Contour Integration: finding residues

[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?

 

[Office_Shredder]

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

 

[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 :)

 

[HallsofIvy]

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