SQU Journal for Science, 2018, 23(1), 68-72  DOI: http://dx.doi.org/10.24200/squjs.vol23iss1pp68-72 

Sultan Qaboos University  

68 

 

Inequalities Concerning the Growth of 
Polynomials 

Bashir A. Zargar* and Ahmad W. Manzoor
 

Department of Mathematics, University of Kashmir, Srinagar-190006, India. *Email: 
bazargar@gmail.com 

ABSTRACT: In this paper we consider a polynomial )( zP  having no zeros in the disk 1|<| z . We investigate the 

dependence of |)(|
1>|=|

zPmax
Rz

 on |)(|
1|=|

zPmax
z

 and obtain a refinement of a famous result due to Rivilin ([5], 

[7]). Our results not only generalize some polynomial inequalities but also refine a result by Aziz [1]. 

 

Keywords: Growth of polynomials, Maximum modulus, Inequalities. 

 بتزايد الدوالعالقة المتراجحات 

   منزورحمد أو  زارجار حمدأشير ب

)( دالةال الورقة هذه في نفترض :صالملخ zP 1 القرص في أصفار لها ليس التي|<| z. 1|)(| المقدار عالقة عن سنبحث>|=| zPmax Rz  على  

|)(|
1|=|

zPmax
z ولكنها بحسف والالد متراجحات لبعض اتعميم ليست نتائجنا إن وريفيلين. أنكيني إلى تعود معروفة نتيجةل تنظيم على والحصول 

 [.1] عزيز نتيجةل نظيمات أيضا

 

 .الحد األقصى، المتراجحات نموذج،  تزايد الدوال  :مفتاحيةالكلمات ال

  

1.  Introduction 

et )( zP  be a polynomial of degree n . Then ([5] or [6], p. 347), for a fixed 1>R , we have  
 

| |= | |=1
(1)| ( ) | | ( ) | .

n

z R z
Max P z R Max P z    

Equality in (1) holds for the polynomial 
n

zzP =)( . 

It was shown by Rivilin ( [5] , [7] ) that if )( zP  is a polynomial of degree n  having no zeros on 1|<| z , then 

(1) can be replaced by  

| |= | |=1
(2)

1
| ( ) | | ( ) | .

2

n

z R z

R
Max P z Max P z

 
  
 

                                                 

Inequality (2) is sharp and equality holds for ,=)(
n

zzP    .||=||   
Aziz [1] has further improved and generalized inequality (2) by proving the following result: 

Theorem A. If )( zP  is a polynomial of degree n  which does not vanish in the disk kz |<|  where 1k , then  

       | |= | |=1 | |=1 (3)
1 1

| ( ) | | ( ) | | ( ) | .
2 2

n n

z R z z

R R
Max P z Max P z Min P z

    
    
   

  

The result is best possible and equality holds for the polynomial ,=)(
nn

kzzP    1.1,|=|=|| k  
 

 

L 



INEQUALITIES CONCERNING THE GROWTH OF POLYNOMIALS 

69 

 

As a generalization of inequality (2), Aziz [1] conjectured the following results.  

 

Conjectured Results. If )( zP  is a polynomial of degree n  which does not vanish in the disk kz |<| , then  

                   

                          
2

| |= | |=1
(4)| ( ) | | ( ) |, < < 1, < 1

1

n n

z r zn

r k
Max P z Max P z k r k

k





    

 and  

           
2

| |= | |=1
(5)| ( ) | | ( ) |, > , > 1.

1

n n

z R zn

R k
Max P z Max P z R k k

k





  

In an attempt to answer inequality (4), Dewan and Hans [4] proved the following partial results. 

 

Theorem B. If )( zP  is a polynomial of degree n , which does not vanish in 1,<,|<| kkz  then for 

1<<<0 rk , 

 

),,(),( 


pM
k

kr
rpM

nn

nn






 
provided |)(| zp  and |)(| zq  attain the maximum at the same point on 1|=| z , where 

1
( ) = ( )

n
q z z P

z
 and .|)(|=),(),,(|=)(|

|=||=|
zPMaxpMrpMzPMax

zrz   

The result is best possible and equality holds for 
nn

kzzp =)( . 
 

Theorem C. If )( zP  is a polynomial of degree n , which does not vanish in 1,<,|<| kkz  then for 

1<<<0 rk   
                           

                                       (6)
1

( , ) ( ,1) ( , )
1 1

n n n

n n

r k r
M p r M p m p k

k k

    
    

    
  

provided |)(| zp  and |)(| zq  attain the maximum at the same point on 1|=| z , where 

1
( ) = ( )

n
q z z P

z
    and   .|)(|=),(

|=|
zPMinkpm

kz
 

The result is best possible and equality in (6) holds for 
nn

kzzP =)( . 
 

In this paper we shall first present the following interesting refinement of Theorem A. 

 

Theorem 1. Let 

 
,......=)(

01

1

1
azazazazP

n

n

n

n




  
be a polynomial of degree n which does not vanish in 1,|<| kkz . Then for 1,>R   

                 | |= | |=1 | |=1 (7)
1

| ( ) | 1 | ( ) | | ( ) |
1 1

n n

z R z z
Max P z R Max P z R Min P z




 
   

 
 

where  

                                                                            

                                                                  
0

0

(8)
| | (| | )

=
| | (| | )

n

n

a R a m

a R a m


 

 
  

| |=1
and = | ( ) | .

z
m Min P z  

The result is sharp and equality in (7)  holds for 1.|=|=||,
2

=)( 


n
z

zP


 

 

Remark 1.1. Here we have replaced k  by   simply not to confuse it with the region for which )( zP  does not 
vanish. Now  



B.A. ZARGAR and A.W. MANZOOR
 

 

70 

 

   

1<
||||

||)|(|
=

0

0

maaR

amaR

n

n






 
It is easy to verify the above inquality for 1<  if  

  

1.>
||

||
0

ma

a

n


 

To show it holds, let |)(|=
1|=|

zPMinm
z

 then |)(| zPm   for 1,|=| z  so that |)(|<|| zPzm
n

  where   

is any real or complex number with 1.|<|   Since )( zP  does not vanish in 1|<| z  the polynomial  

 

 0
.....)(=)(=)( azmamzzPzF

n

n

n
 

 

does not vanish in 1.|<| z Therefore, 1>
0

ma

a

n


 or 1>
||

||
0

ma

a

n


, for every   with 1.|<|   Choosing 

argument of   such that 

 
mama

nn
||||=||  

 
we get 

 
1.|<|,||||>||

0
 maa

n


 
Letting 1||   it follows that 

.||||
0

maa
n


 
Now it is easy to verify that for 1< , 

SO
1 1

< ( 1) <
1 2 1 2

n
n R

R
 

 




 
  

and 

11
<

2

1

2
=

2

1















nnn
RRR

 
which is true. This shows Theorem 1 is an improvement of Theorem A. 

As an application of Theorem 1, we next establish the following result which, in a way, is similar to 

inequality (6).  

Theorem 2. If 
j

j

n

j

zazP 
0=

=)(  is a polynomial of degree n  which does not vanish in 1|| z , then for 

1<0 r , we have  

                  
| |= | |= | |=1

(9)
( 1) 1 (1 )

| ( ) | | ( ) | | ( ) |
1 1

n n

z r z r zn n

r r
Max P z Min P z Max P z

r r

 

 

 
 

 
  

0

0

and
| | (| | )

= .
| | (| | )

n

n

a R a m

a R a m


 

 
 

The result is best possible and equality in (9) holds for the polynomial  
n

zzP =)( , where 1.|=|=||   

 

Lemmas 
For the proof of Theorem 1, we need the following Lemmas. The first Lemma is due to Dubinin [3, Theorem 5]. 

 

Lemma 1. If  

,......=)(
01

1

1
azazazazP

n

n

n

n




  
is a polynomial of degree n  which does not vanish in 1|<| z , then for every 1R   

                            

                                            

0

0

(10)
| | | |

| ( ) |= | ( ) |,| |= 1
| | | |

n

n

a R a
P Rz Q Rz z

R a a




 

where  



INEQUALITIES CONCERNING THE GROWTH OF POLYNOMIALS 

71 

 

 
 zpzzQ n 1/=)(

. 

Equality is attained for the polynomial )( zP  whose zeros lie on the unit circle 1.|=| z  

Our next lemma is due to Aziz and Mohammad [2]. 
 

Lemma 2. If )( zP  is a polynomial of degree n , then for all 1R   and 0 < 2   

 
|)(|1)(|)(||)(|

1|=|
zPMaxRReQReP

z

nii




 
where 

 
1

( ) = ( )
n

Q z z P
z

. 

2. Proofs of Theorems 

Proof of Theorem 1. Let |)(|=
1|=|

zPMinm
z

. Then |)(| zPm   for 1|=| z  so that |)(|<|| zPzm
n

  for 

1|=| z , where   is any real or complex number with 1|<| . 
Since the polynomial  

 
,......=)(

01

1

1
azazazazP

n

n

n

n




  

does not vanish in 1|<| z , an application of Rouches Theorem shows that the polynomial 
n

mzzP )(  does 

not vanish in 1|<| z , so that the polynomial  

              
n

mzzPzF )(=)(  

                                             
1

1 1 0
= ( ) .... ,

n n

n n
a m z a z a z a




      

does not vanish in 1|<| z  for every  , 1.|<|   
Let  

mzQm
z

Pz
z

FzzG
nn

  )(=)
1

(=)
1

(=)(
       

and         
1

( ) = ( ).
n

Q z z P
z

 

Using Lemma 1, it follows that  

0

0

1.for
| | | |

| ( ) | | ( ) |,
| | | |

n

n

z
a R a m

F z G z
R a a m






 


 
  

This implies 

.|)(|
||||

||||
|)(|

0

0 mzQ
maaR

maRa
mzzP

n

nn 



 






 
We now show that for 1|<|  and 1,>R   

                                
0 0

0 0

(11)
| | | | | | (| | )

<
| | | | | | (| | )

n n

n n

a R a m a R a m

R a a m R a a m





   

   
  

Inequality (11) holds if 

 
)|)(|()|(|||)|(|||||

00

22

0
mamaRmaamaaRRa

nnnn
 

 

 
),|(|||)|(|||||||||

00

22

0
mamaRmaamaaRRa

nnnn
 

 
which after a simple calculation, yields  

 
)}.|(|)|{(||||}|)|{(|||

00

2
mamaamamaaR

nnnn
 

 
This implies 1,R which is true. Hence (11) is established. 

Taking in particular 
i

Rez = , where 1>R  and  2<0  , we get 

                           (12)| ( ) | | ( ) |
i n in i

P Re mR e Q Re m
  

      

for every   with 1|<| . Choosing the argument of   in (12) such that 

       
,|||)(|=|)(|

niinni
mRRePemRReP 




 
we get 

         
.|||)(||||)(| mReQmRReP

ini





 



B.A. ZARGAR and A.W. MANZOOR
 

 

72 

 

This gives  

                   | ( ) | | | ( ) | ( ) |, 0 < 2 .
i n i

P Re m R Q Re
 

         (13) 

Letting 1||   in (13), we get 

 | ( ) | ( ) | ( ) |, 0 < 2 .
i n i

P Re R m Q Re
 

        

Adding |)(|



i

ReP  on both sides it follows that 

  
|},)(||)({|)(|)(|1)(




iini
ReQRePmRReP 

 
for all  , 0 < 2   . 

This gives, with the help of Lemma 2, that  

           | |=1 (14)( 1) | ( ) | ( ) ( 1) | ( ) |
i n n

z
P Re R m R Max P z


        

for all   , 0 < 2   . 

From (14), it follows that 

 
| |=1

1
| ( ) | ( 1) | ( ) | ( )

1 1

i n n

z
P Re R Max P z R m

 


 
   

 
 

for all , 0 < 2   , which is equivalent to the desired result.                                                                                   □ 

Proof of Theorem 2. All the zeros of )( zP  lie in 1|| z ; therefore for 1,<0 r  the polynomial )(rzP  has all 

the zeros in 1>
1

||
r

z  . Applying Theorem 1 to the polynomial )(rzP , we obtain 

| |=1 | |=1

1
| ( ) | ( 1) | ( ) | ( 1) .

1 1

n n

z z
Max P rz R Max P rz R m



 
   

 
 

Equivalently, 

 

.|)(|)(
1

1
|)(|1)(

1
|)(|

|=||=|1|=|
zPMinRzPMaxRRzPMax

rz

n

rz

n

z













 

Taking 
r

R
1

= , then for 1<0 r , we obtain 

 

|,)(||)(|
)(1

1

1
|)(|

1)(

1
1|=||=||=|

zPMaxzPMin
r

r
zPMax

r

r
zrzn

n

rzn

n

















 
which proves Theorem 2.                                                                                                                                          □ 

3. Conclusion  

We generalize some polynomial inequalities and refine a previous result on the dependence of 

|)(|
1>|=|

zPmax
Rz

 on |)(|
1|=|

zPmax
z

, where )( zP  is a polynomial having no zeros in the disk 1|<| z .  

References 

1. Aziz, A. Growth of polynomials whose zeros are within or outside a circle. Bulletin of the Australian 
Mathematical Society, 1987, 35, 247-256. 

2. Aziz, A. and Mohammad, Q.G. Simple proof of a Theorem of Erdos and Lax. Proceedings of the American 
Mathematical Society, 1980, 80, 119-122. 

3. Dubinin, V.N. Applications of Schwarz Lemma to inequalities for entire functions with constraints on zeros, 
Journal of Mathematical Sciences, 2007, 143(3), 3069-3075. 

4. Dewan, K.K. and Hans, S. Growth of polynomials whose zeros are outside a circle, Annales Universitates Mariae 
Curie-Sklodowska Lublin-Polonia, LXII, 2008, 61-65. 

5. Rahman, Q.I.  and Schmeisser, G.  Analytic Theory of Polynomials, Oxford University Press, New York, 2002. 
6. Riesz, M. Über einen satz des Herrn Serge Bernstein, Acta Mathematica, 2007, 40(3), 3069- 3075. 
7. Rivilin, T.J. On the maximum modulus of polynomials, American Mathematical Monthly, 1960, 67, 251-253.  
    
 

Received  22 January 2017            

Accepted 26 September 2017