One thing he does not make clear at around 4:35 you are only looking for dominance at the boundary, because on the inside of the function where you are looking for roots, the 1 will eventually take dominance as you get closer and closer to the zero if you are looking at the area. But we only need to prove it for the boundary for the theorem, which is why it is so powerful.
The Ferrero Rouché joke rocks! I couldn't believe my eyes when I saw this new video just now while I was solving a problem using this theorem :)) Thanks a lot!!!
A friend of mine came up to me with this exact setup for a problem some months ago but couldn't even find something related and now you upload it as a casual thing. You're doing god's work
@@reeeeeplease1178 At that step 1 = |z|, not 1 > |z|. Even though the theorem tells you about roots _inside_ the region, the conditions to use the theorem involve z on the _boundary_ of the region.
I remember the exam for undergrad complex variables I took had an application of Rouche, but I couldn't find the perturbing function. It's a nice result, but I have issues using it correctly.
Now this is incredibly interesting to me - and I will have to watch several times before an algorithm kicks in. This is interesting because it (with present days interest rates) the internal value of f.i. an annuity might have several roots. Further complicated by the fact that the interest rate is depending on (proposed) of raise in productivity and inflation - and nothing else. But the practical implication? I don't think it will have one, as economics professors - from experience are mathematically illiterate.
The problem with bankers is that they really don't know, what they are doing (at the best of times). If I only leaned on the math I learned in University - then I would be in bad shape - just like my costudents. Will the situation be corrected? Nope, and definately not by me - I have no desire to be admired by bacteria brains.
I'm also struggling with that because he considered |z|^5 + 1 < 1 + 1 but if |z| = 1, then |z|^5 = 1, and therefore |z|^5 + 1 = 1 + 1 I feel stupid because everyone is accepting it, yet I don't see it clearly how is it true
For Rouché's theorem, the minimum requirement is: - U an open simply connex, subset of C. - f and g meromorphic in U - Gamma simple loop (the boundary of your compact) If |f(z)-g(z)| < |g(z)| in all point of gamma, then: zeroes(f)-poles(f) = zeroes(g) - poles(g) Here, Peyam used a result derives from it that has the particularity of having no poles. (and obviously the criterion of meromorphic functions is respected, as all rational functions verify it). I really encourage people to at least read the Wikipédia article that isn't super long.
@@adityaekbote8498 You can check Dr. Peyam's videos for more clarification but holomorphic is the way for saying a complex number valued function is differentiable. It's basically the limit definition of differentiation for complex numbers. It's insanely powerful in the following result if you can prove a function is complex differentiable once you can prove it is infinitely differentiable and in case you are looking at @PackSciences explanation: meromorphic means the function is complex differentiable EXCEPT at isolated points/poles. Feel free to correct me if I'm wrong but from my viewing of complex analysis videos poles behave like the "dreaded cusp points" that screw everything up in normal calculus/real analysis.
Dear Dr. Peyam. I am not able to understand that in which questions to use complex numbers and where not to. I am in class 11 and we havr learnt complex numbers. While doing quadratics and other polynomials we use it for solutions. But while finding domain and range and in all of Calculus we use, we do not use it. So when do I use it and when not? eg. Say sinx=2 has not real solution but its possible using complex numbers.
@@drpeyam Really?!! خوش بحالتون (Good for you). Please one day talk about your experience with professor Maryam Mirzakhani and also talk about her research and ... . Thank you
@@drpeyam In my opinion talk a bout Professor Mirzakhani and her research can be one of the greatest subjects for your video. And you're one of the best person for doing this, because you're mathematician and also Iranian.
One doubt. I thought we should check for, on lzl= 2 we have roots or not,but instead you checked for lzl=1and subtract that from 3, but the answers will be same cause two different parts(z^5 and 3z^2+1)will never have same mod value on lzl=2 so they can't cancel each other.
I don't understand : in the case where ∣x∣ < 1, if you take, for exemple, x = 0.5, then f(x) = 0.75 and g(x) = 1.03125, so g(x) > f(x) I know I'm wrong, but where ?
One thing he does not make clear at around 4:35 you are only looking for dominance at the boundary, because on the inside of the function where you are looking for roots, the 1 will eventually take dominance as you get closer and closer to the zero if you are looking at the area. But we only need to prove it for the boundary for the theorem, which is why it is so powerful.
Exam writers love this theorem! You just saved so many students from attempting to solve for the roots during the test!
The Ferrero Rouché joke rocks! I couldn't believe my eyes when I saw this new video just now while I was solving a problem using this theorem :)) Thanks a lot!!!
Nice!
The problem is : Prove that all the roots of z^7 - 5 z^3 + 12 = 0 lie between the circle |z| = 1 and |z| = 2
Amazing, it’s so similar to this video!
Actually got an ad for it although i never get candy/... ads
Complex Analysis is so beautiful! Great presentation of a classic result -- thanks Dr. Peyam!
Interesting theorem. The best I was able to say about this is not all the roots are in |z|
A friend of mine came up to me with this exact setup for a problem some months ago but couldn't even find something related and now you upload it as a casual thing. You're doing god's work
It is a mistake at 9:12. It should be 3*z^2=-z^5-1, not 3*z^2=z^5-1. Even though, the conclusion is true.
You've managed to demystify such a simple concept. Thank you
At 4:40 it should've written |z|^5+1
This is one of my favourite results in complex analysis, grar video!!! Also, anyone else notice the fly enthusiast of maths at 4:26 ?
Isnt that step incorrect? How can 3=|3z^2| when 1>|z|
@@reeeeeplease1178 At that step 1 = |z|, not 1 > |z|. Even though the theorem tells you about roots _inside_ the region, the conditions to use the theorem involve z on the _boundary_ of the region.
Love your videos
Love the way you made me understand
Complex analysis was fun :)
I remember the exam for undergrad complex variables I took had an application of Rouche, but I couldn't find the perturbing function. It's a nice result, but I have issues using it correctly.
It’s hard to be honest!
You are really great just cleared all my concepts how to attempt such problems
Technically, there is an explicit formula for the roots of the general quintic polynomial, but it's in terms of Jacobi theta functions.
really? never heard of this before. any links?
@@gasun1274 It's mentioned in the article on Wolfram MathWorld about the quintic equation. UA-cam deletes my comment when I try to post a link to it.
@@gasun1274 Also - on Wikipedia under "quintic function"
Yay complex analysis!
@0:05 Dr. Peyam, I'm not sure if accidental but your phrasing got me thinking of Voldemort memes in my head. hahaha! :)
Accidental haha
How did Cauchy not have his name on this? Did he owe Rouche’ money?
Rouché theorem is kind of the intermediate value theorem for complex functions. :)
Yep, you look at the boundary and know stuff about the 'inside'. Very powerful.
This video is name “ King your videos “
theorem is interesting
Thank you for your efforts
Thank you Dr!, nice videos!
Now this is incredibly interesting to me - and I will have to watch several times before an algorithm kicks in.
This is interesting because it (with present days interest rates) the internal value of f.i. an annuity might have several roots. Further complicated by the fact that the interest rate is depending on (proposed) of raise in productivity and inflation - and nothing else.
But the practical implication? I don't think it will have one, as economics professors - from experience are mathematically illiterate.
The problem with bankers is that they really don't know, what they are doing (at the best of times). If I only leaned on the math I learned in University - then I would be in bad shape - just like my costudents.
Will the situation be corrected? Nope, and definately not by me - I have no desire to be admired by bacteria brains.
At 4:40 you have 3=|3z^2| which is wrong because |z|
Not true, |z| = 1
@@drpeyam It says "CASE 1 |z|
there's nothinf wrong, you need to compare the |f| & |g| at the boundary of the region
I'm also struggling with that
because he considered |z|^5 + 1 < 1 + 1
but if |z| = 1, then |z|^5 = 1, and therefore |z|^5 + 1 = 1 + 1
I feel stupid because everyone is accepting it, yet I don't see it clearly how is it true
Usually in complex analysis, we are concerned with holomorphic functions… is that a necessary condition to this theorem?
Of course
Whats that
For Rouché's theorem, the minimum requirement is:
- U an open simply connex, subset of C.
- f and g meromorphic in U
- Gamma simple loop (the boundary of your compact)
If |f(z)-g(z)| < |g(z)| in all point of gamma, then:
zeroes(f)-poles(f) = zeroes(g) - poles(g)
Here, Peyam used a result derives from it that has the particularity of having no poles. (and obviously the criterion of meromorphic functions is respected, as all rational functions verify it).
I really encourage people to at least read the Wikipédia article that isn't super long.
@@adityaekbote8498 You can check Dr. Peyam's videos for more clarification but holomorphic is the way for saying a complex number valued function is differentiable. It's basically the limit definition of differentiation for complex numbers. It's insanely powerful in the following result if you can prove a function is complex differentiable once you can prove it is infinitely differentiable and in case you are looking at @PackSciences explanation: meromorphic means the function is complex differentiable EXCEPT at isolated points/poles. Feel free to correct me if I'm wrong but from my viewing of complex analysis videos poles behave like the "dreaded cusp points" that screw everything up in normal calculus/real analysis.
Very nice! Thank you!
Thank you very much.
Nice. But how do you prove Rouché’s theorem in general?
At 4:36, I don't see how that can be true. f(0)=0, and g(0)=1
On the boundary, not the whole set
Dear Dr. Peyam.
I am not able to understand that in which questions to use complex numbers and where not to.
I am in class 11 and we havr learnt complex numbers. While doing quadratics and other polynomials we use it for solutions. But while finding domain and range and in all of Calculus we use, we do not use it.
So when do I use it and when not?
eg. Say sinx=2 has not real solution but its possible using complex numbers.
_Ferrero Rocher_
Where a layer of foil defines a boundary? ;)
Omg nice!
Thank You 😀
You're welcome 😊
If there are 5 roots in this circle but you can't find any, do they really exist?
Hahahaha
I see we have a constructivist here lol.
why you dont prove given equation has no zeros on |z|=2
Love from india ❤
Very nice😀
Proof please
留数定理の話かと思ったけどなんか違うね･･･
もちょっと勉強しときます！
this video made this theorem click for me!
Yay!!!
Best sir
Dear Peyam
Have you ever met professor Maryam Mirzakhani?
I did
@@drpeyam Really?!!
خوش بحالتون (Good for you).
Please one day talk about your experience with professor Maryam Mirzakhani and also talk about her research and ... .
Thank you
I only said hello to her
@@drpeyam
It's still great, because I don't have even this experience!
But if it's possible for you, make a video about her research.
Thank you
@@drpeyam In my opinion talk a bout Professor Mirzakhani and her research can be one of the greatest subjects for your video. And you're one of the best person for doing this, because you're mathematician and also Iranian.
How can 3z² has 2 roots?
It has a root of 0 with multiplicity 2
@@drpeyam aha, alright
Soo cool
One doubt. I thought we should check for, on lzl= 2 we have roots or not,but instead you checked for lzl=1and subtract that from 3, but the answers will be same cause two different parts(z^5 and 3z^2+1)will never have same mod value on lzl=2 so they can't cancel each other.
nice
Is this like Descartes Rule of Sign but in the Complex World XD
I never thought of it that way, wow!!!
You're op
Based
Hello Dear Dr Peyam.
No roots but my ...😀😀
Such nice math problem, I like that.
Thank you
I don't understand : in the case where ∣x∣ < 1, if you take, for exemple, x = 0.5, then f(x) = 0.75 and g(x) = 1.03125, so g(x) > f(x)
I know I'm wrong, but where ?