Complex Analysis 20 | Antiderivatives

Please consider to support me if this video was helpful such that I can continue to produce them :)

🙏 Thanks to all supporters! They are mentioned in the credits of the video :)

Thanks to all supporters who made this video possible! They are mentioned in the credits of the video :)

This is my video series about Complex Analysis. I hope that it will help everyone who wants to learn about complex derivatives, curve integrals, and the residue theorem. Complex Analysis has a lof applications in other parts of mathematics and in physics.

x

#ComplexAnalysis
#Analysis
#Calculus
#Mathematics
#curveintegral
#integration

(This explanation fits to lectures for students in their first or second year of study: Mathematics, Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[DOROnoDORO]

It's magical to finally be able to understand this :)

[drewtmacha]

Can't wait for the next one! I'm about to start a PhD program next week, and I'm taking Complex Analysis as my first course. I've started reading through Conway's Functions of One Complex Variable, but seeing your videos brings it to life in my mind much better than struggling through the text!

[malawigw]

This is not complex analysis, it is beatiful analysis

[mehdih]

Yeah! The great comeback of complex analysis!

[Ghetto_Bird]

Beautifully done once again, what a treat!

[GeoffryGifari]

so... just because we have an integral over a closed curve (in which the function has antiderivative), whether or not the result is zero depends on the points enclosed by the curve/domain of the antiderivative?

[individuoenigmatico]

There is something regarding this proof that I can't quite wrap my head around. The proof presented here is that a function f that is holomorphic in an open ball has an antiderivative, and therefore, all integrals of f along closed paths are zero. We already proved in fact in video 20 that if f has an antiderivative in any type of open set U (whether it has holes or not), then all integrals around closed paths contained in U are zero (whether they pass around holes or not).

However, I now have a problem. Your proof seems to generalize easily to any open, path-connected set U (even with holes or isolated singularities). Assuming f is holomorphic in this set U, it seems to me that we could use Goursat's Theorem to prove that f has an antiderivative, and therefore, that ALL integrals around closed paths contained in U are zero (whether or not they pass around holes).

In fact, let's take z_0 € U. Then, let's take any z €U. Because of the openness of U, there is a ball B_ε(z) entirely contained in U. Now, because U is path-connected, there is a path connecting the nearest point - let's call it h - of the ball B_ε(z) to z_0. Since such a path exists, for every point on the path, we can find a ball entirely included in U. In this way, if we want, we can easily transform the path into a segmented path passing through these balls. Then, the proof can continue along your lines. Within the ball B_ε(z) we connect h to z with a straight line and h to any other point in the ball z' with another straight line. Then, we can use this to prove that f has an antiderivative (using Goursat's Theorem because the path passing between h, z, and z' is essentially triangular, and it is entirely contained in the ball B_ε(z) fully contained in U. But then, since f has an antiderivative in U, ALL closed paths within U should have zero integral, irrespective of whether they pass around a hole of U or not.

But this doesn't make sense. There must be some mistake in this "proof" because we know for a fact that the integral of closed paths that pass around holes or isolated singularities are not zero. Yet, I can't pinpoint where the mistake lies. Do you understand my problem?

[enricorossi]

May I ask why is it possible to integrate directly 1/z^2 even though C-{0} is not a simply connected domain? Or am I missing something? Shouldn't one use the Residue theorem?