23 Cauchy Integral Formula II

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23 Cauchy Integral Formula II

Cauchy Integral Formula for general curves We have studied Cauchy’s Integral Formula on a circle:

f(z) = 1 2πi

I

∂∆(a;r)

f(ζ

ζ− zdζ, ∀z ∈ ∆(a; r).

One natural question is: instead of a circle ∂∆(a; r), if we consider any closed curve C inside D, can we have

f(z) = 1 2πi

I

C

f(ζ) ζ− zdζ?

where C is defined by a parametric function φ : [0, 1] → C.

This may not be true. See examples below.

[Example]

❅❅

■

(a) C is given by φ : [0, 2π], θ 7→ φ(θ) = a + e^iθ,

1 2πi

H

C dζ

ζ−a = _2πi¹ R2π 0

de^iθ

e^iθ = _2πi¹ R2π 0

ie^iθdθ

e^iθ = +1.

❅❅❘ (b) C is given by φ : [0, 2π], θ 7→ φ(θ) = a + e^−iθ,

1 2πi

H

C dζ

ζ−a = _2πi¹ R2π 0

de⁻^iθ

e⁻^iθ = _2πi¹ R2π

0 −ie⁻^iθdθ

e⁻^iθ = −1.

❅❅

■❅■❅

(c) C is given by φ : [0, 2π], θ 7→ φ(θ) = a + e^2iθ,

1 2πi

H

C dζ

ζ−a = _2πi¹ R2π 0

de^2iθ

e^2iθ = _2πi¹ R2π 0

2ie^2iθdθ

e^2iθ = +2.

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❅❅❘

❅❅❘ (d) C is given by φ : [0, 2π], θ 7→ φ(θ) = a + e^−2iθ,

1 2πi

H

C dζ

ζ−a = _2πi¹ R2π 0

de⁻²^iθ

e⁻²^iθ = −2.

Therefore, in general, we don’t have a formula such as f(z) = 1

2πi I

C

f(ζ) ζ− zdζ for any closed curve C.

Rotation number Think about f (z) = 1, from above, f (z) = _2πi¹ H

C f(ζ)

ζ−zdζ is always true, which depends on the parametric function of the curve.

Let a ∈ C and C be a closed curve that does not pass through a. The rotation number of C around a is defined by

n(a; C) := 1 2πi

I

C

dζ ζ− a which is an integer.

Remarks

1. In the above examples (a)(b)(c)(d), the rotation numbers of C are n(0; C) = 1, −1, 2, −2.

2. If C is a Jordan curve (no self-intersection point), it excludes the example (c) and (d) so that n(a; C) = ±1. If, in addition, C has counterclockwise orientation, then n(a; C) = 1; otherwise, n(a; C) = −1.

3. If a is outside of the closed curve C, then the rotation number n(0; C) = 0 by Cauchy’s Integral Theorem.

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❅❅❘

a

s

n(a; C)=_2πi¹ H

C dζ ζ−a = 0

4. On December 18, 1811, Gauss wrote to Bessel,

“If we define log x viaR ₁

xdxstarting at x = 1, then arrive at logx having gone around the point x = 0 one or more times or not at all, every circuit adding the constant +2πi or −2πi; thus the fact that every number has multiple logarithms becomes quite clear.” ⁷¹

General Cauchy Integral Formula

Theorem 23.1 (General Cauchy Integral Theorem) Let f be a holomorphic function in a domain D ⊂ C. Let C ⊂ D be a closed curve that is homotopic to a point. Let z ∈ D − C.

Then

n(z; C)f (z) = 1 2πi

I

C

f(ζ) ζ− zdζ.

In particular, if, in addition, C is a Jordan curve with counter-clockwise orientation, then f(z) = 1

2πi I

C

f(ζ) ζ− zdζ.

Proof: Fixing any z ∈ D − C. Let g(ζ) := f(ζ)

ζ− z − f(z)

ζ− z, ∀ζ ∈ D − {z}.

We claim: g(ζ) is holomorphic on D.

If ζ ∈ D with ζ 6= z, g(ζ) is holomorphic at ζ.

To show that g is holomorphic at z.

71R. Remmert, Theory of Complex Functions, Springer, 1991, p.168.

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Let us consider a disk ∆(z; r) ⊂ D and

g(ζ) = f(ζ)

ζ− z − f(z) ζ− z =

P_∞

n=0an(ζ − z)ⁿ− a⁰

ζ− z =

P_∞

n=1an(ζ − z)ⁿ

ζ− a =

∞

X

n=1

an(ζ − z)ⁿ⁻¹,

which is holomorphic at z. Claim is proved. Then applying Cauchy Integral Theorem, H

Cg(ζ)dζ = 0, i.e., I

C

f(ζ) ζ− zdζ −

I

C

f(z) ζ− zdζ =

I

C

f(ζ)

ζ− zdζ − n(z; C)f(z) = 0. The mean value equality and inequality

Theorem 23.2 Let f be a holomorphic function in a domain D ⊂ C with a disk ∆(a, r) ⊂⊂

D. Then

f(a) = 1 2π

Z 2π 0

f(a + re^iθ)dθ.

Consequently,

|f(a)| ≤ sup

∂∆(a,r)|f|.

Proof: By Cauchy integral formula, f(a) = 1

2πi I

∂∆(a,r)

f(ζ)

ζ− adζ = 1 2πi

Z 2π 0

f(a + re^iθ)

(a + re^iθ) − aire^iθdθ = 1 2π

Z 2π 0

f(a + re^iθ)dθ.

Morera’s theorem A complex-valued function f is holomorphic if and only if f is analytic; if and only if f is C¹-smooth satisfying the Cauchy-Riemann equation. Now we have one more equivalent definition.

Theorem 23.3 (Morera’s theorem) Let f be a continuous complex-valued function on a domain D ⊂ C. Then

f is holomorphic ⇐⇒

I

∂∆

f(z)dz = 0, for any triangle ∆ ⋐ D.

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Proof: (=⇒) By Cauchy-Goursat Theorem.

(⇐=) ∀a ∈ D, to show: f is a holomorphic function in ∆(a; r⁰) ⋐ D.

For any point z ∈ ∆(a; r0), we denote by Cz the line segment from a to z.

Cz

s

a

✒ z

❅❅

Γ ■ s

z+ h Let F (z) :=R

Czf(ζ)dζ.

F(z+h)−F (z)

h =

R

Cz+hf −R

Czf h

=

R

Γf

h → f(z) as h → 0.

Since F^′ = f , f is holomorphic in ∆(a; r0).

[Example] Let T be a triangle inside a domain D ⊂ C and w be a point inside this triangle. Let f be a function defined in D and be holomorphic in D − {w}. If f is bounded near w, prove that

I

T

f(z)dz = 0.

Proof: Consider a disk ∆(w, ǫ₀) inside the triangle T . By connecting T with ∂∆(w, ǫ) with a line segment L, we consider a closed curve Tǫ := T + L − ∂∆(w, ǫ) − L. Then by applying Cauchy Integral Theorem, H

Tǫf(z)dz = 0, i.e., I

T

f(z)dz = I

∂∆(w,ǫ)

f(z)dz = lim

ǫ→0

I

∂∆(w,ǫ)

f(z)dz.

Since f is bounded near w, we have |f(z)| ≤ M for any z ∈ ∆(w, ǫ) for any 0 < ǫ ≤ ǫ0 for some ǫ0. Then

| I

∂∆(w,ǫ)

f(z)dz| ≤ M I

∂∆(w,ǫ)|dz| = 2πǫM → 0 as ǫ → 0. Therefore, we obtain

I

T

f(z)dz = lim

ǫ→0

I

∂∆(w,ǫ)

f(z)dz = 0.

In fact by the above argument, we nearly prove the following result.

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Appendix: The Fresnel Integral The integrals RR

0 cost t²dtand RR

0 sin t²dtare called Fresnel integrals after Augustin-Jean Fresnel that are used in optics, which are closely related to the error function ⁷²

Theorem 23.4 For any a ∈ R with |a| ≤ 1, Z ∞

0

e^−(1+ia)²^t²dt= 1 2· 1

1 + ia

√π.

Splitting into the real part and the imaginary part, it gives Z _∞

0

e^(a²^−1)t²cos(2at²)dt = 1 2(1 + a²)

√π,

Z ∞ 0

e^(a²^−1)t²sin(2at²)dt = a 2(1 + a²)

√π, −1 ≤ a ≤ 1.

In particular, when a = 1, we obtain Z ∞

0

cos t²dt= Z ∞

0

sin t²dt= 1 2

r π 2.

Proof: When a = 0, R∞

0 e^−t²dt= ^√₂^π. It suffices to show: I :=R∞

−∞e^−t²dt=√

π, which is so-called Gauss error integral.

Using the polar coordinates x = r cos φ and y = r sin φ, the double integral is evaluated I² =

Z _∞

−∞

e^−x²dx Z _∞

−∞

e^−y²dy= Z Z

R²

e^−x²^−y²dxdy

= Z 2π

0

Z _∞

0

e^−r²rdrdφ= π Z _∞

0

e^−tdt= π.

Assume 0 < a ≤ 1 here. Define f(z) := e^−z².

For a triangle determined by the three points: 0, r and r + ari. Denote by γ1, γ2, γ3 the line segments from the origin to r, from r to r + ira, from the origin to r + ira, respectively.

Then γ₁+ γ₂− γ3 forms a closed curve with counter-clockwise orientation.

72A.J. Fresnel (1788-1827) is a French engineer and physicist.

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By Cauchy Integral Theorem, Z

γ3

f(ζ)dζ = Z

γ1

f(ζ)dζ + Z

γ2

f(ζ)dζ.

We claim that lim

r→∞

R

γ2f(ζ)dζ) = 0. In fact, γ2(t) = r + it for t ∈ [0, ar]. Then

|f(γ²(t))| = |e^−(r+it)²| ≤ e^−r²^+t² ≤ e^−r² · e^rt f or 0 ≤ t ≤ r so that

| Z

γ2

f(ζ)dζ| ≤ Z ar

0 |f(γ²(t))||γ^′2(t)|dt ≤ e^−r² Z r

0

e^rtdt≤ 1 r. Our claim is proved.

Write γ3(t) = (1 + ia)t for t ∈ [0, r]. Then γ^′3(t) = 1 + ia and (1 + ia)

Z _∞

0

e^−(1+ia)²^t²dt= lim

r→∞

Z

γ3

f(ζ)dζ = lim

r→∞

Z

γ1

f(ζ)dζ = Z _∞

0

e^−t²dt= π 2.