Complex$nalytic)unctions with1atural%oundary

ANDREI-FLORIN ALBIŞORU1, DORIN GHIŞA2

1Faculty of Mathematics and Computer Science, Babeş-Bolyai University, Cluj-Napoca, ROMANIA

2Department of Mathematics, Glendon College, York University, Toronto, CANADA

1a2

Abstract: The analytic functions with natural boundaries have been only occasionally mentioned in literature.

They were defined mainly by lacunary power series of Hadamard type, except for the modular function which

is the result of a laborious construction. The case of infinite Blaschke products which cannot be analytically

continued over the unit circle is also known, yet the authors have no knowledge about any study devoted to

these functions. The purpose of this article is to take a closer look upon these functions, to find new techniques

of generating them and to bring this topic into the mainstream study of analytic functions. A special attention is

devoted to the theory of Blaschke products, which is completed with new results related to their boundary behavior,

making possible the study of the Blaschke products with natural boundary. We apply to them the same method

of study as for ordinary infinite Blaschke products obtaining mirror functions with respect to the unit circle. The

working tool is that of the fundamental domains, which are easily revealed by the technique of continuation over

a curve, or lifting of a curve, having its origins in the differential geometry. Graphic illustrations contribute to a

better understanding of the theoretical endeavors.

Key-Words: Blaschke product, the permanence of functional equations, lacunary power series, natural boundary,

denseness theorems, Frostman condition.

Received: May 16, 2024. Revised: October 2, 2024. Accepted: October 23, 2024. Published: November 28, 2024.

1 Introduction

The modular function λ(τ), [1], has been obtained

starting with a domain Ω1bounded by the half-lines

ℜτ=±1,ℑτ≥0,

and the half-circles

τ±1

2=1

2,ℑτ≥0.

By the Riemann mapping theorem, there is a unique

conformal mapping λ(τ)of the domain Ω1onto the

complex plane with a slit alongside the real axis from

−∞ to −1and from 1to +∞such that τ= 0,1,∞

is mapped onto λ= 1,∞,0.

By the Schwartz symmetry principle this mapping

can be extended analytically to the whole upper half

plane, first into symmetric domains Ω2and Ω3of

Ω1with respect to the two half-lines by using the

functional equations

λ(τ+ 2) = λ(τ)

and then into symmetric domains Ω4and Ω5of Ω1

with respect to the two half-circles by using the

functional equation

λτ

1−2τ=λ(τ).

The new half-lines and half-circles obtained as

boundaries can be used to do analytic continuations of

λ(τ)into the symmetric domains with respect to them

and the process can be repeated indefinitely. Finally,

λ(τ)will be defined in the whole upper half-plane.

The real axis becomes a natural boundary of λ(τ),as

it can be seen in Fig. 1a. It can be easily seen that

λ(τ) = λ(τ),ℑτ < 0,

is an analytic function defined in the lower half-plane

which has also as natural boundary the real axis. The

two functions are mirror images one of each other in

the sense that symmetric figures with respect to the

real axis are mapped by the corresponding functions

into symmetric figures with respect to the real axis, as

it can be seen in Fig. 2a and Fig. 2b.

Let us notice that the Möbius transformation

w=i(1 + z)

1−z

maps conformally the unit disk onto the upper half-

plane, carrying the unit circle onto the real axis, hence

λ(w(z)),|z|<1,will have the natural boundary the

unit circle. This affirmation is illustrated in Fig. 1b.

The same function maps the exterior of the unit disk

onto the lower half-plane, hence λ(w(z)),|z|>1,

will have the same natural boundary, namely, |z|= 1.

Reciprocally, if the analytic function f(ζ)has the

unit circle as natural boundary, then the function

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(a) The fundamental

domains of the modular

function

(b) The natural

boundary of the

modular function can

be carried onto the unit

circle

Fig1: The modular function λ(τ)and the function

λ(i1+z

1−z)have the natural boundaries the real axis and

respectively the unit circle

(a) Symmetric

triangles with respect

to the real axis

(b) Images of

triangles are

symmetric with

respect to the real

axis

Fig2: The images by λ(τ)and by λi1+z

1−zof

symmetric figures with respect to the real axis are

symmetric with respect to the real axis

f(z−i

z+i)has as natural boundary the real axis. A whole

class of infinite Blaschke products have the unit circle

as natural boundary. We will deal with them later,

after presenting some new results related to Blaschke

products.

2 A New Look on Blaschke Products

A Blaschke product is an expression of the form

B(z) =

m≤∞

n=1

e−iθnz−an

1−anz,(1)

where

an=rneiθn,0≤rn<1, θn∈R.

When mis finite we have a finite Blaschke product

of degree m. By relation (2), the finite Blaschke

products of degree mare meromorphic functions

in Cwith the zeros anand the poles 1

an, n =

1,2, ..., m. Infinite products (1) may be convergent

or not. Blaschke proved that an infinite product (1) is

convergent in

D={z| |z|<1}

if and only if Σ∞

n=1(1 − |an|)converges. This is

known as the Blaschke condition. It shows that for

a convergent Blaschke product the sequence (an)

cannot have any accumulation point in the disk |z|

<1and it should accumulate to points of the unit

circle fast enough, i.e., the sequence (1−|an|)should

approach 0fast enough to ensure the convergence of

that series. When the Blaschke condition is fulfilled,

the product (1) converges absolutely in |z|<1

and uniformly on compact subsets of the unit disk,

[2], hence B(z)is an analytic function in D. By

definition, for |z|<1,

B(z) = lim

n→∞ Bn(z),

where

Bn(z) =

k=1

e−iθkz−ak

1−akz.

One can easily check that,

Bn(z) = 1

Bn1

z.(2)

It is also known, [2], that

B′(z) = B(z)

∞

n=1

1− |an|2

(an−z)(1 −anz)(3)

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Let us notice that

1− |an|2= (1 − |an|)(1 + |an|)≤2(1 − |an|)

and for every δ > 0there is a finite number Ksuch

that

1 + |an|

|an−z||1−anz|< K if |z−an|> δ,

hence the series (3) converges in the unit disk, except

at the points an.Yet,

B′(an) = lim

z→an

B(z)

z−an

1−akanY

k=n

eiθkan−ak

1−akan

which is a convergent product since Pk=n(1 − |ak|)

converges. Thus, if B(z)satisfies the Blaschke

condition, then its derivative is given by (3) at every

point of the unit disk.

We notice that if B′(ζ)exits for ζ=eiθ, θ ∈R,

then since

1− |an|2

(an−eiθ)(1 −aneiθ )=−e−iθ 1− |an|2

|an−eiθ|2

we have

eiθ

∞

n=1

1− |an|2

(an−ζ)(1 −anζ)<0,

hence B′(ζ)= 0. Now we can formulate the

following proposition.

Proposition 1. For any Blaschke product B(z)

(finite or infinite), the equation B(z) = 1 has distinct

roots, all located on the unit circle, and B′(z)has no

zero on the unit circle.

Indeed, every Blaschke factor

e−iθnz−an

1−anz

maps the unit disk onto itself, the unit circle onto

itself and the exterior of the unit disk onto itself, so

the Blaschke product (1) does the same. Thus the

solutions of the equation B(z) = 1 must be all located

on the unit circle. On the other hand since every

multiple zero of B(z)−1is also a zero of B′(z), [3],

and no zero of B′(z)is located on the unit circle, no

zero of B(z)−1can be multiple zero.

By a theorem of Frostman, [4], [5], if

∞

n=1

1− |an|

|eiθ −an|<∞

(we will call this inequality the Frostman condition at

eiθ), then B(z)has a radial limit at eiθ.

Theorem 1. Suppose that ζ0∈∂D \E, where

Eis the set of cluster points of zeros of the infinite

Blaschke product B(z)and let

Bn(z) =

k=1

e−iθkz−ak

1−akz.

Suppose that α < Arg(ζ0)< β and no point of E

belongs to the arc of the unit circle between eiα and

eiβ.If B(z)satisfies the Frostman condition at eiα

and eiβ, then limn→∞ Bn(ζ0)exists and we can set

B(ζ0) = lim

n→∞ Bn(ζ0).

Proof. We will use the Cauchy integral formula

for ζ0and the domain Dbounded by the rays

{z|Arg(z) = α}and {z|Arg(z) = β}

and the arcs of the circles centered at the origin and of

radius r < 1and 1

rbetween the two rays, such that no

zero of B(z)belongs to D. If γis the boundary of D

oriented counterclockwise, then for every nwe have:

Bn(ζ0) = 1

2πi Z

Bn(ζ)

ζ−ζ0

dζ

2πi "1

Bn(ρeiα)eiαdρ

ρeiα −ζ0

Bn(1

reiθ)i

reiθidθ

reiθ −ζ0

Bn(ρeiβ)eiβdρ

ρeiβ −ζ0

Bn(reiθ)reiθidθ

reiθ −ζ0#.

(4)

By relation (2) we have

Bn1

reiθ=1

Bn(reiθ)

for α≤θ≤β, thus for θ=αand θ=β, after the

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change of variable ρ=1

twe get

Bn(ρeiθ)eiθ dρ

ρeiθ −ζ0

=−

Bn(1

teiθ)eiθ

t2dt

teiθ −ζ0

Bn(ρeiθ )

eiθ

ρ2dρ

ρeiθ −ζ0

hence

Bn(ρeiα)eiαdρ

ρeiα −ζ0

Zr

Bn(ρeiα)eiα

ρeiα −ζ0

Bn(ρeiα)

eiα

ρ2

ρeiα −ζ0

dρ.

Also, for ρ=1

rwe have

Bn(1

reiθ)i

reiθidθ

reiθ −ζ0

Bn(reiθ )

reiθdθ

reiθ −ζ0

and

Bn(ζ0) =

2πi

Zr

Bn(ρeiα)eiα

ρeiα −ζ0

Bn(ρeiα)

eiα

ρ2

ρeiα −ζ0

dρ

−1

2πi

Zr

Bn(ρeiβ)eiα

ρeiβ −ζ0

Bn(ρeiβ )

eiβ

ρ2

ρeiβ −ζ0

dρ

2πi

Bn(reiθ )

reiθdθ

reiθ −ζ0

−1

2πi

Bn(reiθ)reiθidθ

reiθ −ζ0

(5)

Since Bn(z)converges absolutely in |z|<1to

B(z),when taking the limit as n→ ∞ of Bn(ζ0),we

can switch the limit with the integral sign in (4). Also,

by the Frostman condition at eiα and eiβ ,the integrals

from ρ=rto ρ= 1 of B(z)exist for z=ρeiα and

z=ρeiβ.Then taking the limit as n→ ∞ in (4) we

obtain an expression for B(ζ0).Obviously, this is true

for any ζ=eiθ, α ≤θ≤β. 

Corollary 1. Let Ethe cluster points of zeros

of the infinite Blaschke product B(z),be a set of

Lebesgue measure 0. Then |B(eiθ )|= 1 at almost

every point of ∂D \E.

Proof. Indeed, B(z)

Bn(z)being a subharmonic

function in D, [3] (page 248), we have



B(0)

Bn(0)≤1

2πZ

∂D\E

|B(eiθ)|

|Bn(eiθ)|dθ

2πZ

∂D\E

|B(eiθ)|dθ.

Letting n→ ∞, we get

2πZ

∂D\E

|B(eiθ)|dθ = 1,

hence |B(eiθ)|= 1 a.e. in ∂D \E. In other words,

if Eis a set of Lebesgue measure 0,then B(z)maps

∂D \Eonto ∂D. 

Proposition 2. For a Blaschke product B(z)of

degree m, the equation B(z) = 1 has exactly m

distinct roots ζk, k = 1,2, ..., m all located on the unit

circle. They partition the unit circle into mJordan

arcs γksuch that, if we remove one end of each arc, it

is mapped by B(z)one to one onto the unit circle. In

other words, when ztravels once alongside the unit

circle , its image B(z)describes the unit circle m

times.

For any Blaschke product B(z)of degree m, the

equation B′(z)=0has m−1roots, counted

with multiplicities, in the unit disk and other m−1

symmetric roots of these ones with respect to the unit

circle, [6], [7], [8]. There is no root of this equation

on the unit circle. When we perform simultaneous

continuations, [2], by B(z)over the real axis starting

from ζk, k = 1,2, ..., m we obtain mJordan arcs.

These arcs can only meet at the zeros of B′(z),since

the points of intersection of these arcs are branch

points of B(z),[9], i.e., the zeros of B′(z).

When they meet at some zeros of B′(z),then

the values of B(z)at those zeros must be real.

Suppose these values are all real. Then these Jordan

arcs together with the symmetric ones with respect

to the unit circle divide the complex plane into m

fundamental domains of w=B(z). Otherwise, we

need another method to find fundamental domains

for B(z). Such a method has been described in the

literature. Namely, if bkare the zeros of B′(z), we

connect the points B(bk)and the point w= 1 by a

non self-intersecting polygonal line Land we perform

simultaneous continuations above Lfrom the points

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ζk.This time, all the Jordan arcs we obtain in this way

meet at some points bkand every bkis an intersection

point of some of these arcs. Together with their

symmetric arcs with respect to the unit circle, they

bound mfundamental domains of B(z).

It can be easily checked that if B(z)is a finite

Blaschke product, for every z∈Cwe have, [2],

B(z) = 1

B1

z(6)

For the infinite case, when the Blaschke condition

is fulfilled, we set

B(z) = 1

B1

z(7)

for |z|>1.

Theorem 2. Let Ebe the set of cluster points

of the zeros of the infinite Blaschke product B(z).If

∂D \Eincludes an arc of the unit circle, then the

function e

B(z)is a meromorphic continuation of B(z)

to C\E.

Proof. Indeed, for eiθ ∈∂D \Ewe have

eiθ =1

e−iθ =1

eiθ

and by Corollary 2, there is φ∈Rsuch that B(eiθ) =

eiφ,for almost every eiθ ∈∂D \E.

Then, for such eiθ ∈∂D \E, we have

B(eiθ) = 1

B1

eiθ =1

B(eiθ)=1

eiφ =eiφ =B(eiθ),

hence B(z)and e

B(z)coincide a.e. on ∂D \E, and by

the permanence of functional equations, they coincide

everywhere on ∂D\E, which means that indeed ˜

B(z)

is a meromorphic continuation of B(z). We keep the

notation B(z)for this function. It maps the unit disk

onto itself, the exterior of the unit disk onto itself and

∂D\Eonto ∂D. Since the zeros and the poles of B(z)

are symmetric points with respect to the unit circle,

Eis also the set of accumulation points of the poles

of B(z),hence every point of Eis an essential non

isolated singular point for B(z).

We have studied previously the fact that in any

neighborhood of an isolated essential singular point

of an analytic function f(z)there are infinitely many

fundamental domains of f(z).This property should

be true also for the points of E. A proof of this

affirmation for the case where Eis a single point,

E={z0},can be found in [10]. It is based on the

fact that for every partial product Bn(z)of B(z)the

equation Bn(z)=1has exactly ndistinct solutions

ζn,k, k = 1,2, ..., n located all on the unit circle and

counted counterclockwise starting from z= 1.

Theorem 3. Let ζk=eiθkbe the roots of the

equation Bn(z) = 1, for k= 1,2, ..., n. If

ak+1 =rk+1eiαk+1

is a zero of Bk+1(z),and αk+1 =θj,for j=

1,2, ..., n, then the equations Bn(z)=1and

Bn+1(z) = 1 do not have any common root.

Proof: Indeed, for such a root ζwe would have

Bn+1(ζ) = Bn(ζ),

hence

e−iαn+1 ζ−an+1

1−an+1ζ= 1,

e−iαn+1 ζ−rn+1 = 1 −rn+1e−iαn+1 ζ,

thus

−e−iαn+1 ζ(rn+1 + 1) = rn+1 + 1,

hence ζ=eiαn+1 ,which has been excluded.

We conclude that for different values of n, we have

essentially different roots of the equation Bn(z) = 1

and the roots of all these equations form a discrete

set. As there are infinitely many such roots on the unit

circle for n= 1,2, ..., they have at least one cluster

point. 

Theorem 4. Let B(z)be an infinite Blaschke

product satisfying the Blaschke condition and let

Bn(z)be partial products of B(z),n= 1,2, .... If

B(z)admits a meromorphic continuation to C\E,

then every solution of the equation B(z)=1is the

limit of some solutions of the equations Bn(z) = 1.

Proof. Let Pbe the set of the poles of B(z). Then,

by using relations (6) and (7), it can be easily seen

that Bn(z)converges uniformly on compact subsets

of C\(E∪P)to B(z). Let ζn,k, k = 1,2, ..., n, be

the solutions of the equation Bn(z) = 1 and let ζ0be

a solution of the equation B(z) = 1,hence

Bn(ζn,k) = 1 = B(ζ0) = lim

n→∞ Bn(ζ0).

This shows that there is no δ0>0such that

|Bn(ζ0)−Bn(ζn,k)|> δ0for nbig enough,

thus there should be a sequence (ζn,kn)such that

limn→∞ ζn,kn=ζ0.

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Simultaneous continuations over some polygonal

lines through Bn(z)from ζn,k are Jordan arcs which

meet two by two into the zeros of B′

n(z)and provide

a partition of the complex plane into nfundamental

domains Ωn,k of Bn(z)which are mapped by Bn(z)

onto the whole complex plane with some slits

alongside these lines. When anare simple zeros

of B(z),hence of every Bn(z),the domains Ωn,k

contain each one a unique zero ak.If akis a multiple

zero of order jof Bn(z),then the boundaries of j

fundamental domains will contain ak.

The infinite Blaschke product B(z)has a similar

property, if it admits a meromorphic continuation to

C\E. We have the following result.

Theorem 5. The roots of the equation B(z) =

1located in ∂D \Eform a discrete set, having the

cluster points in E.

Proof. Indeed, C\Eis open and by the

permanence of functional equations, if those roots had

a cluster point in ∂D \E, hence in C\E, then B(z)

would be identically equal to 1, which is not possible.

Therefore, the roots of B(z) = 1 can accumulate only

on E.

Let zn,k be the zeros B′

n(z), k = 1,2, ..., n and let

z0be a zero of B′(z).Since

lim

n→∞ Bn(z) = B(z)

uniformly on compact subsets of the unit disk, we

have limn→∞ B′

n(z) = B′(z),for every z,|z|<1,

hence

lim

n→∞ B′

n(z0) = B′(z0) = 0

=B′

n(zn,k) = lim

n→∞ B′

n(zn,k),

which implies, due to the continuity of B′

n(z),that

there is a sequence (zn,kn)such that

lim

n→∞ zn,kn=z0.

As, for every n, B′

n(z)has n−1zeros counted

with multiplicities in the unit disk, B′(z)must

have infinitely many zeros in the unit disk. They

form a discrete set, since otherwise B′(z)would be

identically zero. We can connect the points B(zn),

where znare these zeros, with a non self-intersecting

polygonal line Lstarting from w= 1,like in the finite

case and then by simultaneous continuation from ζk

over L, where ζkare the zeros of B(z)−1,obtain the

boundaries of the fundamental domains of B(z).This

is true for any infinite Blaschke product satisfying the

Blaschke condition and which admits a meromorphic

continuation to C\E.

Theorem 6. The fundamental domains of any

infinite Blaschke product which admits meromorphic

continuation to C\Eaccumulate to some points of E,

in the sense that every neighborhood of such a point

contains infinitely many such domains.

Proof. Indeed, these domains are disjoint

connected open sets. Every one of them has a part in

the unit disk and a symmetric part with respect to the

unit circle, outside the unit disk. Considering those

inside the unit disk, being infinitely many of them,

they should accumulate to some point in the closed

unit disk. This point cannot be in C\E, since in

every one of them, B(z)assumes a given value w0

and then, the points znwith B(zn) = w0would have

a cluster point in C\E, implying B(z)≡w0, which

is absurd. 

Theorem 7. If Eis a discrete set, every point of

Eis a cluster point of zeros of B(z)−1, hence it is an

accumulation point of fundamental domains of B(z).

Proof. Let ζ0∈E. Hence ζ0is an accumulation

point of zeros of B(z)and there is an open arc γof the

unit circle which contains ζ0and not other point of E.

Let (ank)be a sequence of zeros of B(z)convergent

to ζ0and let γnkbe the arcs obtained by continuation

by B(z)over the real axis starting from ank.These

arcs connect ankwith 1

ank

and are orthogonal to the

unit circle, since all have the same image, the real

axis, which is orthogonal to the image by B(z)of the

unit circle, i.e. to the unit circle.

Let us notice that for every ankthere is a

unique circle Cnkpassing through ankand 1

ank

and

orthogonal to the unit circle. The disk bounded by

this circle contains the arc γnksince its image by B(z)

contains the image of that arc. Since

lim

nk→∞ ank=ζ0=lim

nk→∞

ank

the circles Cnkwill contract to ζ0,and therefore will

intersect γ, thus the arcs γnkwill do the same. These

arcs intersect the unit circle at points ζnk,where

B(ζnk) = 1.These points must also accumulate to ζ0,

which proves the theorem. 

As shown in [10], there are occurrences, not quite

unusual as it may seem, where Ecoincides with the

unit circle, in other words the unit circle is the natural

boundary of B(z).

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3 Blaschke Products with Natural

Boundary

We have studied in [10], the Blaschke product B(z)

with the zeros

an,k =1−1

3nekπi

2n−1,(8)

for n= 1,2, ... and k= 1,2, ..., 2n.

Let us notice that

∞

n=1

k=1

(1 − |an,k|) =

∞

n=1 2

3n

= 2,

hence the Blaschke condition for B(z)is fulfilled.

We also notice that for every nthe zeros an,k are

located on a circle centered at the origin and of radius

rn= 1 −1

3n.There are 2nsuch zeros equally

spaced. When n→ ∞, these radii tend to 1and an,k

will reach almost every point of the unit circle. Thus

B(z)has as natural boundary the unit circle.

Obviously, there are infinitely many similar ways

to choose the zeros an,k such that the Blaschke

condition is fulfilled and (an,k)accumulates to every

point of the unit circle, hence there is an infinite

family of Blaschke products with natural boundary

the unit circle.

If we take a subproduct Bα,β (z)of B(z)obtained

by keeping only the zeros an,k located in a sector

E={z|z=ρeiθ,0≤ρ < 1,0≤α≤θ≤β < 2π},

we get a similar Blaschke product having the arc E

as boundary, in the sense that for eiθ ∈E, there is no

limit z→eiθ from Bα,β (z).However, by Theorem

2, Bα,β(z)has an analytic continuation to C\E.

We can construct infinite Blaschke products

having the set Eof the essential singular points

any generalized Cantor subset, [11], of the unit

circle. Such a subset is obtained in the following

way. Suppose that an infinite sequence (αn),0<

αn<1is given. Let us remove from the interval

(0,2π)a closed interval I1,1of length α1.What

remains are two open intervals J1,1and J1,2of lengths

l1,1and l1,2Next, remove from each one of these

last intervals the closed intervals of length α2l1,1

and respectively α2l1,2.What remains are four open

intervals J2,1, J2,2, J2,3and J2,4of lengths equal

respectively to l2,1, l2,2, l2,3and l2,4.Now, we remove

from each one of these intervals a closed interval of

length equal respectively to α3l2,1, ..., α3l2,4and we

continue this process indefinitely.

To every interval Jn,k, k = 1,2, ..., 2n,

corresponds on the unit circle an arc

γn,k ={ζ=eiθ |θ∈Jn,k}.

(a) Mirror images

with respect to the

unit circle

(b) The images by

B(z)and ˜

B(z)of

mirror images are

mirror images

Fig3: The images by B(z)and by B1

zof

symmetric figures with respect to the unit circle are

symmetric with respect to the unit circle

Let rn= 1 −1

22nand let B(z)be the Blaschke

product with the zeros

an,k =rneiαn,k , αn,k ∈γn,k.

Thus, every zero of B(z)belongs to some circle

centered at the origin and of radius rnand located in

the sector determined by γn.We have

∞

n=1

k=1

(1 − |an,k|) =

∞

n=1

22n= 1,

hence the Blaschke condition for B(z)is fulfilled.

Continuation of B(z)can be performed in C\E,

where Eis a perfect set, [11], on the unit circle. We

obtain a meromorphic function in C\E, having the

zeros an,k and the poles 1/an,k.This is a function

which coincide with its mirror in the unit circle, in

the sense that it maps symmetric figures with respect

to the unit circle into symmetric figures with respect

to the unit circle.

Let us notice that Bz−i

z+ihas the natural

boundary the real axis. Fig. 3a and Fig. 3b portray

mirror images of figures by B(z)and e

B(z).

4 Dirichlet Series with Natural

Boundaries

Ageneral Dirichlet series is an expression of the

form:

ζA,Λ(s) =

∞

n=1

ane−λns(9)

where A= (an)is an arbitrary sequence of complex

numbers, an= 0 infinitely many times and

Λ = {0 = λ1≤λ2≤...}

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is a non-decreasing sequence of non-negative real

numbers such that limn→∞ λn= +∞.

If λn=ln n, then e−λns=1

nsand the series (9) is

an ordinary Dirichlet series. If λn=n−1, we obtain

a power series in e−s.Reciprocally, any power series

f(z) =

∞

n=0

(z−z0)n(10)

can be converted into a Dirichlet series

g(s) =

∞

n=0

ane−ns (11)

by the substitution z−z0=e−s.The Hadamard

formula 1

R=lim sup

n→∞

|an|1

n(12)

for the radius of convergence Rof the series (10)

implies that the series (11) converges for ℜs > ln(1

and it diverges for ℜs < ln(1

R).A similar formula

to (12) is known for general Dirichlet series, [12],

namely, if (9) is not convergent for s= 0,then:

σc=lim sup

n→∞

ln 

k=1

ak

λn

>0(13)

If (9) converges for s= 0,then

σc=lim sup

n→∞

λn+1

ln ζA,Λ(0) −

k=1

ak<0(14)

The number σcis called the abscissa of convergence

of the Dirichlet series (9) since ζA,Λ(σ+it)converges

for σ > σcand it diverges for σ < σc.The function

ζA,Λ(s)is an analytic function in the half-plane ℜs >

σc.To every series (9) a Dirichlet series ζA,eΛ(s)can

be associated, in which the sequence (λn)is replaced

by (eλn).It is known that if ζA,eΛ(s)has only isolated

singular points on the imaginary axis, then the series

(9) can be continued analytically to ameromorphic

function in the whole complex plane.Otherwise, the

line ℜs=σcis the natural boundary of ζA,Λ(s).

Examples of Dirichlet series with natural boundaries

can be constructed by using power series with natural

boundaries as in the formula (11).

Theorem 8. To every power series (10) with

natural boundary having the radius of convergence R

corresponds a Dirichlet series (11) having as natural

boundary the line ℜs=ln 1

Proof. Indeed, the abscissa of convergence of the

series (11) is ln 1

R.Suppose that the series can be

Fig4: An illustration of the denseness theorem for

a Hadamard type of power series. This is the image

of the circle |z|= 0.9999.

continued analytically to some swith σ=ℜs <

ln(1

R).Then f(z) = g(z0+e−s)is an analytic

continuation of f(z)to some z=z0+e−σ+it where

e−σ> R, which is absurd, since |z−z0|=Ris the

natural boundary of the series (10). 

Example. To the Hadamard series P∞

n=0 z2n

with the natural boundary the unit circle corresponds

the Dirichlet series P∞

n=0 e−2nshaving the natural

boundary the imaginary axis.

Fig. 4 portrays the denseness property for this

Dirichlet series related to the line ℜs= 0.001.

5 Denseness Theorems for Functions

with Natural Boundaries

Theorem 9. If f(z) = P∞

n=1 anznis a lacunary

series having as natural boundary the circle |z|=

r, then every neighborhood of a point z0=reiα,

where α∈R, contains infinitely many fundamental

domains of f(z).

Proof. Suppose that there is a neighborhood

Vof z0which contains only a finite number of

fundamental domains of f(z).Then some of these

domains must have a part of the boundary on the circle

|z|=r. Let Ωbe such a domain. The function f(z)

maps conformally the domain Ωonto the complex

plane with a slit Lsuch that the boundary ∂Ωof Ωis

carried by f(z)onto L. Then for every common point

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reit, t ∈Rof ∂Ωand the circle |z|=rwe have that

limz→reit f(z)exists and it belongs to L. Thus an arc

of the circle |z|=ris carried by f(z)into Land the

symmetry of f(z)with respect to this arc represents a

continuation of this function outside the disk |z|< r,

which contradicts the hypothesis that |z|=ris the

natural boundary of f(z)and the theorem is proved.



The prototype lacunary series is the Hadamard

series h(z) = Σ∞

n=1z2n.For any integer m > 2

the series Σ∞

n=1zmn,Σ∞

n=1zτ(n)mn,where τ(n)is a

positive arithmetic function, are also lacunary series.

We will call them lacunary series of Hadamard type.

Some other types of lacunary series are at hand, as for

example Σ∞

n=1anzmn,where

lim sup

n→∞

|an|1

mn=1

R= 1.

Besides the fact that all these series have the circle

|z|=Ras natural boundary, they enjoy many other

common properties. One of them is that expressed

by Theorem 9. Based on this theorem, we are now

able to say more about the fundamental domains of a

lacunary series f(z).

Theorem 10. Let f(z)be a lacunary series having

the natural boundary the circle |z|=R. Then f′(z)

has infinitely many zeros accumulating to every point

of |z|=R. They are the knots of an infinite tree

graph with the lives forming a dense set on the circle

|z|=R. The branches of the tree bound domains

where f(z)is injective. If f(z)has real coefficients,

these zeros are two by two complex conjugate.

Proof: Indeed, since the boundary of every

fundamental domain contains at least one branch

point, where f′(z)cancels, and by Theorem 9, there

are infinitely many fundamental domains, the number

of zeros znof f′(z)must be infinite. They cannot

have any cluster point in the disk |z|< R, since

then, by the permanence of functional equations,

f′(z)would be a constant, which is absurd. They

accumulate to every point of the circle |z|=R.

The tree graph is formed by connecting them with

segments of line such that no finite set of these

segments bound a domain. We include every zero

of f′(z)in such a graph. The zeros of f′(z)can

be counted following any counting algorithm of the

knots of a tree. For example, we designate one of

the knots as the root z0of the tree. Then we count

counterclockwise starting from positive real half-axis

the knots directly connected to z0.This is the first

layer of knots. Then we do the same with the knots

directly connected with the first layer knots, which

represent the second layer, and so on. The branches

of the graph end all in points of the circle |z|=R

such that every neighborhood of such a point contains

infinitely many lives, therefore these lives form a

dense set on the circle. Since there are no branch

points of the function between the branches of the

tree, the domains bounded by adjacent branches are

univalence domains of the function. They are not

necessarily fundamental domains.

If f(z)has real coefficients, then so does f′(z). In

addition, f′(z0) = 0 if and only if

f′(z0) = f′(z0) = 0.

Moreover, for any two zeros z1and z2of f′(z),the

segment

z(t) = (1 −t)z1+tz2,0≤t≤1,

is obviously the symmetric with respect to the real

axis of the segment

z(t) = (1 −t)z1+tz2.

We can construct the tree graph such that its branches

are two by two symmetric with respect to the real axis.

Fig. 5 below exhibits such a graph for the Hadamard

series. 

Let us notice that for the Hadamard series we have

h′(z) = 1 + 2z+ 4z3+... and h′(0) = 1.

Thus h(z)is injective in a small disk Dof radius r,

centered at the origin. Moreover, h(z)satisfies the

functional equation h(z) = h(z2) + z. This implies

that limx→1h(x) = +∞.We also have that for zreal,

h(z)is real, which shows that the image by h(z)of the

interval [0,1) is the interval [0,+∞).The mapping

is one to one, since 0≤x1< x2implies h(x1)<

h(x2).

We can normalize every lacunary series such that

it satisfies this property. Suppose that f(z)is an

arbitrary lacunary series such that f(0) = 0 and

f′(0) = 1,thus f(z)is injective in a small disk

of radius rcentered at the origin. For 0<x<

r, we must have f(−x)<0,since otherwise the

injectiveness of f(z)in that disk would be violated.

Let x0be the smallest in absolute value negative

number such that f′(x0)=0.Then f(x)is injective

in the interval (x0, R)and it maps this interval one to

one onto the interval (f(x0),+∞).The point x0is a

branch point for f(z)and we will designate it as the

root of the previously described tree.

Let us notice that the same functional equation

implies that limx→−1h(x) = +∞,hence there must

be a point x0,−1< x0<0such that h′(x0) =

0.The point x0is a branch point of h(z).On the

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other hand, a little computation shows that h′(x)is

a strictly increasing function, hence x0is the only

branch point of h(z)on the real diameter of the

unit circle. Moreover, h(z)maps one to one the

interval (x0,1) and the interval (−1, x0)onto the

interval (h(x0),+∞)and two Jordan arcs γ1and γ2

connecting x0with points ζ1and ζ2of the unit circle

onto (−∞, h(x0)).The arcs γ1and γ2are symmetric

with respect to the real axis.

The functional equation can be seen also as

h(z2) = h(z4) + z2,

hence

h(z) = h(z4) + z+z2,

and in general

h(z) = h(z2n) + z+z2+... +z2n−1.

This last equation implies that

lim

r→1hre2kπi

2n

=lim

r→1hh(r2ne2kπi) + re 2kπi

2n+... +r2n−1ekπii

=h(1) + e2kπi

2n+... +ekπi

=∞,

for k= 1,2, ..., 2n,which shows that for almost every

point ζon the unit circle we have limz→ζh(z) = ∞,

where ztends radially to ζ. The continuation over the

real axis from ζshows that the pre-image by h(z)

of the real axis has at least one component ending

in ζ. Actually, the pre-image of the real axis has

infinitely many components ending in ζsince, by the

Big Picard Theorem, in every neighborhood of ζthere

must be infinitely many points at which h(z)takes

the same real value. On some of these components

limz→ζh(z)is +∞and on others it is −∞.They are

all orthogonal curves to the pre-image of every circle

centered at the origin that they meet, and obviously,

to the unit circle. When a point moves on the pre-

image of a circle centered at the origin between two

consecutive components of the pre-image of the real

axis ending in the same point ζ, its image describes

half of that circle between points with the argument

kπ, k = 0,1,hence on adjacent components of the

pre-image of the real axis ending in ζthe limit +∞

and −∞ of h(z)when z→ζmust alternate.

To find the fundamental domains of h(z)we can

proceed in the following way. Consider the adjacent

component of the pre-image of the real axis to the

real diameter of the circle located in the upper half-

disk which ends in z= 1 and limz→1h(z)=+∞

when zbelongs to that component. There is a point

w1on that component for which h(w1) = h(x0).

Let us do continuation by h(z)over the image of the

segment from x0to z1,where h′(z1) = 0,starting

from w1.This is a Jordan arc connecting w1with

z1whose image by h(z)is the same as the image

of the segment from x0to z1.Let us denote by Ω1

the domain bounded by the two components of the

pre-image of the real axis, the respective segment

and this continuation arc. It can be easily seen

that Ω1is a fundamental domain of h(z),which

is mapped conformally by h(z)onto the complex

plane with a slit alongside the interval (h(x0),+∞)

of the real axis followed by a slit from h(x0)to

h(z1).The component of the pre-image of the interval

(−∞, h(x0)) included in Ω1has the ends in z= 1 and

z=w1,hence limz→1h(z) = −∞,where zbelongs

to this component. The point w1is a branch point of

h(z)on which the two components of the pre-image

of the real axis intersect each other orthogonally. The

value of h(z)at this branch point is real. There

are infinitely many fundamental domains of h(z)in

the upper half disk accumulating to z= 1 having

the same image as Ω1,hence infinitely many branch

points of h(z)in which h(z)has the real value h(x0).

Let us deal now with the component of the pre-

image of the real axis adjacent to the real diameter

of the unit circle located in the upper half disk which

ends in z=−1and such that limz→−1h(z)=+∞

when zbelongs to that component. There is a point

w2on that component such that h(w2) = h(x0).

Let us do continuation by h(z)over the image of

the segment from x0to z2,where h′(z2)=0.We

obtain another Jordan arc connecting x0to w2such

that the image by h(z)of this arc is a slit from h(x0)

to h(z2)and the domain Ω2bounded this arc and

the two components of the pre-image of the real axis

is a fundamental domain of h(z)which is mapped

conformally by h(z)onto the complex plane with a

slit alongside the interval (h(x0),+∞)of the real

axis followed by a slit from h(x0)to h(z2).There are

infinitely many fundamental domains of h(z)in the

upper half disk accumulating to z=−1having the

same image as Ω2,hence another infinity of branch

points of h(z)in which h(z)has the real value h(x0).

We denote by Ω3and Ω4the symmetric domains

of Ω2and Ω1with respect to the real axis. Obviously,

they are also fundamental domains of h(z).With the

second layer of knots, we define similarly a second

layer of fundamental domains, and so on.

A similar construction can be performed for any

normalized lacunary series f(z). When f(z)has non

real coefficients, there is no symmetry with respect

to the real axis, yet the other features are the same.

The geometry of conformal mapping by any lacunary

series is completed by the following two theorems.

Theorem 11. To every lacunary series f(z),

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Fig.5: An idea of infinite tree defining the

fundamental domains of h(z).

having the circle |z|=Ras natural boundary,

corresponds an analytic function g(z)which exists in

the exterior of the that circle, having the same natural

boundary and the zeros R2

an,where anare the zeros

of f(z).If the lacunary series has real coefficients

the symmetric with respect to the circle |z|=R

of the fundamental domains of f(z)are fundamental

domains of g(z).

Proof. Indeed, the function g(z) = fR2

zfor

|z|> R satisfies all these properties. It exists

obviously only for |z|> R, since only there we have

R2

z< R, where fR2

zexists. Moreover, f(an) =

0if and only if gR2

an=f(an) = 0.That g(z)is an

analytic function appears clearly when we replace z

by 1

zin the series of f(z)and then conjugate every

term. We obtain a Laurent series which converges

uniformly on compact subsets of |z|> R, and

therefore it is an analytic function. Moreover, the

transformation z→R2

zcarries the tree defining the

fundamental domains of f(z)into a tree included in

{z| |z|> R}. The arcs of the two trees are symmetric

with respect to the circle |z|=R, therefore they

bound symmetric fundamental domains.

We notice that for |z|> R we have |g(z)|< R

and limz→∞ g(z) = 0.Also gR2

x0=f(x0)and if

f′(x0) = 0,then g′R2

x0= 0,hence the root of the

tree defined by g(z)is R2

x0,where x0is the root of the

Fig.6: Illustration of the density property for the

Dirichlet function ψ(s) = Σ∞

n=0e−2nsof the line s=

0.001 + it,−100 ≤t≤100

tree defined by f(z).

Let f(z) = Σ∞

n=1anzλnbe a lacunary series

having the radius of convergence Rand the natural

boundary the circle |z|=Rand let us define the

Dirichlet series ψ(s) = Σ∞

n=1ane−λnsobtained

from f(z)by the substitution z=e−s.Then ψ(s)

converges for

|e−s|=|e−σ−it|=e−σ< R,

i.e. for σ > ln 1

Rand diverges for σ < ln 1

Moreover, the convergence line s=ln 1

R+it is

the natural boundary of ψ(s).For the Hadamard type

of series f(z),we have R= 1 and therefore the

natural boundary of the corresponding Dirichlet series

ψ(s)is the imaginary axis. For h(z)=Σ∞

n=0z2n,

we have ψ(s)=Σ∞

n=0e−2ns, for which the abscissa

of convergence is σc= 0.Fig. 6 illustrates the

denseness property of the image by ψ(s)of the line

s= 0.001 + it.

Theorem 12 (Denseness Theorem). Let w=

f(z)be an analytic function with the natural boundary

|z|=rand let (wm)be a sequence of points spread

throughout the plane such that two neighboring points

are at a distance less than ϵone of each other for a

given arbitrarily small ϵ. Then for every m0there is

ρ, 0< ρ < r such that the image by f(z)of the circle

Cρ:|z|=ρpasses at the distance less than ϵof every

wmfor m≤m0.

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Fig.7: An illustration of the denseness property for

a Blaschke product with natural boundary. This is the

image of the circle |z|= 0.9999 by B(z).

Proof. We give a constructive proof accompanied

by the illustration Fig. 4 of the case when f(z)is the

Hadamard function and Fig. 7 of the case when f(z)

is a Blaschke product B(z)given by relation (1) with

zeros given by relation (8).

Suppose that a mesh obtained by drawing vertical

and horizontal lines with distance equal to ϵ

3between

adjacent lines is given. The coordinate axes are

supposed to belong to this mesh. Let us denote by

Dmthe eyes of the mesh counted following a spiral

starting from the origin. They are open connected

sets. We can suppose, with no restriction, that wm∈

Dm.Obviously, the distance between neighboring

wmis less then ϵ.

The pre-images of Dmby f(z)is formed by

domains Ωn,m,one in every fundamental domain Ωn

of f(z).By Theorem 5, the domains Ωnaccumulate

to every point of the circle |z|=rand, obviously,

so do the domains Ωn,m.Since f(z)is injective in a

neighborhood of the origin, for small values of ρthe

circle Cρhas the image only in the first four domains

Dm.When ρincreases past of the second knot of

the tree Cρwill intersect a new fundamental domain

Ω2,hence Ω2,1, ..., Ω2,4and in Ω1new domains

Ω1,5, .., Ω1,16.Thus the image of Cρwill pas through

D1, D2, ..., D16.Continuing this way, we reach Dm0

and the image of Cρpasses through every Dm, m ≤

m0.Obviously, it is at a distance less than ϵof every

wm, m ≤m0.

6 Conclusions

The analytic functions with natural boundaries have

been treated up to now as some pathological cases.

No detailed study has been devoted to such functions.

The purpose of this article was to highlight the fact

that these functions are not so unusual as it may seem

and they deserve a better place in the mainstream

study of complex analytic functions. Moreover,

they possess a remarkable property, which is that

of denseness of the image of a line close enough to

the natural boundary. This simulates chaos behavior

in a context different of that which has been well

studied in the literature. We are hinting to a possible

application in quantum physics.

Acknowledgment:

It is an optional section where the authors may write

a short text on what should be acknowledged

regarding their manuscript.

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[1] Albisoru, A.F. and Ghisa, D., Conformal

Self-Mappings of the Fundamental Domains of

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[2] Granath, J., Finite Blaschke products and their

properties, PhD Thesis, University of

Stockholm, 2020.

[3] Ahlfors, L.V., Complex Analysis, International

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Volume 23, 2024

[9] Ahlfors, L.V. and Sario, L., Riemann Surfaces,

Princeton University Press, Princeton, New

Jersey, 1974.

[10] Andreian-Cazacu, C. and Ghisa, D., Global

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London(U.K.), 3-12, 2009.

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[12] Hardy, G.H. and Riesz, M., The General

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Contribution of Individual Authors to the

Creation of a Scientific Article (Ghostwriting

Policy)

The authors equally contributed in the present

research, at all stages from the formulation of the

problem to the final findings and solution.

Sources of Funding for Research Presented in a

Scientific Article or Scientific Article Itself

No funding was received for conducting this study.

Conflict of Interest

The authors have no conflicts of interest to declare

that are relevant to the content of this article.

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WSEAS TRANSACTIONS on MATHEMATICS

DOI: 10.37394/23206.2024.23.83

Andrei-Florin Albişoru, Dorin Ghişa

E-ISSN: 2224-2880

814

Volume 23, 2024