CUBO A Mathematical Journal
Vol.18, No¯ 01, (01–14). December 2016

Uniqueness of meromorphic functions sharing a set in annuli

Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai

Department of Mathematics,

Karnatak University,

Dharwad - 580003, India

renukadyavanal@gmail.com, ashwinimhmaths@gmail.com, madhuramathai@gmail.com

ABSTRACT

The purpose of this article is to investigate the uniqueness of meromorphic functions

sharing a set with counting multiplicity and also with weight 1 in annuli.

RESUMEN

El propósito de este art́ıculo es investigar la unicidad de funciones meromorfas compar-

tiendo un conjunto contando multiplicidad y también con peso 1 en un anillo.

Keywords and Phrases: Annuli, Meromorphic, Sharing Set, Uniqueness.

2010 AMS Mathematics Subject Classification: 30D35, 39A05.



2 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
18, 1 (2016)

1 Introduction and Main Results

We assume the reader is familiar with standard results and notations of Nevanlinna’s theory of

meromorphic functions [4],[10],[11]. The purpose of this paper is to study the uniqueness of mero-

morphic functions in doubly connected domains of complex plane C. By the doubly connected

mapping theorem [1], each doubly connected domain is conformally equivalent to the annulus

{z : r < |z| < R}, 0 ≤ r < R ≤ +∞. We consider only two cases: r = 0, R = +∞ simultaneously and

0 < r < R < +∞. In the latter case, the homothety z "−→ z√
rR

reduces the given domain to the

annulus A =
{
z : 1

R0
< |z| < R0

}
, where R0 =

!

R
r
. Thus, in two cases every annulus is invariant

with respect to the inversion z "−→ 1
z
. Hence, we consider the uniqueness of mermorphic functions

in the annulus A = {z : 1
R0

< |z| < R0}, where 1 < R0 ≤ +∞. We denote by S the subset of distinct

elements in C = C ∪ {∞}. For a meromorphic function f in A, we define

EA(S, f) = ∪a∈S{z ∈ A : fa(z) = 0, counting multiplicities},

E
A
(S, f) = ∪a∈S{z ∈ A : fa(z) = 0, ignoring multiplicities},

where fa(z) = f(z) − a if a ∈ C. We also define

E
A

1(S, f) = ∪a∈S{z ∈ A : all the simple zeros of fa(z)}.

For any constant a, we say that f and g share a Counting Multiplicity(CM), provided that

f − a and g − a have the same zeros with same multiplicities. Similarly, we say that f and g

share a Ignoring Multiplicity(IM), provided that f − a and g − a have the same zeros ignoring

multiplicities.

In 2009, Cao et al.[3] obtained an analog of Nevanlinna’s famous five -value theorem as follows.

Theorem 1.1. Let f1 and f2 be two transcendental or admissible meromorphic functions on the

annulus A =
{
z : 1

R0
< |z| < R0

}
, where 1 < R0 ≤ +∞. Let aj(j = 1, 2, 3, 4, 5) be five distinct

complex numbers in C. If f1, f2 share aj IM for j = 1, 2, 3, 4, 5, then f1(z) ≡ f2(z).

In 2012, Cao and Deng [2] investigated the uniqueness of two meromorphic functions in A sharing

two or three finite sets and obtained the following theorems.

Theorem 1.2. Let f and g be two admissible meromorphic functions in the annulus A. Put S1 =

{0}, S2 = {∞}, and S3 = {w : P(w) = 0}, where

P(w) = awn + n(n − 1)w2 + 2n(n − 2)bw − (n − 1)(n − 2)b2,

where n ≥ 5 is an integer and a and b are two non-zero complex numbers satisfying abn−2 ̸= 1, 2.

If E(S2, f) = E(S2, g) and E(Sj, f) = E(Sj, g)(j = 1, 3), then f ≡ g.

Theorem 1.3. Let f and g be two admissible meromorphic functions in the annulus A. Put

S1 = {∞} and S2 = {w : P(w) = 0}, where P(w) is stated as in Theorem 1.2, and a and b are two

non-zero complex numbers satisfying abn−2 ̸= 2, n ≥ 8 is an integer. If E(S1, f) = E(S1, g) and

E(S2, f) = E(S2, g), then f ≡ g.



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Uniqueness of meromorphic functions sharing a set in annuli 3

In 2013, Xu and Wu [8] obtained the following Theorems.

Theorem 1.4. Let f and g be two admissible meromorphic functions in the annulus A. Let

S = {w ∈ A : P1(w) = 0}, where P1(w) =
(n−1)(n−2)

2
wn − n(n − 2)wn−1 +

n(n−1)
2

wn−2 − c and

c(̸= 0, 1) is a complex number. If EA(S, f) = EA(S, g) and n is an integer satisfying ≥ 11, then

f ≡ g.

Theorem 1.5. Let f and g be two admissible meromorphic functions in the annulus A. If n is

an integer ≥ 7, EA(S, f) = EA(S, g) and Θ0(∞, f) >
3
4
, Θ0(∞, g) >

3
4
, where S is defined as in

Theorem 1.4, then f ≡ g.

Theorem 1.6. Let f and g be two admissible meromorphic functions in the annulus A. If

EA1 (S, f) = E
A
1 (S, g) and n is an integer ≥ 15, where S is defined as in Theorem 1.4, then f ≡ g.

Theorem 1.7. Let f and g be two admissible meromorphic functions in the annulus A. If n is

an integer ≥ 9, EA1 (S, f) = E
A
1(S, g) and Θ0(∞, f) >

5
6
, Θ0(∞, g) >

5
6
, where S is defined as in

Theorem 1.4, then f ≡ g.

The main purpose of this paper is to investigate the uniqueness of meromorphic functions

sharing a set S = {w ∈ A : P(w) = 0}, where P(w) = wn + αwn−m + β, α and β are two non-zero

constants different from the sets considered by Cao and Deng [2] and X.Y.Xu and Z.T.Wu [8]. The

set considered in this paper is more general, as set varies by varying value of m and constants α

and β for a fixed n, where as the set in [2] and [8] are specific for a fixed n.

To prove the main results, we follow the techniques used by X.Y.Xu and Z.T.Wu [8] till the

conclusion part and using different technique, conclusion part is effectively proved as the sharing

set S is different from the set considered in [8].

Theorem 1.8. Let f and g be two admissible meromorphic functions in the annulus A. Let

n > 2m + 8 and m ≥ 2 be integers with n and m having no common factors. Let S = {w ∈ A :

P(w) = 0}, where P(w) = wn + αwn−m + β, α and β are two non-zero constants such that the

algebraic equation wn + αwn−m + β = 0 has no multiple roots. If EA(S, f) = EA(S, g), then f ≡ g.

As inspired by the proof of the Theorem 1.5, Theorem 1.6 and Theorem 1.7, we proved the

following Theorems.

Theorem 1.9. Let f and g be two admissible meromorphic functions in the annulus A. Let n >

2m + 5 and m ≥ 2 be integers. If EA(S, f) = EA(S, g) and Θ0(∞, f) >
3
4
, Θ0(∞, g) >

3
4
, where S

is defined as in Theorem 1.8, then f ≡ g.

Theorem 1.10. Let f and g be two admissible meromorphic functions in the annulus A. Let

n > 2m + 12 and m ≥ 2 be integers. If EA1 (S, f) = E
A
1 (S, g), where S is defined as in Theorem 1.8,

then f ≡ g.

Theorem 1.11. Let f and g be two admissible meromorphic functions in the annulus A. Let

n > 2m+7 and m ≥ 2 be integers. If EA1 (S, f) = E
A
1(S, g) and Θ0(∞, f) >

5
6
, Θ0(∞, g) >

5
6
, where

S is defined as in Theorem 1.8, then f ≡ g.



4 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
18, 1 (2016)

Basic Notations in the Nevanlinna Theory on Annuli

Let f be a meromorphic function in C. We recall the classical notations of the Nevanlinna theory

as follows:

N(R, f) =

∫R

0

n(t, f) − n(0, f)

t
dt + n(0, f) log R,

m(R, f) =
1

2π

∫2π

0

log+ |f(Reiθ)|dθ, T(R, f) = m(R, f) + N(R, f)

Where log+ x = max{log x, 0} and n(t, f) is the counting function of poles of the function f in

{z : |z| ≤ t}.The following are the notations and basic results of Nevanlinna theory on annuli

A = {z : 1
R0

< |z| < R0} for 1 < R < R0 ≤ +∞, which can be found in [5] and [6]. Let

N1(R, f) =

∫1

1
R

n1(t, f)

t
dt, N2(R, f) =

∫R

1

n2(t, f)

t
dt

Where n1(t, f) and n2(t, f) are the counting function of poles of the function f in {z : t < |z| ≤ 1}

and {z : 1 < |z| ≤ t}, respectively. Let

N0(R, f) = N1(R, f) + N2(R, f) , m0(R, f) = m(R, f) + m

"

1

R
, f

#

and

N0(r, f) = N1(R, f) + N2(R, f)

=

∫1

1
R

n1(t, f)

t
dt +

∫R

1

n2(t, f)

t
dt

Where n1(t, f) and n2(t, f) are the reduced counting function of poles of the function f in {z : t <

|z| ≤ 1} and {z : 1 < |z| ≤ t}, respectively.

Finally, we define the Nevanlinna characteristic of f on the annulus A by

T0(R, f) = m0(R, f) − 2m(1, f) + N0(R, f)

In addition, we have

N
(2)

0 (R, f) = N0(R, f) + N
(2

0 (R, f),

N
(2

0 (R, f) = N(R, f) − N
1)

0 (R, f)

where N
1)
(R, f) is the reduced counting function which only counts simple zeros of the function f.

And the Nevanlinna characteristic of f has the following properties.

(i)T0(R, f) = T0
$

R, 1
f

%

,

(ii)max{T0(R, f1 · f2), T0(R,
f1
f2
), T0(R, f1 + f2)} ≤ T0(R, f1) + T0(R, f2) + O(1)

By above properties, the first fundamental theorem on the annulus A is immediately obtained as

follows.



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Uniqueness of meromorphic functions sharing a set in annuli 5

First Fundamental theorem in annuli : Let f be a non-constant meromorphic function on

the annulus A = {z : 1
R0

< |z| < R0}. For 1 < R < R0 ≤ +∞, we have

T0

"

R,
1

f − a

#

= T0(R, f) + O(1) for every fixed a ∈ C.

Definition 1. For every a ∈ C, the reduced deficiency is given by

Θ0(a, f) = 1 − lim sup
R→∞

N0
$

R, 1
f−a

%

T0(R, f)
.

Definition 2. Let f be a meromorphic function on the annulus A = {z : 1
R0

< |z| < R0}, where

1 < R0 ≤ +∞. The function f is called an admissible meromorphic function on the annulus A

provided that

lim sup
R→∞

T0(R, f)

log R
= ∞, 1 < R < R0 = +∞,

or

lim sup
R→R0

T0(R, f)

− log (R0 − R)
= ∞, 1 < R < R0 < +∞,

respectively.

Khrystiyanyn and Kondratyuk[5] obtained the following lemma on the logarithmic derivative on

the annulus A.

Lemma on the logarithmic derivative : Let f be a non-constant meromorphic function on the

annulus A = {z : 1
R0

< |z| < R0}, where R0 ≤ +∞ and let λ > 0. Then,

(I) If R0 = +∞, then m0
&

R, f
′

f

'

= O(log(RT0(R, f))),

for R ∈ (1, +∞) except for the set ∆R such that
∫
∆R

Rλ−1dR < +∞;

(II) If R0 < +∞, then m0
&

R, f
′

f

'

= O
&

log
&

T0(R,f)
R0−R

''

,

for R ∈ (1, R0) except for the set ∆
′
R such that

∫
∆′

R

dR
(R0−R)

λ−1 < +∞.

2 Some Lemmas

For the proof of our main results, we need the following lemmas.

Lemma 2.1. ([3]) Let f be a non-constant meromorphic function on the annulus A = {z : 1
R0

<

|z| < R0}, where 1 < R0 ≤ +∞. Let a1, a2, · · · , aq be q distinct complex numbers in the extended

complex plane C. Then,

(q − 2)T0(R, f) <

q∑

j=1

N0

"

R,
1

f − aj

#

+ S(R, f)



6 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
18, 1 (2016)

(1) If R0 = +∞, then S(R, f) = O(log(RT0(R, f))), for R ∈ (1, +∞) except for the set ∆R such

that
∫
∆R

Rλ−1dR < +∞.

(2) If R0 < +∞, then S(R, f) = O
&

log
&

T0(R,f)
R0−R

''

, for R ∈ (1, R0) except for the set ∆
′
R such

that
∫
∆′

R

dR
(R0−R)λ−1

< +∞.

Lemma 2.2. ([2]) Let f be a non-constant meromorphic function in A, Q1(f) and Q2(f) be two

mutually prime polynomials in f with degree m and n, respectively. Then,

T0

"

R,
Q1(f)

Q2(f)

#

= max{m, n}T0(R, f) + S(R, f),

where S(R, f) is defined as in Lemma 2.1.

Lemma 2.3. ([8]) Let f be a non-constant meromorphic function in A. Then,

N0

"

R,
1

f′

#

≤ N0

"

R,
1

f

#

+ N0(R, f) + S(R, f) + O(1),

where S(R, f) is defined as in Lemma 2.1.

Lemma 2.4. ([8]) Let F and G be admissible meromorphic functions in A satisfying EA(F, 0) =

EA(G, 0) and c1, c2, · · · , cq be q ≥ 2 distinct non-zero complex numbers. If

lim sup
R→∞,R∈I

3N0(R, F) +
∑q

j=1 N
(2)

0

&

R, 1
F−cj

'

+ N0
$

R, 1
F′

%

T0(R, F)
< q

lim sup
R→∞,R∈I

3N0(R, G) +
∑q

j=1 N
(2)

0

&

R, 1
G−cj

'

+ N0
$

R, 1
G′

%

T0(R, G)
< q

Where N
(2)

0 (R, ∗) = N0(R, ∗) + N
(2

0 (R, ∗), N
(2

0 (R, ∗) = N0(R, ∗) − N
1)

0 (R, ∗) and I is some set of R

of infinite linear measure, then

F =
aG + b

cG + d

where a, b, c, d ∈ C are constants with ad − bc ̸= 0.

Lemma 2.5. ([8]) Let F and G be admissible meromorphic functions in A satisfying EA1(F, 0) =

EA1 (G, 0) and let c1, c2, · · · , cq be q ≥ 2 distinct non-zero complex numbers. If

lim sup
R→∞,R∈I

3N0(R, F) +
∑q

j=1 N
(2)

0

&

R, 1
F−cj

'

+ N0
$

R, 1
F′

%

+ 2N
(2

0

$

R, 1
F

%

T0(R, F)
< q

lim sup
R→∞,R∈I

3N0(R, G) +
∑q

j=1 N
(2)

0

&

R, 1
G−cj

'

+ N0
$

R, 1
G′

%

+ 2N
(2

0

$

R, 1
G

%

T0(R, G)
< q



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Uniqueness of meromorphic functions sharing a set in annuli 7

Where N
(2)

0 (R, ∗), N
(2

0 (R, ∗) and I are stated as in Lemma 2.4, then

F =
aG + b

cG + d
,

where a, b, c, d ∈ C are constants with ad − bc ̸= 0.

Lemma 2.6. [6] Let f be a non-constant meromophic function on A = {z : 0 < |z| < ∞}. Let {av}

be any finite collection of complex numbers possibly including ∞. Then,

q∑

v=1

Θ0 (av) ≤ 2.

3 Proof of Theorem 1.1

If P(w) = wn + αwn−m + β, we can get that P(0) = β = c1(̸= 0)

and P(1) = 1 + α + β = c2(̸= 0), and

P ′(w) = wn−m−1 [nwm + α(n − m)] (3.1)

P(w) − c1 = w
n−mQ1(w) (3.2)

Where Q1(w) is a polynomial of degree m and Q1(0) ̸= 0.

P(w) − c2 = (w − 1)Q2(w) (3.3)

Where Q2(w) is a polynomial of degree n − 1 and Q2(1) ̸= 0.

From (3.2) and (3.3), we notice that Q1(w)and Q2(w) have only simple zeros.

Let F = P(f) and G = P(g). Since EA(S, f) = EA(S, g), we get that F and G share the value 0 CM.

From (3.2) and (3.3), we have

N
(2)

0

"

R,
1

F − c1

#

= N0

"

R,
1

F − c1

#

+ N
(2

0

"

R,
1

F − c1

#

≤ 2N0

"

R,
1

f

#

+

m∑

i=1

N0

"

R,
1

f − ai

#

= (m + 2)T0(R, f) + S(R) (3.4)

Where ai(i = 1, 2, · · · , m) are the zeros of Q1(w) in A and S(R) = o{T0(R)}, T0(R) = max{T0(R, f), T0(R, g)}.

And

N
(2)

0

"

R,
1

F − c2

#

= N0

"

R,
1

F − c2

#

+ N
(2

0

"

R,
1

F − c2

#

≤ N0

"

R,
1

f − 1

#

+

n−1∑

j=1

N0

"

R,
1

f − bj

#



8 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
18, 1 (2016)

= nT0(R, f) + S(R) (3.5)

Where bj(j = 1, 2, · · · , n − 1) are the zeros of Q2(w) in A.

From (3.1), we obtain

N0

"

R,
1

F′

#

= N0

"

R,
1

fn−m−1[nfm + α(n − m)]f′

#

≤ N0

"

R,
1

fn−m−1

#

+ N0

"

R,
1

nfm + α(n − m)

#

+ N0

"

R,
1

f′

#

+ S(R)

(3.6)

Using Lemma 2.3, (3.6) reduces to

N0

"

R,
1

F′

#

≤ N0

"

R,
1

f

#

+ N0

"

R,
1

nfm + α(n − m)

#

+ N0

"

R,
1

f

#

+ N0(R, f) + S(R)

≤ (m + 3)T0(R, f) + S(R) (3.7)

From Lemma 2.2, we get

T0(R, F) = nT0(R, f) + S(R) (3.8)

Using (3.4) − (3.8) in Lemma 2.4 and Since n > 2m + 8, we get

lim sup
R→∞,R∈I

3N0(R, F) +
∑2

j=1 N
(2)

0

&

R, 1
F−cj

'

+ N0
$

R, 1
F′

%

T0(R, F)

≤
n + 2m + 8

n
< 2 (3.9)

Similarly, we obtain

lim sup
R→∞,R∈I

3N0(R, G) +
∑2

j=1 N
(2)

0

&

R, 1
G−cj

'

+ N0
$

R, 1
G′

%

T0(R, G)
< 2 (3.10)

Thus, by Lemma 2.4, we have F ≡ aG+b
cG+d

, where, a, b, c, d ∈ C and ad − bc ̸= 0. Since EA(S, f) is

non-empty and EA(S, f) = EA(S, g), we have b = 0, a ̸= 0.

Hence,

F =
aG

cG + d
=

G

AG + B
(3.11)

where A = c
a
and B = d

a
. Now, we consider the following two cases:

Case 1: Suppose A ̸= 0. From (3.11), we notice that every zero of P(g) + B
A

in A has multiplicity

n.



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Uniqueness of meromorphic functions sharing a set in annuli 9

Next, the case 1 is followed by three following subcases:

Subcase 1: Suppose B
A

= −c1. From (3.2), we have

P(g) +
B

A
= gn−m(gm + α) = gn−m(g − a1) · · · (g − am) (3.12)

Where ai(i = 1, 2, · · · , m) are non-zero distinct roots of g. It follows that every zero of g in A has

multiplicity at least m and every zero of g−ai in A has multiplicity of at least n. Then by Lemma

2.1, we have

(m − 1)T0(R, g) ≤ N0

"

R,
1

g

#

+ N0

"

R,
1

g − a1

#

+ · · · + N0

"

R,
1

g − am

#

+ S(R, g)

≤
1

m
N0

"

R,
1

g

#

+
1

n
N0

"

R,
1

g − a1

#

+ · · · +
1

n
N0

"

R,
1

g − am

#

+ S(R, g)

≤

"

1

m
+

m

n

#

T0(R, g) + S(R) (3.13)

Since m ≥ 2 and n > 2m + 8, we arrive at a contradiction.

Subcase 2: Suppose B
A

= −c2. From (3.3), we have

P(g) +
B

A
= (g − 1)(g − b1)(g − b2) · · · (g − bn−1) (3.14)

Where bj ̸= 0, 1 are distinct values. For j = 1, 2, · · · , n − 1, consider

Θ(bj, f) = 1 − lim sup
R→∞

N0(R, bj, f)

T0(R, f)
>

1

2
(3.15)

We can see that P(g) + B
A

has n values satisfying the above inequality. Thus, by Lemma 2.6, we

get a contradiction.

Subcase 3: Suppose B
A

̸= −c1, −c2.

By using the same argument as in Subcase 1 or Subcase 2, we get a contradiction.

Case 2: Suppose A = 0.

If B ̸= 1, then from (3.11), we have F = G
B
; that is

P(f) =
1

B
P(g) (3.16)

From (3.2) and (3.16), we have

P(f) −
c1
B

=
1

B
gn−m(g − a1)(g − a2) · · · (g − am) (3.17)

Since c1
B

̸= c1, from (3.14), it follows that P(f) −
c1
B

has at least n distinct zeros e1, e2, · · · , en.



10 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
18, 1 (2016)

Then by applying Lemma 2.1, we have

(n − 2)T0(R, f) ≤
n∑

i=1

N0

"

R,
1

f − ei

#

+ S(R)

≤ N0

"

R,
1

g

#

+ N0

"

R,
1

g − a1

#

+ N0

"

R,
1

g − a2

#

+ · · · + N0

"

R,
1

g − am

#

+ S(R)

≤ (m + 1)T0(R, g) + S(R) (3.18)

By Applying Lemma 2.4 to (3.16) and from (3.18) and since n > 2m + 8 and m ≥ 2, we arrive at

a contradiction.

Thus, we get A = 0 and B = 1, that is P(f) = P(g)

⇒ fn + αfn−m = gn + αgn−m (3.19)

We set h = f
g
, we substitute f = hg in (3.19), it follows that

gn−m[gm(hn − 1) + α(hn−m − 1)] = 0 (3.20)

If h is a constant. We have from (3.20) that hn − 1 = 0 and hn−m − 1 = 0, which implies h = 1

and hence f ≡ g.

If h is not a constant, then suppose fn ̸≡ gn

Now consider,

gm =
−α (hn−m − 1)

hn − 1
(3.21)

gm =
−α(hn−m−1 + hn−m−2 + · · · + 1)

(hn−1 + hn−2 + · · · + 1)
(3.22)

gm =
−α(h − vn−m−1)(h − vn−m−2) · · · (h − v)

(h − un−1)(h − un−2) · · · (h − u)
(3.23)

where v = exp((2πi)/(n − m)) and u = exp((2πi)/n). Since n and m have no common factors, we

have vj ̸= uk(j = 1, 2, · · · , n − m − 1; k = 1, 2, · · · , n − 1). Suppose that zk is a zero of h − u
k of

order pk. From (3.23), we have pk ≥ m.

Thus

N0

"

R,
1

h − uk

#

≤
1

m
N0

"

R,
1

h − uk

#

≤
1

2
T0(R, h) + O(1) (3.24)



CUBO
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Uniqueness of meromorphic functions sharing a set in annuli 11

By Lemma 2.1 and from (3.24), we obtain

(n − 3)T0(R, h) <
n−1∑

k=1

N0

"

R,
1

h − uk

#

+ S(R, h)

≤
n − 1

2
T0(R, h) + S(R, h)

where S(R, h) is defined as in Lemma 2.1.

Since n > 2m + 8 and m ≥ 2, we arrive at a contradiction and since n and m have no common

factors, we get f ≡ g.

This completes the proof of Theorem 1.8.

Proof of Theorem 1.9 :

Since Θ0(∞, f) >
3
4
and Θ0(∞, g) >

3
4

It follows that

lim sup
R→∞

N0(R, f)

T0(R, f)
<

1

4
, lim sup

R→∞

N0(R, g)

T0(R, g)
<

1

4
(3.25)

By applying (3.25), from Lemma 2.4 and since n > 2m + 5, we deduce

lim sup
R→∞,R∈I

3N0(R, F) +
∑2

j=1 N
(2)

0

&

R, 1
F−cj

'

+ N0
$

R, 1
F′

%

T0(R, F)
(3.26)

<
4

n
lim sup

R→∞,R∈I

N0(R, f)

T0(R, f)
+ lim sup

R→∞,R∈I

(n + 2m + 4)T0(R, f)

nT0(R, f)
< 2 (3.27)

Similarly, we get

lim sup
R→∞,R∈I

3N0(R, G) +
∑2

j=1 N
(2)

0

&

R, 1
G−cj

'

+ N0
$

R, 1
G′

%

T0(R, G)
< 2 (3.28)

Then, from Lemma 2.4, we have F = aG+b
cG+d

, where a, b, c, d ∈ C and ad − bc ̸= 0. Thus, by using

the same argument as that in Theorem 1.8, we can prove the conclusion of Theorem 1.9.

Proof of Theorem 1.10 :

Since EA1 (S, f) = E
A
1(S, g), we have E

A
1(F, 0) = E

A
1 (G, 0)

From (3.1)-(3.3), we can get

N
(2

0

"

R,
1

F

#

=

n∑

i=1

N0

"

R,
1

f − di

#

≤ N0

"

R,
1

f′

#

(3.29)



12 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
18, 1 (2016)

Where di(i = 1, 2, · · · , n) are the distinct zeros of P(w). And from (3.6), (3.29) and by Lemma

2.3, we have

N0

"

R,
1

F′

#

+ 2N
(2

0

"

R,
1

F

#

≤ N0

"

R,
1

f

#

+ N0

"

R,
1

nfm + α(n − m)

#

+ 3N0

"

R,
1

f

#

+ 3N0(R, f) + S(R)

≤ (4 + m)T0(R, f) + 3N0(R, f) + S(R) (3.30)

Since n > 2m + 12, m ≥ 2, T0(R, F) = nT0(R, f) + S(R) and using equations (3.4), (3.5) and (3.30),

we deduce

lim sup
R→∞,R∈I

3N0(R, F) +
∑2

j=1 N
(2)

0

&

R, 1
F−cj

'

+ N0
$

R, 1
F′

%

+ 2N
(2

0

$

R, 1
F

%

T0(R, F)

≤
n + 2m + 12

n
< 2 (3.31)

Similarly, we get

lim sup
R→∞,R∈I

3N0(R, G) +
∑2

j=1 N
(2)

0

&

R, 1
G−cj

'

+ N0
$

R, 1
G′

%

+ 2N
(2

0

$

R, 1
G

%

T0(R, G)
< 2 (3.32)

Thus, from Lemma 2.5, we have F = aG+b
cG+d

, where a, b, c, d ∈ C and ad − bc ̸= 0. Hence, by using

the same argument as that in Theorem 1.8, we can prove the conclusion of Theorem 1.10.

Proof of Theorem 1.11 :

Since Θ0(∞, f) >
5
6
and Θ0(∞, g) >

5
6
,

It follows that

lim sup
R→∞

N0(R, f)

T0(R, f)
<

1

6
, lim sup

R→∞

N0(R, g)

T0(R, g)
<

1

6
(3.33)

By Lemma 2.5, (3.31)- (3.33) and since n > 2m + 7, we deduce

lim sup
R→∞,R∈I

3N0(R, F) +
∑2

j=1 N
(2)

0

&

R, 1
F−cj

'

− N0
$

R, 1
F′

%

+ 2N
(2

0

$

R, 1
F

%

T0(R, F)
(3.34)



CUBO
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Uniqueness of meromorphic functions sharing a set in annuli 13

<
6

n
lim sup

R→∞,R∈I

N0(R, f)

T0(R, f)
+ lim sup

R→∞,R∈I

(n + 2m + 6)T0(R, f)

nT0(R, f)
< 2 (3.35)

Similarly, we get

lim sup
R→∞,R∈I

3N0(R, G) +
∑2

j=1 N
(2)

0

&

R, 1
G−cj

'

+ N0
$

R, 1
G′

%

+ 2N
(2

0

$

R, 1
G

%

T0(R, G)
< 2 (3.36)

Then, from Lemma 2.5, we have F = aG+b
cG+d

, where a, b, c, d ∈ C and ad − bc ̸= 0. Thus, by using

the same argument as that in Theorem 1.8, we can prove the conclusion of Theorem 1.11.

Remark: The method used in this paper to prove the conclusion part of main results can be

applied to the sets of zeros of polynomials containing only three terms including constant term,

but not for more than three terms.

Acknowledgements: The authors are grateful to the referee for his/her keen observations, comments and
valuable suggestions towards the improvement of the present paper. First author is supported by UGC SAP DRS-
III with Ref. No. F.510/3/DRS-III/2016(SAP-I) Dated: 29th Feb. 2016. Second author and third author were
supported by UGC’s Research Fellowship in Science for meritorious Students, UGC, New Delhi. Ref. No.F.7-
101/2007(BSR) and Ref. No. KU/Sch/UGC-UPE/2014-15/894.

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14 Renukadevi S. Dyavanal, Ashwini M. Hattikal, Madhura M. Mathai CUBO
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