EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 4, 2020, 807-813
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

A Private Case of Sendov’s Conjecture

Todor Stoyanov Stoyanov

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

Abstract. In this paper, we prove Sendov’s conjecture, when a polynomial is with real coefficients
and the conjecture is relevant to the zeros, which belong to the set M = D (0, 1) ∩ [D (1, 1) ∪
D (−1, 1)]. We can see it in Figure 1. The conjecture is true for the filled areas.

2020 Mathematics Subject Classifications: 30D20, 30A10

Key Words and Phrases: Zeros, Complex Polynomial, Real Polynomial, Disk, Derivative,
Integral

1. Introduction

The localization of the zeros of the complex polynomials is very important area of the
mathematics. The impossibility to find the zeros of any polynomials using the coefficients
makes every statement here very significant. There exist many conjectures which are not
proved, like Sendov’s conjecture and Obreshkoff’s conjecture. They localize the zeros of
the derivative of many complex polynomial in some areas. Here we present some new
results about the zeros of the derivative of real polynomials.

In the second part “Preliminaries” we define some sets, including the set M . Here we
formulate Sendov’s conjecture.

In the third part “Related results”, we present three statements which are similar
to our theorems. The proofs of Statement 1 and Statement 2, we can see in [3]. If we
put m = 1 in the condition of Statement 2 we obtain Statement 1, which is the famous
Obreshkoff theorem, that can be regarded as a ‘complex version’ of a well-known theorem
due to Laguerre. The proof of Statement 3 we can see in [1].

The main theorems are in the fourth part “Main results”. Theorem 1 is relevant to
the real zeros of the real polynomial r (z). Theorem 2 is relevant to the complex zeros of
the real polynomial r (z). All the results could be connected with the results of [4].

The proofs of these theorems are independent of references. Only [4] is relevant to
them. Many of these results could be applied in [2] and [5].

DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3737

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

https://www.ejpam.com 807 c© 2020 EJPAM All rights reserved.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (4) (2020), 807-813 808

2. Preliminaries

We note:
D (a, r) = {z ∈ C : |z − a| < r} is the open disk with center a and radius r.
D (a, r) = {z ∈ C : |z − a| ≤ r} is the closed disk with center a and radius r.
C (a, r) = {z ∈ C : |z − a| = r} is the circle with center a and radius r.
M = D (0, 1) ∩ [D (1, 1) ∪D (−1, 1)].

Figure 1:

Sendov‘s conjecture: Let us put for n ≥ 2, p (z) =
∏n
k=1 (z − zk), where zk ∈

D (0, 1) , k = 1, 2, . . . , n. Then p′(z) has at least one zero in each of the disks D (zk, 1) , k =
1, 2, . . . n.

3. Related Results

Statement 1(Obreshkoff). Let the zeros zk, k = 1, 2, . . . , n of a polynomial p (z) ∈
C [z] satisfy zk ∈ D (0, 1). Then the zeros z of the polynomial q (z) = nγp (z) + zp

′
(z),

where Re γ ≥ −1
2 , satisfy z ∈ D (0, 1).

Statement 2(Stoyanov). Let the zeros zk, k = 1, 2, . . . , n of a polynomial p(z) ∈
C[z] satisfy zk ∈ D (0, 1). Then if Re γ ≥ −m

2 , m = 1, 2, . . . , n , the zeros z of the

polynomial q (z) = γp (z) +
∑m

k=1
(n−k)!
n! zkp(k)(z), satisfy z ∈ D (0, θ), where

θ =
(

2
1
m − 1

)−1
.

Statement 3(Bojanov). If all the zeros zk, k = 1, 2, . . . n; of a polynomial p(z) ∈
C[z] satisfy zk ∈ D (0, 1) and a is a zero of p(z) of modulus 1, then the derivative p′(z)
has at least one zero in D(a2 ,

1
2).



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (4) (2020), 807-813 809

4. Main results

Case 1.
In this case we consider a polynomial

r (z) = zn + rn−1z
n−1 + · · ·+ r1z + r0,

where rk ∈ R, n ≥ 2, n ∈ N , k = 0, n− 1. The zeros zk of r (z) satisfy the condition

zk ∈ D (−a, 1) , z0 ∈ R, z0 + a = 0, a ∈ (0, 1].

The derivative is

dr

dz
= n (z + a1) (z + a2) · · · (z + al) (z − b1)

(
z − b1

)
· · · (z − bs)

(
z − bs

)
,

where l + 2s = n− 1, ak, bm ∈ D (−a, 1), k = 1, l, m = 1, s, ak ∈ R, k,m ∈ N .

Theorem 1. If z0 = 0 ∈ R is a real zero of r (z), then there exists a zero c of dr
dz which

satisfies c ∈ D (0, 1).

Proof.

Figure 2:

Without loss of generality, we consider Re z0 ≥ 0.
About the root z0 = 0 we assume that in the disk D (0, 1) there is not exists a root

c ∈ dr
dz . Then all the zeros of the derivative ak, bm ∈ D (−a, 1) \ D (0, 1) and we have

r (0)− r (−a) =

∫ 0

−a

dr

dz
dz = −r (−a) ,



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (4) (2020), 807-813 810

i.e.

−r (−a) = n

∫ 0

−a
(z + a1) (z + a2) · · · (z + al) (z − b1)

(
z − b1

)
· · · (z − bs)

(
z − bs

)
dz

where ak > 1, bm = ρme
iϕm , ρm > 1, ϕm ∈

[
π
2 , π

]
, k = 1, l, m = 1, s,

(z − bm)
(
z − bm

)
= z2 − 2ρm cos ϕmz + ρ2m.

Then we obtain
−r (−a) =

= n

∫ 0

−a
(z + a1) · · · (z + al)

(
z2 − 2ρ1 cos ϕ1z + ρ21

)
· · ·

(
z2 − 2ρs cos ϕsz + ρ2s

)
dz

≥ n
∫ 0

−a
(z + 1)l

(
z2 + 2ρ1z + ρ21

)
· · ·
(
z2 + 2ρsz + ρ2s

)
dz

> n

∫ 0

−a
(z + 1)l(z + 1)2sdz = n

∫ 0

−a
(z + 1)n−1dz

= (z + 1)n |0−a = 1n − (1− a)n > a.

We obtain that |r (−a)| > a. But all the zeros zk of r (z) belong to D (−a, 1), i.e.
|r (−a)| ≤ a , because |z0 + a| = 0. This contradiction confirms the Theorem 1.

Case 2.
In this case we consider a polynomial r (z) = zn + rn−1z

n−1 + · · · + r1z + r0, where
rk ∈ R, n ≥ 2, n ∈ N , k = 0, n− 1. The zeros zk of r (z) satisfy the condition zk ∈ D (0, 1),
z0 = aeiθ0 , where a ∈ (0, 1], θ0 ∈

[
0, π2

]
. The derivative is

dr

dz
= n (z − a1) (z − a2) · · · (z − al) (z − b1)

(
z − b1

)
· · · (z − bs)

(
z − bs

)
,

where l + 2s = n− 1, ak, bm ∈ D (0, 1), k = 1, l, m = 1, s, k,m ∈ N .

Theorem 2. If z0 = aeiθ0 is a zero of r (z) belongs to M (in Preliminaries), then there
exists a zero c of dr

dz which satisfies c ∈ D (z0, 1).



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (4) (2020), 807-813 811

Proof.

Figure 3:

Without loss of generality, we consider Re z0 ≥ 0.
Let us assume that in the disk D (z0, 1) there is not exists a root c ∈ dr

dz . Then all the
zeros of the derivative ak, bm ∈ D (0, 1) \ D (z0, 1) \ D (z0,1), because the zero z0 = aeiθ0

of the polynomial must belongs to M . We note with −d, d > 0 : −d = C (z0, 1)
⋂
C (z0, 1)

and let us put v (θ) = aeiθ, θ ∈ [0, θ0] , l + 2s = n− 1,

t (z) =
1

n

dr (z)

dz
=

l∏
p=1

(z + ap)
s∏

p=1

(z − bp)
(
z − bp

)
,

l, s ∈ N [one of these factors could be not existing, i.e. l = 0 or s = 0].
We put

f (θ) =

∫ v(θ)

0

dr

dz
dz = n

∫ v(θ)

0
t (z) dz,

g (θ) = f (θ) .f (θ) .

Let us calculate

dg

dθ
= n

[
t (v (θ))

dv

dθ
f (θ) + t (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.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (4) (2020), 807-813 812

Knowing df
dθ .
∏n−1
p=0 Up = df

dθ .
∏n−1
p=0 Up we have

dg

dθ
=in

f (θ)
n−1∏
p=0

Up − f (θ)
n−1∏
p=0

Up

 ,
d2g

dθ2
=n

2
df

dθ
.

n−1∏
p=0

Up + i
d
∏n−1
p=0 Up

dθ
f (θ)− i

d
∏n−1
p=0 Up

dθ
f (θ)


d2g

dθ2
=n

2n

n−1∏
p=0

|Up|2 −

U0

n−1∑
p=0

∏
j 6=p

Uj

 f (θ)−

U0

n−1∑
p=0

∏
j 6=p

Uj

 f (θ)


d2g

dθ2
=2n

n n−1∏
p=0

|Up|2 −Re

U0

n−1∑
p=0

∏
j 6=p

Uj

 f (θ)

 ,
and consequently

d2g

dθ2
≥ 2n

n−1∏
p=0

|Up|

n n−1∏
p=0

|Up| −

∣∣∣U0
∑n−1

p=0

∏
j 6=p Uj

∣∣∣ . ∣∣f (θ)
∣∣∏n−1

p=0 Up

 .
Since θ ∈ [0, θ0] ⇒ |Up (θ)| ≥ |Up (θ0)| ≥ 1, p = 1, n− 1.
If we assume |f (θ)| =

∣∣f (θ)
∣∣ ≤ a2, then

d2g

dθ2
≥2n

n−1∏
p=0

|Up|
[
na−

(
1 +

∣∣∣∣U0

U1

∣∣∣∣+ · · ·+
∣∣∣∣ U0

Un−1

∣∣∣∣) a2]

≥ 2na
n−1∏
p=0

|Up| [n− (1 + a (n− 1)) a] = 2na
n−1∏
p=0

|Up|
[
n− a− a2 (n− 1)

]
≥ 2na

n−1∏
p=0

|Up| [n− a− a (n− 1)] = 2nan (1− a)
n−1∏
p=0

|Up|.

Then we get d2g
dθ2
≥ 0.

Hence dg
dθ (θ) ≥ dg

dθ (0) = 0. Consequently g (θ0) > g (0), i.e. |f (θ0)| > a, according to
the proof of Theorem 1.

Therefore a < |f (θ0)| ≤ a2, which is impossible. The contradiction proves that
|f (θ0)| > a2, but

f (θ0) =

∫ v(θ0)

0

dr

dz
dz =

∫ z0

0

dr

dz
dz = r (z0)− r (0) = −r (0) ,



REFERENCES 813

i.e. |r (0)| ≤ a2, because z0, z0 are roots of r (z) and |z0|=|z0| = a. Then we have

a2 < |r (0)| ≤ a2.

This contradiction shows that, there exists a zero c of dr
dz , which satisfies c ∈ D (0, 1).

Acknowledgements

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

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