Cauchy’s Theorems I
Ang M.S.
Augustin-Louis Cauchy
1789 – 1857
October 26, 2012
References
Murray R. Spiegel Complex V ariables with introduction to conformal mapping and its applications Dennis G. Zill , P. D. Shanahan A First Course in Complex Analysis with Applications J. H. Mathews , R. W. Howell Complex Analysis For Mathematics and Engineerng
1 Summary
• Cauchy Integral Theorem
Let f be analytic in a simply connected domain D. If C is a simple closed contour that lies in D , and there is no singular point inside the contour, then
ˆ
f (z) dz = 0
C
• Cauchy Integral Formula (For simple pole)
If there is a singular point z0 inside the contour, then
˛
f(z)
dz = 2πj f(z0) z − z0
• Generalized Cauchy Integral Formula (For pole with any order)
˛
f(z)
(z − z0)n
2πj
- dz =
- f
(n−1) (z0)
(n − 1)!
• Cauchy Inequality
- ꢀ
- ꢀ
ꢀ
M · n!
(n)
ꢀ
f
(z0) ≤
rn
• Gauss Mean Value Theorem
ˆ
2π
- (
- )
1f(z0) =
f z0 + rejθ dθ
2π
0
1
2 Related Mathematics Review
2.1 Stoke’s Theorem
- ¨
- ˛
- ¯
- ¯
- ∇ × F · dS =
- F · dr
- Σ
- ∂Σ
(The proof is skipped)
¯
Consider F = (FX , FY , FZ)
-
-
xˆ
∂
yˆ
∂
zˆ
∂
-
-
¯
∇ × F = det
∂x ∂y ∂z
FX FY FZ
Let FZ = 0
-
-
xˆ
∂
yˆ
∂
zˆ
∂
¯
∇ × F = det
∂x ∂y ∂z
FX FY
0
¯
dS = nˆdS , for dS = dxdy , nˆ = zˆ . By zˆ · zˆ = 1 , consider zˆ component only for ∇ × F
-
-
xˆ
∂
- yˆ
- zˆ
∂
- (
- )
- ∂
- ∂FY
∂x
∂FX
∂y
¯
∇ × F = det
=
−
zˆ
∂x ∂y ∂z
FX FY
0
i.e.
- (
- )
- ¨
- ¨
∂FY
∂x
∂FX
∂y
¯
∇ × F · dS =
−
dxdy
- Σ
- Σ
Consider the RHS (let FZ = 0)
- ˛
- ˛
- ˛
¯
F · dr =
- (FX, FY , FZ) · (dx, dy, dz) =
- FXdx + FY dy
- ∂Σ
- ∂Σ
- ∂Σ
Thus
- (
- )
- ¨
- ˛
∂FY
∂x
∂FX
∂y
−
- dxdy =
- FXdx + FY dy
- Σ
- ∂Σ
2.2 Cauchy-Reimann Condition
For f(z) = u(z) + jv(z) = u(x, y) + jv(x, y)
The complex derivative is
f (z0 + h) − f(z0)
- lim
- = f ′ (z0)
hh → 0
h ∈ C
If the limit exists.
2
If the limit exists, limits along real axis should be the same the limit along imaginary axis
f (z0 + h) − f (z0)
∂f (z0)
1 ∂f (z0)
f (z0 + jh) − f (z0)
- lim
- =
- =
- =
- lim
- h
- ∂x
- j
- ∂y
- jh
- h → 0
- h → 0
- h ∈ R
- h ∈ R
i.e. The Cauchy-Riemann Equation
∂f (z0)
1 ∂f (z0)
=
- ∂x
- j
- ∂y
Expand
∂
1 ∂
- [u (x0, y0) + jv (x0, y0)] =
- [u (x0, y0) + jv (x0, y0)]
- ∂x
- j ∂y
Equalize real part and imaginary part
- ∂
- ∂
- u (x0, y0) =
- v (x0, y0)
∂x ∂
∂y
∂
v (x0, y0) = − u (x0, y0)
- ∂x
- ∂y
Or simply as
∂u ∂x
∂v ∂y
=
- ∂v
- ∂u
= −
- ∂x
- ∂y
i.e. If f(z) is analytic, then it fulfill this condition
2.3 ML Inequality
First,
- ˆ
- ˆ
- b
- b
if g(x) ≤ f(x) , then
- g(x)dx ≤
- f(x)dx
- a
- a
This is true if
- sup g(x) ≤ sup f(x)
- inf g(x) ≤ inf f(x)
Then for any function f , it is true that
−|f| ≤ f ≤ |f|
Then apply the inequality above ,
- ˆ
- ˆ
- ˆ
b
−
- b |f(x)|dx ≤
- f(x)dx ≤ b |f(x)|dx
- a
- a
- a
Now is the time to show , for bounded f(z) , i.e. |f(z)| ≤ M ,
- ꢀ
- ꢀ
ˆ
ꢀꢀꢀ
ꢀꢀ
f(z)dz ≤ ML
ꢀ
c
3
Pf.
- ´
- ´
b
By using the b f(x)dx ≤ |f(x)|dx
- a
- a
ꢀ
ˆ
ꢀ
ˆ
ꢀꢀꢀ
ꢀꢀ
f(z)dz ≤ |f(z)|dz
ꢀ
- c
- c
Since f(z) si bounded,
- ꢀ
- ꢀ
- ˆ
- ˆ
- ˆ
- ˆ
ꢀꢀꢀ
ꢀꢀ
- f(z)dz ≤ |f(z)|dz ≤ Mdz = M
- dz = ML
ꢀ
- c
- c
- c
- C
| {z }
L
Therefore
ˆ
f(z)dz| ≤ ML
- |
- |
c
3 The Cauchy Theorem
f(z) = u(x, y) + jv(x, y)
- ˛
- ˛
- ˛
- ˛
- f (z) dz =
- [u (x, y) + jv (x, y)] [dx + jdv] =
- udx − vdy + j
- vdx + udy
- C
- C
- C
- C
3.1 The Real Part
Let FX = u (x, y) FY = −u (x, y) and consider the real part of the integral
- ˛
- ˛
udx − vdy = FXdx + FY dy
C
By the Stoke’ Theorem
- ˛
- ¨
- ¨
udx + (−v)dy =
[(−v)x − uy] dxdy =
[−vx − uy] dxdy
C
Since the function is analytic in the region D , ⇐⇒ the function fulfill Cauchy-Reimann Condition
∂v ∂x
∂u ∂y
= −
Then
- ¨
- ¨
[−vx − uy] dxdy =
[uy − uy] dxdy = 0
´
The real part of C f(z)dz equal zero
4
3.2 The imaginary part
´
Consider the imaginary part of C f(z)dz ,
˛
vdx + udy
C
Let FX be v(x, y) and FY be u(x, y) , and apply Stoke’s Theorem
- ˛
- ¨
- vdx + udy =
- (ux − vy) dxdy
C
∂u ∂x
∂v ∂y
As the function is analytic, so it fulfill Cauchy-Reimann Condition : become zero
=
, thus the integral
Finally, Let f be analytic in a simply connected domain D , for the simple closed contour C that lies in D , the close contour integral equal zero
ˆ
f (z) dz = 0
C
4 The Cauchy Integral Formulas
4.1 Simple Pole
˛
g (z) dz = 2πjg (z0) z − z0
C
Condition • z0 inside C , if z0is outside C , then the integral is just zero • g(z) is analytic in simply connected domain ⇐⇒ domain of g(z) is simply connect closed region, and g(z) fulfill CR-Condition there
Proof. Consider the diagram
g(z)
, the integral of
By concept of path deformation z − z0
along path C that ϵ → 0 along path C is equivalent to the integral
ϵ
- ˛
- ˛
- g(z)
- g(z)
dz = lim
dz
ϵ→0
- z − z0
- z − z0
C
Cϵ
The small circle inside can be expressed as
|z − z0| = ϵejθ 0 ≤ θ < 2π
As the radius ϵ → 0+ , the circle is thus
z − z0 = ϵejθ 0 ≤ θ < 2π
5
Put this back into the integral
- (
- )
- ˛
- ˛
- ˆ
g z0 + ϵejθ
(z0 + ϵejθ) − z0
2π
- (
- )
- g(z)
- g(z)
- dz = lim
- dz = lim
d z0 + ϵejθ
- ϵ→0
- ϵ→0
- z − z0
- z − z0
C
Cϵ
0
- (
- )
- ˆ
- ˆ
2π
(
g z0 + ϵejθ
2π
)
= lim
- jϵejθdθ = jlim
- g z0 + ϵejθ dθ
ϵejθ
- ϵ→0
- ϵ→0
- 0
- 0
- ˆ
- ˆ
- ˆ
- [
- ]
)
- 2π
- 2π
- 2π
(
= j
- limg z0 + ϵejθ dθ = j
- g (z0) dθ = jg (z0)
- dθ = 2πjg (z0)
ϵ→0
- 0
- 0
- 0
∴
˛
g(z)
dz = 2πjg (z0) z − z0
C
4.2 Generalized Cauchy Integral Formula
˛
- f (z0)
- 2πj
- dz =
- 2πj f(n−1) (z0)
(z − z0)n
(n − 1)!
First, the Cauchy Integral Formula for simple pole is
˛
g (z) dz = 2πjg (z0) z − z0
C
(
on both side Not
˛
)
- d
- d
Take
!
- dz0
- dz
˛
g (z)
⇒
- d
- g (z)
- d
- 2πj
- dz = 2πj
- g (z0)
- dz =
g′ (z0)
(z − z0)2
- dz0
- z − z0
- dz0
1!
- C
- C
Repeat,
- ˛
- ˛
g (z)
⇒
- d
- g (z)
(z − z0)2
g (z)
(z − z0)3
- 2πj d
- 2πj
dz =
g′ (z0)
dz =
g” (z0)
(z − z0)3 dz0
- 1! dz0
- 2!
- C
- C
- ˛
- ˛
g (z)
⇒
- d
- 2πj d
- 2πj
dz =
g” (z0)
- dz =
- g
(3) (z0)
(z − z0)4 dz0
- 1! dz0
- 3!
- C
- C
Thus, the general form of Cauchy Integral Formula is
˛
- g (z)
- 2πj
- dz =
- g
(n−1) (z0)
(z − z0)n
(n − 1)!
C
This can be proved using Mathematical Induction.
6
5 Consequences of Cauchy’s Integral Formula
5.1 Cauchy Inequality
The generalized Cauchy Integral Formula is
˛
f (z)
(z − z0)n+1
2πj
- dz =
- f
(n) (z0)
n!
C
Take absolute value
- ꢀ
- ꢀ
ꢀ
ꢀꢀꢀꢀ
ꢀꢀꢀ
- ꢀ
- ꢀ
ꢀ
- ˛
- ˛
- ꢀ
- ꢀ
ꢀꢀ
ꢀꢀ
ꢀꢀ
f (z)