May I ask what do you mean by that ?

 

[shen07]

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

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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?

 

[I like Serena]

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.

 

[alyafey22]

Type 1 is a special case of Type 2 by letting $$k(z) = z-a $$ . Of course we must have that $$h(z) $$ is holomorphic in a neighborhood of $$a$$.

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

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

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

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

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

 

[shen07]

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??

 

[alyafey22]

You could use Type 1 by factorization

$$z^3+1= (z+1)(z^2-z+1)$$

It won't be complicated.

 

[shen07]

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??

 

[alyafey22]

For example :

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

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

Then the Laurent expansion of the function around $$z=0$$ will be

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

Clearly the residue will be $$\frac{-1}{2!}$$

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