Understanding Cauchy’s Theorem and Complex Integrals Explained

welcome back to my video series on complex variables today's video is going to go over cauchy's theorem which is one

of the most important results and complex calculus but before we do that I'm going to briefly go over complex

integrals here's how it works suppose I have a complex function f of Z and I want to integrate that in the

complex plane from a complex number a to another complex number B for real functions like f of X integrating from A

to B would be very simple because when going from A to B you just have to traverse along a straight interval for

complex functions though instead of having a straightforward path along an interval you could go along multiple

paths from A to B when performing your integration each of these paths could give you different results for the

integral so when that sense complex integration is a lot like line integration and that a path has to be

defined for the integration this is all because complex numbers can be thought of as two dimensional numbers and that

they exist on an entire plane instead of just a straight line because complex integration is a lot like line

integration the method of solving complex integration problems is very similar to that for line integration

suppose I want to integrate from A to B along a path see an arbitrary complex function f of Z which is composed of a

real part U and an imaginary part V suppose also that my curve C can be expressed in terms of the parametric

equation in which X and y are both functions of some parameter T which varies from alpha to beta as we traverse

the curve hopefully you're familiar with parametric equations and how they work the integral of f of zdz along the curve

C can be simplified by decomposing F into its real and imaginary parts and doing the same thing for the

differential DZ we can expand out the part inside the integral to get the integral along c of u DX minus the

integral along c of v dy plus the imaginary number times the integral along c of v DX plus the integral along

c of u dy you can now make a change of variables by writing the differential DX as the

product of DX by DT times DT and doing the same for the different hildie why since the integration

variable has now changed at E we need to change the limits of integration accordingly in this case the limits of

integration all become alpha and beta because that's what T varies along it varies from alpha to beta on the curve C

and we're done this is the formula for evaluating the complex integral given a complex function f of Z and a

parametrized path C so if you're given the function f of C to integrate you can find its real and imaginary parts U and

B then you can find the parametric equation of the given curve C and then you can evaluate the integral using this

formula but there's much more to complex integrals than just the simple integration process and this is where we

transition to the real meat of the lecture which is cauchy's theorem cauchy's theorem describes complex

integral specifically it's a theorem about the integral of complex functions F of Z around a closed curve also known

as contour integrals by closed curve I just mean a curve that encloses upon night area specifically cauchy's theorem

states that if I have a complex function f of Z and a closed curve C which satisfy the following three conditions

one is that f is holomorphic or differentiable on and everywhere inside C the second is that C is a simple curve

by which I mean it doesn't cross itself and the third is that C has a finite number of corners so a square is fine

but something crazy like a Weierstrass function isn't if all these three conditions are satisfied then the

contour integral around C of f of zdz is a big fat zero to summarize if f is holomorphic then its contour integral

along a nice enough closed curve will be zero now unlike the little cop-out that I did last time I'm actually going to

man up today and prove this theorem for you the proof is actually pretty simple but in this simple proof I'm going to

impose an additional condition that DF DZ is also continuous now you might say that isn't a holomorphic complex

function supposed to have continuous derivatives in fact isn't it supposed to have derivatives of all orders if you

remember the previous lecture well it is but those facts need to be proven because they came in the form of the

theorem and for that theorem to be proven you actually require cauchy's theorem which we're proving

right here actually required how she's theorem to be true so starting out cauchy's theorem by saying oh with f of

z has a derivative then that derivative must be continuous would be circular logic that's why we've written DF DZ is

continuous as an assumption instead of a fact which follows from that being homework it it's also possible to prove

this theorem without making that assumption but it's much more difficult you want to do that anyway let's begin

as usual we can write a complex function f of Z as a composition of its real part view and its imaginary part beat we can

also write the differential DZ as a composition of its real part DX and its imaginary part dy if we put these

together in our integral of f of Z then we can expand out our expression to end up with the closed integral along c of

udx minus vdy plus I times the closed integral along c of b DX plus u dy and this is where we use a result from your

glory days in vector calculus that result is called greens theorem which says that the line integral along a

simple closed curve of P DX and Q dy equals the double integral over the region enclosed by the curve of DQ DX

minus DP dy provided P and Q are continuous and have continuous derivatives in terms of

two-dimensional vector fields it just means that the line integral of a vector field on a closed loop equals the double

integral of the curl of that vector field over the area enclosed by that mean we can apply greens theorem to both

the first and second integrals in this expression start with the first one we compare it to the P and Q in Green's

theorem then we can see that you hear is like P and negative V here is like Q so when we apply greens theorem you get the

double integral over the area enclosed by C of negative DB DX minus DQ dy now recall that if a function f of Z is

holomorphic then its real and imaginary parts obey the cashier eman relations it then follows that the function we're

using in the proof of cauchy's theorem also obeys the cashier a more relations because after all it's all work

according to one of the initial assumptions the statement since the cow eat Braemar relations

apply D UD y equals negative B B DX since these two partial derivatives are being subtracted in the integrand of I 1

it follows that this integrand becomes 0 which means that this integral I 1 as a whole becomes 0 we can use the same

argument on the integral I 2 when we apply greens theorem on this integral it just becomes the double integral over

the area enclosed by C of D u DX minus DV dy again since the cashier Aemon relations hold true the integrand just

cancelled out and it follows that I 2 is also 0 since I 1 is 0 and I 2 is 0 I 1 plus I 2 is 0 and it follows that the

closed integral along c of f of zdz is 0 plus 0 i which is just 0 and this proves cauchy's theorem it should be noted that

it's also possible to prove cauchy's theorem without assuming the function f has a continuous derivative I mentioned

this earlier as well but in that case the proof is more lengthy and it doesn't really add much to our understanding so

I left that part out anyway that should be it for this video in the next video I'm going to move on to proving a

formula related to contour integrals