Complex Integrals | Contour Integration | Complex Analysis #11

The basics of contour integration (complex integration). The methods that are used to determine contour integrals (complex Integrals) are explained and illustrated with a lot of examples with solutions. You will also learn how to parametrize the standard curves/paths and how to determine when you can ignore the path and only consider the start and end points of the path when determining the contour integral.

But in summary, this video will include concepts as:
► Definition of Contour Integrals
► Definition of a Directed Smooth Curve
► Definition of a Contour
► How to Parametrize the standard curves
► A theorem about Independence of Path
► Methods to solve Contour Integrals
► Tips and Tricks for Contour Integration

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Consider subscribing, liking or leaving a comment, if you enjoyed the video or if it helped you understand the subject. It really helps me a lot.

CONCEPTS FROM THE VIDEO

► Contour Integral
It is like the case with normal integrals, but here the integrand is a complex valued function of a complex variable instead of a real function of a real variable. The complex function is integrated along a specific path and in general the integration will yield different results for different paths (with the same start and end points). This comes from the fact that there are many differents path from one point in the complex plane to another.

One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods. Furthermore if a function f(z) is continous on a directed smooth curve (gamma), and if z(t) (from t=a to t=b) is a parametrization of this curve, then:

Int_(gamma) f(z) dz = Int_a^b f(z(t))*z'(t) dt

► Independence of Path
There are functions that will yield the same result when integrated along different paths which have the same start and end points.

If a function f(z) is continous on a contour (gamma) from z_1 to z_2, and f(z) has an antiderivative F(z) on gamma, then:

Int_gamma f(z) dz = F(z_2) - F(z_1)

so the only thing that matters when calculating the integral is the start and end points of gamma.

► Directed smooth curve
A point set is said to be a smooth curve if it is in range of some continous value function z = z(t) on [a, b] that satisfies the following conditions:

1) z(t) has a continous derivative on [a, b]
2) z'(t) never vanishes on [a, b]
3) z(t) is one to one on [a, b]

But in short this simply means that the curve should not have a any sharp corners and no intersections with itself. A directed smooth curve is a smooth curve with a direction.

► Parametrization
Imagine you start at a time t_0 and start tracing the curve gamma on a graph. At any particular time, a dot is drawn and the locus of dots generated over an interval of time t_0 to t_1 constitutes the curve.

You could say that by using this method we can create a function z as a function of t, such that the curve gamma is in the range of z(t) as t varies between t_0 and t_1. Then we call the function z(t) the parametrization of the curve gamma.

So in short it is a way to describe the curve with the help of the function z(t).

► Contour
A single point or a finite sequence of directed smooth curves, such that the terminal point of one directed smooth curve coincides with the following directed smooth curve.

TIMESTAMPS
00:00 - 00:20 Definition/Theorem Contour Integrals
00:20 - 01:44 Standard Parametrizations
01:44 - 03:42 Theorem Independence of Path
Examples: Compute each Integral
03:42 - 04:52 f(z) = z along a straight line
04:52 - 06:33 f(z) = z along a quarter arc of a circle
06:33 - 09:23 f(z) = z along some weird path
09:23 - 10:05 f(z) = z^bar along two connected paths
10:05 - 12:09 Notes about the most used trap in (pitfall)

ERROR
I missed adjusting for the "+1" term in the second integrand when I determined the primitive in the video at around 11:35.

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[AngelsofDeath]

Had to say thank you, not only for this video but for the entire playlist that has really helped me understand Complex Analysis way better.

[eccedentesiast]

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[iSolarSunrise]

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[elenaclaramaria]

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[ekoi]

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[t.o]

your videos are truly heavensent, please never stop thank you so much!

[Frutyy]

I was wondering about the integral from 0 to 1 of (1+t(1-i))(1+i) because when I integrated it myself I got the anti-derivative to be (1+i)t+t^2. So when i plugged in the values i got 2+i instead of 1 which is what your answer was . Sorry for asking.

[PunmasterSTP]

Another truly incredible video, that touched on so many things I'd heard of before, but didn't fully grasp. Thank you again so much!

[Achrononmaster]

Interesting commentary, a phrase you used piqued my inner philosophy nerd. @4:05 you say you "need to prove" the deformation exists. But the picture is a proof. It is only the formalist mathematical paradigm that says you must turn the picture into some other meaningless symbols. We do not have to accept that paradigm, it is a social construct. Both the Latin/Greek symbols and the picture, are interpreted by a sentient mind imbuing them with meaning due to our social context, so in a more platonic philosophical framework, either one suffices as a proof, one is easier to grasp, the other easier to rigorously defend formally. But there is always a way to translate a good enough picture into a rigorous formal proof, if the implicit spatial dimension is somehow indicated. I get it that visual proofs are hard to find, and diagrams per se can be misleading (not contain the full logic) but in this case of the path deformation I think the picture suffices, one only need do the symbolic algebra to placate a some examiner or editor.

[Amani_Rose]

11:29 How comes we didn't take the derivative of the conjugate?
As in for gamma 2, why is it (1+t(1-i)(1+i)dt instead of (1+t(1-i)(1-i)dt?

If we took the conjugates right after parameterization, wouldn't the derivatives of those give dz bar?

[AbhishekKumar-jggq]

Your videos are very helpful for me in understanding maths deeply once again thanks for this videos love from north india 🇮🇳

[Predator]

Thank you so much, I have a lecturer who refuses to put up worked examples, just the answers and questions, so if you miss any part of the module you have no basis to go off.

[CraftReaper-OmarFaruqTawsif]

one of my biggest regret in this year is not finding this playlist 3 days ago TT

[hamadhalbahrani]

an amazing explanation thank so much I appreciate you effort and giving your time

[Stayawayfrommyname]

Let me begin by thanking you about this amazing series, it has helped me a lot!
Just one thing though, at around 7:20 you say that 3pi/2 <= t <= 0, but shouldn't it be 3pi/2<=t<=2pi? or -pi/2<=t<=0?
again, thank you for your work!

Edit:
I know we will get the same result, but it feels kind of strange implying that 3pi/2 <= 0

[durellodelmattino]

TheMathCoach Hi, I wanted to ask if at 11:45 there were a missed integration of 1dt in the second integral, please correct me if I'm wrong

[KuldeepSingh-tnmb]

At 11:28 why didnt you use congugate of z'(t) with the multiplication of conjugate of f(z(t))?

[loriz]

Most helpful video ever! Learned a lot’

[yousifsalam]

@ 13:33 you said Ln(z) is not defined on the negative real axis including zero, but looking at the restriction for theta, it's -pi(not included to pi(included) so the function is defined at the negative real axis I would say.. can you explain this, please?

[dijkstra]

At 13:04 why is it that it can't handle functions with the argument -pi if it can clearly handle functions with argument pi which is the same thing as -pi. I'm just extremely confused as to why this is the case that you have assumed e^i(-pi) =/= e^i(pi) when they both are the same.