intzero.htm Cauchy's Integral Theorem
Cauchy's Integral Theorem
The fundamental result - Cauchy's Integral Theorem - says roughly:
If C is a simple closed path and w = f(z) is analytic inside and on C, then
⌠
(⎜)
⌡


C 
 f(z)  dz = 0.
There are two common approaches to this result. The first approach uses the Cauchy-Riemann equations and Green's Theorem. The second approach uses less assumptions about the regularity of the derivative f′ and builds up the proof by first considering C to be a simple closed triangle and then approximating the general simple closed path by a simple closed polygonal path.


Cauchy's Integral Theorem using Green's Formula

Theorem. Let C be a simple closed path enclosing a region D. Suppose that on D ∪C, w = f(z) is analytic and that f′ is continuous. Then
⌠
(⎜)
⌡


C 
 f(z)  dz = 0.
 Proof: By the Cauchy-Riemann Equations,
∂f
∂
-
z
 
 = 1
2
 ⎛
⎝ 		∂f
∂x
 + i ∂f
∂y
⎞
⎠ 		=0.
Then by Green's Formula
0 = ⌠
⌡ 		⌠
⌡

D 
 ⎛
⎝ 		∂f
∂x
 + i ∂f
∂y
⎞
⎠ 		 dx  dy
= ⌠
(⎜)
⌡


C 
 f  dy − i ⌠
(⎜)
⌡


C 
 f dx
= −i ⌠
(⎜)
⌡


C 
 f  (dx + i dy)
= −i ⌠
(⎜)
⌡


C 
 f(z)  dz
= 0.


Consequences of Cauchy's Integral Theorem
1.
In Simply Connected Regions Integrals of Analytic functions are Independent of the Path
A region D is simply connected if for every simple closed path C in D, all of the points inside C are also in D. The most important examples of simply connected regions are
•
Circles: {z | |z−z| < R}
•
Half Planes: {z| ℜz > 0}
•
The whole complex plane C
•
Convex regions
Let w = f(z) be analytic in a simply connected region D. Let Z1 and Z2 be two points in D, and take two paths C1 and C2 in D which go from Z1 (initial point) to Z2 (terminal point). Then C1 − C2 can be broken into simple closed paths so that
0 = ⌠
⌡

C1 − C2 
 f(z)  dz
= ⌠
⌡

C1 
 f(z)  dz − ⌠
⌡

C2 
 f(z)  dz,
⌠
⌡

C1 
 f(z)  dz
= ⌠
⌡

C2 
 f(z)  dz.
Thus we define
⌠
⌡ 		Z2

Z1 
 f(z)  dz = ⌠
⌡

C 
 f(z)  dz,
where C is any path in D from Z1 to Z2.


2.
Two Circles Theorem
Let Cϵ be a circle inside a circle CR.

intzpic1.gif

Suppose that f(z) is analytic on the two circles and the region in between the two circles. Then
⌠
(⎜)
⌡


Cϵ 
 f(z)  dz = ⌠
(⎜)
⌡


CR 
 f(z)   dz.
. The proof uses a cut C from the outer circle to the inner circle.

intzpic2.gif

Then
⌠
(⎜)
⌡


CR 
 f(z)  dz − ⌠
(⎜)
⌡


Cϵ 
 f(z)  dz
= ⌠
(⎜)
⌡


CR + C − Cϵ − C  
 f(z)  dz
= 0.


3.
Fundamental Theorem of Calculus Version II
Let w = f(z) be analytic in a simply connected region D. For z ∈ D, define
F(z) = ⌠
⌡ 		z

Z0 
 f(ζ)  dζ.
Then F(z) in analytic in D and F′(z) = f(z).




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On 16 Nov 2013, 21:01.