Dr Penn, at 28:42, there is a much easier way to find f in terms of z. In your formula for f, just replace x with z and y with 0. Then you get f = z - iz^2/2. This is just what you got but without any real thought. Current UA PhD student.
It is very refreshing and nostalgic to see again how easy all this stuff feel know, when at the same time studying for a graduate course on Several complex variables analysis…
I am reading the book Visual Complex Analysis by Tristan Needham and your videos are very helpful to understand some concepts that couldn't clarify from reading. Thank you, you are a great teacher 🙏🏾
21:58 One important semantic point: "del" isn't the Laplacian. Del is the nabla ∇; the Laplacian is "del squared", which is why it gets a different symbol 𝝙.
My American electro-magnetic texts use a triangle (with pointy end down) squared symbol for the Laplacian. However, the Russian literature on atmospheric turbulence uses the triangle (pointy end up) for the Laplacian, just as comrade Penn uses. 🙂
by previous theorem, (f^(-1))’(f(z)) = 1/f’(z) implies f’(z)*(f^(-1))’(f(z)) = 1, the first line at note says f and e^z have inverse relation and the second line can be derived by theorem by knowing (e^z)’=e^z.
My answers for these warm-up questions: Q1 d/dz(tan^(-1)z) = 1/(1+z^2) Q2 (1) f=z^2+C (2) f=-i*z^3+C Thank you professor for your wonderful explanation.
@@AssemblyWizard no. the example is a polynomial so things work out very simply and it is not representative of the general case. you can read e.g. on wikipedia that the harmonic conjugate of a harmonic function is only guaranteed to exist locally (in a neighborhood of each point), but in general it may not be possible to patch these local harmonic conjugates together into a global harmonic conjugate.
why did he say jacobian of f inverse is inverse of jacobian of f and then say those two are equal when the first one is evaluated at f(z) but the second one at z also a_y=-u_y/|f'(z)|²=-(-v_x)/|f'(z)|²=-b_x and I don't understand why he evaluates derivative of f inverse at f(z) and says its equal to a_x+ib_x, without saying where a_x and b_x are evaluated
Dr Penn, at 28:42, there is a much easier way to find f in terms of z. In your formula for f, just replace x with z and y with 0. Then you get f = z - iz^2/2. This is just what you got but without any real thought. Current UA PhD student.
Refreshing complex analysis with this video series is a true joy.
It is very refreshing and nostalgic to see again how easy all this stuff feel know, when at the same time studying for a graduate course on Several complex variables analysis…
I am reading the book Visual Complex Analysis by Tristan Needham and your videos are very helpful to understand some concepts that couldn't clarify from reading. Thank you, you are a great teacher 🙏🏾
21:58 One important semantic point: "del" isn't the Laplacian. Del is the nabla ∇; the Laplacian is "del squared", which is why it gets a different symbol 𝝙.
My American electro-magnetic texts use a triangle (with pointy end down) squared symbol for the Laplacian. However, the Russian literature on atmospheric turbulence uses the triangle (pointy end up) for the Laplacian, just as comrade Penn uses. 🙂
Thank you so much professor for this amazing and insightful video! Your approach is very intuitive and thus easier to understand and apply.
Glad to see you're back.
Can you PLS make a video about the Lambert W function with its all branches in the complex plane*?
W(z)=inverse of z*exp(z)
27:42 should be g(x) not g'(x)
30:25 “slams home the fact”
learning for midsem exam from nit ap.
love from india
NATIONAL INSTITUTE OF TECHNOLOGY ANDHRAPRADESH ( NIT AP) INDIA.
At 14:00 you use the Chain Rule. Have we established the Chain Rule yet? I don't seem to recall.
by previous theorem, (f^(-1))’(f(z)) = 1/f’(z) implies f’(z)*(f^(-1))’(f(z)) = 1,
the first line at note says f and e^z have inverse relation
and the second line can be derived by theorem by knowing (e^z)’=e^z.
Thank God, I missed these uploads last week :0
We missed u bro
My answers for these warm-up questions:
Q1
d/dz(tan^(-1)z) = 1/(1+z^2)
Q2
(1)
f=z^2+C
(2)
f=-i*z^3+C
Thank you professor for your wonderful explanation.
I didn't understand what justifies the existence of a harmonic conjugate ...
good catch! a harmonic function may fail to have a harmonic conjugate, but if the harmonic conjugate exists then it is unique.
@@schweinmachtbree1013 I think the example showed a way to construct the conjugate of any harmonic, no? Therefore a conjugate always exists
@@AssemblyWizard no. the example is a polynomial so things work out very simply and it is not representative of the general case. you can read e.g. on wikipedia that the harmonic conjugate of a harmonic function is only guaranteed to exist locally (in a neighborhood of each point), but in general it may not be possible to patch these local harmonic conjugates together into a global harmonic conjugate.
what is the complete elliptic integral of the second kind, ie, what is the integral function that's derivative produces the integrand
why did he say jacobian of f inverse is inverse of jacobian of f and then say those two are equal when the first one is evaluated at f(z) but the second one at z
also a_y=-u_y/|f'(z)|²=-(-v_x)/|f'(z)|²=-b_x
and I don't understand why he evaluates derivative of f inverse at f(z) and says its equal to a_x+ib_x, without saying where a_x and b_x are evaluated
You surely understand it now, but I am at your same position you were a year ago and I am also lost. Could you explain it to me please?
Harmonic conjugate? More like "Oh man, there are many great"...videos on both your channels!