Volume 9 (2008), Issue 4, Article 119, 9 pp.
INEQUALITIES FOR THE POLAR DERIVATIVE OF A POLYNOMIAL
K.K. DEWAN AND C.M. UPADHYE
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCE
JAMIA M ILIA I SLAMIA (C ENTRAL U NIVERSITY )
N EW D ELHI -110025 (INDIA)
G ARGI C OLLEGE (U NIVERSITY OF D ELHI )
S IRI F ORT ROAD , N EW D ELHI -110049 (INDIA)
c_upadhye@rediffmail.com
Received 17 April, 2007; accepted 15 May, 2008
Communicated by D. Stefanescu
A BSTRACT. In this paper we obtain new results concerning maximum modules of the polar
derivative of a polynomial with restricted zeros. Our results generalize and refine upon the
results of Aziz and Rather [3] and Jagjeet Kaur [9].
Key words and phrases: Polynomials, Inequality, Polar Derivative.
2000 Mathematics Subject Classification. 30A10, 30C15.
1. I NTRODUCTION AND S TATEMENT OF R ESULTS
Let p(z) be a polynomial of degree n and p0 (z) its derivative. It was proved by Turán [11]
that if p(z) has all its zeros in |z| ≤ 1, then
n
(1.1)
max |p0 (z)| ≥ max |p(z)|.
|z|=1
2 |z|=1
The result is best possible and equality holds in (1.1) if all the zeros of p(z) lie on |z| = 1.
For the class of polynomials having all its zeros in |z| ≤ k, k ≥ 1, Govil [7] proved:
P
Theorem A. If p(z) = nv=0 av z v is a polynomial of degree n having all the zeros in |z| ≤ k,
k ≥ 1, then
n
(1.2)
max |p0 (z)| ≥
max |p(z)| .
|z|=1
1 + k n |z|=1
Inequality (1.2) is sharp. Equality holds for p(z) = z n + k n .
Let Dα {p(z)} denote the polar derivative of the polynomial p(z) of degree n with respect to
α, then
Dα {p(z)} = np(z) + (α − z)p0 (z) .
122-07
2
K.K. D EWAN AND C.M. U PADHYE
The polynomial Dα {p(z)} is of degree at most n−1 and it generalizes the ordinary derivative
in the sense that
Dα p(z)
lim
= p0 (z).
α→∞
α
Aziz and Rather [3] extended (1.2) to the polar derivative of a polynomial and proved the
following:
P
Theorem B. If the polynomial p(z) = nv=0 av z v has all its zeros in |z| ≤ k, k ≥ 1, then for
every real or complex number α with |α| ≥ k,
|α| − k
(1.3)
max |Dα p(z)| ≥ n
max |p(z)|.
|z|=1
1 + k n |z|=1
The bound in Theorem B depends only on the zero of largest modulus and not on the other
zeros even if some of them are close to the origin. Therefore, it would be interesting to obtain
a bound, which depends on the location of all the zeros of a polynomial. In this connection we
prove the following:
Theorem 1.1. Let
p(z) =
n
X
n
Y
av z = an (z − zv ),
v
v=0
an 6= 0,
v=1
be a polynomial of degree n, |zv | ≤ kv , 1 ≤ v ≤ n, let k = max(k1 , k2 , . . . , kn ) ≥ 1. Then for
every real or complex number |α| ≥ k,
(1.4)
max |Dα p(z)|
|z|=1
≥ (|α| − k)
n
X
v=1
k
2
1 kn − 1
max |p(z)| + n
min |p(z)| .
k + kv 1 + k n |z|=1
k
k n + 1 |z|=k
Dividing both sides of (1.4) by |α| and letting |α| → ∞, we get the following refinement of
a result due to Aziz [1].
P
Q
Corollary 1.2. Let p(z) = nv=0 av z v = an nv=1 (z − zv ), an 6= 0, be a polynomial of degree
n, |zv | ≤ kv , 1 ≤ v ≤ n, let k = max(k1 , k2 , . . . , kn ) ≥ 1. Then,
n
X
k
2
1 kn − 1
0
(1.5)
max |p (z)| ≥
max |p(z)| + n
min |p(z)| .
|z|=1
k + kv 1 + k n |z|=1
k
k n + 1 |z|=k
v=1
k
Since k+k
≥ 12 for 1 ≤ v ≤ n, Theorem 1.1 gives the following result, which is an improvev
ment of Theorem B.
Q
Corollary 1.3. If p(z) = an nv=1 (z − zv ), an 6= 0, is a polynomial of degree n, having all its
zeros in |z| ≤ k, k ≥ 1, then for every real or complex number |α| ≥ k,
n
1
1
k −1
(1.6)
max |Dα p(z)| ≥ n(|α| − k)
max |p(z)| + n
min |p(z)| .
|z|=1
1 + k n |z|=1
2k
k n + 1 |z|=k
Dividing both sides of (1.6) by |α| and letting |α| → ∞, we obtain the following refinement
of Theorem A due to Govil [7].
Q
Corollary 1.4. If p(z) = an nv=1 (z − zv ), an 6= 0, is a polynomial of degree n, having all its
zeros in |z| ≤ k, k ≥ 1, then
n
1
1
k −1
0
(1.7)
max |p (z)| ≥ n
max |p(z)| + n
min |p(z)| .
|z|=1
1 + k n |z|=1
2k
k n + 1 |z|=k
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 119, 9 pp.
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P OLAR D ERIVATIVE OF A P OLYNOMIAL
3
The bound in Theorem 1.1 can be further improved for polynomials of degree n ≥ 2. More
precisely, we prove the following:
Q
P
Theorem 1.5. Let p(z) = nv=0 av z v = an nv=1 (z − zv ), an 6= 0, be a polynomial of degree
n ≥ 2, |zv | ≤ kv , 1 ≤ v ≤ n, and let k = max(k1 , k2 , . . . , kn ) ≥ 1. Then for every real or
complex number |α| ≥ k,
n
X
k
2
1 kn − 1
(1.8) max |Dα p(z)| ≥ (|α| − k)
max |p(z)| + n
n |z|=1
|z|=1
k
+
k
1
+
k
k
kn + 1
v
v=1
n
k − 1 k n−2 − 1
2|an−1 |
× min |p(z)| +
−
|z|=k
k(1 + k n )
n
n−2
1
+ 1 − 2 |na0 + αa1 | for n > 2
k
and
n
X
2
1 kn − 1
k
(1.9) max |Dα p(z)| ≥ (|α| − k)
max |p(z)| + n
n |z|=1
|z|=1
k
+
k
1
+
k
k
kn + 1
v
v=1
(k − 1)n
1
× min |p(z)| + |a1 |
+ 1−
|na0 + αa1 | for n = 2 .
n
|z|=k
k(1 + k )
k
k
k+kv
≥
1
2
for 1 ≤ v ≤ n, the above theorem gives in particular:
Q
Corollary 1.6. If p(z) = an nv=1 (z − zv ), an 6= 0, is a polynomial of degree n having all its
zeros in |z| ≤ k, k ≥ 1, then for every real or complex number |α| ≥ k,
n
1
1
k −1
(1.10) max |Dα p(z)| ≥ n(|α| − k)
max |p(z)| + n
|z|=1
1 + k n |z|=1
2k
kn + 1
n
k − 1 k n−2 − 1
|an−1 |
−
× min |p(z)| +
|z|=k
k(1 + k n )
n
n−2
1
+ 1 − 2 |na0 + αa1 | for n > 2 ,
k
and
2
1 kn − 1
(1.11) max |Dα p(z)| ≥ (|α| − k)
max |p(z)| + n
|z|=1
1 + k n |z|=1
k
kn + 1
|a1 |(k − 1)n
1
× min |p(z)| +
+ 1−
|na0 + αa1 |, for n = 2 .
n
|z|=k
k(1 + k )
k
n
n−2
Now it is easy to verify that if k > 1 and n > 2, then k n−1 − k n−2−1 > 0. Hence for
polynomials of degree n ≥ 2, the above corollary is a refinement of Theorem 1.1. In fact,
except the case when p(z) has all the zeros on |z| = k, a0 = 0, a1 = 0, and an−1 = 0, the bound
obtained by Theorem 1.5 is always sharper than the bound obtained by Theorem 1.1.
Since
Remark 1. Dividing both sides of inequalities (1.8), (1.9), (1.10) and (1.11) by |α| and letting
|α| → ∞, we get the results due to Jagjeet Kaur [9]. In addition to this, if min |p(z)| = 0 i.e. if
|z|=k
a zero of a polynomial lies on |z| = k, then we obtain the results due to Govil [8].
Finally, as an application of Theorem 1.1 we prove the following:
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 119, 9 pp.
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4
K.K. D EWAN AND C.M. U PADHYE
P
Q
Theorem 1.7. If p(z) = nv=0 av z v = an nv=1 (z − zv ), an 6= 0, is a polynomial of degree n,
|zv | ≥ kv , 1 ≤ v ≤ n, and k = min(k1 , k2 , . . . , kn ) ≤ 1, then for every real or complex number
δ with |δ| ≤ k,
n
X
kv
2
1 − kn
n−1
(1.12) max |Dδ p(z)| ≥ (k − |δ|)k
max |p(z)| + n
m ,
n |z|=1
n)
|z|=1
k
+
k
1
+
k
k
(1
+
k
v
v=1
where m = min |p(z)|.
|z|=k
2. L EMMAS
For the proofs of the theorems, we need the following lemmas.
Lemma 2.1. If p(z) is a polynomial of degree n, then for R ≥ 1
max |p(z)| ≤ Rn max |p(z)| .
(2.1)
|z|=R
|z|=1
The above lemma is a simple consequence of the Maximum Modulus Principle [10].
P
Lemma 2.2. If p(z) = nv=0 av z v is a polynomial of degree n, then for all R > 1,
max |p(z)| ≤ Rn max |p(z)| − (Rn − Rn−2 )|p(0)| for n ≥ 2
(2.2)
|z|=R
|z|=1
and
max |p(z)| ≤ R max |p(z)| − (R − 1)|p(0)| for
(2.3)
|z|=1
|z|=1
n = 1.
This result is due to Frappier, Rahman and Ruscheweyh [5].
Q
Lemma 2.3. If p(z) = an nv=1 (z − zv ), is a polynomial of degree n ≥ 2, |zv | ≥ 1 for
1 ≤ v ≤ n, then
n
Rn + 1
R −1
Rn−2
Rn − 1
max |p(z)| − |a1 |
−
−
min |p(z)|,
(2.4) max |p(z)| ≤
|z|=1
|z|=R≥1
2
n
n−2
2 |z|=1
if n > 2
and
(2.5)
max |p(z)| ≤
|z|=R≥1
R2 + 1
(R − 1)2 R2 − 1
max |p(z)| − |a1 |
−
min |p(z)|, if
|z|=1
2
2
2 |z|=1
n = 2.
The above lemma is due to Jagjeet Kaur [9].
Q
Lemma 2.4. If p(z) = an nv=1 (z −zv ), an 6= 0, is a polynomial of degree n, such that |zv | ≤ 1,
1 ≤ v ≤ n, then
n
X
1
0
(2.6)
max |p (z)| ≥
max |p(z)|.
|z|=1
1 + |zv | |z|=1
v=1
This lemma is due to Giroux, Rahman and Schmeisser [6].
Lemma 2.5. If p(z) is a polynomial of degree n, which has all its zeros in the disk |z| ≤ k,
k ≥ 1, then
2k n
(2.7)
max |p(z)| ≥
max |p(z)|.
|z|=k
1 + k n |z|=1
Inequality (2.7) is best possible and equality holds for p(z) = z n + k n .
The above result is due to Aziz [1].
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 119, 9 pp.
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P OLAR D ERIVATIVE OF A P OLYNOMIAL
5
Lemma 2.6. If p(z) is a polynomial of degree n, having all its zeros in the disk |z| ≤ k, k ≥ 1,
then
kn − 1
2k n
max
|p(z)|
+
min |p(z)|.
(2.8)
max |p(z)| ≥
|z|=k
1 + k n |z|=1
1 + k n |z|=k
The result is best possible and equality holds for p(z) = z n + k n .
Proof of Lemma 2.6. For k = 1, there is nothing to prove. Therefore it is sufficient to consider
the case k > 1.
Let m = min |p(z)|. Then m ≤ |p(z)| for |z| = k.
|z|=k
Since all the zeros of p(z) lie in |z| ≤ k, k > 1, by Rouche’s theorem, for every λ with |λ| < 1,
the polynomial p(z) + λm has all its zeros in |z| ≤ k, k > 1. Applying Lemma 2.5 to the
polynomial p(z) + λm, we get
2k n
max |p(z) + λm| ≥
max |p(z) + λm|.
|z|=k
1 + k n |z|=1
Choosing the argument of λ such that |p(z) + λm| = |p(z)| + |λ|m and letting |λ| → 1, we get
2k n
kn − 1
max |p(z)| ≥
max
|p(z)|
+
min |p(z)|.
|z|=k
1 + k n |z|=1
1 + k n |z|=k
This completes the proof of Lemma 2.6.
Lemma 2.7. If p(z) is a polynomial of degree n and α is any real or complex number with
|α| =
6 0, then
(2.9)
|Dα q(z)| = |nαp(z) + (1 − αz)p0 (z)| for |z| = 1.
Lemma 2.7 is due to Aziz [2].
3. P ROOFS OF THE T HEOREMS
Proof of Theorem 1.1. Let G(z) = p(kz). Since the zeros of p(z) are zv , 1 ≤ v ≤ n, the zeros
of the polynomial G(z) are zv /k, 1 ≤ v ≤ n, and because all the zeros of p(z) lie in |z| ≤ k, all
the zeros of G(z) lie in |z| ≤ 1, therefore applying Lemma 2.4 to the polynomial G(z), we get
n
X
1
0
(3.1)
max |G (z)| ≥
max |G(z)|.
|zv | |z|=1
|z|=1
1
+
v=1
k
Let H(z) = z n G(1/z). Then it can be easily verified that
(3.2)
|H 0 (z)| = |nG(z) − zG0 (z)|,
for |z| = 1.
The polynomial H(z) has all its zeros in |z| ≥ 1 and |H(z)| = |G(z)| for |z| = 1, therefore, by
the result of de Bruijn [4]
|H 0 (z)| ≤ |G0 (z)| for |z| = 1.
(3.3)
Now for every real or complex number α with |α| ≥ k, we have
α
|Dα/k G(z)| = nG(z) − zG0 (z) + G0 (z)
k
0
≥ |α/k||G (z)| − |nG(z) − zG0 (z)|.
This gives with the help of (3.2) and (3.3) that
(3.4)
max |Dα/k G(z)| ≥
|z|=1
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 119, 9 pp.
|α| − k
max |G0 (z)| .
|z|=1
k
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6
K.K. D EWAN AND C.M. U PADHYE
Using (3.1) in (3.4), we get
n
|α| − k X
k
max |Dα/k G(z)| ≥
max |G(z)|.
|z|=1
k
k + |zv | |z|=1
v=1
Replacing G(z) by p(kz), we get
max |Dα/k p(kz)| ≥ (|α| − k)
|z|=1
n
X
v=1
1
max |p(kz)|
k + |zv | |z|=1
which implies
n
α
X
1
0
max np(kz) +
− z kp (kz) ≥ (|α| − k)
max |p(kz)|,
|z|=1
k
k + |zv | |z|=1
v=1
which gives
max |Dα p(z)| ≥ (|α| − k)
|z|=k
n
X
v=1
1
max |p(z)|.
k + |zv | |z|=k
Using Lemma 2.6 in the above inequality, we get
(3.5)
max |Dα p(z)| ≥ (|α| − k)
|z|=k
n
X
v=1
2k n
1
max |p(z)|
k + |zv | 1 + k n |z|=1
n
k −1
+
min |p(z)| .
1 + k n |z|=k
Since Dα p(z) is a polynomial of degree at most n − 1 and k ≥ 1, applying Lemma 2.1 to the
polynomial Dα p(z), we get
max |Dα p(z)| ≤ k n−1 max |Dα p(z)|.
(3.6)
|z|=k
|z|=1
Combining (3.5) and (3.6), we get
max |Dα p(z)|
|z|=1
n
X
k
2
1 kn − 1
≥ (|α| − k)
max |p(z)| + n
min |p(z)|
n |z|=1
n
|z|=k
k
+
|z
|
1
+
k
k
1
+
k
v
v=1
n
X
k
2
1 kn − 1
≥ (|α| − k)
max |p(z)| + n
min |p(z)|
n |z|=1
n
|z|=k
k
+
k
1
+
k
k
1
+
k
v
v=1
which is the required result. Hence the proof Theorem 1.1 is complete.
Proof of Theorem 1.5. Let G(z) = p(kz). Since the zeros of p(z) are zv , 1 ≤ v ≤ n, the zeros
of the polynomial G(z) are zv /k, 1 ≤ v ≤ n, and because all the zeros of p(z) lie in |z| ≤ k,
all the zeros of G(z) lie in |z| ≤ 1, therefore applying Lemma 2.4 to the polynomial G(z) and
proceeding in the same way as in Theorem 1.1, we obtain
(3.7)
max |Dα p(z)| ≥ (|α| − k)
|z|=k
n
X
v=1
1
max |p(z)|.
k + |zv | |z|=k
1
Now let q(z) = z n p z be the reciprocal polynomial
of p(z). Since the polynomial p(z) has
z
all its zeros in |z| ≤ k, k ≥ 1 the polynomial q k has all its zeros in |z| ≥ 1. Hence applying
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 119, 9 pp.
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P OLAR D ERIVATIVE OF A P OLYNOMIAL
7
(2.4) of Lemma 2.3 to the polynomial q kz , k ≥ 1, we get
z k n + 1
z k n − 1 z max q
max q
min q
≤
−
|z|=k
|z|=1
k
2 |z|=1
k
2
k
|an−1 | k n − 1 k n−2 − 1
−
−
,
k
n
n−2
which gives
n
k −1
kn + 1
max |p(z)| ≤
max |p(z)| −
min |p(z)|
|z|=1
|z|=k
2k n |z|=k
2k n
|an−1 | k n − 1 k n−2 − 1
−
−
,
k
n
n−2
which is equivalent to
n
2k n
k −1
(3.8)
max |p(z)| ≥
max |p(z)| +
min |p(z)|
|z|=k
1 + k n |z|=1
1 + k n |z|=k
2|an−1 |k n−1 k n − 1 k n−2 − 1
+
−
.
1 + kn
n
n−2
Using (3.8) in (3.7) we get
n
n
X
1
2k n
k −1
(3.9) max |Dα p(z)| ≥ (|α| − k)
max |p(z)| +
n |z|=1
|z|=k
k
+
|z
|
1
+
k
1 + kn
v
v=1
2|an−1 |k n−1 k n − 1 k n−2 − 1
× min |p(z)| +
−
if n > 2 .
|z|=k
1 + kn
n
n−2
Since Dα p(z) is a polynomial of degree n − 1 and k ≥ 1, from (2.2) of Lemma 2.2, we get
(3.10)
max |Dα p(z)| ≤ k n−1 max |Dα p(z)| − (k n−1 − k n−3 )|Dα p(0)|,
|z|=k
|z|=1
if n > 2 .
Combining (3.9) and (3.10) we have
n
X
k
2
1 kn − 1
max |Dα p(z)| ≥ (|α| − k)
max |p(z)| + n
min |p(z)|
n |z|=1
n
|z|=1
|z|=k
k
+
|z
|
1
+
k
k
1
+
k
v
v=1
n
2|an−1 |
k − 1 k n−2 − 1
+
−
k(1 + k n )
n
n−2
1
+ 1 − 2 |na0 + αa1 |, if n > 2
k
n
X k 2
1 kn − 1
≥ (|α| − k)
max |p(z)| + n
min |p(z)|
n |z|=1
n
|z|=k
k
+
k
1
+
k
k
1
+
k
v
v=1
n
2|an−1 |
k − 1 k n−2 − 1
+
−
k(1 + k n )
n
n−2
1
+ 1 − 2 |na0 + αa1 |, if n > 2
k
which completes the proof of (1.8).
The proof of (1.9) follows on the same lines as the proof of (1.8) but instead of (2.2) and (2.4)
we use inequalities (2.3) and (2.5) respectively. We omit the details.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 119, 9 pp.
http://jipam.vu.edu.au/
8
K.K. D EWAN AND C.M. U PADHYE
Proof of Theorem 1.7. By hypothesis, the zeros of p(z) satisfy |zv | ≥ kv for 1 ≤ v ≤ n such
that k = min(k1 , k2 , . . . , kn ) ≤ 1. It follows that the zeros of the polynomial q(z) = z n p(1/z)
satisfy 1/|zv | ≤ 1/kv , 1 ≤ v ≤ n such that 1/k = max(1/k1 , 1/k2 , . . . , 1/kn ) ≥ 1. On
applying Theorem 1.1 to the polynomial q(z), we get
n
X
2/k n
1
n−1
(3.11) max |Dα q(z)| ≥ k (|α| − 1/k)
max |q(z)|
n |z|=1
|z|=1
1/k
+
1/k
1
+
1/k
v
v=1
1
1/k n − 1
+
min |q(z)| , |α| ≥ .
n
1 + 1/k |z|=1/k
k
Now from Lemma 2.7 it follows that
|Dα q(z)| = |α||D1/ᾱ p(z)| for |z| = 1 .
Using the above equality in (3.11), we get for |α| ≥ 1/k,
(3.12) |α| max |D1/ᾱ p(z)| ≥ k
|z|=1
n−1
n
X
2
kkv
max |p(z)|
(|α| − 1/k)
n |z|=1
k
+
k
1
+
k
v
v=1
1 − kn
+
min |p(z)| .
(1 + k n )k n |z|=k
Replacing
1
α
by δ, so that |δ| ≤ k, we get from (3.12)
|1/δ| max |Dδ p(z)| ≥ k
|z|=1
n−1
n
X
kkv
2
(|1/δ| − 1/k)
max |p(z)|
k + kv 1 + k n |z|=1
v=1
1 − kn
+
min |p(z)| ,
(1 + k n )k n |z|=k
or
n
X
2
1 − kn
kv
max |p(z)| +
min |p(z)| ,
max |Dδ p(z)| ≥ k (k − |δ|)
|z|=1
k + kv 1 + k n |z|=1
(1 + k n )k n |z|=k
v=1
n
X
kv
2
1 − kn
n−1
≥ k (k − |δ|)
max |p(z)| +
min |p(z)| ,
n |z|=1
n )k n |z|=k
k
+
k
1
+
k
(1
+
k
v
v=1
n−1
which is (1.12). Hence the proof of Theorem 1.7 is complete.
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