EIGHT EQUIVALENT PROPERTIES/DEFINITIONS OF ANALYTICFUNCTIONS

(AS PRESENTED IN MY UNLV'S COURSE MAT 709 - COMPLEXFUNCTION THEORY I)

ANGEL S. MULESHKOV

Introduction

The following text is a preliminary version of a chapter in a textbook of Complex Analysis.It needs a lot more work and polishing, but it is made available for students of ComplexAnalysis. Since I believe that the idea of proving the equivalence of the eight de�nitions wasprobably known to the classics of Complex Analysis, this text is not intended as a researchpublication, but only as an original teaching tool that reveals the rich and beautiful natureof analytic functions.

The following eight properties are equivalent and a function f(z) is called analytic, if itsatis�es at least one of these properties. Proving that each of the eight properties implies theother seven is equivalent to proving 56 theorems. The minimal number of theorems neededis eight. However, we succeeded to achieve this by proving twelve theorems, illustrated bythe following chart:

Figure 1.

1

2

The Eight PropertiesThere are two versions for each of the eight properties: One for being analytic at a point

and one for being analytic in a domain. The version for a point will be denoted with ∗ andthe version for a domain will be denoted without ∗.Also, property H has two additionalversions denoted H and H

∗.

Property A*: For a function f(z) and a point z0 in some domain D, there is aneighborhood about point z0 such that

f(z) =+∞∑n=0

an(z − z0)n

for all z in this neighborhood and some sequence {an} of complex numbers.

Property A: For a function f(z) in a domain D, if for every point z0 ∈ D, there isa neighborhood about z0 in D such that

f(z) =+∞∑n=0

an(z − z0)n

for all z in this neighborhood and for some sequence {an} of complex numbers.

Property B*: For a function f(z) and a point z0, the derivative f ′(z) exists insome neighborhood about point z0.

Property B: For a function f(z) in a domain D, the derivative f ′(z) exists every-where in the domain D.

Property C*: For a function f(z) and a point z0, the k − th derivative (∀k ∈ N),f (k)(z), exists in some neighborhood about point z0.

Property C: For a function f(z) in a domain D, the k − th derivative (∀k ∈ N),f (k)(z), exists everywhere in the domain.

Property D*: (Cauchy Theorem) For a given point z0 and every closed piece-wisesmooth contour C in a neighborhood of point z0,

�Cf(z)dz = 0. Also, f(z) is continuous in

this neighborhood.

Property D: (Cauchy Theorem) For every closed piece-wise smooth contour C ina given domain D,

¸Cf(z)dz = 0. Also, f(z) is continuous in D.

Property E*: (Cauchy's Integral Formula) For a given point z0 and any closedpiece-wise smooth contour C in a neighborhood of point z0, as z0 is inside C,

‰C

f(z)

z − z0dz = 2πif(z0)

3

Property E: (Cauchy's Integral Formula) For any closed piece-wise smooth contourC in some domain D and any point z inside of C,

‰C

f(ζ)

ζ − zdζ = 2πif(z)

Property F*: (Independence of Path) If f(z) is continuous in some neighborhoodof z0 containing z1 and z2 and Γ is any piece-wise smooth line in this neighborhood thatconnects z1 and z2. Then

´ z2z1f(z)dz does not depend on Γ.

Property F: (Independence of Path) If f(z) is continuous in domain D and for anyz1, z2 ∈ D, as Γ ⊂ D is any piece-wise smooth line that connects z1 and z2, then

´ z2z1f(z)dz

does not depend on Γ.

Property G*: (Cauchy-Riemann condition) For a given function f(z), z = x + iyand a point z0 = x0 + iy0, let u(x, y) = Ref(z) and v(x, y) = Imf(z). Then u(x, y) andv(x, y) satisfy:

∂u

∂x(x0, y0) =

∂v

∂y(x0, y0),

∂u

∂y(x0, y0) = −

∂v

∂x(x0, y0)

Also, u, v, ∂u∂x, ∂u∂y, ∂v∂x, and ∂v

∂yare continuous at point (x0, y0).

Property G: (Cauchy-Riemann condition) For a given function f(z), z = x+iy anda domain D, let u(x, y) = Ref(z) and v(x, y) = Imf(z). Then u(x, y) and v(x, y) satisfy:

∂u

∂x=∂v

∂y,∂u

∂y= −∂v

∂x

for every point in D. Also, u, v, ∂u∂x, and ∂u

∂yare continuous in D.

Property H*: For a function f(z) = u(x, y) + iv(x, y) and a given point z0 in adomain D where u(x, y) is an arbitrary harmonic function in a neighborhood of point z0(i.e. u satis�es the Laplace PDE, ∂

2u∂x2

+ ∂2u∂y2

= 0, in this neighborhood) and u(x, y) andits �rst and second partial derivatives are continuous in this neighborhood, there exists aharmonic function v(x, y) such that ∂u

∂x= ∂v

∂yand ∂u

∂y= − ∂v

∂xin this neighborhood. If the

value v(x0, y0) = v0 is given for some point (x0, y0) in this neighborhood, the function v isunique. Otherwise there are in�nitely many v with this property as the di�erence of any twoof them is a constant.Property H: For a function f(z) = u(x, y) + iv(x, y) in a domain D where u(x, y)

is an arbitrary harmonic function in D (i.e. u satis�es the Laplace PDE, ∂2u∂x2

+ ∂2u∂y2

= 0, in

D) and u(x, y) and its �rst and second partial derivatives are continuous in D, there exists aharmonic function v(x, y) such that ∂u

∂x= ∂v

∂yand ∂u

∂y= − ∂v

∂xin D. If the value v(x0, y0) = v0

is given for some point (x0, y0) ∈ D, the function v is unique. Otherwise there are in�nitely

4

many v with this property as the di�erence of any two of them is a constant.

Property H∗: For a function f(z) = u(x, y) + iv(x, y) and a given point z0 in a

domain D where v(x, y) is an arbitrary harmonic function in a neighborhood of point z0(i.e. v satis�es the Laplace PDE, ∂

2v∂x2

+ ∂2v∂y2

= 0, in this neighborhood) and v(x, y) andits �rst and second partial derivatives are continuous in this neighborhood, there exists aharmonic function u(x, y) such that ∂v

∂x= −∂u

∂yand ∂v

∂y= ∂u

∂xin this neighborhood. If the

value u(x0, y0) = u0 is given for some point (x0, y0) in this neighborhood, the function u isunique. Otherwise there are in�nitely many u with this property as the di�erence of anytwo of them is a constant.

Property H: For a function f(z) = u(x, y) + iv(x, y) in a domain D where v(x, y)is an arbitrary harmonic function (i.e. v satis�es the Laplace PDE, ∂

2v∂x2

+ ∂2v∂y2

= 0, in D)

and v(x, y) and its �rst and second partial derivatives are continuous in D, there exists aharmonic function u(x, y) such that ∂v

∂x= −∂u

∂yand ∂v

∂y= ∂u

∂xin D. If the value u(x0, y0) = u0

is given for some point (x0, y0) ∈ D, the function u is unique. Otherwise there are in�nitelymany u with this property as the di�erence of any two of them is a constant.

5

Some De�nitions and Lemmas

De�nition 1. (big O) For given functions f(z) and g(z),(1) f(z) = O(g(z)) as z → z0 if and only if ∃c > 0 and ∃δ > 0 such that ∀z (for which

f(z) and g(z) are de�ned), |z − z0| < δ =⇒ |f(z)| ≤ c|g(z)|.(2) f(z) = O(g(z)) as z → ∞ if and only if ∃c > 0 and ∃δ > 0 such that ∀z (for which

f(z) and g(z) are de�ned) satisfying |z| > 1δ, |f(z)| ≤ c|g(z)|.

�

De�nition 2. (small o) For given functions f(z) and g(z),(1) f(z) = o(g(z)) as z → z0 if and only if ∀� > 0, ∃δ(�) > 0 such that ∀z (for which f(z)

and g(z) are de�ned), |z − z0| < δ =⇒ |f(z)| ≤ �|g(z)|.(2) f(z) = o(g(z)) as z →∞ if and only if ∀� > 0, ∃δ(�) > 0 such that ∀z (for which f(z)

and g(z) are de�ned) satisfying |z| > 1δ, |f(z)| ≤ �|g(z)|.

�

De�nition 3. (Contour Integral)(1) A smooth curve Ck is de�ned as

Ck =

{x = x(t)

y = y(t)(1)

where t belongs to some interval Ik for which ẋ(t) and ẏ(t) exist and are not both zero.(2) A piecewise smooth curve C is de�ned as

C =n⋃k=1

Ck (2)

where Ck are smooth and connected curves and some n ∈ N. Also,ˆC

f(z)dz =n∑k=1

ˆck

f(z)dz (3)

(3) For z = x+ iy =⇒ z(t) = x(t) + iy(t) (for t belonging to some interval Ik) and ż(t) =ẋ(t) + iẏ(t) where ż(t) 6= 0 everywhere in Ik, the contour integral of f(z) = u(x, y) + iv(x, y)(dz = dx+ idy = (ẋ(t) + ẏ(t))dt) on a piecewise smooth contour C is de�ned as

ˆC

f(z)dz =n∑k=1

ˆCk

f(z)dz =n∑k=1

ˆCk

(u+iv)(dx+idy) =n∑k=1

( ˆCk

udx−vdy+iˆCk

vdx+udy

)(4)

as the last two integrals in (4) are line integrals of the second kind and

ˆCk

f(z)dz =

ˆCk

udx− vdy + iˆCk

vdx+ udy (5)

=

ˆIk

(u(x(t), y(t))ẋ(t)−v(x(t), y(t))ẏ(t))dt+iˆIk

(v(x(t), y(t))ẋ(t)+u(x(t), y(t))ẏ(t))dt (6)

as the integrals in (6) are de�nite single integrals.

6

Evidently, the following three properties hold

ˆC

(f(z) + g(z))dz =

ˆC

f(z)dz +

ˆC

g(z)dz (7)

ˆC

Af(z)dz = A

ˆC

f(z)dz (8)

ˆC?f(z)dz = −

ˆC

f(z)dz (9)

for any piecewise smooth curve C, as C? is the same curve as C but with opposite directionand any A ∈ C.

Lemma 1. If the derivative of some function f(z) exists at some point z0 (i.e. f′(z0) exists),

then f(z) is continuous at point z = z0.

Proof. Since f ′(z0) exists,

f ′(z0) = limz→z0

f(z)− f(z0)z − z0

= f ′(z0) =⇒f(z)− f(z0)

z − z0− f ′(z0) = o(1)

=⇒ f(z)− f(z0) = (z − z0)(f ′(z0) + o(1)) (10)

As z − z0 → 0, f ′(z0) + o(1)→ f ′(z0). Thus,

f(z)− f(z0) = o(1) =⇒ limz→z0

f(z) = f(z0) (11)

Hence, f(z) is continuous at z = z0. �

Lemma 2. For a given function f(z) in a domain D, let C be a closed piecewise smoothcurve in D, then

˛C

dz = 0 and

˛C

zdz = 0

Proof. Since C is a piecewise smooth curve,

C =n⋃l=1

Cl

where Cl are smooth (1 ≤ l ≤ n).Let zl−1and zl be the end points of Cl (1 ≤ l ≤ n). Since C is closed, z0 = zn.Also let zl = xl+ iyl = z(tl) for (0 ≤ l ≤ n) on Cl. Then z = zl(t) = xl(t)+ iyl(t) and dz =

dzl = ẋl(t)dt+ iẏl(t)dt. Also, zdz = xl(t)ẋl(t)dt− yl(t)ẏl(t)dt+ ixl(t)ẏl(t)dt+ iyl(t)ẋl(t)dt.

7

˛C

dz =n∑l=1

ˆCl

dz =n∑l=1

(

ˆ tltl−1

ẋl(t)dt+ i

ˆ tltl−1

ẏl(t)dt)

=n∑l=1

[xl(t)

∣∣∣∣tltl−1

+ iyl(t)

∣∣∣∣tltl−1

]=

n∑l=1

[xl(tl)− xl(tl−1) + iyl(tl)− iyl(tl−1)

]

=n∑l=1

[xl + iyl − (xl−1 + iyl−1)

]=

n∑l=1

zl −n∑l=1

zl−1

=n−1∑l=1

zl + zn −n−1∑l=1

zl − z0 = zn − z0 = 0 (12)

In a similar fashion,

˛C

zdz =n∑l=1

ˆCl

zdz =n∑l=1

[ˆ tltl−1

x(t)ẋ(t)dt−ˆ tltl−1

y(t)ẏ(t)dt+ i

ˆ tltl−1

[x(t)ẏ(t) + ẋ(t)y(t)]dt

]

=n∑l=1

[1

2x2(t)

∣∣∣∣tltl−1

− 12y2(t)

∣∣∣∣tltl−1

+ ix(t)y(t)

∣∣∣∣tltl−1

]

=n∑l=1

[1

2x2l −

1

2x2l−1 −

1

2y2l +

1

2y2l−1 + ixlyl − ixl−1yl−1

]

=n∑l=1

[(xl + iyl)

2

2− (xl−1 + iyl−1)

2

2

]=

1

2

n∑l=1

z2l −1

2

n∑l=1

z2l−1

=1

2

n−1∑l=1

z2l +1

2z2n −

1

2

n−1∑l=1

z2l −1

2z20 =

1

2z2n −

1

2z20 = 0 (13)

�

Lemma 3. (ML Inequality) For an arbitrary piecewise smooth curve C, where f(z) isbounded on C, i.e. |f(z)| ≤M for every z on C and for some M ≥ 0, then∣∣∣∣ˆ

C

f(z)dz

∣∣∣∣ ≤ML (14)where L is the arc length of C.

Proof. From the de�nition in (6), for t1 < t2∣∣∣∣ ˆC

f(z)dz

∣∣∣∣ = ∣∣∣∣ ˆ t2t1

(u(x(t), y(t))ẋ(t)− v(x(t), y(t))ẏ(t))dt∣∣∣∣ (15)

≤ˆ t2t1

√u2(x(t), y(t)) + v2(x(t), y(t))

√ẋ2(t) + ẏ2(t)dt (16)

≤ Mˆ t2t1

√ẋ2(t) + ẏ2(t)dt = ML (17)

8

�

ProofsHere, only the theorems that involve the properties without ? or − are given and proved.

The theorems that involve the properties with ? or − and their proofs are almost identicalto the ones given below as the domain D is replaced by a neighborhood of a given point z0.

Theorem 1. (D =⇒ F ) If for every closed piecewise smooth contour C in a given domainD, f(z) is continuous in D and

�Cf(z)dz = 0, then for arbitrary points z1, z2 ∈ D and any

piecewise smooth curves C1 and C2 from z1 to z2,´C1f(z)dz =

´C2f(z)dz (independence of

path.)

Proof. Let C1 and C2 be arbitrary piecewise smooth curves in domain D from abitrary pointz1 to z2 (both in D.)

Figure 2.

De�ne C?2 to be the same curve as C2 but starting at z2 and ending at z1 and C = C1∪C?2which is a piecewise smooth contour in D.Thus,

0 =

‰C

f(z)dz =

ˆC1

f(z)dz +

ˆC?2

f(z)dz =

ˆC1

f(z)dz −ˆC2

f(z)dz (18)

Hence,ˆC1

f(z)dz =

ˆC2

f(z)dz (19)

�

Theorem 2. (B =⇒ E) For a given function f(z) in a domain D, if the derivative off(z), f ′(z), exists everywhere in the domain D, then for any arbitrary piecewise smooth

closed contour C in D that contains point z,�Cf(ξ)ξ−z dξ = 2πif(z).

9

Proof.For�C1

dξξ−z , let C1 be the circle parameterized by ξ = z + �e

it =⇒ dξ = i�eitdt fort ∈ [0, 2π) and � > 0 is small enough such that C1 is contained completely in C. Since 1ξ−zis a di�erentiable function of ξ in a domain containing C, C1, and the region between C andC1

‰C

f(ξ)

ξ − zdξ =

�C1

f(ξ)ξ−z dξ =

‰C1

f(ξ)− f(z) + f(z)ξ − z

dξ (20)

=

‰C1

f(ξ)− f(z)ξ − z

dξ +

‰C1

f(z)

ξ − zdξ (21)

=

‰C1

f(ξ)− f(z)ξ − z

dξ + f(z)

‰C1

dξ

ξ − z(22)

Let

I1 =

‰C1

f(ξ)− f(z)ξ − z

dξ, I2 =

‰C1

dξ

ξ − z(23)

Since

limξ→z

f(ξ)− f(z)ξ − z

= f ′(z) =⇒ f(ξ)− f(z)ξ − z

= f ′(z) + o(1), (24)

I1 =

‰C1

(f ′(z) + o(1))dξ =

‰C1

f ′(z)dξ +

‰C1

o(1)dξ (25)

= f ′(z)

‰C1

dξ +

‰C1

o(1)dξ (26)

By Lemma 2,

I1 = f′(z)

‰C1

dξ +

‰C1

o(1)dξ = f ′(z)(0) + o(1) = o(1) (27)

I2 =

‰C1

dξ

ξ − z=

‰C1

i�eit

�eitdt =

ˆ 2π0

idt = 2πi (28)

Thus,

‰C1

f(ξ)

ξ − zdξ = I1 + f(z)I2 = 2πif(z) (29)

Therefore, ‰C1

f(ξ)

ξ − zdξ = 2πif(z) (30)

�

Theorem 3. (E =⇒ C) For a given function f(z) in a domain D, if for any arbitrarypiecewise smooth closed contour C in D and z inside C,

�Cf(ξ)ξ−z dξ = 2πif(z), then the n

th

derivative of f(z), f (n)(z) (∀n ∈ N), exists everywhere in D.

10

Proof. It will be shown by mathematical induction that, if f(z) = 12πi

�Cf(ξ)ξ−z dξ, then

f (n)(z) =n!

2πi

‰C

f(ξ)

(ξ − z)n+1dξ (n ∈ N) (31)

For n = 1

limh→0

f(z + h)− f(z)h

= limh→0

1

2πih

( ‰C

f(ξ)

ξ − z − hdξ −

‰C

f(ξ)

ξ − zdξ)

= limh→0

1

2πih

‰C

f(ξ)(ξ − z)− f(ξ)(ξ − z − h)(ξ − z − h)(ξ − z)

dξ

= limh→0

1

2πi

‰C

f(ξ)

(ξ − z − h)(ξ − z)dξ = lim

h→0

1

2πi

‰f(ξ)

(ξ − z)2(1− hξ−z )

dξ (32)

Since hξ−z = O(h),

hξ−z = o(1)

Using the geometric series

1

1− hξ−z

=1

1−O(h)= 1 +O(h) = 1 + o(1) (33)

Thus,

limh→0

f(z + h)− f(z)h

= limh→0

1

2πi

‰C

f(ξ)

(ξ − z)2(1 +O(h))dξ = lim

h→0

1

2πi

‰C

f(ξ)

(ξ − z)2(1 + o(1))dξ

=1

2πi

‰C

f(ξ)

(ξ − z)2dξ (34)

Hence,

f ′(z) =1!

2πi

‰C

f(ξ)

(ξ − z)2dξ (35)

exists.For n = k, assume f (k)(z) = k!

2πi

�C

f(ξ)(ξ−z)k+1dξ.

Then, using the quotient rule for f(ξ)(ξ−z)k+1 , one gets

f (k+1)(z) = (f (k)(z))′ =k!

2πi

‰C

∂

∂z

( f(ξ)(ξ − z)k+1

)dξ (36)

=k!(k + 1)

2πi

‰C

f(ξ)

(ξ − z)k+2dξ =

(k + 1)!

2πi

‰C

f(ξ)

(ξ − z)(k+1)+1dξ (37)

Thus, by mathematical induction, the nth derivative of f(z) exists for n ∈ N and

f (n)(z) =n!

2πi

‰C

f(ξ)

(ξ − z)n+1dξ (38)

�

Theorem 4. (F =⇒ B) If f(z) is a continuous function in a given domain D and for anytwo points z1, z2 ∈ D, the integral

´ z2z1f(z)dz doesn't depend on the curve in D from z1 to z2

(path independence), then f ′(z) exists everywhere in the domain D.

11

Proof. Fixing a point a ∈ D and letting

g(z) =

ˆC

f(ξ)dξ =

ˆ za

f(ξ)dξ (39)

for some contour C ⊂ D that connects a and z.Since

´ zaf(ξ)dξ does not depend on the path,

g(z) =

ˆ za

f(ξ)dξ (40)

where C is chosen to be the line segment connecting a to z.Thus,

g′(z) = limh→0

g(z + h)− g(z)h

= limh→0

1

h

( ˆ z+ha

f(ξ)dξ −ˆ za

f(ξ)dξ) (41)

= limh→0

1

h

ˆ z+hz

f(ξ)dξ (42)

Let ξ = z + ht for t ∈ [0, 1], so dξ = hdt, then

g′(z) = limh→0

1

h

ˆ 10

f(z + ht)hdt = limh→0

ˆ 10

f(z + ht)dt (43)

Since f(z) is continuous in D, f(z + ht) = f(z + o(1)) = f(z) + o(1).Thus,

g′(z) = limh→0

ˆ 10

(f(z) + o(1))dt = f(z) (44)

Hence, g(z) has a derivative everywhere in domain D and g′(z) = f(z). Hence, g(z)satis�es property B in domain D. It has been shown earlier that B =⇒ E, so g(z) satis�esproperty E. It has also been shown that E =⇒ C. Thus, g(z) satis�es property C. Hence,g(z) has a second derivative g′′(z) and by de�nition g′′(z) = f ′(z). Hence, f ′(z) exists. �

Theorem 5. (B =⇒ G) For a given function f(z) in a domain D, if the derivative off(z), f ′(z), exists everywhere in the domain D, then the Cauchy-Riemann conditions hold forf(z) = u(x, y) + iv(x, y), i.e. ∂u

∂x= ∂v

∂yand ∂u

∂y= − ∂v

∂xin the domain D and u, v, ∂u

∂x, ∂u∂y, ∂v∂x, ∂v∂y

are continuous in D.

Proof. Since f ′(z) exists in D, f(z) is continuous in D according to Lemma 1. Thus, u(x, y)and v(x, y) are continuous in D. It has been shown that B =⇒ E and E =⇒ C, soB =⇒ C. Thus f ′′(z) exists and f ′(z) is continuous. Thus, ∂u

∂x, ∂u∂y, ∂v∂x, ∂v∂y

are all continuous.

f ′(z) is calculated at an arbitrary point z0 = x0 + iy0 ∈ D in two di�erent ways.Horizontally, keeping y = y0 constant and letting x→ x0, z = x0 +h+ iy0 as h→ 0. Then

f ′(z0) = limh→0

f(z)− f(z0)z − z0

(45)

= limh→0

u(x0 + h, y0) + iv(x0 + h, y0)− u(x0, y0)− iv(x0, y0)h

(46)

12

= limh→0

u(x0 + h, y0)− u(x0, y0)h

+ i limh→0

v(x0 + h, y0)− v(x0, y0)h

=∂u

∂x(x0, y0) + i

∂v

∂x(x0, y0)

(47)Vertically, keeping x = x0 constant and letting y → y0, z = x0 + i(y0 + h) as h→ 0. Then

f ′(z0) = limh→0

f(z)− f(z0)z − z0

= limh→0

u(x0, y0 + h) + iv(x0, y0 + h)− u(x0, y0)− iv(x0, y0)ih

(48)

= −i limh→0

u(x0, y0 + h)− u(x0, y0)h

+limh→0

v(x0, y0 + h)− v(x0, y0)h

= −i∂u∂y

(x0, y0)+∂v

∂y(x0, y0)

(49)By equating the real and imaginary parts of (0.39) and (0.41), one gets{

∂u∂x

= ∂v∂y

∂u∂y

= − ∂v∂x

(50)

for an arbitrary point z0 in D. �

Theorem 6. (C =⇒ B) If the kth derivative (∀k ∈ N) of a function f(z), f (k)(z), existseverywhere in a domain D, then the �rst derivative f ′(z) exists everywhere in D.

Proof. This is given for k = 1 �

Theorem 7. (G =⇒ H) For a given function f(z) = u(x, y)+iv(x, y) in a domain D whereu(x, y) and v(x, y) satisfy the Cauchy-Riemann condition (i.e. ∂u

∂x= ∂v

∂yand ∂u

∂y= − ∂v

∂x) and

u(x, y) and its �rst and second partial derivatives are continuous, then u(x, y) and v(x, y)are harmonic (i.e. u(x, y) and v(x, y) satisfy the Laplace PDE).

Proof. From the continuity of u(x, y) and its �rst and second partial derivatives, one gets

4u = ∂2u

∂x2+∂2u

∂y2=

∂

∂x

(∂u

∂x

)+

∂

∂y

(∂u

∂y

)=

∂

∂x

(∂v

∂x

)+

∂

∂y

(− ∂v∂y

)=

∂2v

∂x∂y− ∂

2v

∂y∂x= 0

(51)

4v = ∂2v

∂x2+∂2v

∂y2=

∂

∂x

(∂v

∂x

)+∂

∂y

(∂v

∂y

)=

∂

∂x

(− ∂u∂y

)+∂

∂y

(∂u

∂x

)= − ∂

2u

∂x∂y+

∂2u

∂y∂x= 0

(52)�

Theorem 8. (G =⇒ B) For a function f(z) = u(x, y) + iv(x, y) in a given domain D, ifu(x, y) and v(x, y) satisfy the Cauchy-Riemann condition (i.e. ∂u

∂x= ∂v

∂yand ∂u

∂y= − ∂v

∂x) and

u,v,∂u∂x, ∂u∂y, ∂v∂x, and ∂v

∂yare all continuous everywhere in D, then f ′(z) exists everywhere in D.

Proof. One needs to prove that the directional derivative exists and doesn't depend on thedirection.Let z = x+ iy and ∆z = ρeiθ(so ∆z is in the direction of θ). Then, for �xed θ,

f ′(z) = lim∆z→0

f(z + ∆z)− f(z)∆z

= limρ→0

f(z + ρeiθ)− f(z)ρeiθ

(53)

13

= limρ→0

u(x+ ρ cos θ, y + ρ sin θ) + iv(x+ ρ cos θ, y + ρ sin θ)− u(x, y)− iv(x, y)ρeiθ

(54)

= e−iθ limρ→0

u(x+ ρ cos θ, y + ρ sin θ)− u(x, y)ρ

+ ie−iθ limρ→0

v(x+ ρ cos θ, y + ρ sin θ)− v(x, y)ρ

(55)

= e−iθ(∂u

∂~θ

)+ ie−iθ

(∂v

∂~θ

)= e−iθ

(∂u

∂~θ+ i

∂v

∂~θ

)= e−iθ

(Dθu(x, y) + iDθv(x, y)

)(56)

where Dθf(x, y) denotes the directional derivative of f(x, y) in the direction of θ. To provethe existence of f ′(z), one needs to show that the result in Eq.(46) does not depend on θ.Indeed, since u(x, y) and v(x, y) are continuous and have continuous �rst partial deriva-

tives,

e−iθ(Dθu(x, y) + iDθv(x, y)

)= e−iθ

(∂u

∂xcos θ +

∂u

∂ysin θ + i

∂v

∂xcos θ + i

∂v

∂ysin θ

)(57)

= e−iθ(∂u

∂xcos θ−∂v

∂xsin θ+i

∂v

∂xcos θ+i

∂u

∂xsin θ

)= e−iθ

(∂u

∂x(cos θ+i sin θ)+

∂v

∂x(i cos θ−sin θ)

)(58)

= e−iθ(∂u

∂x(cos θ+ i sin θ) + i

∂v

∂x(cos θ+ i sin θ)

)= e−iθ

(∂u

∂xeiθ + i

∂v

∂xeiθ)

=∂u

∂x+ i

∂v

∂x(59)

Thus,

f ′(z) =∂u

∂x+ i

∂v

∂x(60)

which does not depend on θ. �

Theorem 9. (H =⇒ G) For an arbitrary harmonic function u(x, y) in a domain D (i.e.∂2u∂x2

+ ∂2u∂y2

= 0 in D) such that u and its �rst and second partial derivatives are continuous in

D, there exists another harmonic function v(x, y) which is conjugate harmonic of u(x, y) inD (i.e. ∂u

∂x= ∂v

∂yand ∂u

∂y= − ∂v

∂xin D). v(x, y) is harmonic and continuous in D. The function

v(x, y) is not unique. However, assigning a value at a given point (i.e. v(x0, y0) = v0) makesit unique.

Proof. For some point z0 = x0 + iy0 in D, let C and C′ be any two di�erent piecewise smooth

contours that connect point (x0, y0) with point (x, y) ∈ D. Let C? be the same contour asC ′ but with opposite direction, S = C

⋃C?, P = −∂u

∂y, and Q = ∂u

∂x. Then, according to the

Green's theorem,

‰S

−∂u∂ydx+

∂u

∂xdy =

¨D?

(∂2u

∂x2+∂2u

∂y2)dxdy =

¨D?

0dxdy = 0 (61)

where D? is the domain enclosed by C and C ′.On the other hand,

14

‰S

−∂u∂ydx+

∂u

∂xdy =

ˆC

−∂u∂ydx+

∂u

∂xdy+

ˆC?−∂u∂ydx+

∂u

∂xdy =

ˆC

−∂u∂ydx+

∂u

∂xdy−ˆC′−∂u∂ydx+

∂u

∂xdy

(62)Thus, from Eqs.(51) and (52),

ˆC

−∂u∂ydx+

∂u

∂xdy =

ˆC′−∂u∂ydx+

∂u

∂xdy (63)

so

v(x, y) =

ˆC

−∂u∂ydx+

∂u

∂xdy (64)

Figure 3.

Since v(x, y) does not depend on the curve C, special curve C as given in Fig. 3 isconsidered. Then

v(x, y) =

ˆC

−∂u∂ydx+

∂u

∂xdy =

ˆ (x,y0)(x0,y0)

(−∂u∂ydx+

∂u

∂xdy) +

ˆ (x,y)(x,y0)

(−∂u∂ydx+

∂u

∂xdy) (65)

= −ˆ xx0

∂u

∂ydx+

ˆ yy0

∂u

∂xdy (66)

Hence,

∂v

∂x(x, y) =

∂

∂x

(−ˆ xx0

∂u

∂ydx+

ˆ yy0

∂u

∂xdy

)= −∂u

∂y(x, y0) +

ˆ yy0

∂2u

∂x2dy (67)

= −∂u∂y

(x, y0)−ˆ yy0

∂2u

∂y2dy = −∂u

∂y(x, y0)−

∂u

∂y

∣∣∣∣yy0

= −∂u∂y

(x, y0)−∂u

∂y(x, y) +

∂u

∂y(x, y0) (68)

Thus,

∂v

∂x(x, y) = −∂u

∂y(x, y) (69)

15

Figure 4.

Next, considering another contour C ′ as given in Fig. 4, one obtains

v(x, y) =

ˆC

−∂u∂ydx+

∂u

∂xdy =

ˆ (x0,y)(x0,y0)

−∂u∂ydx+

∂u

∂xdy +

ˆ (x,y)(x0,y)

−∂u∂ydx+

∂u

∂xdy (70)

=

ˆ yy0

∂u

∂xdy −

ˆ xx0

∂u

∂ydx (71)

Hence,

∂v

∂y(x, y) =

∂

∂y

( ˆ yy0

∂u

∂xdy −

ˆ xx0

∂u

∂ydx

)=∂u

∂x(x0, y)−

ˆ xx0

∂2u

∂y2dx (72)

=∂u

∂x(x0, y) +

ˆ xx0

∂2u

∂x2dx =

∂u

∂x(x0, y) +

∂u

∂x

∣∣∣∣xx0

=∂u

∂x(x0, y) +

∂u

∂x(x, y)− ∂u

∂x(x0, y) (73)

Thus,

∂v

∂y=∂u

∂x(74)

Hence, v(x, y) is conjugate harmonic of u(x, y).Also,

∂2v

∂x2+∂2v

∂y2=

∂

∂x

(∂v

∂x

)+

∂

∂y

(∂v

∂y

)= − ∂

2u

∂x∂y+

∂2u

∂y∂x= 0 (75)

as u(x, y) has continuous second partial derivatives. Thus, v(x, y) satis�es the Laplace PDEand is harmonic.Evidently adding a real constant to v(x, y) yields another function v1(x, y) with the same

properties as v(x, y). On the other hand, if v(x0, y0) = v0 is given, the constant and v(x, y)are unique. �

Theorem 10. (E =⇒ A) For a function f(z) in a domain D, if for all closed piecewisesmooth contours in D, f(z) = 1

2πi

�Cf(ξ)ξ−z dξ (Cauchy's Integral formula), then for all z0 ∈ D

there exists a δ > 0 such that for all z satisfying |z − z0| < δ, f(z) =∑+∞

n=0 an(z − z0)n for

16

some sequence {an}+∞n=0 of complex numbers (this series is in fact the Taylor series of f(z)about the point z0.)

Figure 5.

Proof. It is given that

f(z) =1

2πi

‰C

f(ξ)

ξ − zdξ =

1

2πi

‰C

f(ξ)

ξ − z0 − (z − z0)dξ =

1

2πi

‰C

f(ξ)

(ξ − z0)(1− z−z0ξ−z0 )dξ (76)

Since z is in the interior of the disk centered at point z0 with radius less than the minimaldistance from z0 to the boundary of D and ξ is on the boundary of this disk, it follows that| z−z0ξ−z0 | < 1.Thus,

1

1− z−z0ξ−z0

=+∞∑n=0

(z − z0)n

(ξ − z0)n(77)

Hence,

f(z) =1

2πi

‰C

f(ξ)

ξ − z0

+∞∑n=0

(z − z0)n

(ξ − z0)ndξ =

1

2πi

‰C

+∞∑n=0

f(ξ)(z − z0)n

(ξ − z0)n+1dξ (78)

f(z) =+∞∑n=0

(z − z0)n

2πi

‰C

f(ξ)

(ξ − z0)n+1dξ =

+∞∑n=0

(z − z0)n

2πi

(2πi

n!f (n)(z0)

)=

+∞∑n=0

f (n)(z0)

n!(z−z0)n

(79)which is the Taylor series of f(z) about point z = z0. The third equal sign in (69) is justi�edby E =⇒ C. �

Theorem 11. (A =⇒ B) If for a function f(z) in a domain D and every point z0 ∈ D,there is a neighborhood about z0 in D such that f(z) =

∑+∞n=0 an(z − z0)n for all z in this

neighborhood and some sequence {an} of complex numbers, then the derivative of f(z), f ′(z),exists everywhere in D and f ′(z) = g(z) =

∑+∞n=1 nan(z − z0)n−1.

17

Proof. Note that

limx→+∞

x1x = lim

x→+∞eln(x

1x ) = lim

x→+∞e

ln xx = elimx→+∞

ln xx = elimx→+∞

1xx = e0 = 1 (80)

Thus,

limn→+∞

n1n = 1 (81)

Let Rf and Rg denote the radii of convergence of the power series of f(z) and g(z)respectively.Then

1

Rf= lim sup

n→+∞

n√|an| (82)

1

Rg= lim sup

n→+∞

n√|nan| = lim

n→+∞n√n · lim sup

n→+∞

n√|an| = lim sup

n→+∞

n√|an| (83)

Thus, Rf = Rg.Let w = z − z0. Then

f(z + h) =+∞∑n=0

an(z + h− z0)n =+∞∑n=0

an(w + h)n (84)

and

f(z) =+∞∑n=0

an(z − z0)n =+∞∑n=0

anwn (85)

Using the binomial formula

(w + h)n =n∑j=0

(n

j

)wn−jhj = wn + nwn−1h+ o(h) (86)

as h→ 0.Thus,

f(z + h)− f(z)h

=1

h

( +∞∑n=0

an(wn + nwn−1h+ o(h))−

+∞∑n=0

anwn

)=

1

h

+∞∑n=0

an(nwn−1h+ o(h))

(87)

=+∞∑n=0

(annwn−1 + ano(1)) =

+∞∑n=0

annwn−1 + o(1) =

+∞∑n=1

annwn−1 + o(1) = g(z) + o(1) (88)

Hence,∑+∞

n=1 nan(z − z0)n−1 is convergent and

f ′(z) = limh→0

f(z + h)− f(z)h

= g(z) (89)

exists. �

18

Theorem 12. (B =⇒ D) If f ′(z) exists (you may not assume that f ′(z) is continuous anduse Green's theorem) at every point of some domain D, then

�Cf(z)dz = 0 for every closed

piecewise smooth contour C ⊂ D. Also, f(z) is continuous in D.

Proof. According to Lemma 1, since f ′(z) exists in D, f(z) is continuous in D.The grid below consists of L× L squares (L→ +0) and contains the contour C.

Figure 6.

We consider two sets of squares, those entirely inside and those at least partially insidethe curve C.Let R be the domain enclosed in C, M be the number of squares whose points are in C or

on C, and N be the number of squares that have non-empty intersection with R. It is clearthat M ≤ N . Since L → +0, f ′(z) exists in all of these N squares, as long as L is smallenough.By de�nition, a region R has �inner area�, A, if lim

L→+0ML2 = A ≥ 0 and �outer area�, A, if

limL→+0

NL2 = A > 0, as A ≥ A. By de�nition of area i� A = A, then A = A = A is the areaof the region R.The horizontal and vertical lines divide R into N subsets. Call the kth subset Sk, 1 ≤ k ≤

N . zk is the center of the kth square and Ck the boundary of Sk (Ck could be outside of R.)

We observe that cancelation of the integrals on the portions of the sides of the squaresthat are inside of C yields

‰C

f(z)dz =N∑k=1

‰Ck

f(z)dz (90)

limL→+0

NL2 = A =⇒ N = A+ o(1)L2

=⇒ N → +∞ (91)

Since f ′(zk) exists,f(z)−f(zk)

z−zk→ f ′(zk) as z → zk.

Hence, f(z)−f(zk)z−zk

− f ′(zk) = �k(z) = o(1).Then,

19

f(z) = f(zk) + f′(zk)(z − zk) + (z − zk)�k(z) (92)

for z 6= zk. The last equality is also valid for z = zk and hence for all z ∈ SkThus,

‰C

f(z)dz =N∑k=1

˛Ck

[f(zk) + f′(zk)(z − zk) + (z − zk)�k(z)]dz (93)

N∑k=1

(f(zk)− zkf ′(zk))‰Ck

dz +N∑k=1

f ′(zk)

‰Ck

zdz +N∑k=1

‰Ck

(z − zk)�k(z)dz (94)

Hence, by Lemma 2,

‰C

f(z)dz =N∑k=1

‰Ck

(z − zk)�k(z)dz (95)

|‰C

f(z)dz| = |N∑k=1

‰Ck

(z − zk)�k(z)dz| ≤N∑k=1

‰Ck

|z − zk||�k(z)||dz| (96)

Note that

|�k(z)| = o(1) (97)

|z − zk| ≤L√2

(98)

since zk is the center of the kth square.

z = x+ iy =⇒ dz = dx+ idy = ẋdt+ iẏdt =⇒ |dz| =√ẋ2 + ẏ2dt = ṡdt = ds.

Then,

‰Ck

|dz| =‰Ck

ds = 4L (99)

which is the perimeter of each of the squares Sk.Hence,

|‰C

f(z)dz| ≤ maxz∈R|�k(z)|

L√2

4LN = o(1)4L2√

2(A+ o(1)

L2) = o(1) (100)

Thus,

‰C

f(z)dz = 0 (101)

�

�

20

Note: I would like to thank my graduate students Tan Nguyen, Joshua Reagan, and RalphThomas for their contribution to editing/typing of this text.

Angel S. Muleshkov, Ph.D.Associate Professor of Mathematics, UNLV, 1989