Cauchy Integral Theorem

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

f (z) dz = 0

• Cauchy Integral Formula (For simple pole)

If there is a singular point z₀inside the contour, then

dz = 2πj f(z₀) z − z₀

• Generalized Cauchy Integral Formula (For pole with any order)

(z − z₀)ⁿ
2πj

⁽ⁿ⁻¹⁾(z₀)

(n − 1)!

• Cauchy Inequality

M · n!

(z₀) ≤

• Gauss Mean Value Theorem

1f(z₀) =

f z₀+ re^jθdθ
2π

2 Related Mathematics Review

2.1 Stoke’s Theorem

(The proof is skipped)

Consider F = (F_X, F_Y, F_Z)





∇ × F = det

∂x ∂y ∂z

F_XF_YF_Z

Let F_Z= 0




¯
∇ × F = det

∂x ∂y ∂z

F_XF_Y

dS = nˆdS , for dS = dxdy , nˆ = zˆ . By zˆ · zˆ = 1 , consider zˆ component only for ∇ × F




∂x
∂F_X
∂y

¯
∇ × F = det
=

∂x ∂y ∂z

F_XF_Y

∂F_Y
∂x
∂F_X
∂y

∇ × F · dS =

Consider the RHS (let F_Z= 0)

F · dr =

∂F_Y
∂x
∂F_X
∂y

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)

hh → 0

h ∈ C

If the limit exists.
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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)

i.e. The Cauchy-Riemann Equation

∂f (z₀)

1 ∂f (z₀)
=

Expand

Equalize real part and imaginary part



∂x ∂
∂y
∂



v (x₀, y₀) = − u (x₀, y₀)

Or simply as

∂u ∂x
∂v ∂y





i.e. If f(z) is analytic, then it fulﬁll this condition

2.3 ML Inequality

First,

if g(x) ≤ f(x) , then

This is true if

Then for any function f , it is true that

−|f| ≤ f ≤ |f|

Then apply the inequality above ,

Now is the time to show , for bounded f(z) , i.e. |f(z)| ≤ M ,

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f(z)dz ≤ ML

By using the ^bf(x)dx ≤ |f(x)|dx

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f(z)dz ≤ |f(z)|dz

Since f(z) si bounded,

ꢀꢀꢀ
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| {z }

Therefore

f(z)dz| ≤ ML

3 The Cauchy Theorem

f(z) = u(x, y) + jv(x, y)

3.1 The Real Part

Let F_X= u (x, y) F_Y= −u (x, y) and consider the real part of the integral

udx − vdy = F_Xdx + F_Ydy

By the Stoke’ Theorem

udx + (−v)dy =

[(−v)_x− u_y] dxdy =

[−v_x− u_y] dxdy

Since the function is analytic in the region D , ⇐⇒ the function fulﬁll Cauchy-Reimann Condition

∂v ∂x
∂u ∂y

[−v_x− u_y] dxdy =

[u_y− u_y] dxdy = 0

The real part of _Cf(z)dz equal zero
4

3.2 The imaginary part

Consider the imaginary part of _Cf(z)dz ,

vdx + udy

Let F_Xbe v(x, y) and F_Ybe u(x, y) , and apply Stoke’s Theorem

∂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

f (z) dz = 0

4 The Cauchy Integral Formulas

4.1 Simple Pole

g (z) dz = 2πjg (z₀) z − z₀

Condition • z₀inside 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

, the integral of

By concept of path deformation z − z₀

along path C that ϵ → 0 along path C is equivalent to the integral

dz = lim

ϵ→0

The small circle inside can be expressed as

|z − z₀| = ϵe^jθ0 ≤ θ < 2π

As the radius ϵ → 0⁺, the circle is thus

z − z₀= ϵe^jθ0 ≤ θ < 2π

5
Put this back into the integral

g z₀+ ϵe^jθ

(z₀+ ϵe^jθ) − z₀

d z₀+ ϵe^jθ

g z₀+ ϵe^jθ

ϵe^jθ

ϵ→0

dz = 2πjg (z₀) z − z₀

4.2 Generalized Cauchy Integral Formula

(z − z₀)ⁿ

(n − 1)!

First, the Cauchy Integral Formula for simple pole is

g (z) dz = 2πjg (z₀) z − z₀

on both side Not

g^′(z₀)

(z − z₀)²

Repeat,

(z − z₀)²

g (z)
(z − z₀)³

g^′(z₀)

g” (z₀)

(z − z₀)³dz₀

g” (z₀)

⁽³⁾(z₀)

(z − z₀)⁴dz₀

Thus, the general form of Cauchy Integral Formula is

⁽ⁿ⁻¹⁾(z₀)

(z − z₀)ⁿ

(n − 1)!

This can be proved using Mathematical Induction.
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5 Consequences of Cauchy’s Integral Formula

5.1 Cauchy Inequality

The generalized Cauchy Integral Formula is

(z − z₀)ⁿ⁺¹

⁽ⁿ⁾(z₀)

Take absolute value

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Cauchy Integral Theorem

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