Contour Integration: finding residues

1. Aug 27, 2010

Piano man

Hi, I'm stuck on this problem:

$$\int{\frac{1}{z^4+1}}$$

Writing it as a product of its roots, we get:

$$\frac{1}{(z-e^{\frac{i\pi}{4}})(z-e^{\frac{3i\pi}{4}})(z-e^{\frac{5i\pi}{4}})(z-e^{\frac{7i\pi}{4}})}$$

Then applying Cauchy's residue theorem for simple poles:
$$\mbox{Res}(f,c)=\lim_{z\rightarrow c}(z-c)f(z)$$

It's here that I'm stuck - I've got the poles and the function, how do I get the residues in this case?

2. Aug 27, 2010

Office_Shredder

Staff Emeritus
For the limit, the (z-c) that you're multiplying by will cancel with the (z-c) in the denominator and you can just plug in the value for the limit. However, you kind of have a mess left to multiply out. It's a lot easier to use the fact that if $$f(z)=\frac{g(z)}{h(z)}$$ where h(z) has a simple zero at c, the residue at c is $$\frac{g(z)}{h'(c)}$$ if you've seen this before

3. Sep 1, 2010

Piano man

Thanks for that.

Trying both ways, I come out with the same residue for $$c = e^{\frac{i\pi}{4}}$$
of $$\frac{1}{-2\sqrt{2}+2i\sqrt{2}}$$

But the answer I'm looking for is $$\frac{1}{4i\sqrt{i}}$$ because that's what you get when you expand the Laurent series.

So any ideas where I've gone wrong?

Thanks :)

4. Sep 1, 2010

HallsofIvy

Have you considered expanding $4i\sqrt{i}$ in the form a+ bi? You might find that those are the same.