As the integrator got out his pen and began solving the problem, he asked the function. "Are you zero because were integrating along a closed contour? Or are we integrating along a closed contour because you are zero?" The function then replied, "Stand proud. You're strong. But nah I'd win." For unbeknownst to the integrator, the function was not analytic and has a singularity within the contour.. In that moment the function could've saved itself but it didn't know two key things. The first is always bet on integrator. And the second.. is that the integrator knew Residue Theorem
Hello Sir, I wanted to thank you for the high quality throughout this series. You make learning, in this case Complex Analysis, both interesting and progressive due to your rigorous approach. Again, thanks!
I'm a junior in highschool. Keep it up man! Complex analysis in 7th grade is far beyond me. I wish I had your grit to work and perspective to start early.
is it right to say that for a holomorphic function on a complex domain, whenever we do a closed contour integral we either have zero, or (sum of multiples of residues)?
I stuck in a problem. I wonder could you give me a hint? if \omega is a region containing the closed unit disk and f has n simple zeros in the open unit disk D. how can I show Re(f(z)) real part of f has at least 2n zeros on the boundary of the unit disk. Hint is written use Winding number. I appreciate your guidance.
Would it be correct that the contour integral of f(z) being non 0 imply a pole in the interior, and again the reciprocal having a zero in the interior? (assuming reciprocal non 0 on boundary, and all other conditions met)
I think that assumptions only holds true if the contour we consider is true, since an closed contour with no singularities is 0 by Cauchy goursat theorem, but any normal contour that isn't closed can be non-zero without any singularities
I would say, not necessarily? By Cauchy's Theorem we know that if you integrate over a boundary to a domain where the integrand is holomorphic then the result is 0. By contrapositive that would only say that if the result of the integral is not 0, then the integrand is not holomorphic. Now, that could be because there is a singularity, thereby it can't be holomorphic at the singularity. As a matter of fact, integrating the conjugate function f(x+iy) = x-iy does not give 0 if I remember right, and it does not have singularities. It actually is not holomorphic so your hypothesis is unfortunately wrong
As the integrator got out his pen and began solving the problem, he asked the function. "Are you zero because were integrating along a closed contour? Or are we integrating along a closed contour because you are zero?" The function then replied, "Stand proud. You're strong. But nah I'd win."
For unbeknownst to the integrator, the function was not analytic and has a singularity within the contour.. In that moment the function could've saved itself but it didn't know two key things. The first is always bet on integrator. And the second.. is that the integrator knew Residue Theorem
Lobotomy kaisen is everywhere
Are you zero because were integrating along a closed countour ?Or are we integrating along a closed contour because you're zero ?
The first :D
Always bet on the integrator
But the intergrator knows residue theorem
Nah, I'd integrate
Hello Sir, I wanted to thank you for the high quality throughout this series. You make learning, in this case Complex Analysis, both interesting and progressive due to your rigorous approach. Again, thanks!
Thank you for the nice words! And thank you very much for the support :)
You know you are the best person 🎉 to explain math
Btw I am a 7 grader and IMO Iraqi team participant
Thanks!
I'm a junior in highschool. Keep it up man!
Complex analysis in 7th grade is far beyond me. I wish I had your grit to work and perspective to start early.
is it right to say that for a holomorphic function on a complex domain, whenever we do a closed contour integral we either have zero, or (sum of multiples of residues)?
No, the assumptions of the residue theorem need to be satisfied.
is this playlist complete?
See here: tbsom.de/s/ca
I stuck in a problem. I wonder could you give me a hint? if \omega is a region containing the closed unit disk and f has n simple zeros in the open unit disk D. how can I show Re(f(z)) real part of f has at least 2n zeros on the boundary of the unit disk. Hint is written use Winding number. I appreciate your guidance.
Such discussion I want to put into my community forum: tbsom.de/s/community
UA-cam comments are too cumbersome and also miss LaTeX commands.
Would it be correct that the contour integral of f(z) being non 0 imply a pole in the interior, and again the reciprocal having a zero in the interior? (assuming reciprocal non 0 on boundary, and all other conditions met)
I think that assumptions only holds true if the contour we consider is true, since an closed contour with no singularities is 0 by Cauchy goursat theorem, but any normal contour that isn't closed can be non-zero without any singularities
I would say, not necessarily? By Cauchy's Theorem we know that if you integrate over a boundary to a domain where the integrand is holomorphic then the result is 0. By contrapositive that would only say that if the result of the integral is not 0, then the integrand is not holomorphic. Now, that could be because there is a singularity, thereby it can't be holomorphic at the singularity.
As a matter of fact, integrating the conjugate function f(x+iy) = x-iy does not give 0 if I remember right, and it does not have singularities. It actually is not holomorphic so your hypothesis is unfortunately wrong
Do real and complex analysis mean the opposite
No, they just say which number set we use :)
What's the delta before the B_\epsilon?
Boundary of the ball, so the whole thing means the circle :)
@@brightsideofmaths thank you!
Difficult to follow lecturer's english
Sorry about that!