Complex Integrals | Contour Integration | Complex Analysis #11

[@AngelsofDeath]

Had to say thank you, not only for this video but for the entire playlist that has really helped me understand Complex Analysis way better.

[@TheMathCoach]

That is great to hear, I'm just glad that my playlist could help you to understand the subject better! Thank you for the kind words, I appreciate it deeply.

[@iSolarSunrise]

Why is this the last video :( ... Your videos are absolutely amazing so please keep uploading these flawless videos!

[@TheMathCoach]

What a coincidence, this is what I'm doing right now: gyazo.com/8b8b158576f2b853ef5aa08a6c31d43c I'm really glad you are enjoying my content (it really keeps me motivated) and don't worry I'm still determined to make more videos, just need my schedule to let me since I just started my first job 😅

[@eccedentesiast4647]

Ultimately I would say, this underrated channel deserves subscription more! :)

[@TheMathCoach]

I can only agree :) thank you for the kind words!

[@elenaclaramaria8577]

Hi there, I just wanted to say thank you for your amazing videos. Maths is all about creating a good panorama from which you get to choose what you need anytime. And you paint very clear sceneries about every topic. Thanks a lot!

[@TheMathCoach]

Hi, that was quite beautiful and moving, thank you. I'm happy you shared this with me and that my content has been useful for one more person!

[@ekoi1995]

Very accurate penmanship! It's so satisfying!

[@TheMathCoach]

It is one of the things I take very seriously, thank you! I will probably make a video one day where I show my in real life writing, it is even more extreme in my opinion.

[@Fru1tyy]

I was wondering about the integral from 0 to 1 of (1+t(1-i))(1+i) because when I integrated it myself I got the anti-derivative to be (1+i)t+t^2. So when i plugged in the values i got 2+i instead of 1 which is what your answer was . Sorry for asking.

[@TheMathCoach]

It seems like you are completely right, I missed to adjust for the "+1" term in the integrand when I determined the primitive in the video at around 11:35. Good thing you asked since you probably was not alone with thinking that something was wrong :)

[@Predator517]

Thank you so much, I have a lecturer who refuses to put up worked examples, just the answers and questions, so if you miss any part of the module you have no basis to go off.

[@TheMathCoach]

My pleasure to help! I feel you, the step by step solutions are really helpful when trying to grasp a subject!

[@t.o8386]

your videos are truly heavensent, please never stop thank you so much!

[@TheMathCoach]

Glad you like them!

[@CraftReaper-OmarFaruqTawsif]

one of my biggest regret in this year is not finding this playlist 3 days ago TT

[@TheMathCoach]

Ahh that is too bad to hear! Better luck next time!

[@PunmasterSTP]

Another truly incredible video, that touched on so many things I'd heard of before, but didn't fully grasp. Thank you again so much!

[@TheMathCoach]

Glad it was helpful!

[@PunmasterSTP]

@@TheMathCoach Hey I just came across a video my Michael Penn on the Laplace transform and the gamma function ( kzbin.info/www/bejne/qn2Qm3tpg8Z_gdE ). One of the commenters Francesco Santoni brought up a very good point. In the integral of t^a * e^(-st) from 0 to infinity with respect to t, if we make the substitution x = st, then x would become a complex variable since s is complex as well. But then in that case, shouldn't the bounds of the integral change from 0 to infinity to something else, and so we'd need to use a contour integral and perhaps the Jordan lemma to evaluate the integral? I'm curious what your take would be on that situation. Thanks, and thanks again for all of the wonderful videos you've shared!

[@Achrononmaster]

Interesting commentary, a phrase you used piqued my inner philosophy nerd. @4:05 you say you "need to prove" the deformation exists. But the picture is a proof. It is only the formalist mathematical paradigm that says you must turn the picture into some other meaningless symbols. We do not have to accept that paradigm, it is a social construct. Both the Latin/Greek symbols and the picture, are interpreted by a sentient mind imbuing them with meaning due to our social context, so in a more platonic philosophical framework, either one suffices as a proof, one is easier to grasp, the other easier to rigorously defend formally. But there is always a way to translate a good enough picture into a rigorous formal proof, if the implicit spatial dimension is somehow indicated. I get it that visual proofs are hard to find, and diagrams per se can be misleading (not contain the full logic) but in this case of the path deformation I think the picture suffices, one only need do the symbolic algebra to placate a some examiner or editor.

[@TheMathCoach]

I really like this comment and I completely agree with you that "...either one suffices as a proof, one is easier to grasp, the other easier to rigorously defend formally". We always have to remember that they are just different ways to express the *same information* and sometimes one is easier to grasp than the other and different authors/viewers have their own preference about which one to use. In this case, I wanted to stay close to the definition I introduced earlier where we had to find the function but some people would get the point right from the image and the image only. By the way, I guess this comment is about the concept introduced in the "Deformation of Contours" video in this playlist and not this video :)

[@AbhishekKumar-jg7gq]

Your videos are very helpful for me in understanding maths deeply once again thanks for this videos love from north india 🇮🇳

[@TheMathCoach]

Glad you found my videos useful, best of luck with your studies from sweden!

[@individuoenigmatico1990]

In the case of the integral 1/z, you actually can solve the integral even if it passes across the negative real axis, you just cannot use the principal Log, but you can use any other Log that has a cut that does not intersect your curve. In fact all complex logarithmic functions have 1/z as derivative in their domains. For instance Log'(z)=log|z|+iθ where 0

[@TheMathCoach]

Thank you individ for clarifying in the comments!

[@Stayawayfrommyname]

Let me begin by thanking you about this amazing series, it has helped me a lot! Just one thing though, at around 7:20 you say that 3pi/2

[@TheMathCoach]

Stayawayfrommyname I'm happy to help and really glad that you are liking the playlist. Let me get back to your question later today (travelling atm) since I think this is a really observant and interesting question which I would like to answer thoroughly :)

[@TheMathCoach]

Stayawayfrommyname I'm back and you are completly right in your statement, they should be in increasing order even though it all end well since we have periodic argument for the path. I will keep that in mind for my next video about the subject, thanks for the help :) Great name btw!

[@Stayawayfrommyname]

TheMathCoach when you get the time, please keep making videos, I love the motivation you give for everything and I find your voice very relaxing! Thanks, my 12 year old self was pretty funny. Great channel btw :)

[@TheMathCoach]

Ohh I you can be sure I will. It was just that my schedule with working on my *master thesis* and *job applications* have occupied my time for the last 6 months. But rest assured, I will *continue to make videos* and thank you for your kinds words. Always happy to meet someone who appreciate my content and that it can be useful :)

[@leongonzalez5803]

With the limits he used he is integrating in a different path, being this path the 3/4 of circle instead of the desired quarter.

[@hamadhalbahrani8781]

an amazing explanation thank so much I appreciate you effort and giving your time

[@TheMathCoach]

My pleasure and thank you for your kind words!

[@kevincardenas6629]

I can't thank you enough for these videos. Nice work! Also Is it possible that you make a video on Cauchy's Theorem of Integration? please!

[@TheMathCoach]

Glad you found them useful :) I will do that when I continue making videos

[@dijkstra4678]

Woah, Amazing! Thanks!

[@TheMathCoach]

Glad you like it! ^^

[@hadijaffri9856]

You are a life saver

[@TheMathCoach]

Ahoy, happy to be saving another sailor in the ocen!

[@wh17efox]

this video save my life thx

[@TheMathCoach]

It seems like I'm a lifesaver today then :)

[@wh17efox]

@@TheMathCoach my mark at exam was 9/10 thanks to u

[@TheMathCoach]

@@wh17efox that is great to hear! Good job!

[@TheJProducti0ns]

I’m confused on how you got the bounds for the circle example? May you please clarify how you got 3pi/2 and 2pi Btw awesome video!!

[@TheMathCoach]

Think of the path that is drawn as a part of an circle, which can be described with an angle and some radius around a point. The starting point is then located at an angle 3pi/2 and the endpoint at the angle 0pi (or 2pi) when you follow the drawn path.

[@xulq]

this video is really well done

[@TheMathCoach]

Happy to hear that! It took some time to make xD

[@xulq]

never stop teaching people please

[@TheMathCoach]

@@xulq I will try that!

[@ryanmckenna2047]

That was fantastic, thank you.

[@TheMathCoach]

Thank you for the kind words :)

[@loriz6103]

Most helpful video ever! Learned a lot’

[@TheMathCoach]

I'm happy to hear that :)

[@bebarshossny5148]

Man this is so swedish 😂 Loved the video man thanks a lot Subbed

[@TheMathCoach]

[@ranani20]

ikr!! he has the loveliest Swedish accent xDD

[@vicentechomali2047]

Thank you so, so much!

[@TheMathCoach]

Happy to be able to help, hope you have a great day!

[@Amani_Rose]

11:29 How comes we didn't take the derivative of the conjugate? As in for gamma 2, why is it (1+t(1-i)(1+i)dt instead of (1+t(1-i)(1-i)dt? If we took the conjugates right after parameterization, wouldn't the derivatives of those give dz bar?

[@entropiclips]

i hope u keep on making these awesome videos. Nice explanation easy to understand

[@TheMathCoach]

Really happy to hear that you find my content useful and I will continue to make videos when I have the schedule to support it.

[@gaaraofddarkness]

veryy nice. you explained nicely

[@TheMathCoach]

Glad you found it useful :)

[@phoenixk4328]

Sir I have a question. At 11:26 why did u write (1+i)(1-i) instead of the conjugate only which is (1-i)? also Z2 whas Z2=(3-i)+t(1+i), should I take the conjugate of (3-i) too? Thank you very much u r one of the best on KZbin.

[@TheMathCoach]

Hi, happy to help! check the formula to the right, I have to add the derivative when determining the integral which is where the "(1+i)" factor comes from. Where did you get the expression for z2? I don't see it in the video, but I might have missed it.

[@buana9588]

dude you are genius

[@TheMathCoach]

I will just have to add this on my resume now

[@DarkSlayerphlc]

Thank you Professor!!!

[@TheMathCoach]

My pleasure to help!

[@CH-zw2gm]

In the final example why did you use the complex conjugate of z instead of just z? (Great video by the way!)

[@TheMathCoach]

Happy to help out, I used the complex conjugate because that was the integrand that we wanted to determine in that example.

[@jovysu4509]

Still, you haven't explained why the setting started with conjugate instead of normal z, is there a special reason of doing it?@@TheMathCoach

[@SoumalyaPatra07]

Which software are you using for writing? And which devices are you using?

[@TheMathCoach]

I have a video about that on my channel, but in short camtasia (now premiere pro) and a wacom tablet.

[@ashkonnoori1294]

Awesome video

[@TheMathCoach]

Thank you!

[@DargiShameer]

Good explanation

[@TheMathCoach]

Thank you :)

[@comentadorsincero8315]

Thank u so much

[@TheMathCoach]

My pleasure :)

[@KuldeepSingh-tn7mb]

At 11:28 why didnt you use congugate of z'(t) with the multiplication of conjugate of f(z(t))?

[@TheMathCoach]

Hello, The definition states that we can determine the integral by inserting the *parametrization* [z(t)] into the *function* [f(z)] and then takes this times the *parametrization* [z(t)]. The thing here is that our function f(z) = conjugate of z, which means that the first term *f(z(t))* becomes the conjugate of the parametrization but the second factor z'(t) is only the derivate of the parametrization. Hope it clarifices and sorry for the late comment!

[@tanmayprasadkhadsan8142]

Really great series. Btw, I just became your 1000th subscriber!

[@TheMathCoach]

I'm happy that you liked it and thank you for helping me achieve 1000 subscribers, that is a really big milestone for me! Welcome to the channel :)

[@sarah_70]

Hi, TheMathCoach! You explain things so well and I really need your help. I am currently doing Complex Analysis at University and it's destroying me. My lecture notes are really hard to understand. Could you please do a video explaining the important stuff about Affine Maps, Conformal Maps and Power Maps, please? I don’t understand them at all. I am trying to find stuff about this but no one is explaining it properly. Could you please save me?

[@TheMathCoach]

Hello Sarah, I'm happy you find my videos useful and I agree, sometimes lectures notes are just impossible to understand. I might be able to help you out with the Conformal Maps part, since that is a video I have planed to make for a while. However, I'm currently working 24+ hours a day with my master thesis. Therefore, when is your examination for the course? I will see if I can make it before that date, can't promise though :) Best regards,

[@sarah_70]

TheMathCoach, thank you so much for responding to me! I totally understand and I wish you the best of luck with masters! In that case, don’t feel pressured at all to make the video. My exam is either at the start or in the middle of June. :)

[@TheMathCoach]

Thank you and no problem, I always take time to answer all comments. June might just work out, I will let you know if I start working on it :)

[@sarah_70]

Awesome! You’re a legend. :D

[@TheMathCoach]

Hello again, I started gathering material this afternoon for the video we talked about a while ago. It might however take some time for me to finish it (master thesis and work applications takes some time) but I also would like to help you if possible. Hence I wonder if you know when your exam is so I have an approximate deadline? :D

[@yousifsalam]

can you explain how I should think about this integration? can this still bethought of as measuring the area under the curve? I highly doubt it since some functions are independent of path of parametrization.. I'm just confused as to what to think about these integration in terms of their significance and meaning. Thanks a lot for the help!

[@TheMathCoach]

Sorry, I don't really know. You could try to check out this thread maybe: math.stackexchange.com/questions/110334/line-integration-in-complex-analysis/110367#110367

[@inzepinz]

Awesome, thanks

[@TheMathCoach]

My pleasure to help!

[@durellodelmattino3918]

TheMathCoach Hi, I wanted to ask if at 11:45 there were a missed integration of 1dt in the second integral, please correct me if I'm wrong

[@durellodelmattino3918]

Also thanks for the series on complex analysis, It is a huge help and very clear

[@TheMathCoach]

Yes, you are correct there is a note about it in the description "I missed adjusting for the "+1" term in the second integrand when I determined the primitive in the video at around 11:35."

[@TheMathCoach]

Happy to be able to help and glad you find my content useful durello :)

[@MrEzystreet]

Great video series. Shouldn't the product in the integrand between f(z(t)) z'(t) in the first definition be a scalar product? This way in the first example you will have in the integrand [0 times (-i) + (2+2i) times (t(2+2i))]?Cheers!

[@TheMathCoach]

Thanks a lot, I'm glad you are finding the playlist useful! I have been using the normal product denotation since that is used in the definition in my textbook (Fundamentals of Complex Analysis by Saff & Snider) and is the convention I have seen online this far. I would like to argue that we can't use the dot product since this operation only returns a *scalar* (thus in practice we lose the imaginary term 'i'). My reasoning is as follows. In the first example, let us write the two complex numbers on vector form by splitting the complex number (z = a+bi) into its real and imaginary part (z_{vector} = (a,b)). Hence we should get: f(z(t)) = 2t +(2t-1)*i -> f(z(t))_{vector} = (2t, 2t-1), z'(t) = 2+2*i -> z'(t)_{vector} = (2,2). By calculating the dot product of the vectors we get that: f(z(t))_{vector} (*dot product operation*) z'(t) = (2t, 2t-1) * (2,2) = 2t*2 + (2t-1)*2 = 4t+4t-2 = 8t-2. Here we can see that we have lost the imaginary part of the integrand and we will therefore not be able to get the same answer as in the video. Thanks for sharing and asking your question, I'm always open for discussion about my content. Just let me know if I misinterpreted you in any way and furthermore I would love to read into this some more if you could link me some of your sources. Cheers :)

[@MrEzystreet]

Thanks for the reply, loved it. I don't have any source was just a thought seeing how contour integrals are just path integrals.

[@TheMathCoach]

Ahh got it and it is my pleasure, I like to go ham in the comment section with my replies!

[@flameon8185]

In improper integrals, when we say poles above the real axis, does that count for z=0 also?(origin)

[@TheMathCoach]

a pole can exist anywhere in the complex plane, it all depends on the function and the origin is on and not above the real axis :)

[@yousifsalam]

@ 13:33 you said Ln(z) is not defined on the negative real axis including zero, but looking at the restriction for theta, it's -pi(not included to pi(included) so the function is defined at the negative real axis I would say.. can you explain this, please?

[@TheMathCoach]

-pi is the angle that corresponds to all the numbers along the negative part of the real axis and the restriction of theta tells us that theta never can be equal to -pi. The complex logarithmen can't be defined for all the angles on the complex plane, you always need to choose one "branch" = line that it is not defined for and the standard is always the line that makes up all the numbers with the angle -pi.

[@dijkstra4678]

At 13:04 why is it that it can't handle functions with the argument -pi if it can clearly handle functions with argument pi which is the same thing as -pi. I'm just extremely confused as to why this is the case that you have assumed e^i(-pi) =/= e^i(pi) when they both are the same.

[@TheMathCoach]

the argument function is not defined on the whole plane, you have to make a "branch cut", this is normaly done along the negative part of the real axis and therefor the function is not defined for -pi, and even though you can add and/or subtract 2pi when dealing with functions such as sine and cosine, that is not the case when dealing with this function. So in short the input value pi is not the same as the input value -pi when it comes to the argument function.

[@entropiclips]

Do you have any video related to Cauchy's Integral theorem?

[@TheMathCoach]

Nothing right now, but I have planned to cover it in the future (my schedule does not allow for making videos right now) which I intend to do :)

[@Laxury2050]

Good work.However ,I am struggling to see clearly.I am not sure if I am the only one.

[@thomashenstock6133]

At about 11:15 you multiply by the conjugate (1-i) why do you do that in this example but not any of the others?

[@TheMathCoach]

Hello Thomas, I'm not multiplying by the conjugate but I'm simply inserting the conjugate of z (the parameterization) into the integral. The reason I did it in this example here and not in the others is because the integrand in this example is the conjugate of z while in the others the integrand is simply z. Hope this clarifies and if not, let me know so I can elaborate further.

[@aksharasujeendran9987]

i have a doubt, if the initial point is -1 and final point is 0,then do we still write z(t)= -1,t(1+i) 0r id z(t)= -1,t(-1-i)?

[@TheMathCoach]

You always start with the startpoint and end with the endpoint of the curve :) not sure about the z(t) function in your example so can't answere about that. Have a great day!

[@Amani_Rose]

9:31 Oh good, cuz I was not ready to even try and comprehend that squiggly line lol

[@한정우-s5k]

Hi, it may not be relevant, but I’m curious what app is used?

[@TheMathCoach]

Hello, It is absolutely relevant. Adobe Photoshop for the writing (and a lot of patience) and then I record my screen with Camtasia, which is also my editing program. The audio is later recorded over the video by using Audacity and then I match the video and audio files with Camtasia :) Just let me know if you are interested in knowing about my settings in either program.

[@한정우-s5k]

Really? I think it was just an app. thank you for such a good content. If i have questions, I will ask you again. (I want you to create more contents jaja :)

[@TheMathCoach]

No problem, I will hopefully come back and make videos again in the short (I hope atleast) future :)

[@euler0148]

where is the explanation of residue theorem and integrals of cauchy

[@TheMathCoach]

Cauchy's Integral Theorem is going to be covered in the video I release tomorrow and the residue theorem is going to be covered in the video after that (or the one after that since I have to see so it makes sense in the order). I have a new focus and hope that I'm able to release one video each 1-2 weeks.

[@soubej]

PewDiePie of math

[@TheMathCoach]

#Brofist

[@melinahafelt2591]

haha bästa :)) proud SWE ;>

[@TheMathCoach]

So you heard that? :D God jul och Gott nytt år!

[@HDitzzDH]

+TheMathCoach Are you Swedish?

[@TheMathCoach]

I'm, how did you figure it out (is my accent that obvious)?

[@HDitzzDH]

TheMathCoach Indeed it is, could tell immediately :)

[@TheMathCoach]

I thought so xD

[@Ines-dw2ih]

Where are you from?

[@TheMathCoach]

I'm from sweden, how so? :)

[@Ines-dw2ih]

TheMathCoach I like your accent and voice 😊

[@TheMathCoach]

Ohh thank you, I'm really glad to hear that since I struggle with english a lot back in the days :)

[@Ines-dw2ih]

TheMathCoach No you're doing pretty good actually and your lessons are perfectly explained .Good luck !!

[@TheMathCoach]

Thanks for the confident boost and good luck yourself with your studies 😀

[@thequarrymen58]

i get 5/2 + i on last problem

[@TheMathCoach]

that might be correct, I missed adjusting for the "+1" term in the second integrand when I determined the primitive in the video at around 11:35. Did you do that and got your answer after that? :)

[@davidmafeFX]

you are saving my ass

[@TheMathCoach]

happy to be saving that ass

[@trevorsimpson8788]

I don't understand why anybody would dialike this video?

[@TheMathCoach]

I also don't know, kind of weird if you ask me :)