Morera's Theorem and Corollaries -- Complex Analysis 14

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[DeanCalhoun]

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?

[xgx]

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.

[JM-usfr]

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.

[romajimamulo]

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?

[willyh.r.]

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.

[JM-usfr]

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

[jimallysonnevado]

How can we take derivative of F(z) at 3:42? How do we know if it is differentiable/analytic?

[jenny]

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.

[connorfrankston]

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|.

[PunmasterSTP]

Was that a saxophone in the background around 9:30? In any case, thanks again for another wonderful video!

[lexinwonderland]

Hey, I always love your math and I genuinely aspire to emulate you ad a teacher, but also

LOVE THAT SHIRT, GAY RIGHTS
Thanks prof Penn <3 love, a transfem fan of yours

[MacHooolahan]

Pick delta so that the proof is true, then prove the proof? Am lost with this one MP :S