Just wanted to say a quick thank you for all the help throughout various topics around multiple subjects. What I most appreciate about your videos is not only the straightforward, as easy to grasp as possible, approach but also the amount of effort you put in editing your videos to exclude any distractive noises and unnecessary pauses etc. You are doing an admirable job and I can actually see how future teaching could move online with videos like this one allowing me to understand something that my lecturer takes about 1.5 hours to teach. It actually feels like you are saving my time.
These videos are so great, I think the only thing they sometimes lack compared to long university lectures is examples with more advanced complex functions, deriatives, integrals, etc. But as for future online education, you may even learn just as much, if not more, from doing all the examples yourself instead of first seeing a lecturer demonstrate several examples before applying the knowledge to your own work.
These videos have no fat whatsoever. Streamlined, hard hitting stuff. That was Cauchy’s integral theorem in 8 minutes. Earned an instant subscribe from me. Excellent work.
You are an absolute legend, these videos are so useful in making sure my understanding is sound. You have very efficient videos and they are great before an exam. You are a life saver!
Holy shit that's a alot of integrals. Great video btw. Barely understood anything due to the fact I'm only in precalc but I love the format. Its like a cut commentary for math. LOVE it
sir, you know at 0:43 you said "each of these paths [for a line integral in the complex plane] can give you different results", isn't this technically untrue as cauchy's theorem says that, once the appropriate conditions are met, the line integral along a closed loop C is zero must imply that the line integral is path independent? Also, a few things caught my eye. The cauchy riemann equations look a heck of a lot similar to the derivative test (Py=Qx) for determining if a vector field is conservative. I think there is a strong link-is that correct? I think it might be as the cauchy thereom is a condition for conservative vector fields. It would also make sense as I read that there is zero curl and divergence, which is characteristic of a conservative vector field. Thank you very much for these awesome lecture series!
Hi Sean, At 0:43, when I said that "each of these paths [for a line integral in the complex plane] can give you different results", I meant that for a general case (that's why I also used the qualifier 'can' in my statement). For analytic/holomorphic functions, it is true that the paths give the same line integral. However, in general, for any complex function (which could be non-holomorphic), they won't necessarily give the same result. And yes, there is a strong link. A holomorphic complex function is a lot like a conservative vector field, in that its line integral over a nice enough path is 0. Even the derivative test you mention is reminiscent of the Cauchy-Riemann relations (which I write down at 6:23). I might have even mentioned this link in another video. Thank you for the feedback!
Hey, awesome video, great explanations, very helpful! Thank you! If I could make a suggestion, I found that the video moved a little bit fast for me and i had to pause a bit, but other than that awesome job thank you so much!!
I just wanted to ask what do you mean by saying that C has to have a finite number of corners because in the case of a circle, which is similar to a polygon with infinite corners, doesn't that say you can't have smooth curves; any help would be much appreciated.
could you plz explain why function should be holomorphic inside for Cauchy's theorem to holds. why non analytic point matter inside closed curve, the integration is taken along the curve not inside?
Is it a valid proof to say that if we assume that f(z) is some derivative dg/dz then if we integrate f(z) from z to z (in some circular path), we essentially get g(z) - g(z) = 0? Essentially the FTC for Line Integrals? I know that infinite differentiability is something holomorphic functions have, but I don't know if that goes the other way...
Thank you so much for the amazing work! It's been sometime since I have last taken my complex analysis course and your lecture has proven to be extremely helpful in reminding me of the materials I have learned but I have not understood well at that time. Could you please make a video on what it means to 'differentiate' a complex function? Complex function is function from a mapping from another mapping and I have always wondered is there any way I can grasp this notion in terms of 'linear approximation' or some kind of tangent plane-ish concept just as we have those kind of stories for real variable functions.
No problem! Glad you liked it! I think I've discussed differentiation of complex functions in my first video on Complex Analysis: ua-cam.com/video/iUhwCfz18os/v-deo.html Geometrically, it's difficult to come up with a 'tangent plane-ish' concept for complex differentiation. The reason is that a complex function maps a complex number (which is like a 2-D number) to another complex number (another 2-D number). Thus, to create a visualization of a complex function, we would need 4 dimensions (2 + 2), just like how a real function z = f(x,y) would need 3 (2 + 1) dimensions. 4 dimensions are impossible to draw so right now, I'm not sure if I can provide a tangent-plane analogy. Hope that helps!
Could you do a video on the complex formulation of the inverse laplace transform. I understand that if gamma = 0, we uncover the inverse Fourier transform, but I don’t understand how you would compute the indefinite contour integral if gamma is non zero. I love your vids btw.
I was thinking in terms of area as well but if you think that way then remember it is the area under f(z) along the path c, not the area inside the path c if that is what you implying. z=x+yi then you can think of the real parts of f(z) to be perpendicular to your screen, actually f(z) is also two dimensional but for simplification, you would need four dimensions to visualize it. The simplest way to think about is that this is analog to real valued function(x) and the area under the real line from a to b then back to a i.e the integral from a to b plus integral b to a is zero, since the areas cancel out. I was also confused with the fact that the complex function f(z) can have big values on the top of the path c compared to lower path but I think the holomorphic condition forces f(z) to have these nice properties like change of u with respect to x , Ux=Vy etc so when you go round the path c, the integral cancels out. also I think this is not technically the area under f(z) since the value of f(z) cant always be real number, if it was then V would be zero and it wont satisfy the Cauchy equations.
Well, if you are confused about possible larger values of f(z), then you should try to remember that the contour [a->b] is partially with respect to real values, but partially with respect to complex values as well, because unlike on a x,y plane, all point on a contour between two endpoints are defined by two real components (x,y). for all points on the contour between two endpoints on the complex plane, there is one complex component, and one real component. because of the fact that going from a to b involves strictly real components, there is no boundaries on the changes of the complex components. in a sense, as the real components change, the complex components are “let free”, and as long as each point on the contour contains a matching real component of f(z), then the complex component can be (-infty,+infty).
Abstractly, I think I understand. But how can you possibly find an x(t) and a y(t) just like that? Won't it take a long time and a lot of guesswork to find those functions for a random curve?
If you've studied parametric equations, it shouldn't take that long to find those functions. Maybe a simple example will help illustrate the idea and make things seem less complicated: Suppose you have a relatively complicated function like y = x^3 - sin(x), and the curve C is defined from x = 0 to x = pi. A very simple way to find the parametric equations (i.e. express y and x as functions of t) is to just do: x = t y = t^3 - sin(t), where t is on [0,pi] For a circle of radius 1, the parametric equations are actually easier than using the regular y = f(x) form: x = cos(t), y = sin(t), t is on [0, 2*pi] Generally, when solving problems, you're given enough information about C to find its parametric equations. Hope that helps, and let me know if you have any more questions!
I was thinking the following: you can claim that df/dz is continuous simply because f is differentiable inside C as a hypothesis. Since it is differentiable it will also be continuous. The real assumption one has to make here is the one that the partial derivatives of u and v are continuous. Correct me if I'm wrong.
i need to know more about "total derivative becomes partial derivative if it goes to in the integral" i dont mind about fast lecture, because i can keep replaying it
I'm not exactly sure why because the textbooks I've read don't offer much of a reason beyond saying that 'C must have a finite number of corners'. However, my guess is that it has to do with Green's Theorem which I make use of in this proof (at around 5:25). Green's Theorem requires a piecewise smooth curve, which is equivalent to a curve with a finite number of corners. If C had infinitely many corners, then I wouldn't be able to use Green's Theorem or Cauchy's Theorem. Hope that helps!
Great video. Very clearly explained without labouring too much on the background material. Wish I didn't get so distracted by "Cauchy" being mispronounced so much though!
Yup, that's correct. There's a lot of features between complex integration and line integration that are very common (I even mentioned it in the video!).
Just wanted to say a quick thank you for all the help throughout various topics around multiple subjects. What I most appreciate about your videos is not only the straightforward, as easy to grasp as possible, approach but also the amount of effort you put in editing your videos to exclude any distractive noises and unnecessary pauses etc. You are doing an admirable job and I can actually see how future teaching could move online with videos like this one allowing me to understand something that my lecturer takes about 1.5 hours to teach. It actually feels like you are saving my time.
Thank you so much!
These videos are so great, I think the only thing they sometimes lack compared to long university lectures is examples with more advanced complex functions, deriatives, integrals, etc.
But as for future online education, you may even learn just as much, if not more, from doing all the examples yourself instead of first seeing a lecturer demonstrate several examples before applying the knowledge to your own work.
You predicted the future!
These videos have no fat whatsoever. Streamlined, hard hitting stuff. That was Cauchy’s integral theorem in 8 minutes. Earned an instant subscribe from me. Excellent work.
HIS HANDWRITING!
Truly captivating work, have no idea how you smarties do this stuff but I like to watch
You are an absolute legend, these videos are so useful in making sure my understanding is sound. You have very efficient videos and they are great before an exam. You are a life saver!
Glad I could help!
Thank you very much
The only lecture that highlighted the green's theorem
Thank god this isn't in Hindi
i feel you _/\_
lmao TRU
They always say the first few sentences in English too
True kkkkk
True I ******* hate Hindi.
I guess watching these videos is the definition of efficient learning. Thank you!
Really appreciate your videos, simple and elegent explanations!
Thank you so much!!!!!!!!!!!! REALLY HELPFUL
I LOVE THE WHOLE COMPLEX VARIABLE PLAYLIST🥰
the best explanation.
wow the illumination of E&M given to me by complex analysis is grand
Holy shit that's a alot of integrals. Great video btw. Barely understood anything due to the fact I'm only in precalc but I love the format. Its like a cut commentary for math. LOVE it
sir, you know at 0:43 you said "each of these paths [for a line integral in the complex plane] can give you different results", isn't this technically untrue as cauchy's theorem says that, once the appropriate conditions are met, the line integral along a closed loop C is zero must imply that the line integral is path independent?
Also, a few things caught my eye. The cauchy riemann equations look a heck of a lot similar to the derivative test (Py=Qx) for determining if a vector field is conservative. I think there is a strong link-is that correct? I think it might be as the cauchy thereom is a condition for conservative vector fields. It would also make sense as I read that there is zero curl and divergence, which is characteristic of a conservative vector field.
Thank you very much for these awesome lecture series!
Hi Sean,
At 0:43, when I said that "each of these paths [for a line integral in the complex plane] can give you different results", I meant that for a general case (that's why I also used the qualifier 'can' in my statement). For analytic/holomorphic functions, it is true that the paths give the same line integral. However, in general, for any complex function (which could be non-holomorphic), they won't necessarily give the same result.
And yes, there is a strong link. A holomorphic complex function is a lot like a conservative vector field, in that its line integral over a nice enough path is 0. Even the derivative test you mention is reminiscent of the Cauchy-Riemann relations (which I write down at 6:23). I might have even mentioned this link in another video.
Thank you for the feedback!
What do you write on? Your handwriting’s incredibly neat and clear just like the information you present. Subbed
Thanks for subbing! I write on a writing tablet connected to my PC on smoothdraw.
Hey, awesome video, great explanations, very helpful! Thank you! If I could make a suggestion, I found that the video moved a little bit fast for me and i had to pause a bit, but other than that awesome job thank you so much!!
No problem, glad to help! I've slowed down my more recent videos a bit, so hopefully that should address things!
I just wanted to ask what do you mean by saying that C has to have a finite number of corners because in the case of a circle, which is similar to a polygon with infinite corners, doesn't that say you can't have smooth curves; any help would be much appreciated.
That's the best explanation. Thank you!
Great stuff! 😊😊😊
Thank you for explaining so clearly
I liked your quick response
you are awesome
Fantastic series!
Salute to you .Thanks for the video😁
thank you alot, i have finally understood this theorams, i look forward to learn more from you.
Safdar Ansari No problem! Glad you found it useful!
Thank you for your lecture. What kind of program do you use to write letters and graph?
very clear and helpful. thank you so much.
No problem and glad you liked it!
could you plz explain why function should be holomorphic inside for Cauchy's theorem to holds. why non analytic point matter inside closed curve, the integration is taken along the curve not inside?
what is the use of holomorphicity of f on the curve C
great video!
Is it a valid proof to say that if we assume that f(z) is some derivative dg/dz then if we integrate f(z) from z to z (in some circular path), we essentially get g(z) - g(z) = 0? Essentially the FTC for Line Integrals? I know that infinite differentiability is something holomorphic functions have, but I don't know if that goes the other way...
No, that's not a valid proof, but it is a hypothesis.
Thanks for the great lecture.
Thank you so much for the amazing work! It's been sometime since I have last taken my complex analysis course and your lecture has proven to be extremely helpful in reminding me of the materials I have learned but I have not understood well at that time. Could you please make a video on what it means to 'differentiate' a complex function? Complex function is function from a mapping from another mapping and I have always wondered is there any way I can grasp this notion in terms of 'linear approximation' or some kind of tangent plane-ish concept just as we have those kind of stories for real variable functions.
No problem! Glad you liked it!
I think I've discussed differentiation of complex functions in my first video on Complex Analysis: ua-cam.com/video/iUhwCfz18os/v-deo.html
Geometrically, it's difficult to come up with a 'tangent plane-ish' concept for complex differentiation. The reason is that a complex function maps a complex number (which is like a 2-D number) to another complex number (another 2-D number). Thus, to create a visualization of a complex function, we would need 4 dimensions (2 + 2), just like how a real function z = f(x,y) would need 3 (2 + 1) dimensions. 4 dimensions are impossible to draw so right now, I'm not sure if I can provide a tangent-plane analogy. Hope that helps!
Could you do a video on the complex formulation of the inverse laplace transform. I understand that if gamma = 0, we uncover the inverse Fourier transform, but I don’t understand how you would compute the indefinite contour integral if gamma is non zero. I love your vids btw.
i have a doubt ..does cauchys theorem imply that area of contour is zero ? how can that be?
I was thinking in terms of area as well but if you think that way then remember it is the area under f(z) along the path c, not the area inside the path c if that is what you implying. z=x+yi then you can think of the real parts of f(z) to be perpendicular to your screen, actually f(z) is also two dimensional but for simplification, you would need four dimensions to visualize it. The simplest way to think about is that this is analog to real valued function(x) and the area under the real line from a to b then back to a i.e the integral from a to b plus integral b to a is zero, since the areas cancel out. I was also confused with the fact that the complex function f(z) can have big values on the top of the path c compared to lower path but I think the holomorphic condition forces f(z) to have these nice properties like change of u with respect to x , Ux=Vy etc so when you go round the path c, the integral cancels out. also I think this is not technically the area under f(z) since the value of f(z) cant always be real number, if it was then V would be zero and it wont satisfy the Cauchy equations.
Well, if you are confused about possible larger values of f(z), then you should try to remember that the contour [a->b] is partially with respect to real values, but partially with respect to complex values as well, because unlike on a x,y plane, all point on a contour between two endpoints are defined by two real components (x,y). for all points on the contour between two endpoints on the complex plane, there is one complex component, and one real component. because of the fact that going from a to b involves strictly real components, there is no boundaries on the changes of the complex components. in a sense, as the real components change, the complex components are “let free”, and as long as each point on the contour contains a matching real component of f(z), then the complex component can be (-infty,+infty).
@shreyasinha it is not area of contour it is area beneath f(z) along curve C
So a contour integral is really just a line integral?
This is good thanks very much
Marvelous💯💯
please upload a video on residue theorem
I already have! It's right here:
ua-cam.com/video/YWIseo5LwgQ/v-deo.html
Great video! One question though. Is holomorphic the same as analytic ?
Thank you! Yes, it is! See: en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions
Abstractly, I think I understand. But how can you possibly find an x(t) and a y(t) just like that? Won't it take a long time and a lot of guesswork to find those functions for a random curve?
If you've studied parametric equations, it shouldn't take that long to find those functions. Maybe a simple example will help illustrate the idea and make things seem less complicated:
Suppose you have a relatively complicated function like y = x^3 - sin(x), and the curve C is defined from x = 0 to x = pi. A very simple way to find the parametric equations (i.e. express y and x as functions of t) is to just do:
x = t
y = t^3 - sin(t), where t is on [0,pi]
For a circle of radius 1, the parametric equations are actually easier than using the regular y = f(x) form:
x = cos(t), y = sin(t), t is on [0, 2*pi]
Generally, when solving problems, you're given enough information about C to find its parametric equations. Hope that helps, and let me know if you have any more questions!
thankyou so much.but I would like to see some solved examples please
I was thinking the following: you can claim that df/dz is continuous simply because f is differentiable inside C as a hypothesis. Since it is differentiable it will also be continuous. The real assumption one has to make here is the one that the partial derivatives of u and v are continuous. Correct me if I'm wrong.
You are best.. Thanks a lot lot lot...
Glad you liked it!
Sir , how can we visualise this, can you make it more apparent.
i need to know more about "total derivative becomes partial derivative if it goes to in the integral" i dont mind about fast lecture, because i can keep replaying it
Thank you prof.
Why did you require condition 3?
I'm not exactly sure why because the textbooks I've read don't offer much of a reason beyond saying that 'C must have a finite number of corners'. However, my guess is that it has to do with Green's Theorem which I make use of in this proof (at around 5:25). Green's Theorem requires a piecewise smooth curve, which is equivalent to a curve with a finite number of corners. If C had infinitely many corners, then I wouldn't be able to use Green's Theorem or Cauchy's Theorem. Hope that helps!
Great video. Very clearly explained without labouring too much on the background material. Wish I didn't get so distracted by "Cauchy" being mispronounced so much though!
how does it pronounced? I think it is correct
It helped me lot.Thank you for making those videos.
@@duastan2087 Coh-shee
Thank you!
doesn't the region have to be simply connected
I believe I mention that in the video (e.g. in 3:39 when I write that the curve is simple and has a finite number of corners).
thank you sir
No problem! Glad you found it useful!
thanks sir....
then why the integrals of 1/z is 2i*pi
Hey guys i want to advance my learnings. where should i start ?
You should start by subscribing to my channel!
@@FacultyofKhan done
I think the curve should be piece wise smoth
Love it !
U can use complex integration to solve line integral problems and u can use line integral concepts to prove results in complex analysis
Yup, that's correct. There's a lot of features between complex integration and line integration that are very common (I even mentioned it in the video!).
شكرا جزيلا
elegant proof -- worth a bitcoin!
Thanks
it would be nice to see a proof without Green's theoreme
We will need 4d for plotting these graphs
nice
👍👍👍👍👍👍👍👍👍👍👍👍
with more numerical
So with examples you mean? Ah okay, in that case, I'll do that later when I get to example problems in Complex Variables.
Are you a Khan Academy faculty
Nope, just an independent creator.
You really butchered Cauchy's name lmao
You're saying Cauchy wrong.
How is Cauchy pronounced again? I hear cow shit
🤡