Residues - complex analysis confusion.

[Zukias]

Hi! I am having a hard time with residues. :( I understand the formal definition of a residue, but anything past that and I am struggling. My course lecturer has a very confusing way of organising the course material and it is very hard to comprehend so was wondering if anyone could help with these questions or direct me to another source which would help explain what to do?

part i: Calculate res(f, 0) of an even meromorphic function f which has a pole at 0. (...the order of the pole is not given.)

part ii: Can anything be deduced about res(f,0) for an odd meromorphic function f with a pole at 0? Justify your answer. (...again, the order of the pole isn't given.)

Part iii:
Find \int^{\infty}_{-\infty}\frac{cosx}{x^2 + 2x + 4}\,dx and \int^{\infty}_{-\infty}\frac{sinx}{x^2 + 2x + 4}\,dx

Any help would be much appreciated. :)

 

[chisigma]

Hi! I am having a hard time with residues. :( I understand the formal definition of a residue, but anything past that and I am struggling. My course lecturer has a very confusing way of organising the course material and it is very hard to comprehend so was wondering if anyone could help with these questions or direct me to another source which would help explain what to do?

part i: Calculate res(f, 0) of an even meromorphic function f which has a pole at 0. (...the order of the pole is not given.)

Wellcome on MHB Zukias!...

Let's start from the part I, that is absolutely not trivial!... let's suppose that f(*) can be represented as Laurent series around z=0...

$\displaystyle f(z) = \sum_{n = - \infty}^{+ \infty} a_{n}\ z^{n}\ (1)$

In this case the residue of f(*) in z=0 is the coefficient $a_{-1}$...

Kind regards

$\chi$ $\sigma$

 

[DreamWeaver]

Welcome on MHB Zukias!...

Seconded! A very warm welcome to the forum, Zukias! :D (Heidy)

 

[Zukias]

Thanks for the welcomes (drink)

Wellcome on MHB Zukias!...

Let's start from the part I, that is absolutely not trivial!... let's suppose that f(*) can be represented as Laurent series around z=0...

$\displaystyle f(z) = \sum_{n = - \infty}^{+ \infty} a_{n}\ z^{n}\ (1)$

In this case the residue of f(*) in z=0 is the coefficient $a_{-1}$...

Kind regards

$\chi$ $\sigma$

Thanks for replying :) and I should have specified the reason I am struggling with part i. I know that an even function is a function like cos(x) which is symmetric about the y-axis and I know the laurent expansions of cos(x) and maybe a few other individual even functions, but I don't know where to start in such a general case as in the question.

 

[alyafey22]

You have to use two facts

$$f(-x)=f(x)$$

And that

$$f(x)=\sum^\infty_{n=-\infty} a_n x^n $$

Now use fact 1 to compare coefficients of $x^{-1}$ in the Laurent expansion.

 

[Zukias]

You have to use two facts

$$f(-x)=f(x)$$

And that

$$f(x)=\sum^\infty_{n=-\infty} a_n x^n $$

Now use fact 1 to compare coefficients of $x^{-1}$ in the Laurent expansion.

Wouldn't the coefficients of $x^{-1}$ in the LE's of f(x) and f(-x) be the same, since if f(x) = f(-x), then surely they'd have identical Laurent expansions.

Also if $b_n$ are the coefficients for the Laurent Expansion of f(-x), since the 1/z term has odd power between the two series of f(x) and f(-x), $$a_{-1}/z = -b_{-1}/z$$, implying $$a_{-1} = -b_{-1}$$

And the only constant which is equal to the negative of itself is zero, so res(f,0)=0 ? Hope I'm on the right tracks... :p

 

[Prove It]

This may help you with the third part.

 

[alyafey22]

Wouldn't the coefficients of $x^{-1}$ in the LE's of f(x) and f(-x) be the same, since if f(x) = f(-x), then surely they'd have identical Laurent expansions.

The coefficients of odd terms will not be the same they will differ in sign. So all odd terms will vanish leaving only even terms which is necessary for the function to be even.

Also if $b_n$ are the coefficients for the Laurent Expansion of f(-x), since the 1/z term has odd power between the two series of f(x) and f(-x), $$a_{-1}/z = -b_{-1}/z$$, implying $$a_{-1} = -b_{-1}$$

And the only constant which is equal to the negative of itself is zero, so res(f,0)=0 ? Hope I'm on the right tracks... :p

Yes, but don't use a different symbol because this will create confusion but I guess you have the concept. Remember ,also, that the coefficients might be complex but the same conclusion follows from equality of complex numbers.