# Homework Help: Residue Theorem

1. Mar 30, 2009

### latentcorpse

for the function $f(z)=\frac{ze^{iz}}{z^4+\alpha^4}, \alpha>0$

what are the residues of the poles in the upper half plane

so i factorised the denominator to $(z^2+i \alpha^2)(z^2-i \alpha^2)$
my problems are:

(i)but then i wasn't sure how to characterise this the z^2 had me confused as to whether these were simple poles or dobule poles?

(ii)also the question said "poleS" in the upper half plane making me think there was more than 1???

(iii) i dont know how to proceed without being able to get the denominator into the form of (z-a)(z-b) or something along those lines

can anybody help me here?

2. Mar 31, 2009

### gammamcc

Keep factoring (factorizing [sic]). Hint: i=exp(i*pi/2).

3. Mar 31, 2009

### latentcorpse

i get $z^2+i \alpha^2=(z+ie^{\frac{i \pi}{4}} \alpha)(z-ie^\frac{{i \pi}{4}} \alpha)=(z-\alpha)(z+\alpha)$
so poles at +/-alpha

the other term in the denominator gives poles at +/-i alpha.

so we have four simple poles and then it should be easy to get residues etc.

would there have been an easier way to factorise this that would have avoided dealing with exponentials?

4. Mar 31, 2009

### HallsofIvy

Ahhh, $\alpha$ is NOT equal to $ie^{\frac{i\pi}{4}}\alpha$. Why would you think it was?
$$ie^{\pi/4}= i(\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2})= -\frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}$$

The fourth roots of $-\alpha$ lie on a circle in the complex plane with radius $\alpha^{1/4}$ and center at 0. Since the argument of -1 is $\pi$ the arguments of the fourth roots of $-\alpha$ are $\pi/4$, $3\pi/4$, $5\pi/4$, and $7\pi/4$. The first two of those are in the upper half plane.

5. Mar 31, 2009

### latentcorpse

my bad. i was still working with $e^{\frac{i \pi}{2}}$ and so just used i*i and i*(-i). a lesson in taking your time. ok so then i just evalueate the residues of simples poles as usual i.e.

now if i let $Q=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ then the residue at Q can be obtained by

$\lim_{z \rightarrow Q} \frac{(z-Q)(ze^{iz})}{z^4+\alpha^4}$ yes?

secondly,
could i also have found the poles by considering the fourth roots of $-\alpha^4$ and using de Moivre then like youy did above? (or would it be the fourth roots of $\alpha^4$ - i'm inclined to go with the first but would like to double check)

thanks.

6. May 6, 2009

### latentcorpse

surely multiplying by z-Q on the top is going to leave a whole bunch of crap on the denominator and its going to get very messy!