Recently, Will Chen made a much more comprehensive video overviewing Polya vector fields on his channel: th-cam.com/video/itEqPTJpxUo/w-d-xo.html Please check it out if you want to improve your intuition about complex integrals. I think his work does the concept justice.
Thank you! Until now, I was solving complex integrals without having any idea what they meant. This model also helped me develop some intuition of why Cauchy's theorem works
It is a shame that this great video has such low volume. I use to watch TH-cam on volume level 6-8 not to disturb the house too much. Here I am still barely hearing anything at volume level 30. That is scarily loud compared to the regular video on TH-cam. Now all I can think about is not to forget to turn the volume down again before the video is over. Oh no! ADVERTS!!!!
It's wonderful how by looking at a complex function I can recognize it - looks similar to its real counterpart . Haven't seen such depiction of complex functions as vector fields before. The resemblance between real and complex functions is remarkably beatiful.
Great explanation! The example Polya fields in the end where a great touch, my only complaint it would be better if you didn't redraw the axes every time they transitioned. Also you could leave them on for a bit longer. Thanks for uploading!
Nice video! This is definitely something I'd wanted to see spelled out in animation so I'm glad you did it. A nitpick for your next video: ß (a "long s" and a regular "s" glued together) is not β (a Greek letter related to "b").
This is a reasonable question to ask, as it seems more intuitive to visually construct the vector field by drawing a vector pointing directly from the input 'z' to the output 'f(z)'. However, if we do that, we cannot create the corresponding Polya vector field using the "trick" described around the 5:00 minute mark. If f(z) maps two complex numbers to the same point and we use the method described in the video to visualize f(z), the vectors projecting from both of these points will look identical. If we use the method you describe, the vectors projecting from both of these points will not look the same.
@@lemmaxiom so actually correct vector field is that one that i described. You used another vector field because it required for integration. Sorry seems im wrong, in video you noticed that we translate vector to construct __required__ vector field. Thats fully correct because required vector field could be arbitrary :) My thoughts are only correct if we talk about vector field for complex function.
My Complex Analysis professor recommended "Complex Analysis by Donald Marshall", but in our undergrad class he thought it was too advanced for us, and made us learn instead from "Saff E.B., Snider A.D. -Fundamentals of Complex Analysis with Applications to Engineering, Science, and Mathematics". I downloaded both books for free (illegally) from Library Genesis (libgen dot rs). Also, if you're not familiar with the Fundamental Theorem of Algebra and its proof using Complex Analysis, you should check out the Numberphile video on it here th-cam.com/video/shEk8sz1oOw/w-d-xo.html
7:59 .... I can see why the real parts cancel out .... I do not see how we get the magnitude 2*pi*i ... |f(z)| * 2*pi*i makes a little more sense to me ...
This is correct intuition - what's interesting here though, and I admit I didn't explain this in the video, is that the size of the circle we draw (in fact, it doesn't even have to be a circle in this example; it could be any simple closed curve around the origin, which is a fact I regret not clarifying and expanding upon) "cancels out" the average magnitude of the function along the curve. There is a reason for this that I do not explain in the video, namely that the function in question is _holomorphic_ on all of the plane save the origin. If we were to draw another circle concentric with and twice the circumference of the one I illustrated (the circle in the video is assumed to have radius 1), then the value of |f(z)| along each point of _that_ circular curve would be _less on average_ than it is along the smaller circular curve, while the larger circumference of the bigger circle in this case "balances" the integral's value to again yield 2*pi*i. This works for _any_ circle (again, even any simple closed curve!) enclosing what's called the "pole" of the particular function at the origin; the value of the integral _remains the same_ no matter what size our circle is: 2*pi*i. I used circles as my simply closed curves here because it becomes obvious where the factor of pi is coming from. If I had more time and depth of understanding when I made this video, I would likely have tried articulating these ideas more carefully. Maybe I'll eventually produce a more prudent presentation of this concept in the future, because there is more to say about it than I did. I hope this helps - I want to leave you with this lemma, because I think you'll be able to see why it's true with the visualizations from this video in mind: en.wikipedia.org/wiki/Estimation_lemma
@@lemmaxiom Thank you for your kind response. You get an A+ for an intuitive demonstration of why the real part is zero. As for the imaginary being = 2*pi*i due to the presence of a pole, I was aware of that only because of a memorized formula. Using the Estimation lemma, and noticing from symmetry that every point on the contour circle would seem to have the same magnitude, I pick a suitable z, such as z=1 + 0i, and then integrating around a circle, I get 2*pi for a magnitude. Then I just wing it and say that there must be an i multiplied in to point things in the right direction....
The idea is that the flux is positive if it "flows out" of a simple closed contour, and negative if it "flows into" it. Because the standard direction of integration around a simply connected domain in the complex plane is _counter-clockwise_ , flux ought to be defined as it is here in order for "internal" and "external" flow to be negative and positive, respectively.
You say this is the integral of a complex number, but to me it just looks like the gradient of a vector field. Given that in some contexts, vectors can be multiplied to give complex numbers (while the two remain distinct), f(z) could also just be f(v), which could theoretically generalize it to R^3 or even beyond. (Though in R^3 the integral would give a quaternion rather than a complex number.)
This video is completely inaudible. Max volume on everything, video player, computer, speakers, and I can just barely hear that there's any speaking at all.
Recently, Will Chen made a much more comprehensive video overviewing Polya vector fields on his channel:
th-cam.com/video/itEqPTJpxUo/w-d-xo.html
Please check it out if you want to improve your intuition about complex integrals. I think his work does the concept justice.
jesus CHRIST, this is AMAZING!!!!! Thank you for elucidating the Polya Vector Field -- I finally understand it!!!
Thank you!
Until now, I was solving complex integrals without having any idea what they meant. This model also helped me develop some intuition of why Cauchy's theorem works
This is the third of your videos I've watched, on the same day! This could get habit-forming - wait; it already has! Thanks.
It is a shame that this great video has such low volume. I use to watch TH-cam on volume level 6-8 not to disturb the house too much. Here I am still barely hearing anything at volume level 30. That is scarily loud compared to the regular video on TH-cam. Now all I can think about is not to forget to turn the volume down again before the video is over. Oh no! ADVERTS!!!!
Yup - I definitely need to fix up some stuff on this vid. I had very little time to refine it and it definitely shows.
@@lemmaxiom don’t worry. I like your video enough that I am planning on watching it with headphones as soon as time permits.
It's wonderful how by looking at a complex function I can recognize it - looks similar to its real counterpart . Haven't seen such depiction of complex functions as vector fields before. The resemblance between real and complex functions is remarkably beatiful.
7:59 is a revelation.
A nicely produced video that explains this method of visualising a complex integral very well. Thanks
More and more, please!!! Thanks for these amazing videos.
Thanks for the great video. What software you use to make this presentation?
Thank you - I used Adobe Premiere Pro for just about everything.
The list of your videos proves that you have a great interest in visualization
Omg I legit never thought I'd see the day
Great explanation! The example Polya fields in the end where a great touch, my only complaint it would be better if you didn't redraw the axes every time they transitioned. Also you could leave them on for a bit longer. Thanks for uploading!
Awesome video! Thank you!
What an impresive and high quality content, you should upload more often, delightful . +1 sub
great job, thanks you so much, cheers from Argentina
Nice video! This is definitely something I'd wanted to see spelled out in animation so I'm glad you did it.
A nitpick for your next video: ß (a "long s" and a regular "s" glued together) is not β (a Greek letter related to "b").
Oops, you're right - looks like I used *Eszett* - added to the corrections!
Thank you. These subject is beyond my understanding but can connect by Visible graph connect to the each individual connect formulae.
Amazing
Pla make more such amazing videos
Also I have some thoughts how intuitively translate integration real function to complex function.
Thank you
i^Z
✨👍✨
1:20 - why dont we draw new vector from top of blue to top of violet? function transforms 1+i to 0.5(1-i)
This is a reasonable question to ask, as it seems more intuitive to visually construct the vector field by drawing a vector pointing directly from the input 'z' to the output 'f(z)'.
However, if we do that, we cannot create the corresponding Polya vector field using the "trick" described around the 5:00 minute mark. If f(z) maps two complex numbers to the same point and we use the method described in the video to visualize f(z), the vectors projecting from both of these points will look identical. If we use the method you describe, the vectors projecting from both of these points will not look the same.
@@lemmaxiom so actually correct vector field is that one that i described. You used another vector field because it required for integration.
Sorry seems im wrong, in video you noticed that we translate vector to construct __required__ vector field. Thats fully correct because required vector field could be arbitrary :)
My thoughts are only correct if we talk about vector field for complex function.
Muy interesante, voy a ver si lo puedo graficar en Python
Bro can you please suggest me a complex book i want to do research in complex analysis
My Complex Analysis professor recommended "Complex Analysis by Donald Marshall", but in our undergrad class he thought it was too advanced for us, and made us learn instead from "Saff E.B., Snider A.D. -Fundamentals of Complex Analysis with Applications to Engineering, Science, and Mathematics". I downloaded both books for free (illegally) from Library Genesis (libgen dot rs). Also, if you're not familiar with the Fundamental Theorem of Algebra and its proof using Complex Analysis, you should check out the Numberphile video on it here th-cam.com/video/shEk8sz1oOw/w-d-xo.html
Thank you Sir, this video is very informative. Would like to see such more videos. I am from INDIA. Please make more videos.
Thank you, that's great to hear! I will definitely work on more videos when I can.
7:59 .... I can see why the real parts cancel out .... I do not see how we get the magnitude 2*pi*i ... |f(z)| * 2*pi*i makes a little more sense to me ...
This is correct intuition - what's interesting here though, and I admit I didn't explain this in the video, is that the size of the circle we draw (in fact, it doesn't even have to be a circle in this example; it could be any simple closed curve around the origin, which is a fact I regret not clarifying and expanding upon) "cancels out" the average magnitude of the function along the curve. There is a reason for this that I do not explain in the video, namely that the function in question is _holomorphic_ on all of the plane save the origin.
If we were to draw another circle concentric with and twice the circumference of the one I illustrated (the circle in the video is assumed to have radius 1), then the value of |f(z)| along each point of _that_ circular curve would be _less on average_ than it is along the smaller circular curve, while the larger circumference of the bigger circle in this case "balances" the integral's value to again yield 2*pi*i. This works for _any_ circle (again, even any simple closed curve!) enclosing what's called the "pole" of the particular function at the origin; the value of the integral _remains the same_ no matter what size our circle is: 2*pi*i.
I used circles as my simply closed curves here because it becomes obvious where the factor of pi is coming from. If I had more time and depth of understanding when I made this video, I would likely have tried articulating these ideas more carefully. Maybe I'll eventually produce a more prudent presentation of this concept in the future, because there is more to say about it than I did.
I hope this helps - I want to leave you with this lemma, because I think you'll be able to see why it's true with the visualizations from this video in mind: en.wikipedia.org/wiki/Estimation_lemma
@@lemmaxiom Thank you for your kind response. You get an A+ for an intuitive demonstration of why the real part is zero. As for the imaginary being = 2*pi*i due to the presence of a pole, I was aware of that only because of a memorized formula.
Using the Estimation lemma, and noticing from symmetry that every point on the contour circle would seem to have the same magnitude, I pick a suitable z, such as z=1 + 0i, and then integrating around a circle, I get 2*pi for a magnitude. Then I just wing it and say that there must be an i multiplied in to point things in the right direction....
You sound kinda like Cr1tical, I'm assuming that wasn't intentional, but I'd love to know if it was
Not intentional - but I do think I hear it just a tad as well now.
3:20 I don't know why it's so much harder to see in the flux example than in the work one
The idea is that the flux is positive if it "flows out" of a simple closed contour, and negative if it "flows into" it. Because the standard direction of integration around a simply connected domain in the complex plane is _counter-clockwise_ , flux ought to be defined as it is here in order for "internal" and "external" flow to be negative and positive, respectively.
@@lemmaxiom no, like, harder to see the colored regions
@@romajimamulo Ah, got it. I'll make sure to fix up some little things like that if I get the chance.
I can barely hear anything but good video!
You say this is the integral of a complex number, but to me it just looks like the gradient of a vector field. Given that in some contexts, vectors can be multiplied to give complex numbers (while the two remain distinct), f(z) could also just be f(v), which could theoretically generalize it to R^3 or even beyond. (Though in R^3 the integral would give a quaternion rather than a complex number.)
This video is completely inaudible. Max volume on everything, video player, computer, speakers, and I can just barely hear that there's any speaking at all.