EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 12, No. 2, 2019, 649-653
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Some estimates below the modulus of integrals of some
polynomials in the complex plane

Todor Stoyanov Stoyanov
1 University of Economics, Department of Mathematics bul. Knyaz Boris I 77, Varna 9002,
Bulgaria

Abstract. In this paper, we make some estimates below the modulus of some integrals in the
complex plane. Our aim is to prove the Conjecture1, which we could see in [2–4]. The proof of the
conjecture appears the Corollary.

2010 Mathematics Subject Classifications: 30A10

Key Words and Phrases: Zeros, complex polynomial, real polynomial, disk, derivative, integral

1. Introduction

In papers [2–4], we consider the Conjecture 1: If ak ≥ 0, ak ∈ R, Then we assert∣∣∣∣∣∣∣
eiϕ∫
0

n∏
k=1

(x+ ak) dx

∣∣∣∣∣∣∣ ≥
1

n+ 1
,

for arbitrary natural n, ϕ ∈
[
0, π2

]
. There exists a connection between this conjecture and

Conjecture2: If φk ∈
[
π
2 , π

]
, then∣∣∣∣∣∣

0∫
−1

(x+ 1)
n∏
k=1

(
x− eiφk

)
dx

∣∣∣∣∣∣ ≥ 1

n+ 2
.

Both conjectures are very important for the proofs of some famous conjectures, like
Sendov’s and Obreshkoff’s ones. A possible connection between both conjctures appears
[5]. Here we shall extend this problem (Conjecture1): what kind of set L satisfies this
assertion, i.e. if ak belongs to the set L, then the upper inequality is true. The results
related with the Conjecture 1, we observe in Theorem 1 , Theorem 2. In Theorem 4 we
generalize and prove the extended conjecture. We can see the results of Theorem 1 in

DOI: https://doi.org/10.29020/nybg.ejpam.v12i2.3372

Email address: todstoyanov@yahoo.com (T. S. Stoyanov)

http://www.ejpam.com 649 c© 2019 EJPAM All rights reserved.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 12 (2) (2019), 649-653 650

[2, 4]. Such one of Theorem 2 could be seen in [3]. Many authors use some modulus of
some integrals in the complex plane for various estimates in their works. For example we
can see how Bojanov and Rahman in [1] use this method. These estimates are explored
for the localization of the zeros of some polynomials. The results are useful in the (open)
problems of [6–9].

2. Related Results

Theorem 1. Let k = 1, 2, ...n, n ∈ N, ak ∈ [0, 1]ϕ ∈
[
0, π2

]
. Then the function∣∣∣∣∣∣∣

eiϕ∫
−1

x
n
Π
k=1

(x+ ak) dx

∣∣∣∣∣∣∣ ≥
1

n+ 2

for n = 1, 2, 3.

Theorem 2. let k ∈ N, a ∈ R, a ∈ [0, 1] . Then the function∣∣∣∣∣∣
i∫
0

x (x+ a)k dx

∣∣∣∣∣∣ ≥ 1

n+ 2
.

3. Preliminaries

We note:
D (a, r) = {z ∈ C : |z − a| < r} is the open disk with center a and radius r.
D (0, r) = {z ∈ C : |z − a| ≤ r} is the closed disk with center a and radius r.
A = {z ∈ C, Rez ≤ 0} is the left semiplane.

4. Main Results

Theorem 3. We consider a polynomial r(z) = zn−1 + rn−1z
n−1 + ... + r1z + r0. where

rk ∈ R, n ≥ 1, n ∈ N, k = 0, n− 1. The zeros zk of r (z) satisfy the condition Rezk ≤ 0. If

a ≥ 0, then I = n
a∫
0

r (z) dz ≥ an.

Proof. Let r (z) = (z + a1) (z + a2) ... (z + a1) (z − b1)
(
z − b1

)
... (z − bs)

(
z − bs

)
, where

1+2s = n−1, ak ≥ 0, bm ∈ C, k = 1, l,m = 1, s, ak ∈ R, k,m ∈ N. and bm = ρme
iϕm, ρm ≥

0, ϕm ∈
[
π
2 , π

]
, (z − bm)

(
z − bm

)
= z2 − 2ρm cosϕmz + ρ2m ≥ z2. Then

n
a∫
0

r (z) dz = n
a∫
0

(z + a1) (z + a2) ... (z + a1) (z − b1)
(
z − b1

)
... (z − bs)

(
z − bs

)
d ≥

n
a∫
0

zn−1dz = an.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 12 (2) (2019), 649-653 651

Theorem 4. We consider a polynomial r (z) = zn−1 + rn−1z
n−2 + ... + r1z + r0, where

rk ∈ R, n ≥ 1, n ∈ N, k = 0, n− 1. The zeros zk of r (z) satisfy the condition zk ∈
A \D (z0, a) \D (z0, a) , z0 = aeiθ0 , where a ≥ 0, θ0 ∈

[
0, π2

]
. Then

I =

∣∣∣∣∣∣n
z0∫
0

r (z) dz

∣∣∣∣∣∣ ≥ an.
Proof. Let us put v (θ) = aeiθ, θ ∈ [0, θ0] , I + 2s = n− 1,

r (z) =
l
Π
p=1

(z + ap)
s
Π
p=1

(z − bp)
(
z − bp

)
,

l, s ∈ N (one of the factors could be not existing, i.e.,l = 0 or s = 0).

We put f (θ) = n
v(θ)∫
0

r (z) dz, g (θ) = f (θ) .f (θ) .

Let us calculate

dg

dθ
= n

[
r (v (θ))

dv

dθ
f (θ) + r (v (θ))

dv

dθ
f (θ)

]
,

dv

dθ
=
daeiθ

dθ
= iaeiθ,

and if we put

U0 = v (θ) = aeiθ, Up = v (θ) + ap, p = 1, l, Ul+2p+1 = v (θ)− bp,

Ul+2p+2 = v (θ)− bp, p = 0, s− 1.

Knowing
df

dθ
.Πn−1

p=0Up =
df

dθ
.Πn−1

p=0Up,

we have
dg

dθ
= in

[
f (θ)

n−1
Π
p=0

Up − f (θ)
n−1
Π
p=0

Up

]
,

d2g

dθ2
= n

[
2
df

dθ

n−1
Π
p=0

Up + i
dΠn−1

p=0Up

dθ
f (θ)− i

dΠn−1
p=0Up

dθ
f (θ)

]
,

d2g

dθ2
= n

[
2n

n−1
Π
p=0
|Up|2 −

(
U0

n−1
Σ
p=0

Π
j 6=p

Uj

)
f (θ)−

(
U0

n−1
Σ
p=0

Π
j 6=p

Uj

)
f (θ)

]
,

d2g

dθ2
= 2n

[
n
n−1
Π
p=0
|Up|2 −Re

(
U0

n−1
Σ
p=0

Π
j 6=p

Uj

)
f (θ)

]
,



T. S. Stoyanov / Eur. J. Pure Appl. Math, 12 (2) (2019), 649-653 652

and consequently

d2g

dθ2
≥ 2n

n−1
Π
p=0
|Up|

nn−1Π
p=0
|Up| −

∣∣∣∣U0

n−1
Σ
p=0

Π
j 6=p

Uj

∣∣∣∣ . ∣∣f (θ)
∣∣

Πn−1
p=0Up

 .
If we note

B = {A \D (z, a) \D (z, a)} , B0 = {A \D (z0, a) \D (z0, a)}

and since
θ ∈ [0, θ0] =⇒ B0 ⊂ B, i.e., |Up (θ)| ≥ a, p = 1, n− 1.

If we assume
|f (θ)| =

∣∣f (θ)
∣∣ ≤ an,

then
d2g

dθ2
≥ 2n

n−1
Π
p=0
|Up|

[
naan−1 −

(
1 +

∣∣∣∣U0

U1

∣∣∣∣+ ...+

∣∣∣∣ U0

Un−1

∣∣∣∣) .an]

≥ 2na.Πn−1
p=0 |Up|

[
nan−1 −

(
1 +

a (n− 1)

a

)
.an−1

]
= 0.

Then
d2g

dθ2
≥ 0.

Hence
dg

dθ
(θ) ≥ dg

dθ
(0) = 0.

Consequently g (θ0) > g (0) , i.e., |f (θ0)| > an, according to the proof of Theorem 3.
Therefore an < |f (θ0)| ≤ an, which is impossible. The contradiction proves the Theorem
4.

Corollary. If in the condition of Theorem 4, we put a=1, and s=0, i.e., all the zeros of
r(z) are real and negative, then we get that the Conjecture 1 is true.

Acknowledgements

This research was financed from University of Economics of Varna research grants No.
19 2018-04-27



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