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Complex Analysis 35 | Application of the Residue Theorem
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- čas přidán 8. 11. 2022
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This is my video series about Complex Analysis. I hope that it will help everyone who wants to learn about complex derivatives, curve integrals, and the residue theorem. Complex Analysis has a lof applications in other parts of mathematics and in physics.
#ComplexAnalysis
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(This explanation fits to lectures for students in their first or second year of study: Mathematics, Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)
I know that was not the point of the video, but leaving sin(pi/6) like that and not replacing it with 1/2 left me hanging with a feeling of emptiness...
Sorry! :)
Mr. I cannot thank you enough, your video series for Measure Theory and Complex Analysis saved me from failing my Analysis III exam. Keep up the good work!!
You're very welcome!
Thank you for your videos. They have been so so so helpful in my undergraduate studies
thank you, really detailed explained, helped me a lot.
best video on this ever, way better than my lectures!
Glad you think so! :)
Thank you.
such a good play list. please consider Lie groups, Lie algebra, Haar measure.
Thank you for this series, I learned a lot !! It would be great if you could consider extending it by covering some more advanced topics (residue at infinity, conformal maps, Rouché's theorem...) :)
Great suggestion! I will definitely do that!
very nice videos! Thanks for the explainations! I just do not understand, how you find so quick the residues for z1,z2 and z3.
Thanks you very much! Did you watch the previous videos about residues?
Hello. Your videos are very useful. I thank you for all the efforts you put in them.
I have a request: can you make a video series on Fourrier and Laplace transforms? Thank you.
Great suggestion!
Hi, I have a problem understanding Rouche's theorem, but I couldn't find any videos on it in your videos. I was wondering if you have covered that topic. Thank you!
Great suggestion!
Amazing!
Thanks :)
on the last part when we discard the cos( ), do we implicitly take Re( ) of the entire contour integral? so that the real part of a contour integral must equal the integral we want?
The contour integral *is* what we want, no real-part-taking required. He's saying the result of evaluating all the sines and cosines will be real at the end regardless (otherwise, something's gone wrong). In other words, notice how there's an i out front and, by Euler's formula, exp(it) = cos(t) + i sin(t). This means i exp(it) = i cos(t) - sin(t), so if we're expecting a real result out of all the sines and cosines being added there, then all the cosines better go away - so just make them go away by ignoring them because they all will cancel out anyways.
@@whalep thanks!
@@GeoffryGifari You can definitely take the real part on both sides because the left-hand side will not change. It is just a matter of taste how you structure the calculation.
Großartig. Endlich verstanden. Klausur hoffentlich gerettet. Jetzt hoffe ich das es drankommt, das Gegenteil von vor diesem Video.
Good luck :)
Perhaps not the right place to ask this question you might think, but as the people who are watching this are so up to date with your work, they're likely to be someone in the same shoes as myself....
Having finished a degree 10 years ago, and not being proud of my result, I am attempting to learn things again.
Without exams to sit, there is seemingly no pressure to get it done, and if I'm honest I am failing. I can't but help think things like, why am I even doing this? What am I trying to prove.
Has anyone any advice, maybe even the content creator himself. Thanks in advance.... am I alone in this I wonder?
I guess, the very fact that you have no exams to sit, actually frees you up to learn what YOU want, not what the SYLLABUS wants. So, you may not have learned this particular topic as well as it ought to be, but if it is as good as you'd like yourself to be, why not?
5:44 you are a mathematician so I wanted to ask you if it's correct practice to write the inequality as
I would never use inequalities with the O notation. Maybe you give an additional meaning for this??
But how do we calculate the roots so quickly? Any videos you recommend? Also, I really love your videos!
I have videos about calculating complex roots: tbsom.de/s/ov
Why we take only upper half of the circle?
You could also choose the lower half.
Thanks for your videos! I have an issue with the formula for simple poles (h(z) /g(z) ) used. When can we use it? I tried solving with the formula lim z-z0 [(z-z0) f(z)] to calculate the limit and the answers are different. Also tried it for the improper integral for f(x) = x^2/(1+x^4) and had a different answer also.
The formula for simple poles I referred to from the video is res(h/g, zi) = h(z) /g'(z)
Yes, you need a simple pole for this :)
This directly follows from the general formula to compute the residue at a pole. Note that if z_0 is a simple pole of h/g, then g(z_0)=0 hence lim_{z\to z_0} (z-z_0) h(z)/g(z) = lim_{z\to z_0} d/dz h(z)/[g(z)-g(z_0)/(z-z_0)] = h(z_0)/g'(z_0)
Why did not you compute the value of sin(π/6) at the end?
여기서 한국인을 보네 ㅋㅋ
@@user-km1os8oq5e 안녕하세요. :)
Oh, I didn't think that this was important for this demonstration. I wanted to show things that can be generally used to calculate similar integrals.
@@brightsideofmaths I see. Thank you for replying.
@@brightsideofmaths Good decision. Such calculations would just distract us from the central points.
Where can I find more info on standard estimates? I am lost on that part
It's discusses in this series :)
Is this the final part of this series or will there be more? :)
At the moment, the series has an end here but I will continue it next year!
05:03 Why is there no Pi in the power? Since its a halfcircle the power should be i*pi*t right?
I think it does not affect the estimation however.
It depends what domain your t has, doesn't it?
@@brightsideofmaths thanks for your answer!
do you mean because of i*pi*t=i*t' for some t' from domain?
If that:
But how do you know t' is inside [-R,R]?
For example t=R, then pi*t=t' isnt inside the domain.
@@MrChicken1joe t does not have to be inside [-R,R] for the delta curve :)
@@brightsideofmaths ah I can see, thanks!!
Complex analysis seems awesome but what does real analysis do?
Thanks! Real Analysis is calculus together with the theory :)
@@brightsideofmaths ah ok. So if we know calculus already, it’s not gonna really add any cool integral techniques like this, just proofs?
That depends what you already know. You should definitely watch it :D@@darcash1738
bro is not solving for a 8 munit
What?
@@brightsideofmaths now that i got your attraction, we love you :D