The residue theorem -- Complex Analysis 22

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  • Опубліковано 19 гру 2024

КОМЕНТАРІ • 27

  • @ingobojak5666
    @ingobojak5666 2 роки тому +15

    5:33 I think the explanation given both verbally and by the drawing is incorrect. The condition |z_j - z_k| >= epsilon does not ensure that "none of the bubbles intersect". It only ensures that no other pole lies within the bubble around any one pole. For the equal sign, a different pole could be on the epsilon-boundary though. It probably would be safest here to say |z_j - z_k| > 2 * epsilon with double the distance and a greater sign. Then the explanation as given is right.

    • @Tehom1
      @Tehom1 2 роки тому

      Good catch.
      He could actually prove it with intersecting bubbles by integrating around just the intersecting region to add it back in since it got subtracted twice, and similar for multi-intersections, but it would be more complex.

  • @thomashoffmann8857
    @thomashoffmann8857 2 роки тому +5

    At 29:15.. The quotient rule would require to square the demoninator. So it should be (z-3i)^4 instead of ^2

    • @melontusk7358
      @melontusk7358 2 роки тому

      This should be pinned to the top. I was wondering why the derivative seemed weird.

  • @romajimamulo
    @romajimamulo 2 роки тому +6

    15:49 I think that's the wrong order? Multiply first, then the derivative

  • @Noam_.Menashe
    @Noam_.Menashe 2 роки тому +6

    One of my favorite and most useful theorems in applied math.

    • @mcqueen424
      @mcqueen424 2 роки тому +1

      This isn’t applied math

    • @Noam_.Menashe
      @Noam_.Menashe 2 роки тому +1

      @@mcqueen424 Complex variables are used in physics.

    • @navierstokes2356
      @navierstokes2356 2 роки тому +1

      @@Noam_.Menashe in physics you use topolgy and that doesn't mean that topology is applied math....

    • @navierstokes2356
      @navierstokes2356 2 роки тому +1

      @@Noam_.Menashe also you and computer scientist use Abstract Algebra, and that doesn't imply that Abstract Algebra is applied math...

    • @navierstokes2356
      @navierstokes2356 2 роки тому

      @@mcqueen424 true

  • @JM-us3fr
    @JM-us3fr 2 роки тому +11

    Are we going to explore how these results get applied to more unusual functions, like the gamma function or zeta function? Also, maybe this is outside the scope of this course, but will we look at how to understand transformations like the Laplace, Fourier, and Mellin transformations?

  • @WhattheHectogon
    @WhattheHectogon 2 роки тому +3

    Right up there with Cantor's Theorem as my favorite theorems from undergrad :)

  • @QP9237
    @QP9237 3 дні тому

    Wished you would have included the winding number based Residue theorem for the more general case

  • @euclid3044
    @euclid3044 2 роки тому

    this series is gamechanging oml

  • @TimHaloun
    @TimHaloun 2 роки тому +3

    So I should think of this as basically this is extending the original cauchy integral formula? Instead of the integration around the boundary being zero for a function wholly analytic inside, we can now integrate over functions with singularities and the residues are the 'correction' terms? Sometimes I wish the prof would step back and give some wider context.

  • @JM-us3fr
    @JM-us3fr 2 роки тому +2

    Here's my warm up solutions. I haven't double-checked them, so feel free:
    For the integral of sin(z)/(z^2-1)^2, I got pi*i*(cos(1)-sin(1)) (the two poles had the same residue),
    For the integral of z^2/cos(z), I got 0 (the two poles had opposite residue),
    For the integral of sin(z)^6/(z-pi/6)^3, I got 3*pi*i/16.

    • @Windolon
      @Windolon 2 роки тому +1

      The last integral should be 21*pi*i/16. This is because the residue of the function at z=pi/6 is 21/32. I've checked in Mathematica.

    • @DendrocnideMoroides
      @DendrocnideMoroides Рік тому

      @@Windolon Yes that is correct

  • @uguraltntas2630
    @uguraltntas2630 Рік тому

    Durmak için güzel bir yer😃Greetings from Turkey🖐

  • @edcoad4930
    @edcoad4930 2 роки тому

    Totally agree on how exciting messy integrals get with complex analysis. Possibly one of the best short cuts ever..note: as a physicist, maths is a hammer and I know how to use it but don't always care why it works! 😉

  • @s4623
    @s4623 2 роки тому

    21:39 wouldn't it be easier to do the residue at infinity just like the previous example because it wouldn't be an essential singularity at infinity?

  • @sergpodolnii3962
    @sergpodolnii3962 2 роки тому

    32:30: It is just an imaginary circle :)

  • @Eis461
    @Eis461 11 місяців тому

    18:04 you mean clockwise

  • @synaestheziac
    @synaestheziac 2 роки тому +1

    Am I right to assume that shirt says “and that’s a good place to stop” in serious different languages?

  • @kirkwilliams1609
    @kirkwilliams1609 2 роки тому

    Well, given what was said at 30:25, I'd claim that the z-junior residual function implies that a nested bubble-recognition proof is necessary to even imperfectly understand the exectual defenestrational. flirty and fun nature of complex analysis. My cousin Ron punctured his disk last fortnight, I