Morera's Theorem and Corollaries -- Complex Analysis 14
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- Опубликовано: 7 фев 2025
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love the shirt!
I feel like there needs to be some grand theorem of complex analysis, that just says “With whatever necessary assumptions, the following are equivalent:
1)f is analytic
2)f is holomorphic
3)f satisfies the Cauchy-Riemann equations
4)f is harmonic
5)f is locally conformal
6)The contour integral of f vanishes on every rectangle
7)Whatever others I can’t remember
For the first proof, I am confused as to where exactly it was used that the integral over dR is 0 for any rectangle in D. What makes that condition necessary for the proof given?
Yeah, that confused me too. I *think* the answer is that because that condition implies path independence, then our definition of F as that antiderivative is well-defined.
@@Alex_Deam He didn't explain this which I think was a pretty big mistake. The integral is well defined regardless of path independence because he stated what the path is. The condition that the integral over a rectangle is 0 was used to get that F(z+dz)-F(z) is equal to the integral between z and z+dz. If you draw a picture of what is happening you can notice that this equality only holds if the integral is 0 over the rectangle formed by the paths on the Y axis to z and z+dz.
The condition is necessary for the step at 9:16, and the logic was brushed over. See my other comment for an explanation!
Me too... still confused...
Thank you for your generosity sharing math college concepts and problems. Regarding complex analysis, could you condense the key concepts with guided problems in one video? I wanna have a quick refresher of it. Thanks in advance.
Morera's theorem is usually formulated with triangles. The proof is much cleaner (see "green" Rudin) and avoids the problem at 5:30 already pointed in one of the comments. Also by waltzing around this problem, it becomes unclear where the assumption of the theorem comes into play.
I am a bit puzzled why people feel it necessary to present homegrown proofs on youtube, while better proofs are available in classic texts.
Less puzzling, but more dangerous is the approval of this by so many commenters, some of whom are students and are not yet able to distinguish between good and bad proofs.
If a "proof" is a proof, it cannot be a bad proof. :-) (It can be difficult to understand, unaesthetic (always a matter of taste), unnecessarily complicated, etc.).
At 9:16 I think you mean the "shortest path is length |Delta z|." Since the length of _any_ path will give an upper bound, it suffices to choose the shortest for the tightest bound.
Technically some logic was brushed over at 9:16. The result ends up being correct but this is where the assumption of vanishing on any rectangle needs to be invoked.
From the definition of the path at the start of the proof, we know that the path length is Delta x + Delta y (note it should read | dw | after using the triangle inequality for integrals around 9:00). This is smaller or equal to sqrt(2) |Delta z| (equality holds for Delta x = Delta y) which is smaller than 2 |Delta z|. The fact that integrals on closed rectangular contours vanish was used at the start to write F(z+Delta z) - F(z) as a single integral over a rectangular path between z and z + Delta z.
It would be nice to compare Morera’s theorem to Cauchy’s integral theorem. In the former the domain need not be simply connected. A counterexample is therefore given by f(z) = 1/z which is analytic on the punctured plane, but any closed contour integral surrounding the origin does not vanish. If the domain is simply connected then the converse of Morera’ s theorem holds and is exactly given by Cauchy’ s theorem plus the fact that any analytic function is continuous.
Very nice! However, I am taking some issue with the justification of your step at 9:16. Technically, from the definition of F, the integral path consists of three linear contours, one from z to Re(z), one from Re(z) to Re(z+delta z), and one from Re(z+ delta z) to z+delta z. These have lengths Im(z), Re(delta z), and Im(z+delta z), respectively.
At this point, you can proceed by introducing an auxiliary path to complete the rectangle, and then invoke the assumption about the contour vanishing on any closed rectangle to demonstrate equivalence to an integral over a second contour that is confined to a |delta z| neighborhood of z. Then you can apply continuity to bound the integrand, and the path length of the contour is |Re(delta z)|+|Im(z+ delta z)-Im(z)|, which is the same as |Re(delta z)|+|Im(delta z)|. This is indeed bounded by 2|delta z|.
5:30 why can we do this? If delta z is only in the imaginary axis, it checks out, but if it has a real component, don't we have the one along the real axis and then the one along the imaginary axis doesn't overlap?
The difference is a boundary integral of a rectangle which vanishes by assumption. Because of this assumption and the definition of the path, you can apply the same rules for combining limits as you would for real analysis.
Was that a saxophone in the background around 9:30? In any case, thanks again for another wonderful video!
How can we take derivative of F(z) at 3:42? How do we know if it is differentiable/analytic?
If we can show that the limit in 3:42 exist, then F(z) is differentiable and its derivative F'(z) = that limit.
Hey, I always love your math and I genuinely aspire to emulate you ad a teacher, but also
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Thanks prof Penn
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Pick delta so that the proof is true, then prove the proof? Am lost with this one MP :S
The goal is to show f has an antiderivative F, and that F is analytic. It suffices to show F'(z)=f(z) for all z, and since f is continuous, by the definition of analytic, F would be analytic. Since the derivative of an analytic function is also analytic, f must be analytic.
To show F'(z)=f(z) he needs to show F'(z)-f(z)=0, which he expresses as the integral at 6:35 . He chooses epsilon arbitrarily and a delta is guaranteed with the property he describes because that is the formal definition of continuity i.e. "for every epsilon>0, there exists a delta>0 such that for all w satisfying |w-z|z."