Complex Analysis Overview
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- Опубликовано: 12 сен 2024
- In this video, I give a general (and non-technical) overview of the topics covered in an elementary complex analysis course, which includes complex numbers, complex functions, the Cauchy-Riemann equations, Cauchy’s integral formula, residues and poles, and many more! Watch this video if you want to know what complex analysis is like
When we study mathematics, the biggest obstacle is fear. Your videos defeat the enemy. Great!
I want a full lecture series on complex analysis by you
"This is just the start of heaven and there's more heaven waiting for you."
I love it!
“Before I was invented” it’s true he’s a robot!
good joke!!
Oh boi, a 36 minute video from the Doc.
"Prepares popcorn, softdrink and a sheet of paper to take useful notes".
My favourite class in undergrad...brings back wonderful memories.
20:47 I think, it is the Liouville Theorem. F bounded and differentialable in *whole complex plane*, then F is constant. Marvelous video!
Correct!
I just got rick rolled in a complex analysis video, nice!
You explain everything to him in such a beautiful way. I have never seen such a professor❤️❤️
Love you sir..
I haven't even looked at analysis, but this video is going to help me once I (hopefully) get to complex analysis! Thanks Dr. Peyam!
By the way, I enjoy your videos. You have great enthusiasm and present the concepts very well....I bet the classes you teach are both fun and educational!
Never gonna give YOU up Dr P
Hi, Dr. Peyam. I discovered your channel last week, and my last week has been PEYAMAZING because of it. Your teaching is so enthusiastic that one is almost forced to have a great time. While I've always been intrigued by complex analysis, I never knew anything about it. This video excites me, because what you showed me was awesome, and you said there's a lot more! I look forward to working my way through your videos. Thank you for making them. A part two to this complex analysis overview would be so cool!
Thanks so much!!!! :)))
I wish my lectures taught like this. I'm doing a resit in complex analysis and I have to rely on this. Kenyan education system needs prayers
+Dr. Peyam
A few remarks:
7:00:
A function f:D→C (D being the domain of f) is not called holomorphic at z=z0 if it is merely differentiable at z=z0. A function being holomorphic is a way stronger condition.
The definition is: f is holomorphic at z=z0 (ϵD) if there exists an open neighborhood U of z=z0 in D (hence, z0 must be an interior point of D) such that for each uϵU f is differentiable at z=u.
As a counterexample, f(z)=|z|^2 is differentiable at z=0, which can be easily shown by the definition. However, f is NOT differentiable for every z≠0 (consider taking the limit along "rays and circles through z").
Hence, although f is differentiable for z=0, it is not holomorphic for z=0 (since for every z≠0 in a neighborhood of z it is not differentiable).
20:50:
This is the so-called Liouville's theorem. The prerequisite is that f be an entire function, that is that f has domain the whole of C and that f be holomorphic at every zϵC.
33:30:
Again, we are missing the conditions here.
Here, f and g should be holomorphic on an open, bounded set U and the condition |g|
I remember how excited I got the first time a plotted different power series on the complex plane and saw that they extended just as well off the real axis as they did on it. It also makes you go "ohhh, that's why they call it a **radius** of convergence"
Thank you for this clear overview. This is one of the few videos I saw on complex analysis and I realy understood what you siad.
Your videos are great Dr Payam. Please do more complex analysis, it’s EPIC
There’s a complex analysis playlist
Dr Peyam Just found it, HEAVEN😍
I love to watch you happly talking about maths!
"I love math so does he!"
Thank you for the video! It is very clear and I can see your passion in teaching! Please have more videos!!!
Thanks so much 🙂
Needed that daily dose of Rick Astley. Thank you Peyam! LOL
Very nice overview! Thank you, this was exactly the overview I needed.
One question: for Rouché's theorem, is it required that |g|
On the contour!
Dr. Peyam's Show Thanks for answering, thought so, but good to know. TFAE: Peyam, Fun, Quality, Insight.
May you get incredible regard of it for teaching us for free! ❤ from Pakistan
Do we have split complex analysis, hypercomplex analysis or surreal analysis?
If you ever use hypercomplex or surreal analysis I‘ll give you i Dollar
If you ever use hypercomplex or surreal analysis I‘ll give you i Dollar
Ii was a real pleasure to watch your video, keep it up mat. Share the knowledge through the web.
Love your enthusiasm, thank you for your contribution to the subject!
Such an amazing video of complex analysis overview 🔥I can totally feel your zeal for math👍
Hey there Peyam! Quick question for you. Is there a good way to describe the action of the complex exponential WITHOUT using the power series definition? Maybe geometrically. The reason for why I ask this is because Rudin defines the real exponential b^x for a base b > 1 and a real number x as the supremum of all b^r where r is a rational number and r
Indeed there is! If you multiply a number r e^i(theta), it's like rotating it by an angle theta, then dilating it by a factor of r
Great video. I enjoyed every second of it. Very useful. Thank you.
Love you Dr. Peyam you are the absolute best!!!!
at 34:22 |g| is not less equal to |f| near about 0, I think that 2 should be added in f instead of g so that it will take care the required inequality...
By the way very nice revision of the whole subject in a single lecture...
Please do such on abstract algebra, ring theory, real analysis, ODE, PDE, optimization etc.
We love to watch your lectures.... Thanks a lot.
I think for cauchy-goursat integral thm that the region has to be simply connected..
Yep
I have a complex analysis exam today ... Thank you
I hope you did well! ❤️
@@ryelyndesch1018 Not at all, it was a disaster xD But thank you for the kind words, it's a pleasure nontheless !
I don't remember complex analysis being so entertaining.
Great videos! Really glad I discovered channel.
For integrals over closed curves, if f is a constant function f : z -> z0, my intuition tells me that the integral of f over any curve, closed or not, is just the length of the path multiplied by z0, which is not 0 in the general case.
And as it is a pretty simple function, i expect it to have an antiderivative...
Something must be wrong there, maybe because going back to the starting point actually makes you count backwards in some way... Maybe I'll figure it out in the second half of the video
Think of it more like F(b) - F(a), where F is an antiderivative of f. If b = a, then you get 0
i have to admit i have never messed around with complex analysis yet, so my intuition is poor atm.
When integrating f over a close curve "starting" and "ending" at γ(a) = γ(b), you want to compute the integral of f(γ(t)) from t=a to t=b, and γ(a) = γ(b) but a b ? I'm kind of confused..
Isn't F the antiderevative of f∘γ? i don't understand the concept of an antiderivative being used to integrate when integration actually depends on the path you take ahah. In real analysis, you don't have this problem, but there...
I definitely want to learn more about complex analysis. It pops up quite often in your videos, and others', and almost everytime i have to wait until the video goes back to something i understand ;) Unfortunately next year we will only redo integrals, and begin series... I already know so much about them thanks to you, papa flammy and bprp :DD
The contour integral of a complex function is equal to 2πi times the sum of residues inside the contour sounded to me like Amperes Circuital law in Magnetism which occurs when you use the magnetic field vector instead of an arbitrary vector function in Stokes Theorem.
And it turns out this residue integral result is a consequence of generalising the Stokes Theorem.
My hunch proved to be correct after all...
At 8:30
I thought that the Cauchy-Riemann Equations worked as an implication i.e. If f(z) = U + iV is holomorphic, then f(z) satisfies the Cauchy-Riemann Equations. But I don't think that the converse is necessarily true.
The converse is true, but it is harder to prove if you do not have the assumption that U and V are continuously differentiable.
You are right. Cauchy-Riemann equations can be considered as a sufficient condition for a function f= u+iv to be holomorphic, only if u and v are continuously differentiable in sense of real functions.
So if U and V satisfy the Cauchy-Riemann equations, then U + iV is holomorphic. Otherwise, it's inconclusive. Right?
Xander Gouws U and V are assumed to be two real functions. So, there is no need to check Cauchy-Riemann equations for them. They only need to have continuous differentiations.
Ye when I say 'U and V satisfy Cauchy-Riemann Equations' I mean that dU/dx = dV/dy and dU/dy = - dV/dx.
i just subscribed, you are gifted passionate teacher!
Every time I think "why does nobody talk about log base z of x" and though it is technically trivial by the log definitions log_z(x)=log(x)/log(z)=log_z(y)*log_y(x), it's still significant because all the time people are told that, much like the contents of log, that the base must be positive and additionally not 1
Love you sir...
Great Video! Love your enthusiasm..! for ironing better ;)
Maybe all those surprising facts about complex functions wouldn't be so surprising if we could truly see their geometry as is , in 4D I mean.Maybe the function being fully defined by a single strip in its domain is the most intuitive fact a 4D being has conceived 😛
thanks Dr. Peyam for sharing
I feel like you need some sort of condition for Laurent series. You can't just pick any random function from C to C and expect it to have that form.
Holomorphic
PackSciences PackSciences PackSciences But the way he says it he makes it seem like that condition is not rquired. "Even if a function is not differentiable" 23:22
I think it needs to be holomorphic except for isolated singularities.
@@seanfraser3125 That sounds right.
I made a course on complex analysis so I think I can predict the highlights but I will leave it running in the background for those sweet, sweet views
I have a question, you said the an integral over a closed curve will always be 0 if f(z) has an AD.
wouldn't it be more accurate to say that it will be 0 if f(z) has a non-multi-valued AD?
Does this mean that I can finally understand how to do half of your other videos?
Yeah 😂😂😂
I am not familiar with the differential notation given in the system of differential equations.
Do you mean the following?
dU/dx = dV/dy
dU/dy = -dV/dx
also what is the Z naught r thing at 18:10 ?
I am a little over my head here, but I enjoyed it and was mostly able follow what is going on. Thank you for the introduction. Seriously interesting stuff.
Yeah, and z0 is a given point
Thank you so much for this!!!
best video ever
Seriesly impressive :-)
and this is the easy stuff for this dude
Books you like for undergraduate complex analysis?
Brown and Churchill, stein and Shakarchi
Please part 2!!!
Riemann surfaces?
I love this guy, he's adorable! I wish he was my math teacher. But, thanks to the internet, he is!
Wait! A Lebesque integration, or a what? An abstract integration? Is there a third kind that's not Riemann or Lebesque?
I have never been rickrolled on a math lecture... Yet
22:25 Missed an opportunity to talk about the ray of convergence. 1/(1-z) does no equal to the sum from n=0 to infinity of z^n.
PackSciences does not*
Sorry, english is are will was no my main language
thank you for the summary!
Great overview, thanks
Nothing too complex for Dr Peyam!
At 26:26 the integral of 1/z-zo raised to the n disappears because its anti derivative exists
Does that mean that the anti derivative is holomorphic ? Or is differentiable? Because analytic functions dont have singularity but these functions do have singularity why don't we consider them as analytic ? Am I missing something??
@Alejo Sanchez makes sense
Thanks for the reply
Wonderful overview
You're the man!
Please make overview on vector calculus
thank you very much!
So do all of parts from 2 onwards only work if your function ‘f’ is a complex function (I.e. ‘z’ has a non-zero imaginary part)?
I also had this question. It seems that a real-valued function cannot be holomorphic (unless it is a constant function).
When you explain it is so understandable like straght way and when my teacher explains it's always a curly way🥴
there is a dz right?
Could you derivate x!...?
Zeta function fun!
Factorial is an operator reserved to integers:
It goes from N to N.
How do you define derivative of a non-continuous function?
If you are talking about the extension x! """""="""""" Gamma(x+1), you need to use the digamma function, but it's a really interesting result. (and please don't use factorial operator :'( )
It's gamma function, sorry!
Anyways, for integers gamma(n) = (n-1)!
It extends to the complex numbers
gamma(z) = integral from 0 to inf of (x^(z-1) *e^-x dx)
Javier Gonzalez Differentiate, not derivate
@Deepto Chatterjee : People understund when people will said derivate. Also dériver is WAAAAY smoother to say than differentiating
In contour integration who we know the appropriate contour we use ? A circle or semi circle or rectangle or or or .... ?
It doesn’t matter provided that they have the same singularities in the inside (unless I forgot some exception)
Taking z real then in the interval (0, 1) z(exp 5) is no greater than z(exp 2) + 2
Could you try this:- definite integral of (x^100-1)/lnx with respect to x ranging from o to 1
Im not Dr. Peyam, but i`ll try take on your challenge lol. Also i dont think you can integrate something with Natural log as an exponent and denominator, unless you got some random trick.
also identity theorem?
Thanks thanks
23:20 Im dead😂
13:33 RICK ROLL ALERT !
Hi, Dr. Peyam! 6:50 Holomorphic is not technically the same as differentiable. Differentiability only applies to a single point. Holomorphicity applies to the entire domain of a function (i.e., differentiable at all points in the domain). I guess that's not so important though. bleh bleh bleh (Dracula). Also, for Rouche's Thm, I didn't see you state that |g| < |f| on the boundary. You made it seem like it had to be true for the entire region. Hey, he Rick Rolled me!!!!!! :P
I know NOTHING about complex analysis, i have only heard of the theorems and some other helpful techniques for integration/differentiation, but im sure the guy wasnt going to state everything detail by detail, its just an overview of how undergrad complex analysis is for the most part. Looks really fun tho tbh!
we say that a function is differentiable if it is differentiable in every point of its domain
R U ok foxy?
Is there a way to contact you?
No
I've never liked it when people say that i² = -1 because that gives two solutions. I prefer i = √(-1)
Sure but its good to have i squared=-1 since that doesnt over complexify things to understand. Plus, he expects us to atleast know calculus and a bit of real analysis, we should all know what i is equal to.
+Dominic Boggio
First of all, he did not state i² = -1 as an equation, but as a property of i.
The standard way of introducing complex numbers (as Dr. Peyam also did in the video) is by introducing them as z = x + yi (where x,y ϵ R); i being some magical identity (number) with the property that i² = -1. But this makes no sense at all! It merely gives some kind of representation of a complex number, including the symbol "+", though nothing is said about the meaning of "+". Since i is not a member of R (since i² = -1), it cannot represent the "+" as defined for the real numbers. So what to make of it?
There is a (way) better way to introduce complex numbers however, in which the definition of i is crystal clear!
Consider the set of 2-tuples (x,y) where x,y ϵ R. As a set, this is equal to R². Now let us define two operators "+" and "*" (addition and multiplication) on it as follows:
- define addition of 2-tuples by:
(a,b)+(c,d) = (a+c,b+d)
- define multiplication of 2-tuples by:
(a,b)*(c,d) = (ac-bd,ad+bc)
It is not difficult to prove that this set of 2-tuples, together with the operators "+" and "*" as defined above, turn this set into a field. Let us call this set (together with the operations) C.
We can now VERY precisely define the element "i": i=(0,1). From the definition of multiplication, it follows that i² = (0,1)*(0,1) = (-1,0).
R can be embedded in C in a natural way by identifying xϵR with (x,0)ϵC. Note that this natural embedding extends the addition and multiplication as already defined on R, while preserving it for elements "identified" as elements of R.
Using the definition of "i" as given above, the real element -1 is identified with i², i.e. i² = -1.
Finally, how to get to the "standard" representations of zϵC as z = x + yi? Well, I think you already guessed it: by writing (read: identifying) z = (x,y) (as an element of C) with (the "sloppy" notation) z = x + yi! It turns out (purely by means of the definitions above) that the (initially meaningless) notation z = x + yi completely coincides with the 2-tuple definition if we separate reals and multiples of i by means of "+" (which still remains kind of meaningless), together with the standard rules to add/multiply numbers (while treating "i" as a separate entity).
Hope this helps.
what is inf*i? just infinity? Or maybe some sort of imaginary infinity?
It’s not infinity, it would be indeed some imaginary infinity. If you think of C as R2, then it would be (0,inf)
Just as far but in a different direction?
@@JayAbel there would be 8 infinities
Laurent series? Or do you mean MacLaurin series?
Laurent
@@drpeyam Thanks but i've never heard of these series.
Normal, it’s mainly covered in a complex analysis course
i don't know how to feel about being rick rolled like that πm
#BetterthanAhlfors
13:47 was very funny
Bet you can't do a whole video saying 'zed' instead of 'zee'.
Portsmouth FC most english people can't
Jorge C. M. .....actually, most English speaking people say say "zed". It is only Americans who say "zee".
Remlat Zargonix That's what he said lol
I interchange zed and zee pronunciations, and I'm American ;)
we say zed here and I am Canadian but British people also say zed.
👍🏻
ur cute ;)
I really wise you would stop using all caps.
Unfortunately I can’t, because the only alternative for me is to write in cursive, which I promise you nobody can’t read
Still love your videos though.