# Recent content by beefcake24

1. ### Complex Analysis Residues at Poles

Ohh haha that was stupid. Thanks!
2. ### Complex Analysis Residues at Poles

Homework Statement Find the residue at each pole of zsin(pi*z)/(4z^2 - 1) Homework Equations An isolated singular point z0 of f is a pole of order m if and only if f(z) can be written in the form: f(z) = phi(z)/(z-z0)^m where phi(z) is analytic and nonzero at z0. Moreover, Res(z=z0) f(z)...
3. ### Complex Analysis Integration with Sin and Cos

Yeah, me too. I re-read that section in the book and it wasn't really helpful. Im pretty stumped, I think I'm just going to ask my professor how to solve it and hope there are no problems like this on the final.
4. ### Complex Analysis Integration with Sin and Cos

Oh, good point. I'm a little fuzzy on meromorphic functions, I'll re-read that part of the book and see if it provides any insight. Thanks for your help jackmell, your input has been really helpful.
5. ### Complex Analysis Integration with Sin and Cos

Crap that didn't work at all. Anyway, it's supposed to be the integral from -p to -1 of sin(i[ln(2r) + i*pi]/(8r^3 - 1) dr + the contour integral of f(z) around the unit circle - the integral from -p to -1 of sin(i[ln(2r) + 3*i*pi]/(8r^3 - 1) dr + the contour integral of f(z) around Cp.
6. ### Complex Analysis Integration with Sin and Cos

Ok, so I have this so far: \int_-p^-1 \! \frac{sin(i[ln(2r)+i∏]}{8r3-1} \, \mathrm{d} r + \oint_C f(z) dz - \int_-p^-1 \! \frac{sin(i[ln(2r)+3i∏]}{8r3-1} \, \mathrm{d} r + \oint_Cp f(z) dz = 2∏i*Res f(z) at z = 1/2 Does this look right? One more thing, what do you do with the contour...
7. ### Complex Analysis Integration with Sin and Cos

But then the singularity z = 1/2 is on the branch cut. So we can just make the branch cut on the negative x axis, with z behaving like e^(i*pi) along the top and behaving like e^(3i*pi) on the bottom right?
8. ### Complex Analysis Integration with Sin and Cos

Ok, so after factoring, we can see that there's a simple pole at z = 1/2. Also, f is undefined at z = 0, but analytic at every point in a neighborhood of z = 0. We also have a branch along the positive x axis, with z behaving like 1 on top and e^(2∏i) when approaching on the other side. So...
9. ### Complex Analysis Integration with Sin and Cos

Homework Statement Compute the integral from 0 to 2∏ of: sin(i*ln(2e^(iθ)))*ie^(iθ)/(8e^(3iθ)-1) dθ (Sorry for the mess, I don't know how to use latex) Homework Equations dθ=dz/iz sinθ = (z - z^(-1))/2i The Attempt at a Solution So I tried to change it into a contour integral of a...
10. ### Integrate over complex contour

Perfect, that was just what I needed to finish the problem. Thanks!
11. ### Complex analysis integration

Haha yeah, I definitely worded that incorrectly, just edited my original post. Thanks for the reply, that's exactly what I needed!
12. ### Converting complex power series into a function

Hey guys, sorry for sending out so many questions so fast. I just discovered this site, and it looks great. Plus, I have my first complex analysis midterm tomorrow, so I'm pretty stressed (you'd think after 4 years of math/econ/computer science you'd get used to it but there's nothing like the...
13. ### Complex analysis integration

Evaluate the integral of f over the contour C where: f(z) = 1/[z*(z+1)*(z+2)] where C = {z(t) = t+1 | 0 <= t < infinity} Over this contour, is f a real valued function? z(t) just maps t to the t+1, so it seems as if the contour is a real-valued continuous function, and f does not have any...
14. ### Integrate over complex contour

The contour is just the unit circle, but starting at (0,i) when t=0 instead of (0,0) and traversed clockwise. Does this mean I can just take the negative of the integral around the unit circle? I know traversing a path in the opposite direction just changes the sign of the integral, and I...
15. ### Integrate over complex contour

I don't know, we haven't covered the residue theorem yet, and we do not need to know the residue theorem for our midterm, so I think we should be able to solve this without it.