
021804, 08:28 AM #1
 Posts
 454
Cauchy's Integral Theorem
I would like to understand this theorem better.
If f(z) is analytic in a simply connected domain D, then for every simple closed path C in D we have
[CONTOUR INTEGRAL]f(z)dz=0
The simply closed path is also called a contour.
Now if we have a function which is entire in its domain (is analytic everywhere) say exp(z) or cos(z) or even sec(z), then the contour integral is zero. Right? So what is special about this? Is it that we are considering the 'complex' version of an INDEFINITE integral?
I have done complex integration before with given endpoints (like the integral of z^n from 1+i to 0 as an example), but now Mr Cauchy is telling me that the integral of z^n is actually zero? Obviously I am not getting this idea of 'taking paths' through the complex plane and thus evaluating contour integrals. Could someone (probably lethe ) give me an explanation or forward me to a site.
Cheers.

021804, 09:36 AM #2
it becomes clear when you learn when this theorem doesn't hold. the contour integral around any closed curve for a function that is not necessarily holomorphic inside that curve need not be zero.
for example, let f(z) be a holomorphic function, then make a new function, f(z)/(za), and perform the countour integral around the point a. obviously, this new function is not holomorphic in this domain (it has a pole at a), and so the integral is nonzero. in fact, you know what the value is, it is f(a).
what is happening here? well, the CauchyRiemann conditions on the real and imaginary parts of f make this countour integral actually a restatement of Green's theorem. remember Green's theorem? the line integral around a closed curve of Pdx+Qdy is equal to the double integral over the disk enclosed by the curve of dQ/dx  dP/dy.
but this last expression vanishes by the CauchyRiemann condition, for a holomorphic function.
this is also very much analogous to Gauss' Law. did you learn Gauss' Law in electrostatics? the integral of the electric field over the surface is equal to the integral of the divergence of the electric field over the volume enclosed by that surface. the divergence is only nonzero at places where there are charges, which have electric fields like 1/r^2, which is singular. the only thing that can make the flux nonzero is the presence of singularities inside the surface you are integrating.
so you have the same thing here for the Cauchy integral. if the function is holomorphic then it cannot have a poles, and so the line integral vanishes.

021804, 09:40 AM #3
also, the requirement that D be simply connected is very important. a meromorphic function (a holomorphic function everywhere but for a point where it has a simple pole) can be considered a holomorphic funcition on the region with the point of its pole removed.
but such a region is no longer simply connected.
the relationship is this: Poincaré's lemma says that on a simply connected region, every closed form is an exact form. the integral around a closed loop of an exact form has to vanish, by the fundamental theorem of calculus (what is F(b)F(a) when the endpoints coincide, as they do for a closed curve?)
so the presence of a pole is equivalent to restricting to a holomorphic function on a nonsimply connected region

021804, 01:34 PM #4
 Posts
 36
haha, this brings to mind a joke (that's actually not very funny):
Q: What's the contour integral around Western Europe?
A: Zero, because all the Poles are in Eastern Europe!
Addendum: Actually, there ARE some Poles in Western Europe, but they are removable!

021904, 06:18 AM #5
 Posts
 454
this is also very much analogous to Gauss' Law. did you learn Gauss' Law in electrostatics? the integral of the electric field over the surface is equal to the integral of the divergence of the electric field over the volume enclosed by that surface. the divergence is only nonzero at places where there are charges, which have electric fields like 1/r^2, which is singular. the only thing that can make the flux nonzero is the presence of singularities inside the surface you are integrating.
for example, let f(z) be a holomorphic function, then make a new function, f(z)/(za), and perform the countour integral around the point a. obviously, this new function is not holomorphic in this domain (it has a pole at a), and so the integral is nonzero. in fact, you know what the value is, it is f(a). what is happening here? well, the CauchyRiemann conditions on the real and imaginary parts of f make this countour integral actually a restatement of Green's theorem. remember Green's theorem? the line integral around a closed curve of Pdx+Qdy is equal to the double integral over the disk enclosed by the curve of dQ/dx  dP/dy.
So for the contour integral of f(z) which is holomorphic (analytic) at each point in and on the contour C, the value of the integral is independent of path along the contour between the two endpoints.
With that sorted we can write f(z) = u+iv and dz=dx+idy. Then
[contour integral]f(z)dz=[integral between a and b](u+iv)(dx+idy)=
Denote S by the integral...
S(udx+iudy+ivdx+i^2vdy)=S(udx  vdy) + i S(vdx + udy)
Now you said that Green's Theorem allows us to express line integrals as double integrals  specifically if two realvalued functions together with their firstorder partial derivatives are continuous throughout the closed region. Which they are! Then
S (Pdx + Qdy) = SS (Q_x  P_y)dA
Then
S f(z)dz = SS (v_x  u_y)dA + i SS (u_x  v_y)dA
But we know that from the CR equations that u_x = v_y and u_y = v_x then the right hand expression is zero. Wow!! That really does work! Thankyou for pointing me to Green's Theorem. That helped. Awesome stuff

021904, 08:43 AM #6
or another theorem you can use from vector calculus:
remember conservative force fields? ∫<b}F</B>*ds is independent of path iff <b>F</b>=∇f for some function f.
∫<b>F</b>*ds=∫F<sub>1</sub>dx+F<sub>2</sub>dy
we will have <b>F</b>=∇f iff ∂F<sub>1</sub>/∂y=∂F<sub>2</sub>/∂x (which is equivalent to the equality of mixed partial derivatives)
apply this to the real and complex part of the contour integral, and you obtain the CauchyRiemann conditions.Last edited by lethe; 021904 at 09:10 AM.

021904, 10:12 AM #7
 Posts
 686
Originally Posted by oxymoron

021904, 10:23 AM #8
you know.... i hope that my illustration to other theorems of calculus helped put this into context for you, but you should realize that all i have done is change the context
in other words, the fact that the Cauchy integral is zero is just another statement of Stoke's Theorem, so all i did was show you how it is related to a few versions of stoke's theorem (which is itself a generalisation of the fundamental theorem of calculus, which states that antiderivatives are integrals)
someone just asked on this forum <i>why</i> antiderivatives are the same thing as integrals, and well there are some nice plausibility arguments, none of them really felt like a convincing <i>why</i>.
sometimes, the best answer is, as JamesR said, "because the theorem says so!"

021904, 07:58 PM #9
 Posts
 454
Lethe, the 'analogies' you showed me where great. It is as though they are intertwined  all using the same basic properties (which I guess they are  FTOC).
Errandir. After all the conversations I am finally starting to understand that it must equal zero. Thankyou guys for helping me.
But there is more...
Let C denote the circle z  z_0 = R taken counterclockwise. Use the parametric representation z = z_0 + Re^iq, (where pi <= q <= pi ) for C to derive the following integration formula:
[contour integral over C] dz / (z  z_0) = 2[pi]i
If I can show this then I can complete my understanding of Cauchy's Integral Theorem.

021904, 08:16 PM #10Originally Posted by oxymoron
dz = iRe^iq*dq.
You also have zz_0 = Re^iq, so that your contour integral becomes:
[contour integral over C] dz / (z  z_0) = [integral 0 > 2PI]iqRe^iq/(Re^iq)dq
= i[integral 0 > 2PI]dq = 2PI i

022004, 06:45 AM #11Originally Posted by Electric Jaguar
There was a plane from "air poland" that was arriving to the states. Then the pilot said on his microphone that on the right side, they can see the Statue of Liberty. So all the passengers moved to the right side and look by the windows. Suddenly the plane crashed.
Q. Why did the plane became unstable and crashed?
A. There were too many poles on the right side of the plane.

022004, 06:08 PM #12
 Posts
 454
1100f, thankyou for the solution.
With your answer now let f(z) = 1 in the equation so that
f(z)/(z  z_0) is what we are trying to integrate. Then substituting
f(z) = f(z_0) + [f(z)  f(z_0)]
we have
[contour integral] f(z)dz/(z  z_0) = f(z_0) [C integral] dz/(z  z_0) + [C integral] (f(z)  f(z_0))dz/(z  z_0)
The first term on the right in the above equation equals f(z_0)2[pi]i (that is why I needed that help)
The last term on the right in the above equation equals zero since it is everywhere analytic except at z_0 and by the principle of deformation of path we can choose a smaller circle around which we integrate without any problem. Then since f(z) is analytic and thus continuous there is an epsilon greater than zero and we should be able to find a delta greater than zero such that f(z)  f(z_0) < epsilon for all z in the disk  z  z_0  < delta. By prescribing the radius of our smaller circle p we have
 (f(z)  f(z_0)) / z  z_0  < epsilon/p
By the ML inequality we have the inequality integral < 2[pi]epsilon.
Now since epsilon can be chosen so that it is arbitrarily small, it follows that the last integral must be zero.
Similar Threads

By FrankExchangeOfViews in forum Physics & MathLast Post: 050506, 06:44 AMReplies: 5

By oxymoron in forum Physics & MathLast Post: 012204, 10:24 AMReplies: 17

By wet1 in forum Astronomy, Exobiology, & CosmologyLast Post: 040702, 10:46 PMReplies: 0

By Caleb in forum Religion ArchivesLast Post: 090101, 09:09 PMReplies: 17

By CTtheory in forum Pseudoscience ArchiveLast Post: 120899, 09:47 AMReplies: 0
Bookmarks