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Poopsilon
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Ok I can do the integral and see that it is equal to 2∏i, but thinking about it in terms of 'adding up' all the points along the curve I feel like every every point gets canceled out by its antipode, e.g. 1/i and -1/i.
Poopsilon said:Ok I can do the integral and see that it is equal to 2∏i, but thinking about it in terms of 'adding up' all the points along the curve I feel like every every point gets canceled out by its antipode, e.g. 1/i and -1/i.
One way of explaining his point is that in a Riemann sum ##\sum_k \frac{1}{z_k} \Delta z_k##, it's not just the 1/zk that's equal in magnitude and opposite in direction on the opposite side of the circle. ##\Delta z_k## is also equal in magnitude and opposite in direction. So if the contribution to the integral from one short segment of the circle is 1/z Δz, then the contribution from the "opposite" segment is (-1/z)(-Δz)=1/z Δz, i.e. exactly the same.Poopsilon said:Office_Shredder could you expand on what you said, maybe with special attention to our 1/z case, as from my POV it's this very fact that direction matters which I'm using in my argument that intuitively the integral should be zero, and not 2∏i.
I would say that it does. The contribution from opposite sides are the same in this case too, for precisely the same reason.Poopsilon said:But Fredrik, why doesn't this argument hold for f(z)=z?
It's precisely because it's either clockwise the whole way through or counter-clockwise the whole way through that that Δz has "opposite" values on opposite sides.Poopsilon said:And furthermore, I was under the impression that our direction of integration is not changing, it's either counter-clockwise or clockwise, and thus positive or negative accordingly.
But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.
How is it "only perceived"? The Δz clearly has the same magnitude but opposite sign on the opposite side. That's not a matter of perception.Poopsilon said:But that is only perceived change in direction, you don't multiply by an extra negative just because you're down at 5pi/4.
Why would it be? Office_Shredder and I have only used this observation about contributions from opposite segments to explain why the opposite sign of 1/z on opposite sides doesn't automatically make the integral zero.Poopsilon said:But then why wouldn't the integral of z over the unit circle also be non zero?
The theorem we need is the one that says that if B and C are two contours with the same endpoints, and B can be continuously deformed to C without going outside an open set on which f is analytic, then ##\int_B\, f(z)dz=\int_C\, f(z)dz##. That is a pretty deep theorem, but it's surprisingly easy to prove (see e.g. Saff & Snider). It follows almost immediately that if C is a closed contour, and f is analytic on an open set that contains C, then ##\int_C\, f(z)dz=0##.Poopsilon said:Clearly the reason z is zero and 1/z is not is that 1/z has a singularity inside the unit circle. The reason for this is I feel for some deep reason involving the proof of Cauchy's Integral Theorem and the proof of Pullbacks for integrals.
Poopsilon said:In fact upon performing the appropriate pullback I obtain the integral i∫dt from 0 to 2∏. So in this case I now do seem to be simply adding up the points along the curve. I'm not sure what this means..
@lugita: yes that seems like a good way to phrase it.
If ##\{z_0,\dots,z_n\}## are points on the circle, thenPoopsilon said:Fredrik, could you explain to me exactly what you mean by Δz.
Bacle2 said:But I think we can see that the net contributions of antipodes do not cancel out
(we can ignore the contribution of r, since it cancels out). While it is true that:
1/z= 1/eiθ =e-iθ =cos(-θ)+isin(-θ)=cosθ-isinθ
and, for the antipode ei(θ+∏)= -cosθ+isinθ ,
The contributions of θ=0 must cancel out that of both θ=∏. But the contribution of ∏
must also cancel out that of 2∏.
Office_Shredder said:This is incorrect. When integrating from 0 to 2pi the angles 0, pi and 2pi are the only angles for which you get this happening. And assuming your Riemann sum is a left or right hand sum the angles 0 and 2pi won't both show up
The integral of 1/z over a unit circle is also known as the contour integral of complex function 1/z. It represents the area enclosed by the unit circle in the complex plane.
The integral of 1/z over a unit circle is not zero because the function 1/z is not continuous at z=0, which is the center of the unit circle. This causes the integral to have a non-zero value.
The unit circle is used in this integral because it is a simple and commonly used contour in complex analysis. It allows for the evaluation of complex integrals using the Cauchy Integral Formula.
The integral of 1/z over a unit circle is an important concept in complex analysis as it is used to evaluate complex integrals and to prove the Cauchy Integral Theorem, which states that the value of a complex integral depends only on the values of the function inside the contour.
Yes, the integral of 1/z over any contour that encloses the singularity at z=0 will be non-zero. This is because the function 1/z is not continuous at z=0, which means it cannot be integrated along a path that passes through this point.