i find it incredible that i pay a lot of money to attend a top ranked uni and be taught this, but end up learning all of it for free on youtube. this is amazing, thank you.

@covingtonkua9404

4 жыл бұрын

can't relate more

@FighterAceee94

3 жыл бұрын

And here I thought the reason I wasn't taught this at uni was because my uni is free (paid by taxpayers)

@mettataurr

3 жыл бұрын

More teachers need to learn from this neon-focused study

@SouravBagchigoogleplus

3 жыл бұрын

I also attend a top ranked University. That's why our professor assumed that we learnt these simple things in high school.

@technoguyx

3 жыл бұрын

You don't pay to learn -- you're expected to learn yourself. The point of university is to meet people and find opportunities. And it should be free for anyone capable of entering it.

Thank you Dr. Khan. My proffessor finished up complex analysis within 1.5 month and we tought it was impressive. But with you I understood the course within a night. Please continue teaching us because you are better than any proffessor I know.

This is not just a lecture on complex variables and residues. This is a service to the human race. Years back when I took complex variables at Stanford, it was near impossible to understand Cauchy's theorem, residues, and their uses. It took me an extremely long time to understand what this lecture communicated in 15 minutes. Current students from the University fo Toronto, which prides itself as one of the worlds top 20 post some horror stories about their freshmen courses in calculus, physics and YES even pedestrian chemistry and biology.

After hours of confusion with my terrible textbook (fischer), you have cleared everything up for me in less than 15 mins, thank you

Oh my god. This is great. Why I haven't discovered you earlier 👍I just loved your presentation style speaking style. You have earned a fan 🙏

WOW, I'm so astonished by the fact that the contents and explanations are laconic, comprehensive and engaging, excellent in all aspects! Thank you so much for this marvellous course!

Wow! I have watched this video before. At first, I haven't understood it, but then I watched the whole series (with skipping the proofs) and it is very clear. Thank you sir, your explanation is to the point, and very clear!

Thank you SO much. I have an exam on complex function theory in a few days and my mind had gone totally blank! Amazing, clear, thorough presentation. Thanks for saving my degree!

This series is pretty cool. Concise, formal and clear, thanks!!

Brilliant, thank you for posting. It's like being in a dark room and having the light switched on.

@georgedelgado1995

5 жыл бұрын

alex zorba 🕺🏻

Absolutely Clean And Amazing Explanation, Thank you

This video perfectly shows why I love math !! Greatly appreciated. Keep it up.

Thank you really much. The explanation is so well done and easy to understand. Looking forward to learn more from your videos!

Just to inform you sir You are doing a great work for learners like us. Thank you

This video is exactly what I was telling my friend I wanted from my professor but wasn't getting. Thanks!

Concepts so beautifully explained

I was looking for this video all over the internet, nice explanation👍👍👍

Great work! I wish you would continue with complex analysis making more videos about the theory and examples.

You, my sir, are what would generally be known as a LEGEND.

Hope you are getting the recognition you deserve. Great material. Thank you very much for your effort.

@FacultyofKhan

Жыл бұрын

Much appreciated!

Thank you so much for this high quality video!!

Your videos are extremely helpful! Thanks!

i loved this lecture .... all clear to me now .. thank you soooo much

these videos are sooo beautiful and entertaining

@FacultyofKhan

7 жыл бұрын

Thank you so much!

@sabariprasad4239

5 жыл бұрын

how it is entertaining...

just wanted to say that you're amazing and thank you :)

I would like to see an example of the residue theorem where a pole lies on a boundary (not inside, not outside). Great videos by the way!

@FacultyofKhan

7 жыл бұрын

Interesting, I can try that in a future video for sure, and thank you for the kind words!

@uhbayhue

2 жыл бұрын

4 years late, but I got a solution for you! Suppose a pole z0 lies on the real axis while you consider a contour in the upper half-plane. Then, you approximate z0 as z0+ i*p, in the limit as p approaches 0. That is, z0 = lim (p --> 0) [ z0 + i*p]. The addition of i*p removes the pole from the boundary and pushes it into the upper half plane. You can equivalently let z0 go to z0 - i*p to push it into the lower half plane. So replace every instance of z0 with z0 +/- i*p, simplify, and take the limit at the end. This way, you can tackle poles on boundaries of contours.

Very clear explanation. Thank you!

Thankyou so much for posting this video!!!!

Much thanks for proving technic 3. My book does the same by saying it's easy to identify that without show me the proof. Now have the complete picture of computing residues.

Really good work! Thank you so much for this

Thx for this explanation. Made a lot of things clear for me!

The massive dusty tome I was reading on this has been hard going but your lectures made it easy and enjoyable to understand

Awesome Lecture Thank you so much sir

Thank you so much for your great vidios! Sometimes they are life saving.

@FacultyofKhan

7 жыл бұрын

No problem! I appreciate the kind words!

What a beatiful lecture!

Excellent lecture series man!

It is very helpful. Thanks a lot for it.

Thank you so very much for this :)

Thanks so much Khan. You just saved me hours trying to revise Complex Calculus. Your videos are so concise and easy to understanding. Keep these videos coming :D !

@FacultyofKhan

6 жыл бұрын

Thank you, and I will!

Great channel. Thank you so much!

@FacultyofKhan

7 жыл бұрын

Glad you like it!

Very nice demonstration.

I might be missing something here, and technically you are correct, but surely for a function f(z) having a pole of order n, it is unnecessary to multiply the Laurent Series for f about the pole α by (z - α) to any exponent greater than n. Indeed (as demonstrated in your last example) for such a function f, with Res(f, α) the residue of f, Res(f, α) = (1/(n-1)!) lim(z → α) d^(n-1)/dz^(n-1) [(z - α)^n f(z)] For m > n, all that we achieve is more work in determining the residue, as fun as repeated differentiating may be.

Thank you so much, life saver

AMAZING. many thanks

Wonderfully clear description. Huge thanks. What drawing tool are you using?

Great video thank you so much

i like the speed of explaining

Thanks a lot! I will pass my finals hopefully.

Really great content

Wow...your video was very helpful...but can you tell me the tool of making this video? I mean can you tell me the software you are using for making this video?

this i steh first math video i needed to put at 0.75 because you are god dam fast, BTW: nice explanation

Amazing videos!!

@FacultyofKhan

7 жыл бұрын

Glad you like them!

give thanks to you from korea teacher you made me keep dreaming

ayyyyy you the G my man

You are an angel.

Love it!

Very good lesson

Does the next video on complex variables exist yet? If not, please consider this a request :) great video by the way!

@FacultyofKhan

7 жыл бұрын

Doesn't exist yet, but I'm making one. Thank you for the kind words!

Thank you

5:19 When z^2 - 1 is factored, wouldn't that add a second (z - 1 ) term into the denominator?

Amazing

I want to make similar videos. How do you make them. What app?? Or tricks?

I think I'm being very dim but I don't get the first example. The taylor series of the sin(z) is the analytic part right? I dont get how dividing through by 1/z^2 makes it the principal part. I'm clearly not getting how to construct the Laurent series. Does anyone know of any pages that go through step by step how to get the laurent series without the taylor series shortcut?

"The proof is very simple and left it as an exercises" LOL I am preparing the midterm and all my roommate are looking at me like I am a fool

Hello! Can you recommend any books related to this topic? Thank You

thank you

Your proof works if the set of singularities is discrete topologically, because only then you can draw small disjoint circles around them. Are they always like this? Can the set of singularities have an accumulation point, for example?

I like the lecture it is very

at 4.02, could you explain why it stops at j=1 for a simple pole ?

You're the best. I wish you could offer a Coursera or EdX course. Are you a faculty at a University?

@FacultyofKhan

6 жыл бұрын

Nope, not a faculty. I'm a university student.

@marcelasiqueira6658

4 жыл бұрын

Most professors at universities do NOT teach so good.

GOAT!!!

Why is it that in the third technique we do not have to the take the limit as z -> z_0 as we did when z_0 was a simple pole?

Valeu! Thanks!!

Exam in 2 days. Thank you for this

@FacultyofKhan

4 жыл бұрын

All the best!

allow me, but this is fu*king nice!!

ur a legend

Can I use rule three for every residue function?

Please help us to find poles and residue of 1/sin²z

Another important technique: If you have a function of the form f(z)=p(z)/q(z), where q(z) has a zero at z=a but its first derivative is nonzero at z=a, and p(a) is nonzero, the residue at z=a will be equal to p(a)/q'(a).

Hi! Thanks for the nice video! I have a question, for the case of finding the residue for a pole of higher than 1 order, let's take as an example a pole of 2nd order. Wouldn't it give the same result and also be simpler if the residue would be calculated as lim z-> z0 { (z-z0)^2 * F(z)} instead of doing so many derivatives? it means that it would apply to a pole of any order n>=1 as follows: R[z0]= lim z-> z0 { (z-z0)^n * F(z)} with n = the order of the pole. What is the difference? Thanks!

@mbrusyda9437

2 жыл бұрын

No, plugging z0 to the series gives b_n not the residue b_1

Cant thank you enough

5:27 why is z = i is the only simple pole? how about z=-i, z=1 and z=-1? cosz for those zeros are also analytic and they all appear once in the denominator.

well , if lim (z-z_0)f(z) when z tends to z_0 tends to infinity , z_0 also could be a essential singularity but if there is a finite n such the lim((z-z_0)^n)f(z) is finite, then z_0 is nesesarily a pole

How do you know if something has a pole ?. Is there a video that you have on this concept ?

@FacultyofKhan

9 күн бұрын

I define poles here (I'd suggest watching the videos before this one on the playlist so you have a stronger idea of what's happening): kzread.info/dash/bejne/Yo6Dr82mqZDHhNI.html

Please make a video on branch point, too.

@FacultyofKhan

5 жыл бұрын

Working on it!

Sir can u provide video to find resume(sin1/z-1,1) plz sir

I don't really see why for a pole of order n we can't just multiply with (z -zo)^n. Souldn't it also give us the residue at zo? Obviously what I say is wrong because it doesn't give the same result but I can't find out why. EDIT: I just got it if we multiply by (z-zo)^n we would actually get bn and not b1

This is so useful.. Thank you! you really help me. I'm student of physics so.. If you need some help in whatever you want.. just send me a message :).

@FacultyofKhan

6 жыл бұрын

Thank you for the feedback and the offer! I don't need help right now but if I do, I'll be sure to make an announcement!

Why are we only finding b1 using technique 3 , what about the other bn values for the residue ?

@mogustew

3 жыл бұрын

Only the coefficient b_1 is called the residue.

Hi, how is the cosine of z continuous and holomorphic at i ? Can someone explain.

@MrBorderlands123

4 жыл бұрын

Use the Cauchy-Riemann equations to show that cos(z) is an analytic function.

How to solve fourier series and transforms

How about residue of (z-Sin z)/z?

Is this your natural voice or is the voice computer generated?

@FacultyofKhan

4 жыл бұрын

A bit of both :) I usually edit the voice afterwards to make the audio cleaner.

i love you

@georgedelgado1995

5 жыл бұрын

Nick Cooley 🍎...even without a soul

How to find residues of trigonometric functions

Damn you're cool bro

✌✌

Why b1 = Re(Z0) ?

@FacultyofKhan

6 жыл бұрын

Because that's the definition of the residue (the coefficient of the 1/(z-z0) term in the Laurent expansion).

cool

godsend