Advances in Pure Mathematics, 2011, 1, 23-27
doi:10.4236/apm.2011.12006 Published Online March 2011 (http://www.SciRP.org/journal/apm)
Inequalities for the Polar Derivative of a Polynomial
Gulshan Singh1, Wali Mohammad Shah2, Yash Paul1
1Bharathiar University Coimbatore, Tamil Nadu, India
2Department of Mathematics, Kashmir University, Srinagar, India
Received February 20, 2011; revised March 1, 2011; accepted March 5, 2011
Abstract
If

=0
:n
j
j
j
Pz az is a polynomial of degree n, having all its zeros in ,zK 1K, then it was pro-
vied by Aziz and Rather [2] that for every real or complex number

with K  ,


=1z
MaxD Pz






=1
1z
n
nK
M
axP z
K
 
. In this paper, we sharpen above result for the poly nomials

Pz of degree >3n.
Keywords: Polynomial, Inequality, Polar Derivative
1. Introduction
Let

=0
:= n
j
j
j
Pz az
 be a polynomial of degree n and

P' z its derivative, then
 
=1 =1zz
M
axPznMaxP z
 (1)
Inequality (1) is a famous result due to Bernstein and
is best possible with equality holding for the polynomial

=n Pz z

, where

is a complex number.
If we restricted ourselves to a class of polynomial
having no zeros in <1z, then the above inequ ality can
be sharpened. In fact, Erdös conjectured and later Lax [6]
proved that if

0Pz in <1z, then
 
=1 =1
2
zz
n
M
axPzMaxPz
 (2)
On the other hand, it was proved by Turán [10] that if

Pz is a polynomial of degree n having all its zeros in
1z, then
 
=1 =1
2
zz
n
M
axPzMaxPz
 (3)
The inequalities (2) and (3) are also best possible and
become equality for polyno mials which have all zeros on
1z.
For the class of polynomials having all the zeros in
zK, Malik [7] (See also Govil [5]) proved that if

Pz is a polynomial of degree n having all zeros lie in
zK, then
 
=1=1 , if 1,
1
zz
n
MaxPzMaxP zK
K


(4)
where as Govil [5] showed that
 
=1=1 , if 1
1
zz n
n
MaxPzMaxP zK
K

 (5)
Both the inequalities are best possible, with equality in
(4) holding for
 
=n Pzz K and in (5) the equality
holds for the p olynomial


=nn
P
zzK.
Let


DP z

denote the polar derivative of the
polynomial


Pz of degree n with respect to

, then



 
=DP znP zz Pz



 .
The polynomial


DP z

is of degree at most 1n

and it generalizes the ordinary derivative in the sense that
 
=.
lim DP zPz

  
Aziz and Rather [2] extended (5) to the polar deri-
vative of a polynomial and proved the following:
Theorem 1: If the polynomial

=0
:= n
j
j
j
Pz az
 has
all its zeros in zK

, 1K, then for every real or
complex number

with
K
 ,




=1 =1
1
zz
n
nK
M
axDP zMaxPz
K

 
 (6)
In this paper, we prove th e following result which is a
refinement as well as generalization of Theorem 1.

G. SINGH ET AL. 24
Theorem 2: Let 
=0
:= n
j
j
j
P
zaz
, 00
n
aa  be a po-
lynomial of degree >3n, having all its zeros in zK

,
1K, then for every real or complex number

with
K
 ,



 









1
=1=1 =
2
2
1
21
1
1
11 121
2, if >3.
123
n
n
zzzK
n
nn
n
K
nK a
MaxDP zMaxP zMinP zK
nn
K
KnKK nK
an
nnn n














 
 
 


 
 
 
 


(7)
Remark 1: For =1
K
, Theorem 2 provides a refi-
ment of a theorem proved by Shah [9].
Remark 2: For >1
K
, and for >1y,



11
1
y
KyK
yy

 

 and


1
y
K
y
 are both increa-
sing functions of y and so the expressions






2
11 121
123
nn
KnKK nK
nnn n


 
 
 


 
 
 
 

and



11
n
KK
n










are always non-negative so that for polynomials of
degree >3n, Theorem 2 is an improvement of Theo-
rem 1.
Dividing both sides of (7) by

and letting  ,
we get the following:
Corollary 1: Let 
=0
=n
j
j
j
P
zaz
, 00
n
aa  be a po-
lynomial of degree >3n, having all its zeros in zK

,
1
K
, then


  









1
=1=1 =
2
2
1
21
1
1
11 121
2, if >3.
123
n
n
zzzK
n
nn
n
K
a
n
MaxPzMaxP zMinP zK
nn
K
KnKK nK
an
nnn n












 
 
 


 
 
 
 


(8)
2. Lemmas
We need the following lemmas.
Lemma 1: Let


Pz be a polynomial of degree n,
then for 1R.
 
==1
.
n
zR z
M
axPzRMaxPz
The above lemma is a simple consequence of the
maximum modulus principle [8].
Lemma 2: If

=0
:= n
j
j
j
Pz az
, 0
n
a, is a poly-
nomial of degree n having all its zeros in 1z, then
 

=1=1 =1
2
zzz
n
M
axPzMaxPzMinP z
.
This lemma is due to Aziz and Dawood [1].
Lemma 3: If

=0
:= n
j
j
j
Pz az
 is a polynomial of
degree n having no zeros in 1z, and


=1
=z mMin Pz, then for 1R and >3n,




 











=1
2
111
20
,1
221
11 121
0123
nnn
z
nn
RRR
P
MPRMax PzmR
nn
RnRR nR
Pnnn n














 
 
 



 
 
 
 

.
The above result is a special case of a result due to
Dewan, Singh and Mir [4, Theorem 1] with =1K and
=1

.
Remark 3: Here we note that for the proof of this
result an additional hypothesis that

00P is required.
A simple counter example in this case is

=n Pz z.

G. SINGH ET AL. 25
3. Proof of Theorem 2
Since

Pz has all its zeros in zK, therefore
 
=Gz PKz has all its zeros in 1z and hence by
applying lemma 2 to the polynomial

Gz, we get
  

=1=1=1 .
2
zzz
n
M
axGzMaxG zMinG z
 (9)
Let

1
=n Hz zGz



. Then it can be easily verified
that
 


=, for =1.HznGzzGzz

 (10)
The polynomial


H
z has all its zeros in 1z
and
 
=
H
zGz for 1z

, therefore, by result of
a de Bruijn [3]


for =1Hz Gzz

 (11)
Now for every real or complex nu mber α with
K
 ,
we have
  
 
=
K
DGznGzzG zGz
K
GznGzzGz
K







For this, we get by using (10) and (11)
 
=1 =1zz
K
K
M
axD G zMaxGz
K

 
 (12)
Using (9) in (12), we get



 

=1=1 =1
2
zzz
K
Kn
M
axDG zMaxGzMinG z
K

 
.
Replacing

Gz by

PKz, we have





||=1 =1 =1
2z
zz
K
nK
M
axDP KzMaxP KzMinPKz
K

 
 .
This gives
 


 

=1 =1 =1
2
zzz
nK
M
axnPKzzKPKzMaxP KzMinP Kz
KK





 

 .
Equivalently



 

===
.
2
zKzK zK
nK
M
axDP zMaxP zMinP z
K

 
 (13)
Since the polynomial P(z) has all its zeros in
zK, 1K. If

1
=n QzzP z



be the reciprocal
polynomial of


Pz. Then the polyn omial z Q
K



has
all its zeros in 1z. Hence applying lemma 3 to the
polynomial z
Q
K



, 1
K
, we get













1
==1=1
2
2
11 1
21
221
11 121
2123
nn n
n
zK zz
nn
n
KK K
a
zzz
MaxQMax QMin QK
KKKnn
KnKK nK
annn n





 
 






 



 
 
 


 
 
 
 

This in particular gives






 









1
=1 ==
2
2
11 1
21
1
22
11 121
2123
nn n
n
zzKzK
nn
nn
n
KK K
a
MaxP zMaxP zMinP zK
nn
KK
KnKK nK
annn n





 









 
 
 


 
 
 
 


G. SINGH ET AL. 26
which is equivalent to











 





1
==1=
2
2
11
24
1
1
111
11 121
4
123
1
nn
n n
n
zK zzK
nn n
nn
n
n
n
KK
a
KK
MaxP zMaxP zMinP zK
nn
KK K
KnKK nK
Kannn n
K








 







 
 
 


 
 
 
 

(14)
Using (14) in (13), we get












 




 
1
==1=
2
2=
11
24
{ 1
21
111
11 121
4
123
1
nn
n n
n
zKz zK
nn n
nn
n
nzK
n
KK
nK a
KK
MaxDP zMaxP zMinP zK
Knn
KK K
KnKK nK
KaMinPz
nnn n
K














 
 
 


 
 
 
 


, if >3.n
Equivalently



 









11
==1=
2
2
1
21
1
1
11 121
2, if >3.
123
n
n
n
zKz zK
n
nn
n
K
nKK a
MaxDP zMaxP zMinP zK
nn
K
KnKK nK
an
nnn n























 
 
 


 
 
 
 


(15)
Since

DP z

is a polynomial of degree 1n

and 1K, therefore by using Lemma 1, we get




1
==1
n
zK z
M
axDP zKMaxDP z


 (16)
Combining (16) and (15) we have



  









1
=1=1 =
2
2
1
21
1
1
11 121
2||, if >3.
123
n
n
zzzK
n
nn
n
K
nK a
MaxDP zMaxPzMinP zK
nn
K
KnKK nK
an
nnn n














 
 
 


 


 
 


This completes the proof of Theorem 2.
4. Acknowledgements
The authors are extremely grateful to the referee for his
valuable su ggestions.
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