Thanks a million for starting with such a basic example and running it all the way thru, especially with the subtle suggestions your intuition may guide you. This is excellent teaching.
It is typically a reflection of your teacher and your teachers skills at providing context about what you are doing so you do not stick in memorized facts inappropriately or fail to memorize key facts. Teachers of advanced math are the worst offenders, going slow as a snail on trivial stuff then assuming non trivial stuff is obvious and flying through it teaching nothing. What is the context of this comment? Who got a C or D?
I had one math class in third year undergrad university that I got 20% or so on the unit test and my mark was average. Nobody got above 45% on it. All smart students but the professor was useless. It was horrible, total waste of time and resources. I think it was an advanced calculus class involving partials, it was not complex analysis. I could see why missing the fact that complex roots are periodic would cost a lot of marks but not really mean that the student has no idea what they are doing.
For calculating the integral of /z following gamma1, you can notice that /z = z exp(-iπ/4). You just need multiply your first result. For the second integral, z = /z following gamma2 and /z = -z following gamma3 😉
In the theory of functions of a complex variable, analytical functions and non-analytical functions are distinguished. The integral of an analytic function between two points does not depend on the path (trajectory) of integration, for a non-analytic function it does. Conditions of analyticity of the function f(z)=u(x,y)+i v(x,y) in the sense of Cauchy-Riemann equations: ∂u/∂x=∂v/∂y and ∂u/∂y= - ∂v/∂x. Function f(z)=z =x+iy is analytic. The function f(z)=z* =x-iy is not analytic. No miracles.
But can “analyticity” explain why the real part of the integral of z* is always the same, even though the imaginary part varies? Seems like the whole theory is a waste of time if not.
@@edwardlulofs444 well the explanation given in this current video handles it just fine; the real part of the integral is x dx+y dy. And this can be antidifferentiated to be (x^2+y^2)/2, so whatever path you take the real part will always be (x^2+y^2)/2 at the endpoint minus (x^2+y^2)/2 at the start point.
@@noahtaul f(z)=z is analytic (it is easy to check the fulfillment of the Cauchy-Riemann equations). For it, there is the antiderivative F(z)=ʃ zdz=z^2/2 +c. ʃ (from point z1=0 to point z2=1+i) zdz= F(z2)-F(z1)= (1+i)^2/2 -0 =i, regardless of the integration path. For the function f(z)=z*= x-iy the Cauchy-Riemann condition is not satisfied : ∂u/∂x =1 ≠ ∂v/∂y =-1. This function is not analytical. The result of integration depends on the trajectory. Perhaps your question concerns the calculation of ʃ (by γ1 +γ2) z*dz = ʃ (by γ1)z*dz+ ʃ (by γ2)z*dz ? After parametrization of the segment γ1 : z*= t-i*0 =t, z=t+i*0=t, 0≤t ≤1 (that is, the imaginary part of z* is zero, and the real part varies from 0 to 1), dz=dt => ʃ (by γ1)z*dz= =ʃ(from 0 to 1) t*dt =1/2. Here, on the interval y2, the real part is 1, and the imaginary part varies from 0 to 1: z*=1 -i*t, z=1+i*t, dz=i*dt, 0≤t ≤1. ʃ (by y2)z*dz = ʃ(from 0 to 1)(1-i*t)*idt =i+1/2.
just using the cauchy-riemann differential equations allows you to easily calculate whether a given function is path-independent or not which is always nice to remember
Having watched this after seeing that the complex conjugate of a baseline complex number does not satisfy the Cauchy-Riemann differential equations summarizes why I expected the complex conjugate to be path dependent, yielding a different answer over different paths. If it is not holomorphic in differential form, we shouldn't expect it to be holomorphic in integral form due to the fundamental theorem of calculus. This lack of complex analytic behavior necessarily implies that the complex conjugate term is not continuously differentiable. Therefore, there is a specific path dedicated to its integration instead of multiple paths. The final part of this video is particularly nice, because it's essentially a proof of the fundamental theorem of complex variable calculus.
Question for michael penn. Q: denote (n/p)=+,- 1 as legendre symbol for quadratic residues or non residue. Consider the sum (a(a+1)/p) as 'a' runs from 1 to p-2. Show that this sum equals to '-1'.
We can proof easily that gamma 1 = gamma2 + gamma3 First, notice that each complex number from the path gamma1 has the real part equal to the imaginary part. Then, we see that gamma2 is indeed describing the real parts of the complex numbers of gamma1, and the gamma3 is describing the the imaginary parts of the complex numbers of gamma1, so if we take gamma2(t) + gamma3(t) it's indeed equal to gamma1(t).
Hi! Love your content, quick question: for gamma2 and gamma3, why can you have both run from 0 to 1? Wouldn't that represent two separate paths where gamma2 goes from 0 to 1 and the gamma3 from 1 to 1+i at the same time and same rate? For a single path wouldn't it be gamma2 = 2t and runs from t = 0 to 1/2, and gamma3 = 1+i(2t-1) and runs from t = 1/2 to 1?
The idea is that no matter how fast you go, if you integrate while traveling the same path, you will get the same answer. This is just a consequence of the chain rule: if we replace t with anything else, say f(u) where u ranges over some interval, then we’re integrating g(f(u))*f’(u) du since dt=f’(u) du, but this is just the original integral using u-sub.
@@noahtaul for sure the integral is rate independent, and so he can just set the interval to [0,1] which is easier to work with. But isnt having both gamm2 two and gamma3 on the same interval going to cause a well-defined issue? Like for all values of t the gamma2/gamma3 path is at two positions: the one from gamma2 and the one from gamma3. So shouldnt the intervals be something like [a,b] and [b,c], which in my og comment i chose to be [0,1/2] and [1/2,1] for ease due to the rate independence you mention?
@@minamagdy4126 not annoyed, just curious. My understanding was that gamma2 and gamma3 were already a piecwise function of the path, that's why i questioned them having the same domain, it results in a multi-valued function. And i get that the interval itself doesnt really matter, because it's just a parameter, but it's not obvious to me that the integral calculations are valid when the path function isnt necessarily continuous/differentiable
Nice video as always I guess there are more families of functions that have path independence? And what about those that are path dependent? Are there special paths I can take that all yield the same value anyway?
A continuous complex valued function defined on all of C has path independent integrals if and only if it has a complex derivative. The "if" part of that statement is a consequence of Cauchy's integral theorem (complex derivative of F exists-> F has path independence integrals) and the "only if" part is a consequence of Morera's theorem (path independence -> the function has a complex derivative).
People forgets that complex functions are surfaces. :P People forgets too that complex functions requires paths since the Green's theorem happens everywhere over these surfaces. Paths are inevitable and watching out for those poles. :P
sqrt(x) in general is wonky in that it's supposed to be an inverse function, but the function that it's an inverse of has multiple inputs that give the same output. In C it "only" has 2 for any given output, but other systems give infinite solutions to y^2 = x for any given x.
Thank you,professor.From this video i recollected the course of complex analisys that i studied in university.Only one question:what complex integral means.I mean ,definite integral of real fuction means square.
I am not well-versed in complex integrals, so is it reasonable to think of the conjugate function in C being like the absolute value in R. In the sense that a real integral of abs(x) with an interval including 0 doesn't play nice would be analogous to an integral of the conjugate not playing nice over the complex plane because the path might cross the complex axis?
I’m not sure what you mean by the “complex axis” but it’s actually the *real* axis where the conjugate function does something special, since the conjugate of any real number is itself. But, at least to my intuition, taking the conjugate of a complex number seems analogous not to taking the absolute value, but instead to taking the opposite of a real number, since each of f(x)=-x and g(z)=z\ is its own inverse and it corresponds to a reflection geometrically. And I think this is precisely the issue: intuitively, we shouldn’t expect a reflection to behave in a path-independent way; to make this rigorous, one needs to look at the Cauchy-Riemann equations.
In the complex realm, it's usually best to start with a purely algebraic perspective on how functions and operators work. There's not always an easy geometric understanding, and even when there is, it's not always a perfect analogy.
Thanks a million for starting with such a basic example and running it all the way thru, especially with the subtle suggestions your intuition may guide you. This is excellent teaching.
I ABSOLUTELY LOVE his reaction at 3:33; very Human !
it gives a man hope and dreams
very human design 😀👍
i want a compilation of all the 3:32 moments 😂
i think his students are supposed to cut those bloopers out, but often miss a few. i've seen him do it a lot and it's p funny lmao
14:09 Getting a C or D on a test does not describe you as a person. A test score is simply a happening, not a person. Have a good day ☀️
It is typically a reflection of your teacher and your teachers skills at providing context about what you are doing so you do not stick in memorized facts inappropriately or fail to memorize key facts.
Teachers of advanced math are the worst offenders, going slow as a snail on trivial stuff then assuming non trivial stuff is obvious and flying through it teaching nothing.
What is the context of this comment? Who got a C or D?
In complex analysis, everybody gets a C or D on their tests. Usually it's e^(2 pi i t) or |z|
I had one math class in third year undergrad university that I got 20% or so on the unit test and my mark was average. Nobody got above 45% on it. All smart students but the professor was useless. It was horrible, total waste of time and resources. I think it was an advanced calculus class involving partials, it was not complex analysis.
I could see why missing the fact that complex roots are periodic would cost a lot of marks but not really mean that the student has no idea what they are doing.
For calculating the integral of /z following gamma1, you can notice that /z = z exp(-iπ/4). You just need multiply your first result. For the second integral, z = /z following gamma2 and /z = -z following gamma3 😉
So you made integrals more *complex* …
Complex calculus is actually simple.
not funny
jk
In the theory of functions of a complex variable, analytical functions and non-analytical functions are distinguished.
The integral of an analytic function between two points does not depend on the path (trajectory) of integration, for a non-analytic function it does.
Conditions of analyticity of the function f(z)=u(x,y)+i v(x,y) in the sense of Cauchy-Riemann equations:
∂u/∂x=∂v/∂y and ∂u/∂y= - ∂v/∂x.
Function f(z)=z =x+iy is analytic.
The function f(z)=z* =x-iy is not analytic.
No miracles.
This is shown in his other channel on complex analysis.
But can “analyticity” explain why the real part of the integral of z* is always the same, even though the imaginary part varies? Seems like the whole theory is a waste of time if not.
@@noahtaul I've never seen such an explanation. The usual procedure in physics is to ONLY consider analytic functions.
@@edwardlulofs444 well the explanation given in this current video handles it just fine; the real part of the integral is x dx+y dy. And this can be antidifferentiated to be (x^2+y^2)/2, so whatever path you take the real part will always be (x^2+y^2)/2 at the endpoint minus (x^2+y^2)/2 at the start point.
@@noahtaul f(z)=z is analytic (it is easy to check the fulfillment of the Cauchy-Riemann equations).
For it, there is the antiderivative F(z)=ʃ zdz=z^2/2 +c.
ʃ (from point z1=0 to point z2=1+i) zdz= F(z2)-F(z1)= (1+i)^2/2 -0 =i, regardless of the integration path.
For the function f(z)=z*= x-iy the Cauchy-Riemann condition is not satisfied : ∂u/∂x =1 ≠ ∂v/∂y =-1.
This function is not analytical. The result of integration depends on the trajectory.
Perhaps your question concerns the calculation of ʃ (by γ1 +γ2) z*dz = ʃ (by γ1)z*dz+ ʃ (by γ2)z*dz ?
After parametrization of the segment γ1 : z*= t-i*0 =t, z=t+i*0=t, 0≤t ≤1
(that is, the imaginary part of z* is zero, and the real part varies from 0 to 1), dz=dt =>
ʃ (by γ1)z*dz= =ʃ(from 0 to 1) t*dt =1/2.
Here, on the interval y2, the real part is 1, and the imaginary part varies from 0 to 1:
z*=1 -i*t, z=1+i*t, dz=i*dt, 0≤t ≤1.
ʃ (by y2)z*dz = ʃ(from 0 to 1)(1-i*t)*idt =i+1/2.
Reminds me of state functions Pchem where chemical properties are path-independent and depend only on the initial and final points of the system
This is a really nice little introduction!
Just want to say your videos are fantastic, particularly regarding format and clarity of exposition.
Actually whenever integrating over a domain in which the function is holomorphic one can say that the result is path-independent, right?
yeah
not within a nonlinear field potential
Complex Analysis has always been fun to me because it IS complex.
Thank you, professor.
just using the cauchy-riemann differential equations allows you to easily calculate whether a given function is path-independent or not which is always nice to remember
The CR equations can't tell you that integrating 1/z is path dependent while integrating 1/z^2 is path independent.
@Scuffed Physics Yup.
Having watched this after seeing that the complex conjugate of a baseline complex number does not satisfy the Cauchy-Riemann differential equations summarizes why I expected the complex conjugate to be path dependent, yielding a different answer over different paths. If it is not holomorphic in differential form, we shouldn't expect it to be holomorphic in integral form due to the fundamental theorem of calculus. This lack of complex analytic behavior necessarily implies that the complex conjugate term is not continuously differentiable. Therefore, there is a specific path dedicated to its integration instead of multiple paths.
The final part of this video is particularly nice, because it's essentially a proof of the fundamental theorem of complex variable calculus.
This helps a lot thanks Professor penn
Can you make a series about discrete mathematics, and thank you.
Question for michael penn.
Q: denote (n/p)=+,- 1 as legendre symbol for quadratic residues or non residue. Consider the sum
(a(a+1)/p) as 'a' runs from 1 to p-2. Show that this sum equals to '-1'.
do you assume p to be prime?
incredible video!
This is the famous "proof by example"
We can proof easily that gamma 1 = gamma2 + gamma3
First, notice that each complex number from the path gamma1 has the real part equal to the imaginary part. Then, we see that gamma2 is indeed describing the real parts of the complex numbers of gamma1, and the gamma3 is describing the the imaginary parts of the complex numbers of gamma1, so if we take gamma2(t) + gamma3(t) it's indeed equal to gamma1(t).
At 2:46 why use 0 to 1 , kindly give me some englintment.
6:17 We got the same answer i-ther way
In the "Suggest a problem" section, there is no way to upload a picture, just plain text. So how would you recieve a geometry problem?
Put the pic on the web or find it on the web and type a link to it.
Turn it into SVG and copy the code?
ASCII pictures, obviously :^)
complex integral be like
"im built different"
Hi! Love your content, quick question: for gamma2 and gamma3, why can you have both run from 0 to 1? Wouldn't that represent two separate paths where gamma2 goes from 0 to 1 and the gamma3 from 1 to 1+i at the same time and same rate? For a single path wouldn't it be gamma2 = 2t and runs from t = 0 to 1/2, and gamma3 = 1+i(2t-1) and runs from t = 1/2 to 1?
The idea is that there are two timers in series. We go through gamma2, then reset the timer to then go through gamma3.
The idea is that no matter how fast you go, if you integrate while traveling the same path, you will get the same answer. This is just a consequence of the chain rule: if we replace t with anything else, say f(u) where u ranges over some interval, then we’re integrating g(f(u))*f’(u) du since dt=f’(u) du, but this is just the original integral using u-sub.
@@noahtaul for sure the integral is rate independent, and so he can just set the interval to [0,1] which is easier to work with. But isnt having both gamm2 two and gamma3 on the same interval going to cause a well-defined issue? Like for all values of t the gamma2/gamma3 path is at two positions: the one from gamma2 and the one from gamma3. So shouldnt the intervals be something like [a,b] and [b,c], which in my og comment i chose to be [0,1/2] and [1/2,1] for ease due to the rate independence you mention?
@@Brettorini If that is annoying you so much, we can just combine gamma2 and gamma3 into one piecewise function that goes from 0 to 2, with the 0
@@minamagdy4126 not annoyed, just curious. My understanding was that gamma2 and gamma3 were already a piecwise function of the path, that's why i questioned them having the same domain, it results in a multi-valued function. And i get that the interval itself doesnt really matter, because it's just a parameter, but it's not obvious to me that the integral calculations are valid when the path function isnt necessarily continuous/differentiable
Nice video as always
I guess there are more families of functions that have path independence?
And what about those that are path dependent? Are there special paths I can take that all yield the same value anyway?
A continuous complex valued function defined on all of C has path independent integrals if and only if it has a complex derivative. The "if" part of that statement is a consequence of Cauchy's integral theorem (complex derivative of F exists-> F has path independence integrals) and the "only if" part is a consequence of Morera's theorem (path independence -> the function has a complex derivative).
People forgets that complex functions are surfaces. :P
People forgets too that complex functions requires paths since the Green's theorem happens everywhere over these surfaces. Paths are inevitable and watching out for those poles. :P
Yes, but the integral of z(bar)dz(bar) is path independent
Excellent!
3:28 Missed the cut
You just define at the integral over the banana from a to b, and call it Christmas.
Here we are told that in calculus course you need to forget what you learnt in complex numbers and say √(-1) is undefined 🤔
sqrt(x) in general is wonky in that it's supposed to be an inverse function, but the function that it's an inverse of has multiple inputs that give the same output. In C it "only" has 2 for any given output, but other systems give infinite solutions to y^2 = x for any given x.
Thank you,professor.From this video i recollected the course of complex analisys that i studied in university.Only one question:what complex integral means.I mean ,definite integral of real fuction means square.
I am not well-versed in complex integrals, so is it reasonable to think of the conjugate function in C being like the absolute value in R. In the sense that a real integral of abs(x) with an interval including 0 doesn't play nice would be analogous to an integral of the conjugate not playing nice over the complex plane because the path might cross the complex axis?
I’m not sure what you mean by the “complex axis” but it’s actually the *real* axis where the conjugate function does something special, since the conjugate of any real number is itself. But, at least to my intuition, taking the conjugate of a complex number seems analogous not to taking the absolute value, but instead to taking the opposite of a real number, since each of f(x)=-x and g(z)=z\ is its own inverse and it corresponds to a reflection geometrically. And I think this is precisely the issue: intuitively, we shouldn’t expect a reflection to behave in a path-independent way; to make this rigorous, one needs to look at the Cauchy-Riemann equations.
I don't think it has anything to do with crossing an axis. Seems to me it's still not path-independent even if you stay in one quadrant.
question.... the last part of the chain rule for gamma 3 should be -idt?
I was tripped up by this as well, but don’t confuse the domain (path from 1 to 1+I) with the function on the range.
Are analytical functions path independent?
If you watch the video on his MathMajor channel about complex integration (Complex Analysis #12), you might find the answer you’re looking for…
@@ConManAU in video #3, right now!
In my day, we called this PATH INDEPENDENCY.
I would have expected complex integrals to involve computing the hypervolume under the 2D are defined by the end points.
In the complex realm, it's usually best to start with a purely algebraic perspective on how functions and operators work. There's not always an easy geometric understanding, and even when there is, it's not always a perfect analogy.
Michael's videos are beta than just about anything else out there...
No, I dont think I will.