Recent content by beefcake24

Ohh haha that was stupid. Thanks!

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) =...

Yeah, me too. I re-read that section in the book and it wasn't really helpful. I am 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.

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.

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.

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...

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?

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...

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...

Perfect, that was just what I needed to finish the problem. Thanks!

Haha yeah, I definitely worded that incorrectly, just edited my original post. Thanks for the reply, that's exactly what I needed!

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...

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...

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...

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.