Recent content by mancini0

Thank you. When x <0, ψ1 = Aeik1x+Be-ik1x When 0<x<L, ψ2 = Cekx+ De-kx When x> L , ψ3 = Feik1x+Ge-ik1x Now I can see how that the A term represents the incident wave, the B term represents the reflected wave, The F term represents the transmitted wave, and the G term must be zero...

Homework Statement "Use the semi - infinite well potential to model a deuteron, a nucleus consisting of a neutron and a proton. Let the well width L be 3.5 x 10^-15 meters and V - E = 2.2 MeV. Determine the energy E, and determine how many excited states there are."Homework Equations Since V >>...

I see. Thank you. The follow up asks me if f is bounded for |z|<= N. Since f is bounded and continuous, f is constant by Liouville's Theorem. Doesn't this mean f(z) is bounded for all z then?

Homework Statement Hi everyone. I must show that if f is a continuous function over the complex plane, with limit as z tends to infinity = 0, then f is in fact bounded. The Attempt at a Solution Since f is continuous and lim z --> infinity f(z) = 0, by definition of limit at infinity I know...

I meant the integral of z^n dz, which should be 0 by the problem statement. Somebody get this man a Field's Medal! Thanks for your help.

Okay, for example suppose n = -2. Then integrating (e^i*theta)^-2 from 0 to 2pi gives me zero, as desired. I suppose I could use mathematical induction from here on out?

Yikes. I made an elementary mistake in taking the derivative of e^i*theta. Then the integral reduces to the integral of i d(theta) from theta = 0 to theta = 2pi, which is 2(pi)i.

Thank you. The fact that theta and r are not analytic follow from the fact that theta and r are functions of z = x+iy. That thought was bothering me. When n = -1, my problem is that I calculate the integral of z^-1 dz to in fact be zero, when z = re^i*theta. But a previous problem showed me that...

Shucks... the exponential is analytic everywhere, so I can use Cauchy's Integral Theorem. :0

Homework Statement Hello everyone. I'm trying to finish the following problem: Show that \intz^n dz = 0 for any closed smooth path and any integer not equal to -1. [If n is negative, assume that γ does not pass through the origin, since otherwise the integral is not defined.] Homework...

Okay, thank you very much for your help thus far. By De Moivre's Theorem I would take the nth power of the modulus and multiply the argument by n. But why would I do this on e^2ix? I thought De Moivre's theorem only applies when a complex number is raised to an integer?

Hmmm, I think it hit me...if x = -iy, then z=x+iy implies z = -iy +iy = 0 is the root.

We haven't discussed the complex logarithm, that comes tomorrow! I don't see how my above solution shows sin(z) is always real valued.

e^2(iz) = e^2i(x+iy) = e^(2ix -2y) = e^2ix / e^2y since subtraction in the exponents corresponds to division of base. The above = 1 when e^2ix = e^2y i.e, when 2ix = 2y, when ix = y x = -iy

Hmmm... so the roots are when e^(2iz) - 1 = 0, i.e, when e^(2iz) = 1. e^(2iz) = e^(iz) * e^(iz) = (cos(z) + isin(z) ) * (cos(z) + isin(z)) 1= cos^2(z) - sin^2(z) + i(sin(z)cos(z)) Recognizing that cos^2(z) - sin^2(z) = cos(2z), I can rewrite the above as 1 = cos(2z) + isin(z)cos(z) Now...