# Some Bounds for the Polar Derivative of a Polynomial

Authors: Jiraphorn Somsuwan; Keaitsuda Maneeruk Nakprasit
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

---

## Abstract

The polar derivative of a polynomialp
(
z
) of degree n with respect to a complex number α is a polynomial n
p
(
z
)
+
α
-
z
p
′
(
z
), denoted by D
α
p
(
z
). Let 1
≤
R
≤
k. For a polynomial p
(
z
) of degree n having all its zeros in z
≤
k, we investigate a lower bound of modulus of D
α
p
(
z
) on z
=
R. Furthermore, we present an upper bound of modulus of D
α
p
(
z
) on z
=
R for a polynomial p
(
z
) of degree n having no zero in z
<
k. In particular, our results in case R
=
1 generalize some well-known inequalities.


---

## Body

## 1. Introduction

One of the interesting and fruitful subjects in geometry of polynomials is the geometrical relation between the modulus of a complex polynomial on a circle and the position of zeros of this polynomial inside or outside this circle. Many propositions in the area of polynomial inequalities are presented by Bernstein-type inequalities. Many results are derived from Bernstein’s inequality. We start with a result due to Bernstein (see [1]). Let p
(
z
) be a polynomial of degree n
. Then according to the well-known result known as Bernstein’s inequality, we have(1)
max
z
=
1
⁡
p
′
z
≤
n
max
z
=
1
⁡
p
z
.The equality holds if and only if p
(
z
) has all its zeros at the origin.For the class of polynomials having no zero inz
<
1, Lax [2] proved that if p
(
z
) is a polynomial of degree n having no zero in z
<
1, then(2)
max
z
=
1
⁡
p
′
z
≤
n
2
max
z
=
1
⁡
p
z
.The result is best possible and equality holds for p
(
z
)
=
λ
+
μ
z
n
,
λ
=
μ.On the other hand, Turán [3] showed that, for a polynomial having all its zeros in z
≤
1,(3)
max
z
=
1
⁡
p
′
z
≥
n
2
max
z
=
1
⁡
p
z
.Inequality (3) is best possible and becomes equality for polynomials which have all its zeros on z
=
1.An extension of inequality (3) was shown by Govil [4] that, for a polynomial p
(
z
) of degree n having all its zeros in z
≤
k
,
k
≥
1,(4)
max
z
=
1
⁡
p
′
z
≥
n
1
+
k
n
max
z
=
1
⁡
p
z
.Inequality (4) is best possible and equality holds for a polynomial p
(
z
)
=
z
n
+
k
n.In 1999, Dewan et al. [5] proved the following result.Theorem 1 (see [5]).
Letp
(
z
)
=
a
0
+
∑
ν
=
μ
n
a
ν
z
ν
,
1
≤
μ
≤
n, be a polynomial of degree n such that p
(
z
) has no zero in z
<
k
,
k
>
0
. Then, for 0
<
r
≤
R
≤
k,(5)
max
z
=
r
⁡
p
z
≥
k
μ
+
r
μ
k
μ
+
R
μ
n
/
μ
max
z
=
R
⁡
p
z
+
1
-
k
μ
+
r
μ
k
μ
+
R
μ
n
/
μ
min
z
=
k
⁡
p
z
.Inequality (5) is best possible and becomes equality for a polynomial p
(
z
)
=
(
z
μ
+
k
μ
)
n
/
μ, where n is multiple of μ.Thepolar derivative of a polynomial p
(
z
) of degree n with respect to a complex number α, denoted by D
α
p
(
z
), is defined by(6)
D
α
p
z
=
n
p
z
+
α
-
z
p
′
z
.The polynomial D
α
p
(
z
) is of degree at most n
-
1 and it generalizes the ordinary derivative in the sense that lim
α
→
∞
⁡
(
D
α
p
(
z
)
/
α
)
=
p
′
(
z
).In 1998, Aziz and Rather [6] established the extension of inequality (4) to a polar derivative of a polynomial.Theorem 2 (see [6]).
Letp
(
z
) be a polynomial of degree n having all its zeros in z
≤
k
,
k
≥
1.
Then, for every real or complex numberα with α
≥
k,(7)
max
z
=
1
⁡
D
α
p
z
≥
n
α
-
k
1
+
k
n
max
z
=
1
⁡
p
z
.In 2008, Dewan and Upadhye [7] proved the following theorem.Theorem 3 (see [7]).
Letp
(
z
) be a polynomial of degree n having all its zeros in z
≤
k
,
k
≥
1.
Then, for every real or complex numberα with α
≥
k,(8)
max
z
=
1
⁡
D
α
p
z
≥
n
α
-
k
1
1
+
k
n
max
z
=
1
⁡
p
z
+
1
2
k
n
k
n
-
1
k
n
+
1
min
z
=
k
⁡
p
z
.Furthermore, Dewan et al. [8] proved the following result.Theorem 4 (see [8]).
Ifp
(
z
)
=
a
0
+
∑
ν
=
μ
n
a
ν
z
ν
,
1
≤
μ
≤
n, is a polynomial of degree n having no zero in z
<
k
,
k
≥
1, then, for every real or complex number α with α
≥
1,(9)
max
z
=
1
⁡
D
α
p
z
≤
n
1
+
s
0
α
+
s
0
max
z
=
1
⁡
p
z
-
α
-
1
m
,where m
=
min
z
=
k
⁡
p
z and s
0
=
k
μ
+
1
μ
/
n
a
μ
/
a
0
-
m
k
μ
-
1
+
1
/
μ
/
n
a
μ
/
a
0
-
m
k
μ
+
1
+
1.Let1
≤
R
≤
k
. In this paper, we investigate a lower bound of D
α
p
z on z
=
R for a polynomial p
(
z
) of degree n having all its zeros in z
≤
k and an upper bound of D
α
p
z on z
=
R for a polynomial p
(
z
) of degree n having no zero in z
<
k.


## 2. Main Results

Theorem 5.
Letp
(
z
)
=
a
0
+
∑
ν
=
μ
n
a
ν
z
ν
,
1
≤
μ
≤
n
,
a
0
≠
0, be a polynomial of degree n having all its zeros in z
≤
k
,
k
≥
1
. Then, for every real or complex number α with α
≥
k and 1
≤
R
≤
k,(10)
max
z
=
R
⁡
D
α
p
z
≥
n
R
n
-
1
α
-
k
R
n
R
n
+
k
n
k
μ
+
1
k
μ
+
R
μ
n
/
μ
max
z
=
1
⁡
p
z
+
k
n
R
n
+
k
n
1
-
k
μ
+
1
k
μ
+
R
μ
n
/
μ
+
1
2
k
n
k
n
-
R
n
k
n
+
R
n
m
,where m
=
min
z
=
k
⁡
p
z.Proof.
LetF
(
z
)
=
p
(
R
z
)
. Observe that F
(
z
) has all its zeros in z
≤
k
/
R.
Applying Theorem3 to F
(
z
), we obtain that(11)
max
z
=
1
⁡
D
α
/
R
F
z
≥
n
α
-
k
R
R
n
R
n
+
k
n
max
z
=
1
⁡
F
z
+
R
n
2
k
n
k
n
-
R
n
k
n
+
R
n
min
z
=
k
/
R
⁡
F
z
.Applying the relations max
z
=
1
⁡
D
α
/
R
F
z
=
max
z
=
R
⁡
D
α
p
z, max
z
=
1
⁡
F
z
=
max
z
=
R
⁡
p
z, and min
z
=
k
/
R
⁡
F
z
=
min
z
=
k
⁡
p
z into inequality (11), we have that(12)
max
z
=
R
⁡
D
α
p
z
≥
n
R
n
-
1
α
-
k
1
R
n
+
k
n
max
z
=
R
⁡
p
z
+
1
2
k
n
k
n
-
R
n
k
n
+
R
n
min
z
=
k
⁡
p
z
.
Letq
(
z
)
=
z
n
p
(
1
/
z
¯
)
¯
=
a
0
¯
z
n
+
∑
ν
=
μ
n
a
ν
¯
z
n
-
ν.
Sincep
(
z
) has all its zeros in z
≤
k, q
(
z
) has no zero in z
<
1
/
k.
LetQ
(
z
)
=
q
(
z
/
(
k
2
)
). Then Q
(
z
) is a polynomial of degree n having no zero in z
<
k. Applying Theorem 1 to Q
(
z
) with r
=
1, we obtain that(13)
max
z
=
1
⁡
Q
z
≥
k
μ
+
1
k
μ
+
R
μ
n
/
μ
max
z
=
R
⁡
Q
z
+
1
-
k
μ
+
1
k
μ
+
R
μ
n
/
μ
min
z
=
k
⁡
Q
z
.By maximum modulus principle (see [9], p. 128), we obtain that(14)
max
z
=
1
⁡
Q
z
=
1
k
2
n
max
z
=
1
⁡
p
k
2
z
¯
¯
≤
1
k
2
n
max
z
=
k
2
⁡
p
k
2
z
¯
¯
≤
1
k
2
n
max
z
=
R
⁡
p
z
,
max
z
=
R
⁡
Q
z
=
R
n
k
2
n
max
z
=
R
⁡
p
k
2
z
¯
¯
≥
R
n
k
2
n
max
z
=
1
⁡
p
z
,and min
z
=
k
⁡
Q
z
=
1
/
k
n
min
z
=
k
⁡
p
z.
Combining these relations with inequality (13), we obtain that(15)
max
z
=
R
⁡
p
z
≥
k
μ
+
1
k
μ
+
R
μ
n
/
μ
R
n
max
z
=
1
⁡
p
z
+
1
-
k
μ
+
1
k
μ
+
R
μ
n
/
μ
k
n
min
z
=
k
⁡
p
z
.This inequality together with inequality (12) yields(16)
max
z
=
R
⁡
D
α
p
z
≥
n
R
n
-
1
α
-
k
R
n
R
n
+
k
n
k
μ
+
1
k
μ
+
R
μ
n
/
μ
max
z
=
1
⁡
p
z
+
k
n
R
n
+
k
n
1
-
k
μ
+
1
k
μ
+
R
μ
n
/
μ
+
1
2
k
n
k
n
-
R
n
k
n
+
R
n
m
,where m
=
min
z
=
k
⁡
p
z.Remark 6.
Inequality (10) is best possible and the equality holds for p
(
z
)
=
z
-
k when R
=
1 and α
≥
k
≥
1 or R
=
n
≥
2 and α
≥
k
=
n.Remark 7.
In caseR
=
1, inequality (10) reduces to(17)
max
z
=
1
⁡
D
α
p
z
≥
n
α
-
k
1
1
+
k
n
max
z
=
1
⁡
p
z
+
1
2
k
n
k
n
-
1
k
n
+
1
min
z
=
k
⁡
p
z
.This means that Theorem 5 generalizes Theorem 3.Theorem 8.
Ifp
(
z
)
=
a
0
+
∑
ν
=
μ
n
a
ν
z
ν
,
1
≤
μ
≤
n, is a polynomial of degree n having no zero in z
<
k
,
k
≥
1, then, for every real or complex number α with α
≥
R and 0
<
r
≤
R
≤
k,(18)
max
z
=
R
⁡
D
α
p
z
≤
n
1
+
s
0
α
R
+
s
0
k
μ
+
R
μ
k
μ
+
r
μ
n
/
μ
max
z
=
r
⁡
p
z
-
α
R
-
1
+
α
R
+
s
0
k
μ
+
R
μ
k
μ
+
r
μ
n
/
μ
1
-
k
μ
+
r
μ
k
μ
+
R
μ
n
/
μ
m
,where m
=
min
z
=
k
⁡
p
z and s
0
=
k
/
R
μ
+
1
(
μ
/
n
)
R
μ
a
μ
/
a
0
-
m
(
k
/
R
)
μ
-
1
+
1
/
(
μ
/
n
)
R
μ
a
μ
/
a
0
-
m
(
k
/
R
)
μ
+
1
+
1.
The equality holds for a polynomialp
(
z
)
=
z
+
k, where α
∈
R with α
≥
1.Proof.
LetF
(
z
)
=
p
(
R
z
)
. Observe that F
(
z
) has no zero in z
<
k
/
R.
Applying Theorem4 to F
(
z
), we obtain that(19)
max
z
=
1
⁡
D
α
/
R
F
z
≤
n
1
+
s
0
α
R
+
s
0
max
z
=
1
⁡
F
z
-
α
R
-
1
min
z
=
k
/
R
⁡
F
z
,where s
0
=
k
/
R
μ
+
1
(
μ
/
n
)
|
R
μ
a
μ
|
/
|
a
0
|
-
min
z
=
k
/
R
⁡
F
z
(
k
/
R
)
μ
-
1
+
1
/
(
μ
/
n
)
|
R
μ
a
μ
|
/
|
a
0
|
-
min
z
=
k
/
R
⁡
F
z
(
k
/
R
)
μ
+
1
+
1
.
We apply the relationsmax
|
z
|
=
1
⁡
D
α
/
R
F
z
=
max
z
=
R
⁡
D
α
p
z, max
z
=
1
⁡
F
z
=
max
z
=
R
⁡
p
z, and min
z
=
k
/
R
⁡
F
z
=
min
z
=
k
⁡
p
z into inequality (19) to obtain(20)
max
z
=
R
⁡
D
α
p
z
≤
n
1
+
s
0
α
R
+
s
0
max
z
=
R
⁡
p
z
-
α
R
-
1
m
,where m
=
min
z
=
k
⁡
p
z and s
0
=
k
/
R
μ
+
1
(
μ
/
n
)
|
R
μ
a
μ
|
/
|
a
0
|
-
m
(
k
/
R
)
μ
-
1
+
1
/
(
μ
/
n
)
|
R
μ
a
μ
|
/
|
a
0
|
-
m
(
k
/
R
)
μ
+
1
+
1.
Theorem1 implies that(21)
max
z
=
R
⁡
p
z
≤
k
μ
+
R
μ
k
μ
+
r
μ
n
/
μ
max
z
=
r
⁡
p
z
-
k
μ
+
R
μ
k
μ
+
r
μ
n
/
μ
-
1
min
z
=
k
⁡
p
z
.This inequality together with inequality (20) yields(22)
max
z
=
R
⁡
D
α
p
z
≤
n
1
+
s
0
α
R
+
s
0
k
μ
+
R
μ
k
μ
+
r
μ
n
/
μ
max
z
=
r
⁡
p
z
-
α
R
+
s
0
k
μ
+
R
μ
k
μ
+
r
μ
n
/
μ
-
1
+
α
R
-
1
m
,where m
=
min
z
=
k
⁡
p
z and s
0
=
k
/
R
μ
+
1
(
μ
/
n
)
|
R
μ
a
μ
|
/
|
a
0
|
-
m
(
k
/
R
)
μ
-
1
+
1
/
(
μ
/
n
)
|
R
μ
a
μ
|
/
|
a
0
|
-
m
(
k
/
R
)
μ
+
1
+
1.
Next, we show that the equality holds for a polynomialp
(
z
)
=
z
+
k, where α
∈
R with α
≥
1
. Let α
∈
R with α
≥
1
. Consider p
(
z
)
=
z
+
k; we obtain that max
z
=
R
⁡
D
α
p
z
=
α
+
k
. Observe that the right-hand side of inequality (18) becomes |
α
|
+
k
. Furthermore, α
+
k
=
α
+
k by condition of α
. Therefore, the equality holds for a polynomial p
(
z
)
=
z
+
k.Remark 9.
In caser
=
R
=
1, inequality (18) reduces to(23)
max
z
=
1
⁡
D
α
p
z
≤
n
1
+
s
0
α
+
s
0
max
z
=
1
⁡
p
z
-
α
-
1
m
,where m
=
min
z
=
k
⁡
p
z and s
0
=
k
μ
+
1
(
μ
/
n
)
|
a
μ
|
/
|
a
0
|
-
m
k
μ
-
1
+
1
/
(
μ
/
n
)
|
a
μ
|
/
|
a
0
|
-
m
k
μ
+
1
+
1.
That is, inequality (18) generalizes inequality (9) in Theorem 4.



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