How to find the residue of a complex function

[Redwaves]

Homework Statement

Find the residue $$f(z) = \frac{z^2}{(z^2 + a^2)^2}$$

Relevant Equations

$$Res f(± ia) = \lim_{z\to\ \pm ia}(\frac{1}{(2-1)!} \frac{d}{dz}(\frac{(z \pm a)^2 z^2}{(z^2 + a^2)^2}) )$$

Hi,
I'm trying to find the residue of $$f(z) = \frac{z^2}{(z^2 + a^2)^2}$$
Since I have 2 singularities which are double poles.
I'm using this formula
$$Res f(± ia) = \lim_{z\to\ \pm ia}(\frac{1}{(2-1)!} \frac{d}{dz}(\frac{(z \pm a)^2 z^2}{(z^2 + a^2)^2}) )$$

then,
$$\lim_{z\to\ \pm ia} \frac{d}{dz}(\frac{z^2}{z^2 + a^2})$$

At this point, I don't get the correct answer which is $$\pm \frac{1}{4ai}$$

 

[fresh_42]

Let me see. With the formula from Wikipedia I get
$$
\operatorname{Res}_{+ i a }=\lim_{z \to + i a} \dfrac{d}{dz}\left(\dfrac{z^2}{(z+ia)^2}\right)\, , \,\operatorname{Res}_{- i a }=\lim_{z \to - i a} \dfrac{d}{dz}\left(\dfrac{z^2}{(z-ia)^2}\right)
$$
which is not what you have.

 

[Redwaves]

Let me see. With the formula from Wikipedia I get
$$
\operatorname{Res}_{+ i a }=\lim_{z \to + i a} \dfrac{d}{dz}\left(\dfrac{z^2}{(z+ia)^2}\right)\, , \,\operatorname{Res}_{- i a }=\lim_{z \to - i a} \dfrac{d}{dz}\left(\dfrac{z^2}{(z-ia)^2}\right)
$$
which is not what you have.

So I should replace a with ia and get the Res separately for + and -?

 

[fresh_42]

So I should replace a with ia and get the Res separately for + and -?

Yes. The formula is
$$
\textstyle \operatorname {Res}_{c}f=\tfrac{1}{\left(n-1\right)!}\lim_{z\rightarrow c}\dfrac{d^{n-1}}{d z^{n-1}} [(z-c)^{n}f(z)]
$$
and
$$
f(z)=\dfrac{z^2}{(z^2+a^2)^2}=\dfrac{z^2}{[(z+ia)(z-ia)]^2}=\dfrac{z^2}{(z+ia)^2(z-ia)^2}
$$
Alternatively, you could perform a long division and calculate the Laurent series, or use the integral definition with appropriate paths.

 

[Redwaves]

Speaking of Laurent series. I have a hard time to use it. Every videos of examples I watch they uses taylor or mclaurin series as Laurent series. I mean I though that the Laurent series of a complex function was given by $$f(z) = \sum_{n=-\infty}^{\infty} a_n (z-c)^n$$

 

[fresh_42]

Yes. But that does not say anything about $a_n=0$. They do not have to be infinite at either end. A polynomial is also a Laurent series, although a rather simple one.

 

[Redwaves]

Yes. But that does not say anything about $a_n=0$. They do not have to be infinite at either end. A polynomial is also a Laurent series, although a rather simple one.

I meant in general

 

[fresh_42]

In general, a Laurent series is a formal power series seen as a function: $f(z)=g(z)+h(1/z)$ with Taylor series for $g$ and $h$. They include polynomials, and Taylor series, which include Maclaurin Series. Specific examples are usually only at one end infinite and start or end at a certain final index.

Theoretically, however, Laurent series are of the form $\sum_{n=-\infty }^\infty a_n(z-c)^n$ without any information about the coefficients $a_n$.

 

[Redwaves]

So basically, we never use the form $$
\sum_{n=-\infty }^\infty a_n(z-c)^n
$$

 

[fresh_42]

So basically, we never use the form $$
\sum_{n=-\infty }^\infty a_n(z-c)^n
$$

We use it. We just rarely have functions with all $a_n \neq 0.$

 

[Redwaves]

Is it the same as Taylor's series without the derivative.
c is the point of a singularity, right?

 

[fresh_42]

Is it the same as Taylor's series without the derivative.
c is the point of a singularity, right?

Yes. The derivatives and faculties in the Taylor series are the coefficients in the Laurent series. Laurent series are a generalization of Taylor series.

 

[Redwaves]

$$a_n$$ is the Cauchy's integral formula, why we didn't simply use that?

 

[fresh_42]

See post #4. But I think the formula you suggested is the easier calculation. We only have to differentiate a quotient once.

 

[Redwaves]

I think I understand a bit, but I really need to know how to use Laurent series. What makes me really confused is the general formula with Cauchy, It seems like there are a lot of way to find a Laurent series. Some use geometric series as Laurent series. It's probably not clear what I mean.