the complex derivative is strange...

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[@Phlosioneer]

It should be noted that the limits shown are a necessary but NOT sufficient condition. It’s not enough to only consider linear paths towards 0 in a complex limit. There are functions that have a defined and equal limit for all lines through the origin, but on one specific curve (e.g. a parabola) they have a different limit. With some work, you can show that all “pathological” functions like that will also be discontinuous elsewhere, sidestepping the need to take a limit and completing the proof. It’s still a huge landmine for anything else related to complex limits, though.

[@viccaboson4064]

Oh, yes!!! Necessary but NOT sufficient!!! This was my first reaction when Michael said "if and only if"!

[@e.d.8924]

It is even more *basic* than that , here is a quote of a comment I just wrote for this video: "The hidden assumption in this video is that if the derivative of a complex function along the Real Path is identical to the derivative of the same complex function along the Imaginary Path, then they are both identical to the derivative along *any* other path! But is that assumption correct? Let's think of a Real function of two variable x and y i.e. Q(x,y): If one proves that the derivative along the x-axis is identical to the derivative along the y-axis, does that means that they are both identical to the derivative along any other direction/path?" So *even* if we consider only paths which are straight lines, this prove is still not enough!

[@Phlosioneer]

@@e.d.8924​​⁠ considering orthogonal lines is actually *almost* enough for all lines. All you need to do is multiply the original function f by a constant but unknown factor `c*i + d` before decomposing. That rotates and skews the function by arbitrary amounts, allowing the real and imaginary decomposition to represent any arbitrary pair of lines in the original coordinates (like a vector basis). If you do multiply by `c*i + d`, I *think* it falls out/cancels out by the time you take partial derivatives.

[@e.d.8924]

@@Phlosioneer Thank you! After all if these years I finally fully understand the justification for Cauchy-Riemann equations, and as a bonus I understand why the complex function case is different from the case of the real function of 2 variables in the sense that in the complex case two orthogonal directions are indeed enough for the derivative to be identical along any straight line! It's because only in the complex case we can use a complex parameter (what you called 'c*i+d') in order to multiply it with the function so that the all coordinates system will rotates.

[@raptor9514]

Any complex differentiable function is analytic, so any real differentiable function which is not analytic will not be complex differentiable

[@robvdm]

Yeah I was surprised by the “everything” point, there are plenty of very “nice” functions that aren’t analytic. The bump function is a good example of this.

[@dang-x3n0t1ct]

when you're so early there's no sponsorblock segment yet.

[@karanaima]

You skip either 60 or 90 seconds and voilà, no necessity for an external tool when you still have a brain

[@dang-x3n0t1ct]

@@karanaima but it's convenient isn't it?

[@jellymath]

Isn't the money he is paid dependent on the amount of watch time in the sponsoring segments?

[@olasdorosdiliusimilius2174]

​@@karanaimaWay better to have that done automatically, while also not missing 4 seconds potentially.

[@kennnnn]

​@@jellymath Remember, you dont owe content creators anything.

[@markolson4660]

Huh. So *that's* where the Cauchy-Riemann relations come from. I've wondered for almost fifty years. (My complex analysis prof simply stated them and dove in, and I never subsequently made use of complex analysis to have a need to a deeper understanding.) Thanks!

[@paulshin4649]

Another way to see this (which I find conceptually quite pleasant) is to note that the complex numbers can be viewed naturally as a unital subalgebra of real linear endomorphisms of C = R^2 (setting a complex number equal to multiplication by that number), and via this identification, the complex derivatives of a partial map C->C at a point in its domain of definition are precisely the total derivatives which happen to themselves be complex numbers. Via the standard identification of real linear endomorphisms of C with the algebra of 2x2 real matrices, the Cauchy-Riemann equations state simply this very fact above: a partial map C->C at a point in the domain of definition's topological interior is complex-differentiable if and only if it has a complex total derivative.

[@brian8507]

​@paulshin4649 ok buddy. Ur just flexing ur math vocabulary at this point lol. Remember true genius is the ability to explain tricky stuff like that to a child. Like one time a kid came to me and asked why all the whole numbers added up equal -1/12. Well instead of flexing how smart I was. I gave them cool allegorical stories about walking around a path in a park which loops back on itself will eventually bring u back to zero.... unless there was a singularity in the middle of park... then ur phase changes by 2*pi*I And then using thus little fact u can kinda plant seeds in their brain so that maybe one day they can understand analytic continuation and hankle contours and bernoulli numbers. Idk.... math has a barrier that is turning people away... and what u wrote just encapsulates that

[@paulshin4649]

​​​@@brian8507 Sorry if what I wrote came across as rude, that was not my intention - admittedly I do have a tendency to dot all the i's and cross all the t's. Perhaps a better explanation is as follows: there is a natural way to view complex numbers as 2x2 arrays of real numbers, and under this identification, a complex-valued function of a complex variable is complex-differentiable at pretty much any point of interest precisely when it has a complex total derivative there. The mantra that I keep in mind is this: "The C-R equations are just the assertion that complex derivatives are complex total derivatives."

[@kenleybrown8330]

@@paulshin4649 this is a solid explanation with reduced mathematical jargon, kudos!

[@FrostDirt]

​@@brian8507 but the original commenter probably isn't a kid, lol

[@100_IQ_EQ]

One of the very few videos I have understood completely. Michael Penn speaks slowly and explains immediately the possible doubt points like he does when he explains why u(x,y) instead of just u(x). A great teacher. If only others took so much pain and emphasized on students' understanding and not completing syllabus.

[@goodplacetostop2973]

25:12 Homework 26:00 Good Place To Stop

[@homerthompson416]

00:00 start of video 26:36 end of video

[@miikavihersaari3104]

27:01 - 01:22i outside the video axis

[@pierreabbat6157]

e^(-1/x²) is real differentiable at 0, but not complex differentiable at 0 (it goes to infinity on the imaginary path). e^(-1/x⁴) is not complex differentiable at 0 either. The derivative goes to 0 on both real and imaginary paths, but to ∞ on the diagonals.

[@vascomanteigas9433]

It is an essential singularity.

[@numbers93]

saw on stackexchange: f(x) = |x|^2 is differentiable (everywhere), but f(z) = |z|^2 is not holomorphic -- the complex derivative is only well-defined at z = 0. You can pretty much get any value for the complex derivative anywhere else by choosing your path wisely.

[@danieleferretti9117]

An analogous video about quaterionic derivation would be great!

[@christopherneufelt8971]

These things are of Satan! Repent and return to Jesus and Cantor!

[@jounik]

Deriving derivation in a noncommutative division ring is tantamount to hoping the inevitable sign errors cancel.

[@christopherneufelt8971]

@@jounik And behold a White Horse: another Algebra approaches the mystical candidate.

[@mosquitobight]

Would it be possible to derivate quaternions as three sets of partial complex derivatives?

[@christopherneufelt8971]

​@@mosquitobight Aside of humor now. The partial complex derivatives should be in this case coupled equations in order to make this sort of thing? In addition, there will be a need for a Variable to bridge them? P.S. I participated once in an MRI image processing project where it was involving such objects, however in the section of data acquisition (magnetic sensors and ADC), since I am not mathematician, though I did classes on some advanced math (some pure algebra and transformations).

[@GreenMeansGOF]

The complex conjugate function and the real part function are just the identity on R. So the R version is f(x)=x which IS differentiable.

[@iabervon]

That comes down to the question of whether two functions of a real variable can be considered different if they have all of the same values but follow different rules that only affect complex numbers. Michael is working in the formalism where, if you write something as a function of a real variable, you forget the extra rules, and then when you go to a complex variable, you get the one that is differentiable, because that's how you're figuring out what the rules for complex numbers are from the values for real numbers.

[@hyperduality2838]

The complex derivative is conformal as it is independent of angle at a point -- Cauchy Riemann equations. Space/time symmetries are dual to Mobius maps -- stereographic projection. Real is dual to imaginary -- complex numbers are dual. "Always two there are" -- Yoda.

[@kilianklaiber6367]

This is a really good explanation. I had to learn this way back in university and used it so seldom that I had to look it up everytime it was necessary.

[@iabervon]

There's a function with a real derivative everywhere except one point, but no complex derivative anywhere, at least in the most obvious extension (the absolute value, although extending it differently gives some fun functions not differentiable at paths or with weird branch cuts). It's interesting that the point at 0 for 1/x is no big deal with complex numbers, despite that.

[@tomkerruish2982]

You can do way better than that. You can construct a function that is infinitely differentiable (smooth) on all of R but analytic nowhere. See "Non-analytic smooth function" on Wikipedia for an example.

[@thierrypauwels]

What I remember of it, is that you can combine both equations at the end to get an equation for f, without having to use its real and imaginary parts: f_x = u_x + i v_x = v_y - i u_y = - i (u_y + i v_y) = -i f_y. And you get finally the single condition f_x + i f_y = 0.

[@e.d.8924]

Such a *sad* thing! This video ignores something that almost every textbook also ignores: The hidden assumption in this video is that if the derivative of a complex function along the Real Path is identical to the derivative of the same complex function along the Imaginary Path, then they are both identical to the derivative along any other path! *But* is that assumption correct? Let's think of a *Real* function of two variable x and y i.e. Q(x,y): If one proves that the derivative along the x-axis is identical to the derivative along the y-axis, does that means that they are both identical to the derivative along any other direction/path?

[@miloszforman6270]

I don't think that this is ignored in textbooks. If you have a real-valued function, this can only have a complex derivative if it is constant.

[@cauchym9883]

In order to solve the homework one needs to know how \partial f / \partial bar z is defined in the first place. It is defined as 1/2*(\partial f / \partial x + i* \partial f / \partial y). With the notation in the video this can be written as 1/2*( u_x +iv_x +i*(u_y+i*v_y)). From there it is straightforward to see this simplifies to 0 when the C-R Differential equations hold. You can look up the term "Wirtinger derivatives" for a motivation for that definition.

[@ianmathwiz7]

2:19 Wouldn't the complex conjugate function be an example of a function that's R-differentiable but not C-differentiable? Restricted to R, the complex conjugate is just the identity function, which is obviously differentiable, but it's the canonical example of a non-differentiable function on C.

[@burk314]

Yes, his statement of R-diff -> C-diff needs a really big asterisk. Even in the cases where it is true, it is only true for the correct extension. In your case, the identity function on R must be extended into the identity function on C. In fact, we can prove that no other extension of the R-identity can be differentiable on all of C.

[@jamesmcadory1322]

This was so well explained thanks so much!

[@MooImABunny]

2:15 if you take g(x) to be one of those smooth on R but problematic on C, like g(x) = {e^-1/x, x > 0; 0, x ≤ 0} or g(x) = e^-1/x² and integrate them f(x) = ∫ g(x)dx I'm pretty sure you'd get a function that's differentiable on R but not on C

[@reeeeeplease1178]

Or functions with poles off the real line like 1/(x^2 +1)

[@Bodyknock]

Note though that the definition "f(x) = e⁻ˣ if x

[@deinauge7894]

e^(-1/|sin(x)|) is differentiable to any degree on R, but not on C. But it includes the |x| function, so it's not a better example than Re(x)

[@frankstengel6203]

@@deinauge7894 Simply exp(-1/x^2) if x≠0 and 0 if x=0 smooth on R (all derivatives at 0 are equal to 0) but has an essential singularity in 0 if considered as a function of a complex variable.

[@nikitakipriyanov7260]

The function for which we were checking Cauchy-Riemann equations was actually f(z) = z³-2z²

[@Dravignor]

Is this the start of a complex analysis series?

[@iabervon]

There's a Complex Analysis series on his other channel, MathMajor. He doesn't really do videos that build on each other here.

[@seitanarchist]

I definitely have to take issue with your "everything* diff. in R is diff. in C" statement. The counterexamples aren't exactly sparse. You need to start with a real analytic function to be able to extend it to a complex differentiable function.

[@celkat]

Very nice video. However it's not clear to me why constructing two certain limit approaches to be equal (along the real line by setting b=0, and along the imaginary line by setting a=0) is sufficient to conclude that the limit exists for *all* approaches of h-->0 ...

[@Pootycat8359]

Several years ago, I attended a lecture on "Fractional Calculus," which really blew my mind! For those who don't know what it is (I didn't, & I have an engineering degree), here it is. You take the formula for determining the derivative of an integer order, and replace "n!" with the "Gamma Function + 1" of the order. The Gamma Function is defined for ALL numbers: fractions, irrational, & COMPLEX.

[@miloszforman6270]

I can't see any n! in most formulas for derivatives. So what are you talking about?

[@letsmakeit110]

i bet this is how a pet cat or dog feels when watching human shows. like i dont really know what's happening but i like the sounds and colors.

[@scottmiller2591]

I recall that there are complex functions that exist and satisfy the Cauchy-Riemann equations, but are not differentiable. This requires using a curved path for you limit, rather than a = c*b, where c is a constant. Since differentiability requires the limit works over ALL paths, not just straight lines, this is not complex differentiable. Does this ring a bell?

[@miloszforman6270]

I don't think that such a function exists if the partial derivatives exist in a complete neighborhood of z and fulfill the CR-equations. I think that this was once mentioned in a lecture on complex analysis, but the professor did not bother to show this.

[@talastra]

I couldn't do this, but I followed the explanation better than usual.

[@miloszforman6270]

From my point of view, this is somehow missing the one central feature of complex differentiability, namely that this is a kind of _linear approximation._ Linear here means that it is compatible with the algebraic structure of the complex numbers as a _field._ So instead of f'(a) = lim [h->0] (f(a+h) - f(a)) / h we should better write f(a+z) = f(a) + z*f'(a) + R(a, z) with some "small remainder function" R(a, z) where R(a, z) = o(|z|), that is, lim [z->0] R(a, z)/z = 0. This may look more complicated on first sight, but the central characteristic of complex differentiability is this f(a+z) ≈ f(a) + z*f'(a) feature. For example, from this we can immediately see that f(z) := conjugate(z) is not differentiable, as the conjugate means mirroring on the x-axis, while multiplication with a constant means rotating and stretching.

[@mathboy8188]

Exactly how I think of it too. Also, the CR equations fall out if you think of f : C --> C as Calc 3 multivariable f : R^2 --> R^2 to get f(x,y) ~= f(x0,y0) + Df(x0,y0) (x-x0, y-y0), where Df(x0,y0) (x-x0, y-y0) is now considered a 2x2 matrix acting on an R^2 vector. In C multiplication looks like ( a + bi )( c + di ) = (ac-bd) + (ad+bc) i. So to view C as R^2 introduces this vector multiplication in R^2: (a,b) times (c,d) = (ac-bd, ad+bc) In order for a 2x2 matrix | p q | | s t | to act on EVERY vector (x,y) in R^2, in a way that's equivalent to a single complex number a+bi complex-multiplying that R^2 vector that's now considered as x + iy, then must have | p q | |x| | s t | |y| = | ax-by | | bx+ay | For that to be true for all x & y requires a = p, -b = q, b = s, and a = t. Thus for 2x2 matrix to act on every R^2 vector in a way that's equivalent (when switching between C and R^2 perspectives) to being complex multiplication by a single complex number, that 2x2 matrix must look like | p q | | s t |, but with p = t (=a) and s = -q (=b), so the matrix must look like | p q | | -q p | if it's to correspond to complex multiplication by a single complex number ( which will be p - i q). (Obviously get to the same conclusion if view complex numbers in polar form, as that's a magnitude times a rotation matrix.) If f : R^2 --> R^2 by f(x,y) = u(x,y) + i v(x,y), then Df(x0,y0) is the matrix | u_x u_y | | v_x v_y | evaluated at (x0,y0) (which I'll drop hereafter as it's cumbersome notation), where the underscores indicates partial derivatives. Thus for that f to represent multiplication by a single complex number, f ' (x0+iy0), must have | u_x u_y | | p q | | v_x v_y | = | -q p | for some p and q, which says u_x = v_y and u_y = - v_x, i.e. the CR Equations. Moreover get that f ' (x0+iy0) = p - iq = u_x - i u_y = u_x + i v_x.

[@miloszforman6270]

@@mathboy8188 Well yes, multiplication with a complex number a+ib means multiplication with some matrix |a -b| |b a| if we consider complex numbers as two-dimensional vectors, and the CR-equations for the derivative follow quite naturally. I would prefer to put the minus symbol to the upper right, otherwise I would be multiplying with a - ib. Which is of course only a formal deviation. This matrix clearly can't be any matrix, otherwise it won't be compatible with complex multiplication, that is: with the algebraic structure of ℂ. Of course one could also investigate complex functions which are only derivable as functions ℝ² --> ℝ². But that's not the topic of complex analysis. As mentioned earlier, this aspect of the complex derivative as a linear approximation is missing in the video, and it is missing in many textbooks. I just looked at "Ahlfors", an old but renowned textbook on complex analysis: it's just the same. And that's why they conclude: "The complex derivative is strange", which is the title of this video. It's rather strange that they find it strange.

[@mathboy8188]

@@miloszforman6270 It's rather strange that they find it strange. - Agreed. And it's always, in my experience, presented as some artifact of needing to get the same value however you approach the point in the limit - and while that's true, it's hiding the essential point. I mean, you also need the same value however you approach the point in the limit just thought of as a map in the R^2 --> R^2 case, so that's obscuring what's actually special here - that it's the best linear approximation in z, exactly as you said.

[@miloszforman6270]

@@mathboy8188 One other aspect of the "linear approximation" approach is the fact that the product rule and the chain rule come out of it quite straight forward and naturally. That's certainly far from being new, but I frequently noticed that this point is somehow neglected in lectures and textbooks. Not sure about all of them, of course. For example f(a+z) ≈ f(a) + z*f'(a) g(a+z) ≈ g(a) + z*g'(a) h(z) := f(z)*g(z) h(a+z) ≈ h(a) + z*f(a)*g'(a) + z*g(a)*f'(a) ≈ h(a) + z* ( f(a)*g'(a) + g(a)*f'(a) ) for "small" z. We can neglect the fourth term z²*f'(a)*g'(a), as it has a higher order z factor. This applies not only to complex derivatives, but also to the standard real one, as well as its generalizations for multivariable functions.

[@mathboy8188]

​@@miloszforman6270 YES! I'm actually in the process of outlining my own Calculus 1-3 courses. I intend to do it in two passes - the first without even limits, but instead relying entirely on the "self evident" (ahem... hand-waving) notion of linear approximations. It's absolutely the right insight and intuition for using Calculus (which is what students will ultimately need), and having already outlined it, I know for a fact that following just that is enough to get you through the whole of undergrad Calculus (including multivariable calculus, where you simply start by understanding graphs of planes). (Your derivation being exactly one of the bits already sitting in my outline.) And then a second pass done with more rigor, full rigor, compared to typical Calculus 1-3, where hopefully exposure to the ideas in the first pass will provide a better insight into what those epsilons and deltas are doing. I've felt the courses I've seen and taught always threw the students somewhere in the middle between insight and technical proof competence. I think those should be emphasized separately, with the intuition part being more important to all but future math majors. Calculus throws so much at students so fast, and invariably devolves into "knowing calculus" defaulting to just meaning grinding through calculations of limits, integrals, and derivatives, which is really one of the least important aspects for them to learn (though it still matters, of course). What seems to always be missing is an emphasis on _This is the way you should be thinking about it - and it's the way you WILL be thinking about it once you fully understand it._

[@shawon265]

9:35 For d/dz[z*] and d/dz[Re(z)], I think instead of using z=a+bi, polar format z=re^(iθ) would have been better. Because, r→0 would have meant h→0.

[@Phlosioneer]

It would be a more complete and correct limit, but it would also be a nightmare to manipulate and wouldn’t result in the partial-derivative based answer. It’s not actually solvable with generic f. You’d need to know the specific form of f to solve it. You’re right that a+bi is not sufficient here, but unfortunately the limit isn’t analyzable otherwise. The actual, full proof has to show undefined curved-path limits are impossible for other reasons. Curved paths with arbitrary complexity are just too difficult to handle.

[@eartphoze]

Its like the mona lisa vs Declaration of independence Wire this up like a stringer breadboard, X=0, 0:08(if statement) 1:09(A.I) 2:10(BJ) 3:11(CK) 4:12(DL) 5:13(EM) 6:14(FN) 7:15(GO) 8:16(HP) 9:17(IO) 10:18(JR) 11:19(KS) 12:20(LT)(ends at 12) A> H = 8, I > M= 5, N >R= 5, I[H=8]=[E=5](8,5) [I

[@eveeeon341]

For the checking of differentiability of complex functions, my initial reaction was that "why do you only have to check real and imaginary paths, what about the other ones?" I think the answer is that any path can be constructed as a linear combination of real and imaginary paths, therefore, you can do the same proof with any two paths that aren't along the same line in the complex plane. This got me thinking how you might you prove this all in one go, and my initial thought was to use a kind of symmetry, where rather than a->0 then b->0 for real and imaginary respectively, keep the a+bi->0 and then compare to the same where you swap a and b (b+ai->0), and you should get the same result. While you technically are taking two different limit paths, it should only require one calculation. I'm not sure if more can be said about the symmetry, but this kind of approach appealed to me.

[@Phlosioneer]

Take a function F such that it is everywhere zero except the parabola a^2=b, where it equals 1. Along any linear limit, the line will cross the curve at either two points, one point, or never; therefore the limits can ignore those anomalies and converge to 0. But for the limit along the parabola’s curved path, the function is always 1, so the limit at a+bi=0 is also 1. A contradiction. It’s an extremely tough problem; while this function example is obviously discontinuous, you can construct similar functions that are not obviously discontinuous. It turns out that such functions must necessarily be discontinuous at an infinite number of points, but it’s a lot of work to prove that.

[@magma90]

2:30 Any real analytic function can be extended to the complex numbers and be differentiable, however if the real valued function is not analytic, it cannot be extended to a differentiable complex function

[@egoreremeev9969]

One way to look at complex differentiation is to consider what kind of "variable" is needed to encompass "all degrees of freedom" of functions on 2d space but with coordinates as complex numbers, i.e. (x,y) ~ z= x+iy. Eith that all functions on C must be of form f(z,z*). Although it doesn't directly show how and why, but in this notation df/dz only has "meaning" if df/(dz*) = 0, which is equivalent to Cauchy-Riemann relations

[@richardbrewer2937]

Great, brought back memories, as soon as Michael mentioned Cauchy Riemann Conditions

[@danuttall]

The complex magnitude function ( |z| ) can not be differentiated for the same same reason as the other given examples (different limits along the real and imaginary axes). The equivalent real function is the absolute value ( |x| ), which has a cusp at x=0, so is differentiable anywhere except at x=0. So they have similar, but different restrictions.

[@mtiganik]

Thank you. Very interesting video. Good to watch

[@DataRae-AIEngineer]

Your content is so good and well explained. I wish I knew about you when I was doing analysis in grad school lol.

[@tomholroyd7519]

This super interesting topic about how restrictive analyticity is --- I would love to see how this works with dual numbers, dx is nilsquare, no limits

[@user-pr6ed3ri2k]

Weird how all the simple and important complex functions aren't differentiable

[@ChaineYTXF]

This is very, very good. Thank you

[@cameronspalding9792]

@ 5:54 you forgot the i in front of the b in the denominator

[@Geenimetsuri]

What happens to the limits if we "restrict" h to a specific number type? There is no good a priori reason to assume h would have the same the type as the parameter. So we could have lim h->0 for Z where h would be in C, R or even C+, Q, Z, N ,etc.. where Z could be of type anything as well. Most would default to being illegal, but not all 🤔

[@Alan-zf2tt]

Thank you for sharing 🌟

[@Don-ev5ov]

Thank you. I think Tristan Needham said the partial derivatives condition corresponds to a differential "square."

[@cameronspalding9792]

@ 5:02 the complex conjugate on real numbers is equivalent to the identity function

[@max.caimits]

In the last equality: which independent variables does the partial derivative wrt z-bar assume? Which variables are considered constant? I don't understand this notation.

[@ominollo]

Nice refresh 😊

[@mannixshowell7522]

youve in the past done a few videos constructing algebras with weird properties e.g. where 1/0 is defined as an element. I was wondering if you would do something similar for terrence howards 1*1=2 nonsense? Just for shits and giggles and given its newfound social viral-ness, I was wondering if you could make some "terrence group" with numbers that are consequential from 1*1=2, and maybe a consequential redefinition of addition/multiplication if necessary

[@rogerlie4176]

An extra exercise as homework: What is the function in Michael's last example expressed in z?

[@MichaelRothwell1]

f(z) =(x³-2x²-3xy²+2y²)+i(-y³+3x²y-4xy) =(x³-3xy²-iy³+3ix²y)+(-2x²+2y²-4ixy) (grouping terms of same order together) =(x³+3ix²y-3xy²-iy³)+(-2x²-4ixy+2y²) (ordering each group in descending powers of x) =(x³+3ix²y-3xy²-iy³)-2(x²+2ixy-2y²) =(x+iy)³-2(x+iy)² =z³-2z²

[@moonwatcher2001]

Awesome Mike ❤

[@kirbo722]

If both the limits for the derivative of f(z) are the same, doesn't that mean there still might be some other "path" towards the origin that can be not equal to the result of the other 2 limits? Don't multivariable limits work like that? Pls help🙏

[@miloszforman6270]

You have to take care what you are defining. Of course other paths might give different limits. But that implies that this function is not differentiable within a complete neighborhood of z. It's sufficient for complex differentiability if the partial derivatives exist in such a neighborhood and fulfill the CR-equations. You don't even have to presuppose that these derivatives have to be continuous, as this follows from their mere existence. It gets tricky here, and I think textbooks usually rely on the continuity of the partial derivatives.

[@hu5116]

Great video! You’re a great instructor! I got all of it ;-)

[@moskitomoskitani]

Hello sir Is there a function that gives the same answer when we derive from real and imaginary paths and different answers in other paths? How to calculate other paths? I have watched your complex analysis course and learned a lot from your teaching Thank you very much incerely, one of your students from Iran

[@gegebenein.gaussprozess7539]

Well. f(z) is differentiable in (0,0) iff ALL paths going to this point yield the same limit (the differential quotient). But here we have chosen two SPECIAL paths, one along the real axis and one along the imaginary axis. We didn't consider all paths. Can somebody please help me with this?

[@chrishetzler6724]

He is showing the function is NOT differentiable in C. It is sufficient to show that two paths have different limits in order to demonstrate that not all paths have the same limit. Those two paths (along Re and Im) are the easiest ones to check.

[@gegebenein.gaussprozess7539]

@@chrishetzler6724 Thank you for your answer❤! But I think you misunderstood me. I understand all the examples and the arguments when a complex function is NOT differentiable, by choosing two paths (here the the real and the imaginary axes) that deliver different limits of the differential quotient ((f(z+h) - f(z))/h, h going to zero. This is clear to me. What I don't understand: He must show that if you use ALL possible paths leading to z then and only then the function is differentiable see 13:49. He doesn't show this. Instead, he shows that when you choose two special paths ( again real and imaginary axis) and the two limits coincide, then the C-R-equations follow. But he chooses two special paths! He should take all different paths to z for showing differentiability, shouldn't he? Thank you for your patience. (Sorry for my formulations, but English is not my mother tongue). PS. Meanwhile, two other comments raised the same question, @themibo899 and @normanstevens4924.

[@Debg91]

The Cauchy-Riemann equations are necessary and sufficient conditions for differentiability, but in this video he only showed the right implication, as the converse is more tricky to prove. But you can show that taking those two paths is equivalent to taking any two paths.

[@gegebenein.gaussprozess7539]

@@Debg91 Thank you very much!! You answered my question. Now, would you be able and willing to give me a source (textbook, video) to see a proof of the "tricky part". Or to a proof of your last sentence. I thank you wholeheartedly.

[@chrishetzler6724]

⁠​​⁠ Ok, I see what you’re getting at now based on the time stamp. Yes, good question, I wasn’t even thinking about the fact that he was only using two paths at that point in the video.

[@ontoverse]

So, from someone who knows plenty of math, but almost no analysis and in particular very little complex analysis-- is there some reason to not simply treat complex numbers as a 2-vector and use multivariate calculus (ie differentials instead)? What do we gain from a "complex derivative" compared to a gradient on the complex plane?

[@APaleDot]

The complex derivative is more restrictive than the multivariate derivative. The multivariate derivative exists as long as all the partial derivatives exist, and in general it's expressed as a matrix (gradient is 1xn matrix). The complex derivative on the other hand can always be expressed as a complex number. Since complex numbers can only express uniform scaling and rotation, the complex derivative must be a rate of stretching/rotation near a point. It follows that if a function has a complex derivative, then that function preserves angles locally (because uniform scaling and rotation preserve angles). We call such functions "holomorphic". So, a function can only have a complex derivative if the function looks like a transformation that preserves angles near every point. If you imagine a little real and imaginary differential extending out from a point, those differentials must remain at right angles to each other after the transformation. They are coupled together. This coupling is expressed in the Cauchy-Riemann equations: U_x = V_y, U_y = - V_x, which you may notice look suspiciously like a rotation by 90 degrees (swap the components and negate one). In multivariate calculus, no such coupling is required. Each axis is independent of each other and the derivative is therefore an arbitrary linear transformation. The arbitrariness means that many more functions are differentiable, but there's less you can conclude from such differentiability.

[@jongraham7362]

I'm going to ask an off the wall question. Is there any relationship between the curl = 0 in line integrals, and the Cauchy Riemann equations?

[@miloszforman6270]

Yes, of course. uᵧ = -vₓ means that uᵧ + vₓ = curl (u(z), v(z)) = 0, and this also implies that any complex diffentiable function has an antiderivative, which is at least defined in some neighborhood of z.

[@jongraham7362]

@@miloszforman6270 Thank you! I've always wondered. I'm not at the level to fully comprehend that yet. Does this mean that if the curl is not zero that a complex function is not differentiable?

[@jongraham7362]

@@miloszforman6270 Actually, what I am more curious about is- is there any relationship between path independence of a line integral in vector calculus and whether a complex function is differentiable or not?

[@HuangfanHelloWorld]

I think it will be better to mention df / dz = \partial f / \partial z when f is complex differentiable at a point.

[@silassheriff7868]

Nice Video! Next: complex Integration

[@Pootycat8359]

See my comment about "Fractional Calculus." If the order of the derivative is negative, it is the integral. Of course, I'm sure the math nerds will have a much more complex (pun intended) way of doing it.

[@Khashayarissi-ob4yj]

With luck and more power to you. hoping for more videos.

[@dimastus]

I think You should state the homework more precise. Because I didn't realise at first what do you even mean. But in the end, if I understood correctly you want us: 1) to look at function g(x,y)=u(x,y) + i*v(x,y), then express it as one complex function f(z, z') //I will designate comjugation with apostrophe I you don't mind// 2) So f(z, z') = g( (z+z')/2 , (z-z')/(2*i) ) 3) And now take partial derivative with respect to z' getting precisely Cauchy-Riemann equations from real and imaginary part of resultant expression.

[@chalkchalkson5639]

21:51 wait, what about a function like 0 if Re(z) = Abs(z) or Im(z) = Abs(z) and 1 otherwise? Ie the function is constant along the two axes and constant everywhere else, but not the same constant. Then the jacobian of the component wise notation is all 0s at z=0, so in particular the main diagonal is the same and it is anti symmetric. So the cauchy riemann condition is fulfilled, but the limit doesn't converge for any path, in fact it diverges for all other paths. Is there a missing assumption here? Or am I missing something?

[@miloszforman6270]

I would say that the partial derivative uᵧ is not defined on the real axis, while uₓ is not defined on the imaginary axis. So uᵧ and uₓ do not exist for all z within any neighborhood of 0, which means that you can't apply the CR-equations at z=0.

[@Cor97]

Is "taking the real part" not a special kind of function? I mean, like "taking only positive numbers" in a real valued function, or like "numbers that are closer to an odd integer than to an even integers.

[@100_IQ_EQ]

It is. So?

[@Cor97]

@@100_IQ_EQ . That means that both with real numbers and with complex numbers there are plenty of non-differentiable functions. So in this respect complex differentiation is not so much different from real diffentiation as initially claimed.

[@asmithgames5926]

It seems like it would make more sense if there were real and imaginary derivatives, which could be different, as long as all path directions of h -> 0 (from all possible complex angles) ended up with a set of values which were "planar" - being able to be defined by multiplying the directional vector by a 2x2 complex matrix that would represent the plane. In such a system, the complex conjugate or Re(x) would be differentiable, I'm pretty sure. TL;DR: the complex derivative should give you a 2x2 complex plane representing the planar relationship of the real and imaginary parts, then you could calculate this for almost any function.

[@tomkerruish2982]

Look up Wirtinger derivatives. They might be what you're looking for.

[@asmithgames5926]

@@tomkerruish2982 Just did. Thank you

[@bendunselman]

If lim Re(h) / h is multivalued, 0 and 1, how does it follow it does not exist. How does multivaluedness imply nonexistence? I would rather think that nonvaluedness would imply nonexistence.

[@bjornfeuerbacher5514]

A limit of a function at a point does only exist if it has the _same_ value, however you approach the point. That's how a limit is _defined_.

[@mekbebtamrat817]

Nice derivation of Cauchy Riemann

[@paperstars9078]

21:40 what if only the partial derivates exist? how would an example look like in this case?

[@miloszforman6270]

f(z) = conjugate(z) would be such an example. u(z) = Re(z) v(z) = -im(z) uₓ = 1 uᵧ = 0 vₓ = 0 vᵧ = -1 so uₓ ≠ vᵧ which means that the CR-equations are not fulfilled.

[@davidblauyoutube]

f(z) = z^3 - 2z^2

[@MichaelRothwell1]

f(z) =(x³-2x²-3xy²+2y²)+i(-y³+3x²y-4xy) =(x³-3xy²-iy³+3ix²y)+(-2x²+2y²-4ixy) (grouping terms of same order together) =(x³+3ix²y-3xy²-iy³)+(-2x²-4ixy+2y²) (ordering each group in descending powers of x) =(x³+3ix²y-3xy²-iy³)-2(x²+2ixy-2y²) =(x+iy)³-2(x+iy)² =z³-2z²

[@ClarkPotter]

If this were any other topic, he'd be on 1.25-1.5x speed.

[@robfielding8566]

why is h not a vector length?

[@robfielding8566]

using clifford algebra for the vectors: e1 e1 = e2 e2 = 1. e1 e2 = -e2 e1. e1 and e2 are orthogonal vectors in space, their multiplication anti-commutes. (e1 e2) = i (e1 e2)(e1 e2) = -(e1 (e2 e2) e1) = -1. ie: i^2=-1. f:vector2 a:vector2 # finite vector in any direction da:vector2 # infinitesimal vector in any direction a = a1 e1 + a2 e2 da = da1 e1 + da2 e2 (da1)^2 = 0 # infinitesimal scalar length (da2)^2 = 0. # infinitesimal scalar length # vector with infinitesimal coefficiens squares to 0 (da)^2 = (da1 e1 + da2 e2)^2 = (da1^2 + da2^2) + (da1 da2 (e1 e2) + da2 da1 (e2 e1)) = (da)^2 = (da1 e1 + da2 e2)^2 = (da1^2 + da2^2) + (da1 da2 (e1 e2) - da1 da2 (e1 e2)) = 0 # so, if f is a^2 f = a^2 # df is the differential... df = (a + da)^2 - a^2 = a^2 + a da + da a + (da)^2 - a^2 = a da + da a // careful because a and da do not commute, because they are multiplied vectors = (a1 e1 + a2 e2)(da1 e1 + da2 e2) + (da1 e1 + da2 e2)(a1 e1 + a2 e2) = (a1 da1 + a2 da2) + (a1 da2 (e1 e2) - a2 da1 (e1 e2)) + (da1 a1 + da2 a2) + (a2 da1 (e1 e2) - a1 da2 (e1 e2)) = a1 da1 + a2 da2 + a1 da2 (e1 e2) + -a2 da1 (e1 e2) + da1 a1 + da2 a2 + a2 da1 (e1 e2) + - a1 da2 (e1 e2) = a1 da1 + a2 da2 + a1 da1 + a2 da2 = 2 a1 da1 + 2 a2 da2 # da1 > 0. (da1)^2 = 0. take da2=0 for this partial df/da1 = 2 a1 + 2 a2 da2/da1 = 2 a1 # da2 > 0. (da2)^2 = 0. take da1=0 for this partial df/da2 = 2 a1 da1/da2 + 2 a2 = 2 a2

[@dang-x3n0t1ct]

I thought he was gonna talk about a complex order derivative

[@paperstars9078]

just what I needed for my exam in 2 days! anyone want to reommend their favourite complex analyiss vids on this channel? I won't have time to go through every single one before my test...

[@cycklist]

Love the thumbnail.

[@6cef]

1:38 so cool.

[@tuongnguyen9391]

Is this wertinger calculus 😢😢😢

[@PleegWat]

Asking if a function in R is differentiable in C is the wrong question. The function is defined on R. lim_b→0 (f(z+bi)-f(z))/bi has no value because z+bi is not in the domain of f. To properly answer the question we first need to specify how we are extending f onto the complex plane. For example, if we set F(z)=f(z) if z is a real number and F(z)=f(z₀)+f'(z₀)(z-z₀), then F(z) is an extension of f onto C which is differentiable in z₀, regardless of which f we started with, as long as f is differentiable in z₀.

[@roberttelarket4934]

I love things that are complex and strange!!! Simplicity and normalcy are boring!!!

[@yuseifudo6075]

This doesn't make any sense in the context of the video. Complex is a term we used and doesn't mean it's really "complex" and "strange"

[@jonahansen]

@@yuseifudo6075 Thus proving you are a total math nerd (it's a compliment).

[@yuseifudo6075]

@@jonahansen Thanks lol!

[@roberttelarket4934]

@@yuseifudo6075: It makes plenty of sense!

[@moskitomoskitani]

@@yuseifudo6075 Is there a function that gives the same answer when we derive from real and imaginary paths and different answers in other paths? How to calculate other paths?

[@normanstevens4924]

CauchyRiemann equations do no prove differentiability.

[@moskitomoskitani]

Is there a function that gives the same answer when we derive from real and imaginary paths and different answers in other paths? How to calculate other paths?

[@bjornfeuerbacher5514]

@normanstevens: Huh? As far as I know, they _do_ prove it, and e. g. the ProofWiki agrees with that. Why do you think otherwise?

[@moskitomoskitani]

@@bjornfeuerbacher5514 I'm sorry My English is not good and I am not a math student but I am very interested I have seen in a video that a teacher said that Cauchy - Riemann conditions are necessary but not sufficient In this video, it is said that in order to be derivable, all paths must give the same answer, but only two paths are explained Is there a function that two real and imaginary paths give the same answer and one of the other paths give a different answer?

[@akashpremrajan9285]

@@bjornfeuerbacher5514 If you mean the Cauchy-Riemann_Equations article in ProofWiki, then that does NOT say what you think it says. You are forgetting one of the premises there: u and v must be differentiable when viewed as functions of R2 to R. Even if partial derivatives exist everywhere and satisfy cauchy-riemann everywhere, even there the functions u and v can fail to be differentiable (in the multivariable sense of f: R2->R) at a particular point, and it would NOT be complex differentiable at that point. You can also look at looman-menchoff theorem, you need at least that f(z) be continuous in an open set. Then cauchy-riemann in the entire open set, and complex differentiable in that same domain, are equivalent.

[@bjornfeuerbacher5514]

@@akashpremrajan9285 Thanks for pointing that out, I indeed missed that part.