2011-06-03, 05:25 AM
(This post was last modified: 2011-06-04, 08:24 AM by 2147483647.)
I'm not sure what you're trying to say here. Cauchy's formula (the first path integral over the unit circle):
∮1/z dz = 2πi
is indeed correct. The point here is that it should be 2πi all the time because of the deformation theorem, but for whatever reason, I'm getting 0 whenever I'm not integrating over the unit circle.
I believe you're talking about ∇arctan(y/x), the vortex vector field. I don't think ∇arctan(y/x) is applicable here, because z∈ℂ, but x,y∈ℝ. In both situations, the winding number appears partially due to the singularity at (0,0), but note that for z∈ℂ, singularities don't work the same way as they do in ℝ. This can be shown by the following integral:
∮1/z² dz = 0
z = e^(iθ
dz = ie^(iθ dθ
∮1/z² dz = ∫ ie^(iθ / (e^(iθ)^2 dθ, [0,2π]
= ∫ ie^(iθ / e^(2iθ dθ, [0,2π]
= ∫ ie^(-iθ dθ, [0,2π]
= -e^(-iθ [0,2π]
= 0
Even though 1/z² has a singularity of order two at z=0, the result of ∮1/z² dz is still 0, because 1/z² is holomorphic. Which is the other thing that's confusing to me. If z has two singularities, why is it holomorphic? Actually, why is 1/z^n, where n∈ℤ \{1} holomorphic? It almost seems to defy the residue theorem, which suggests that I should just count up the number of singularities and multiply by 2πi to obtain the closed contour integral over the entire Re-Im plane, kinda like in here.
∮1/z dz = 2πi
is indeed correct. The point here is that it should be 2πi all the time because of the deformation theorem, but for whatever reason, I'm getting 0 whenever I'm not integrating over the unit circle.
I believe you're talking about ∇arctan(y/x), the vortex vector field. I don't think ∇arctan(y/x) is applicable here, because z∈ℂ, but x,y∈ℝ. In both situations, the winding number appears partially due to the singularity at (0,0), but note that for z∈ℂ, singularities don't work the same way as they do in ℝ. This can be shown by the following integral:
∮1/z² dz = 0
z = e^(iθ
dz = ie^(iθ
dθ∮1/z² dz = ∫ ie^(iθ
/ (e^(iθ)^2 dθ, [0,2π]= ∫ ie^(iθ
/ e^(2iθ dθ, [0,2π]= ∫ ie^(-iθ
dθ, [0,2π]= -e^(-iθ
[0,2π]= 0
Even though 1/z² has a singularity of order two at z=0, the result of ∮1/z² dz is still 0, because 1/z² is holomorphic. Which is the other thing that's confusing to me. If z has two singularities, why is it holomorphic? Actually, why is 1/z^n, where n∈ℤ \{1} holomorphic? It almost seems to defy the residue theorem, which suggests that I should just count up the number of singularities and multiply by 2πi to obtain the closed contour integral over the entire Re-Im plane, kinda like in here.