ON THE MAXIMUM MODULUS OF
POLYNOMIALS. II
Maximum Modulus of
Polynomials
M. A. QAZI
Department of Mathematics
Tuskegee University
Tuskegee, Alabama 36088, USA
EMail: qazima@aol.com
M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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Received:
15 February, 2007
Accepted:
23 August, 2007
Communicated by:
N.K. Govil
2000 AMS Sub. Class.:
30D15, 41A10, 41A17.
Key words:
Polynomials, Inequality, Zeros.
Abstract:
Pn
ν
Let f (z) :=
be a polynomial of degree n having no zeros in
ν=0 aν z
the open unit disc, and suppose that max|z|=1 |f (z)| = 1. How small can
max|z|=ρ |f (z)| be for any ρ ∈ [0 , 1)? This problem was considered and solved
by Rivlin [4]. There are reasons to consider the same problem under the additional assumption that f 0 (0) = 0. This was initiated by Govil [2] and followed
up by the present author [3]. The exact answer is known when the degree n is
even. Here, we make some observations about the case where n is odd.
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Contents
1
Introduction
3
2
Statement of Results
7
3
An Auxiliary Result
10
4
Proofs of Theorems 2.1 and 2.2
13
Maximum Modulus of
Polynomials
M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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1.
Introduction
For any entire function f let
M (f ; ρ) := max |f (z)|
|z|=ρ
(0 ≤ ρ < ∞) ,
and denote by Pn the class of all polynomials of degree at most n. If f ∈ Pn , then,
applying the maximum modulus principle to the polynomial
Maximum Modulus of
Polynomials
f ∼ (z) := z n f (1/z) ,
M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
we see that
(1.1)
M (f ; r) = rn M (f ∼ ; r−1 ) ≥ rn M (f ∼ ; 1) = rn M (f ; 1)
(0 ≤ r < 1) ,
where equality holds if and only if f (z) := cz n , c ∈ C, c 6= 0. For the same reason
(1.2)
M (f ; R) = Rn M (f ∼ ; R−1 ) ≤ Rn M (f ∼ ; 1) = Rn M (f ; 1)
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(R ≥ 1) .
Rivlin [6] proved that if f ∈ Pn and f (z) 6= 0 for |z| < 1, then
n
1+r
(1.3)
M (f ; r) ≥ M (f ; 1)
(0 ≤ r < 1),
2
P
where equality holds if and only if f (z) := nν=0 cν z ν has a zero of multiplicity n
on the unit circle, that is, if and only if c0 6= 0 and |c1 | = |p0 (0)| = n|c0 |.
The preceding inequality was generalized by Govil [2] as follows.
Theorem A. Let f ∈ Pn . Furthermore let f (z) 6= 0 for |z| < 1. Then,
n
1 + r1
(0 ≤ r1 < r2 ≤ 1) .
(1.4)
M (f ; r1 ) ≥ M (f ; r2 )
1 + r2
Here again equality holds for polynomials of the form f (z) := c (1 + eiγ z)n , where
c ∈ C, c 6= 0, γ ∈ R.
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The next result which is also due to Govil [2] gives a refinement of (1.4) under
the additional assumption that f 0 (0) = 0.
P
Theorem B. Let f (z) := nν=0 cν z ν 6= 0 for |z| < 1, and let c1 = f 0 (0) = 0. Then
for 0 ≤ r1 < r2 ≤ 1, we have
n (
n−1 )−1
1 + r1
(1−r2 )(r2 −r1 )n 1+r1
(1.5) M (f ; r1 ) ≥ M (f ; r2 )
1−
.
1 + r2
4
1+r2
Improving upon Theorem B, we proved (see [3] or [5, Theorem 12.4.10]) the
following result.
P
Theorem C. Let f (z) := nν=0 cν z ν 6= 0 for |z| < 1, and let λ := c1 /(nc0 ). Then
n
1 + 2|λ|r1 + r12 2
(1.6)
M (f ; r1 ) ≥ M (f ; r2 )
(0 ≤ r1 < r2 ≤ 1) .
1 + 2|λ|r2 + r22
Note. It may be noted that 0 ≤ |λ| ≤ 1.
If n is even, then for any r2 ∈ (0 , 1], and any r1 ∈ [0 , r2 ), equality holds in (1.6)
Maximum Modulus of
Polynomials
M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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for
f (z) := c(1 + 2|λ|eiγ z + e2iγ z 2 )n/2 ,
c ∈ C, c 6= 0, |λ| ≤ 1, γ ∈ R.
By an argument different from the one used to prove Theorem C, we obtained in
[4] the following refinement of (1.6).
P
Theorem D. Let f (z) := nν=0 cν z ν 6= 0 for |z| < 1, and let λ := c1 /(nc0 ). Then,
for any γ ∈ R, we have
n
1 + 2|λ|r1 + r12 2
iγ
iγ
(1.7)
|f (r1 e )| ≥ |f (r2 e )|
(0 ≤ r1 < r2 ≤ 1) .
1 + 2|λ|r2 + r22
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Again, (1.7) is not sharp for odd n. The proof of (1.7) is based on the observation
that for 0 ≤ r < 1, we have
r<
n
n
f 0 (r)
=n−<
≤n−
,
f (r)
1 − r ϕ(r)
1 + r |ϕ(r)|
where
f 0 (z)
zf 0 (z) − nf (z)
is analytic in the closed unit disc, and max|z|=1 |ϕ(z)| ≤ 1. Since ϕ(0) = −λ,
a familiar generalization of Schwarz’s lemma [7, p. 212] implies that |ϕ(r)| ≤
(r + λ)/(λr + 1) for 0 ≤ r < 1, and so if 0 ≤ r1 < r2 ≤ 1, then
Z r 2
n
1 + 2|λ|r2 + r22 2
f 0 (r)
|f (r2 )| = |f (r1 )| exp
<
dr ≤ |f (r1 )|
,
f (r)
1 + 2|λ|r1 + r12
r1
ϕ(z) :=
which readily leads us to (1.7).
It is intriguing that this reasoning works fine for any even n, and so does the one
that was used to prove Theorem C, but somehow both lack the sophistication needed
to settle the case where n is odd. We know that when n is even, the polynomials
which minimize |f (r1 )|/|f (r2 )| have two zeros of multiplicity n/2 each. However,
n/2 6∈ N when n is odd, and so the form of the extremals must be different in the
case where n is even.
Q.I. Rahman, who co-authored [4], had communicated with James Clunie about
Theorem D years earlier, and had asked him for his thoughts about possible extremals when n is odd and c1 is 0. In other words, what kind of a polynomial f of
odd degree n would minimize |f (r)|/|f (1)| if
!
n
n
Y
X
f (z) :=
(1 + ζν z)
|ζ1 | ≤ 1, . . . , |ζn | ≤ 1 ;
ζν = 0 ?
ν=1
ν=1
Maximum Modulus of
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M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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Generally, one uses a variational argument in such a situation. In a written
P note, Clunie remarked that, in the case of odd degree polynomials, the condition nν=1 ζν = 0
is much more difficult to work with than it is in the case of even degree polynomials,
and proposed to check if
(1.8)
1 + r3
|f (r)|
≥
|f (1)|
2
for 0 ≤ r ≤ 1 if n = 3 and f 0 (0) = 0
Maximum Modulus of
Polynomials
and
M. A. Qazi
3
1+r 1+r
|f (r)|
≥
|f (1)|
2
2
2
for 0 ≤ r ≤ 1 if n = 5 and f 0 (0) = 0 .
He added that if above held, it would seem reasonable to conjecture that if n =
2m + 1, m ∈ N, and f 0 (0) = 0, then
m−1
1 + r3 1 + r2
|f (r)|
for 0 ≤ r ≤ 1 .
(1.9)
≥
|f (1)|
2
2
We shall see that (1.8) does not hold at least for r = 0. The same can be said
about (1.9).
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2.
Statement of Results
Let λ ∈ C,
≤ 1. We shall denote by Pn,λ the class of all polynomials of the form
P|λ|
n
f (z) := ν=0 cν z ν , not vanishing in the open unit disc, such that c1 /(nc0 ) = λ.
Thus, if f belongs to Pn,λ , then
!
n
n
Y
X
f (z) := c0
(1 + ζν z)
|ζ1 | ≤ 1, . . . , |ζn | ≤ 1 ;
ζν = n λ .
ν=1
Maximum Modulus of
Polynomials
ν=1
M. A. Qazi
Let us take any two numbers r1 and r2 in [0, 1] such that r1 < r2 . Then by (1.7), for
any real γ, we have
n
1 + 2|λ|r2 + r22 2
|f (r2 eiγ )|
≤
(0 ≤ r1 < r2 ≤ 1) .
|f (r1 eiγ )|
1 + 2|λ|r1 + r12
iγ
vol. 8, iss. 3, art. 72, 2007
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iγ
In addition, we know that the upper bound for |f (r2 e )|/|f (r1 e )| given by the
preceding inequality is attained if the degree n is even, and that it is attained for
a polynomial which has exactly two distinct zeros, each of multiplicity n/2 and of
modulus 1. When it comes to the case where n is odd, this bound is not sharp. What
then is the best possible upper bound for |f (r2 eiγ )/|f (r1 eiγ )| when n is odd; is the
bound attained? If the bound is attained, can we say something about the extremals?
We shall first show that
|f (r2 eiγ )|
(2.1)
Ωr1 ,r2 ,γ := sup
: f ∈ Pn,λ
|f (r1 eiγ )|
is attained. For this it is enough to prove that for any c 6= 0 the polynomials
f ∈ Pn,λ : f (r1 eiγ ) = c
form a normal family of functions, say Fc (for the definition of a normal
Q family see
[1, pp. 210–211]). In order to prove that Fc is normal, let f (z) := a0 nν=1 (1 + ζν z),
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where |ζ1 | ≤ 1, . . . , |ζn | ≤ 1. Then |f (z)| ≤ |a0 | 2n for |z| = 1 whereas |c| =
|f (r1 eiγ )| ≥ |a0 | (1 − r1 )n . Hence
2n
|c| ,
max |f (z)| ≤
|z|=1
(1 − r1 )n
and so, by (1.2), we have
2n
|c| Rn
(f ∈ Fc ) .
(2.2)
max |f (z)| ≤
|z|=R>1
(1 − r1 )n
Since any compact subset of C is contained in |z| < R for some large enough R,
inequality (2.2) implies that the polynomials in Fc are uniformly bounded on every
compact set. By a well-known result, for which we refer the reader to [1, p. 216],
the family Fc is normal. Hence Ωr1 ,r2 ,γ , defined in (2.1), is attained. This implies
that
|f (r1 eiγ )|
(2.3)
ωr1 ,r2 ,γ := inf
: f ∈ Pn,λ
|f (r2 eiγ )|
is also attained.
Given r1 < r2 in [0 , 1] and a real number γ, let E = E(n; r1 , r2 ; γ) denote
the set of all polynomials f ∈ Pn,λ for which the infimum ωr1 ,r2 ,γ defined in (2.3)
is attained. Does a polynomial f ∈ Pn,λ necessarily have all its zeros on the unit
circle? We already know that the answer to this question is “yes" for even n, we
have yet to find out if the same holds when n is odd. The following result contains
the answer.
Theorem 2.1. P
For λ ∈ C, |λ| ≤ 1 let Pn,λ denote the class of all polynomials of the
form f (z) := nν=0 cν z ν , not vanishing in the open unit disc, such that c1 /(nc0 ) =
λ. Given r1 < r2 in [0 , 1] and a real number γ, let E = E(n; r1 , r2 ; γ) denote
the set of all polynomials f ∈ Pn,λ for which the infimum ωr1 ,r2 ,γ defined in (2.3) is
attained. Then, any g ∈ E must have at least n − 1 zeros on the unit circle.
Maximum Modulus of
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M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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The theoretical possibility that a polynomial g ∈ E may not have all its n zeros
on the unit circle can indeed occur in the case where n is odd. This is illustrated by
our next result.
P
Theorem 2.2. Let f (z) := 3ν=0 cν z ν 6= 0 for |z| < 1, and let c1 = 0. Then, for
any real γ, we have
(2.4)
|f (0)|
4
≥
iγ
|f (ρ e )|
4 + 4ρ2 + ρ4
(0 < ρ ≤ 1) .
M. A. Qazi
For any given ρ ∈ (0 , 1] equality holds in (2.4) for constant multiples of the polynomial
!
!
p
p
ρ + i 4−ρ2 −iγ
ρ − i 4−ρ2 −iγ ρ fρ (z) := 1−
ze
1−
ze
1+ z .
4
4
2
Remark 1. Inequality (2.4) says in particular that (1.8) does not hold for r = 0. In
(1.8) it is presumed that the lower bound is attained by a polynomial that has all its
zeros on the unit circle. Surprisingly, it turns out to be false.
The following
result
is a consequence of Theorem 2.2. It is obtained by choosing
γ such that f ρ eiγ = max|z|=ρ |f (z)|.
P
Corollary 2.3. Let f (z) := 3ν=0 cν z ν 6= 0 for |z| < 1, and let c1 = 0. Then
(2.5)
|f (0)| ≥
4
max |f (z)|
4 + 4ρ2 + ρ4 |z|=ρ
The estimate is sharp for each ρ ∈ (0 , 1].
Maximum Modulus of
Polynomials
(0 < ρ ≤ 1) .
vol. 8, iss. 3, art. 72, 2007
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3.
An Auxiliary Result
√
Lemma 3.1. For any given a ∈ [0 , 1/2], b := 1 − a2 and β ∈ R, let
fa,β (z) := 1 + (a + ib)zeiβ 1 + (a − ib)zeiβ 1 − 2azeiβ .
Then, for any ρ ∈ [0 , 1] and any real θ, we have
fa,β ρeiθ ≤ fa,β −ρe−iβ = 1 + (1 − 4a2 )ρ2 + 2aρ3 .
Proof. It is enough to prove the result for β = 0. The case a = 1/2 being trivial, let
a ∈ (0, 1/2). We have
2
fa,0 ρeiθ 2 = 1 + aρeiθ 2 + b2 ρ2 e2iθ 1 − 4aρ cos θ + 4a2 ρ2
2
= 1 + 2aρeiθ + ρ2 e2iθ 1 − 4aρ cos θ + 4a2 ρ2
2
= e−iθ + 2aρ + ρ2 eiθ 1 − 4aρ cos θ + 4a2 ρ2
2
= 1 + ρ2 cos θ + 2aρ + i −1 + ρ2 sin θ
× 1 − 4aρ cos θ + 4a2 ρ2
= 1 − 2ρ2 + 4a2 ρ2 + ρ4 + 4aρ + 4aρ3 cos θ + 4ρ2 cos2 θ
× 1 − 4aρ cos θ + 4a2 ρ2
2
= 1 − 1 − 4a2 ρ2 + 4a2 ρ6 + 4aρ3 3 − ρ2 + 4a2 ρ2 cos θ
+ 4 1 − 4a2 ρ2 cos2 θ − 16aρ3 cos3 θ.
So, fa,0 ρeiθ ≤ |fa,0 (−ρ)| for all real θ if and only if
aρ(3 − ρ2 + 4a2 ρ2 )(1+cos θ)−(1 − 4a2 )(1−cos2 θ)−4aρ(1+cos3 θ) ≤ 0 ,
Maximum Modulus of
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M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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that is, if and only if
aρ(3 − ρ2 + 4a2 ρ2 ) − (1 − 4a2 )(1 − cos θ) − 4aρ(1 − cos θ + cos2 θ) ≤ 0 .
To prove this latter inequality, we may replace cos θ by t, set
A(t) := aρ 3 − ρ2 + 4a2 ρ2 − 1 + 4a2 − 4aρ + 1 − 4a2 + 4aρ t − 4aρt2
and show that A(t) ≤ 0 for −1 ≤ t ≤ 1. First we note that
Maximum Modulus of
Polynomials
M. A. Qazi
2
2
A(−1) ≤ A(1) = {−1 − (1 − 4a )ρ } aρ < 0 ,
and so, we may restrict ourselves to the open interval (−1, 1).
Clearly, A0 (t) vanishes if and only if t = (1 − 4a2 + 4aρ)/(8aρ) which is inadmissible for ρ ≤ (1 − 4a2 )/(4a). So, if ρ ≤ (1 − 4a2 )/(4a), then A0 (t) is positive
for all t ∈ (−1, 1) since A0 (0) is; and A(t) ≤ A(1) ≤ 0.
Now, let ρ > (1 − 4a2 )/(4a). Since A00 (t) = −8aρ < 0, the function A must
have a local maximum at t = (1 − 4a2 + 4aρ)/(8aρ). However,
1 − 4a2 + 4aρ
A
= aρ(3 − ρ2 + 4a2 ρ2 ) − 1 + 4a2 − 4aρ
8aρ
(1 − 4a2 + 4aρ)2 (1 − 4a2 + 4aρ)2
+
−
8aρ
16aρ
3
2
= −{aρ + (1 + aρ )(1 − 4a )}
(1 − 4a2 )2 + 16a2 ρ2 + 8aρ(1 − 4a2 )
+
16aρ
(1 − 4a2 )2 1
+ (1 − 4a2 )
= −(1 + aρ3 )(1 − 4a2 ) +
16aρ
2
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1
1 − 4a2
3
= −
+ aρ +
(1 − 4a2 )
2
16aρ
1
1 − 4a2
+ aρ3 (1 − 4a2 ) since ρ >
<−
4
4a
< 0.
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M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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4.
Proofs of Theorems 2.1 and 2.2
Q
Proof of Theorem 2.1. Let g(z) := c0 nν=1 (1 + ζν z). Suppose, if possible, that
|ζj | < 1 and |ζk | < 1, where 1 ≤ j < k ≤ n. Now, consider the function
ψ(w) :=
{1 + (ζj − w)r1 eiγ }{1 + (ζk + w)r1 eiγ }
,
{1 + (ζj − w)r2 eiγ }{1 + (ζk + w)r2 eiγ }
which is analytic and different from zero in the disc |w| < 2δ for all small δ > 0.
Hence, its minimum modulus in |w| < δ cannot be attained at w = 0. This means
that if gw is obtained from g by changing ζj to ζj − w and ζk to ζk + w, then, for all
small δ > 0, we can find w of modulus δ such that
g r eiγ g r eiγ 1
w 1
<
.
iγ
iγ
gw (r2 e ) g (r2 e ) Maximum Modulus of
Polynomials
M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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This is a contradiction since gw ∈ Pn,λ for |w| < min{1 − |ζj | , 1 − |ζk |}.
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Proof of Theorem 2.2. We wish to minimize the quantity |f (0)|/|f (ρeiγ )| over the
class P3,0 of all polynomials of the form
!
3
3
Y
X
f (z) := c0
(1 + ζν z)
|ζ1 | ≤ 1, |ζ2 | ≤ 1, |ζ3 | ≤ 1,
ζν = 0 .
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ν=1
ν=1
Given ρ ∈ (0 , 1] and γ ∈ R, let
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mρ,γ := inf
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|f (0)|
: f ∈ P3,0
|f (ρeiγ )|
.
As we have already explained, mρ,γ is attained, i.e., there exists a cubic f ∗ ∈ P3,0
such that
|f ∗ (0)|
= mρ,γ .
|f ∗ (ρeiγ )|
In fact, there is at least one such cubic f ∗ with f ∗ (0) = 1. By Theorem 2.1, the
∗
∗
cubic
Q3 f must have at least two zeros on the unit circle. In other words, if f (z) :=
ν=1 (1+ζν z), then at most one of the numbers ζ1 , ζ2 , and ζ3 can lie in the open unit
disc. Thus, only two possibilities need to be considered, namely (i) |ζ1 | = |ζ2 | =
|ζ3 | = 1, and (ii) |ζ1 | = |ζ2 | = 1, 0 < |ζ3 | < 1.
Case (i). Since ζ1 + ζ2 + ζ3 = 0, the extremal f ∗ could only be of the form
f ∗ (z) := 1 + z 3 e3iβ , β ∈ [0, 2π/3], and then we would clearly have
(4.1)
1
|f ∗ (0)|
≥
∗
iγ
|f (ρe )|
1 + ρ3
(0 < ρ ≤ 1, γ ∈ R) .
Maximum Modulus of
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M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
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Case (ii). This time, because of the condition ζ1 + ζ2 + ζ3 = 0, the extremal f ∗
could only be of the form
f ∗ (z) := 1 + (a + ib)zeiβ 1 + (a − ib)zeiβ (1 − 2azeiβ ) ,
√
where 0 < a < 1/2, b = 1 − a2 and β ∈ R. Then, for any real γ and any
ρ ∈ (0 , 1], we would, by Lemma 3.1, have
(4.2)
|f ∗ (0)|
1
4
≥
min
=
.
|f ∗ (ρeiγ )| 0<a<1/2 1 + (1 − 4a2 )ρ2 + 2aρ3
4 + 4ρ2 + ρ4
Comparing (4.1) and (4.2), we see that if f ∈ P3,0 , then
|f (0)|
4
≥
iγ
|f (ρe )|
4 + 4ρ2 + ρ4
which proves (2.4).
(0 < ρ ≤ 1, γ ∈ R) ,
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References
[1] L.V. AHLFORS, Complex Analysis, 2nd edition, McGraw–Hill Book Company,
New York, 1966.
[2] N.K. GOVIL, On the maximum modulus of polynomials, J. Math. Anal. Appl.,
112 (1985), 253–258.
[3] M.A. QAZI, On the maximum modulus of polynomials, Proc. Amer. Math. Soc.,
115 (1992), 337–343.
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M. A. Qazi
vol. 8, iss. 3, art. 72, 2007
[4] M.A. QAZI AND Q.I. RAHMAN, On the growth of polynomials not vanishing
in the unit disc, Annales Universitatis Mariae Curie - Sklodowska, Section A, 54
(2000), 107–115.
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[5] Q.I. RAHMAN AND G. SCHMEISSER, Analytic Theory of Polynomials, London Math. Society Monographs New Series No. 26, Clarendon Press, Oxford,
2002.
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[6] T.J. RIVLIN, On the maximum modulus of polynomials, Amer. Math. Monthly,
67 (1960), 251–253.
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[7] E.C. TITCHMARSH, The Theory of Functions, 2nd edition, Oxford University
Press, 1939.
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