EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 13, No. 3, 2020, 663-673
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Some Localization of the Zeros of the Third Derivative
of a Complex Polynomial in the Disks or Generalized

Cardioid Interiors

Todor Stoyanov Stoyanov

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

Abstract. In this paper, we localize the zeros of the third derivatives of a complex polynomial in
some sets. These sets are relevant to the first, second and third derivative of the polynomial, and
they are respectively disks and cardioid interiors or generalized cardioid interiors. Here, for the
first time we consider generalized cardioids, as the areas of zeros.

2020 Mathematics Subject Classifications: 30D20

Key Words and Phrases: Zeros, Complex Polynomial, Disks, Cardioid Interiorities

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, Obreshkoff’s conjecture. The assertions localize the zeros
of the derivative of the any complex polynomial in some areas. Here we present some new
results about the zeros of the derivative of the complex polynomials. Theorem 1 could be
seen in [3]. Theorem 3, Theorem 4 and Theorem 5 we can see in [4]. Their results could
be applied for the localization of the zeros of the derivative of the polynomials- these are
Theorem 3 and Theorem 4. In Theorem 5 we localize the zeros of the second derivative of
the complex polynomial. Theorem 6 and the Corollary could be used everywhere in the
fields of mathematics, independently of their application here in Theorem 7. Theorem 7
appears the main result of the article.

For the first time here, we consider a generalized cardioid. We see that the roots of
polynomial must belong to the generalized cardioid interiorities, created by the zeros of the
given polynomial. Many of these results could be applied for the solving of the unproved
conjectures. Especially that could be seen in [1]. Other possibilities are [2, 5] and [6].

DOI: https://doi.org/10.29020/nybg.ejpam.v13i3.3736

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

https://www.ejpam.com 663 c© 2020 EJPAM All rights reserved.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 664

2. Preliminaries

We note:
D (a, r) = {z ∈ C : |z − a| < r} is the open disk.
D (a, r) = {z ∈ C : |z − a| ≤ r} is the closed disk.
C(a, r) – the open cardioid interior – how to define it: after translation t, t (a) = r ∈ R

and then rotation with angle ϕ = −arg a; a ∈ C. Then coordinates must satisfy(
x2 + y2 − 2rx

)2
< 4r2(x2 + y2).

Figure 1:

C(a, r) – the closed cardioid interior.
Cq(a, r) – the open generalized cardioid interior – how define it: after translation

t, t (a) = r ∈ R and then a rotation with angle ϕ = −arg a; a ∈ R. The coordinates
must satisfy (

x2 + y2 − 2rx
)
< 4q2r2

(
x2 + y2

)
Cq(a, r) – the closed generalized cardioid interior.
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

Theorem 1. 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) = γp (z) + zp′(z), where Re γ ≥ −n

2 ,



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 665

satisfy z ∈ D(0, 1).

Proof. Let z be such that q(z) = 0 and p(z) 6= 0. Then

q (z)

p(z)
= γ +

z

z − z1
+ · · ·+ z

z − zk
= 0.

Hence

q (z)

p(z)
=γ +

z
2 −

z1
2 + z

2 + z1
2

z − z1
+ · · ·+

z
2 −

zn
2 + z

2 + zn
2

z − zk
= 0.

q (z)

p(z)
=γ +

n

2
+

1

2

[
(z + z1) (z − z1)
|z − z1|2

+ · · ·+ (z + zn) (z − zn)

|z − zn|2

]
= 0.

Therefore

Re
q (z)

p(z)
= Re γ +

n

2
+

1

2

[
|z|2 − |z1|2

|z − z1|2
+ · · ·+ |z|

2 − |zn|2

|z − zn|2

]
= 0.

If we assume z /∈ D(0, 1), then we obtain Re γ > 0 , when is impossible.

Theorem 2. 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
(
a
2 ,

1
2

)
.

Proof. Let p (z) = (z − a)q(z). If we denote by z1, z2, . . . , zn−1 the zeros of q(z) and
by w1, w2, . . . wn−1 those of p′(z), then in the non-trivial case q (a) 6= 0 we obtain

n−1∑
k=1

Re
a

a− wk
=Re

ap′′(a)

p′(a)
== 2Re

q′(a)

q(a)

=2
n−1∑
k=1

Re
a

a− zk
≥ 2

n− 1

2
= n− 1.

Here we essentially, that

p
′
(z) = (z − a) q

′
(z) + q (z) ,

p
′′

(z) = (z − a) q
′′

(z) + 2q (z) ,

and |a| = 1.
Hence Re a

a−wk
≥ 1 for some k, (1 ≤ k ≤ n− 1). That means

Re
a
2 −

wk
2 + a

2 + wk
2

a− wk
=Re

(
1

2
+

1

2

a+ wk
a− wk

)



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 666

=Re

[
1

2
+

1

2

(a+ wk) (a− wk)
|a− wk|2

]
=

1

2
+

1

2

|a|2 − |wk|2

|a− wk|2
≥ 1,

i.e. |a− wk|2 + |wk|2 ≤ |a|2 which confirms that wk ∈ D
(
a
2 ,

1
2

)
.

Theorem 3. If all the zeros of a polynomial p(z) ∈ C[z] are zk, k = 1, 2, . . . , n. Then
for each zero w of the derivative p′(z) exists some k0 ∈ N, 1 ≤ k0 ≤ n, such that

w ∈ D
(
zk0
2 ,
|zk0 |
2

)
.

Proof. Let w ∈ C be such that p
′
(w) = 0. We except the trivial case p (w) = 0, which

confirms the assertion. Then

w
p
′
(w)

p(w)
=

w

w − z1
+

w

w − z2
+ · · ·+ w

w − zn

=
w
2 −

z1
2 + w

2 + z1
2

w − z1
+

w
2 −

z2
2 + w

2 + z2
2

w − z2
+ · · ·+

w
2 −

zn
2 + w

2 + zn
2

w − zk

=
n

2
+

1

2

[
(w + z1) (w − z1)
|w − z1|2

+ · · ·+ (w + zn) (w − zn)

|w − zn|2

]
= 0.

Therefore

Re w
p
′
(w)

p(w)
=
n

2
+

1

2

[
|w|2 − |z1|2

|w − z1|2
+ · · ·+ |w|

2 − |zn|2

|w − zn|2

]
= 0.

We put

αk =
|w|2 − |zn|2

|w − zn|2
, k = 1, 2, . . . n.

Then α1 + α2 + · · · + αn = −n. Consequently there exists k0 ∈ N, 1 ≤ k0 ≤ n such

that αk0 ≤ −1. It means that |w|2 + |w − zk0 |
2 ≤ |zk0 |

2, i.e. w ∈ D
(
zk0
2 ,
|zk0 |
2

)
.

Theorem 4. If all the zeros of a polynomial p(z) ∈ C[z] are zk, k = 1, 2, . . . , n. Then
for each zero w of the derivative p′(z) exists some k0 ∈ N, 1 ≤ k0 ≤ n, such that

w /∈ D
(
zk0
2 ,
|zk0 |
2

)
.

Proof. Let w ∈ C be such that p
′
(w) = 0. We except the trivial case p (w) = 0, which

confirms the assertion.
Further we repeat the proof of Theorem 3 and we get α1 + α2 + · · · + αn = −1.

Consequently, there exists k0 ∈ N, 1 ≤ k0 ≤ n such that αk0 ≥ −1. It means that

|w|2 + |w − zk0 |
2 ≥ |zk0 |

2, i.e. w /∈ D
(
zk0
2 ,
|zk0 |
2

)
.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 667

Theorem 5. If all the zeros of the polynomial p(z) ∈ C[z] are zk, k = 1, 2, . . . , n. Then
for each zero t of the second derivative p′′(z) exists some, k0 ∈ N, 1 ≤ k0 ≤ n, such that

t ∈ C
(
zk0
2 ,
|zk0 |
2

)
.

Proof. We denote by t the zero of the second derivative, i.e. p
′′

(t) = 0.
According to Theorem 3 there exists such a zero w of the derivative p′(z), that t ∈

D
(
w
2 ,
|w|
2

)
. For this zero w of the derivative p′(z), according to Theorem 3, there exists

a zero of the polynomial p (z)− zk0 , such that w ∈ D
(
zk0
2 ,
|zk0 |
2

)
.

In order to find the geometric places of the points t, let us take w on the boundary

of the disk D

(
zk0
2 ,
|zk0 |
2

)
and t on the boundary of the disk D

(
w
2 ,
|w|
2

)
. For better

understanding, let us take zk0 ∈ X, and arg w = α, arg t = α+ β.

Figure 2:

Here we have α, β ∈
[
0, π2

]
. Let us put |zk0 | = a. Then

x =a cos α cos β cos (α+ β) ,

y =a cos α cos β sin (α+ β),

x2 + y2 =a2 cos2α cos2β,(
x2 + y2 − ax

)2
=
(
a2 cos2α cos2β − a2 cos α cos β cos (α+ β)

)2
=a4 cos2α cos2β sin2α sin2β ≤ a2a2 cos2α cos2β = a2(x2 + y2).

This means that our geometric place belongs to C

(
zk0
2 ,
|zk0 |
2

)
.



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 668

Figure 3:

4. Main results

Lemma 1. If x, y ∈ [0, 1], then the function h (x, y) = x
√

1− y2 + y
√

1− x2 ≤ 1.

Proof. If we fix y ∈ [0, 1], then

t (x) = h(x, y) = x
√

1− y2 + y
√

1− x2.

∂t

∂x
=
√

1− y2 − xy√
1− x2

=

√
1− x2 − y2+x2y2 − xy

√
1− x2

,

where x 6= 1.

The function l(x) =

√
1− x2 − y2+x2y2 − xy = 0, when 1 − x2 − y2 = 0 ⇐⇒ x =√

1− y2. We know

l(x) > 0 with x <
√

1− y2,

l(x) < 0 with x >
√

1− y2,

l (0) =
√

1− y2 > 0,

l (1) = − y < 0.

Then h(x, y) ≤ t
(√

1− y2
)

= 1− y2 + y2 = 1, which confirms the lemma.

Theorem 6. Let x, y, z ∈ [0, 1]. Then the function f (x, y, z) = xy
√

1− z2 +yz
√

1− x2 +

zx
√

1− y2 satisfies f (x, y, z) ≤ 2
√
3

3 .



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 669

Proof. Let us calculate

∂f

∂x
= y

√
1− z2 + z

√
1− y2 − xyz√

1− x2
,

∂f

∂y
= x

√
1− z2 + z

√
1− x2 − xyz√

1− y2
,

∂f

∂z
= x

√
1− y2 + y

√
1− x2 − xyz√

1− z2
.

We want x 6= 1, y 6= 1, z 6= 1. We need to solve the system

y
√

1− z2 + z
√

1− y2 − xyz√
1− x2

= 0,

x
√

1− z2 + z
√

1− x2 − xyz√
1− y2

= 0,

y
√

1− x2 + x
√

1− y2 − xyz√
1− z2

= 0.

If x 6= 0, y 6= 0, z 6= 0, the system is

√
1− z2
z

+

√
1− y2
y

=
x√

1− x2
,

√
1− z2
z

+

√
1− x2
x

=
y√

1− y2
,

√
1− x2
x

+

√
1− y2
y

=
z√

1− z2
.

We put √
1− x2
x

= p,

√
1− y2
y

= q,

√
1− z2
z

= s. (1)

Then we have

q + s =
1

p
, (2)

s+ p =
1

q
, (3)

p+ q =
1

s
. (4)

After the subtraction (2) – (3), we get q + 1
q = p+ 1

p , i.e. p = q or p = 1
q .



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 670

Analogously q = s or q = 1
s , and s = p or s = 1

p .

Because of (1), the only possibility is p = q = s and therefore p+ p = 1
p , i.e. p =

√
1
2 ,

i.e.

x = y = z =

√
2

3
.

The critical point is M
(√

2
3 ,
√

2
3 ,
√

2
3

)
.

We have:

∂2f

∂x2
=

−yz
(1− x2)

3
2

,
∂2f

∂y2
=

−zx
(1− y2)

3
2

,
∂2f

∂z2
=

−xy
(1− z2)

3
2

∂2f

∂x∂y
=
√

1− z2 − zy√
1− y2

− zx√
1− x2

,

∂2f

∂y∂z
=
√

1− x2 − xy√
1− y2

− xz√
1− z2

,

∂2f

∂z∂x
=
√

1− y2 − yx√
1− x2

− zy√
1− z2

.

Then for the matrix

H =


∂2f
∂x2

∂2f
∂x∂y

∂2f
∂x∂z

∂2f
∂y∂x

∂2f
∂y2

∂2f
∂y∂z

∂2f
∂z∂x

∂2f
∂z∂y

∂2f
∂z2

 ,

we get

H (M) =

−2
√

3 −
√

3 −
√

3

−
√

3 −2
√

3 −
√

3

−
√

3 −
√

3 −2
√

3

 = +
√

3

−2 −1 −1
−1 −2 −1
−1 −1 −2

 .

The determinants:

D1 = |−2| = −2 < 0, D2 =

∣∣∣∣−2 −1
−1 −2

∣∣∣∣ = 3 > 0, D3 =

∣∣∣∣∣∣
−2 −1 −1
−1 −2 −1
−1 −1 −2

∣∣∣∣∣∣ = −4 < 0.

That means: M is a local maximum f (M) = 2
√
3

3 . On the plane z = 0 we have r(x, y) =

f(x, y, 0) = xy ≤ 1. On the plane z = 1 we have h(x, y) = f(x, y, 1) = x
√

1− y2 +
y
√

1− x2 ≤ 1, according to the Lemma 1. Because f is a symmetric function and defined

in the cube [0, 1]3, therefore f ≤ 2
√
3

3 .



T. S. Stoyanov / Eur. J. Pure Appl. Math, 13 (3) (2020), 663-673 671

Corollary 1. If α, β, γ ∈
[
0, π2

]
, then the function

g (α, β, γ) = sin α sin β cos γ + sin α sin γ cos β + sin β sin γ cos α ≤ 2
√

3

3
.

Proof. If in the conditions of Theorem 6 we put x = sin α , y = sin β, z = sin γ, and
we obtain that

g (α, β, γ) = f (x, y, z) ≤ 2
√

3

3
.

Theorem 7. If all the zeros of the polynomial p(z) ∈ C[z] are zk, k = 1, 2, . . . , n. Then
for each zero v of the third derivative p′′′(z) exists some k0 ∈ N, 1 ≤ k0 ≤ n, such that

v ∈ Cq
(
zk0
2 ,
|zk0 |
2

)
, where q = 2

√
3

3 .

Proof. We denote by v zero of the third derivative, i.e. p′′′ (v) = 0.
According to Theorem 3 there exists such a zero t of the second derivative p′′(z), that

v ∈ D
(
t
2 ,
|t|
2

)
. For this zero t according again to Theorem 3, there exists a zero w of the

polynomial p′ (z), such that t ∈ D
(
w
2 ,
|w|
2

)
.

And again for this zero w of the derivative p′ (z), there exists a zero zk0 of the polyno-

mial p(z), such that w ∈ D
(
zk0
2 ,
|zk0 |
2

)
.

In order to find the geometric places of the points t, let us take w on the boundary

of the disk D

(
zk0
2 ,
|zk0 |
2

)
and t on the boundary of the disk D

(
w
2 ,
|w|
2

)
. For better

understanding, let us take zk0 ∈ X, and arg w = α, arg t = α+ β, arg v = α+ β + γ.

Figure 4:

Here we have α, β, γ ∈
[
0, π2

]
.



REFERENCES 672

Let us put |zk0 | = a. Then

x = a cos α cos β cos γ cos (α+ β + γ) ,

y = a cos α cos β cos γ sin (α+ β + γ) .

Let us calculate

ax = a2 cos α cos β cos γ [cos (α+ β) cos γ − sin (α+ β) sin γ ]

= a2 cos α cos β cos γ [cos α cos β cos γ − sin α sin β cos γ

− sin α sin γ cos β − sin β sin γ cos α ] .

Obviously

x2 + y2 = a2 cos2α cos2β cos2γ.

Then we have

x2 + y2 − ax = a2 cos α cos β cos γ [sin α sin β cos γ + sin α sin γ cos β

+ sin β sin γ cos α ] .

But

sin α sin β cos γ + sin α sin γ cos β + sin β sin γ cos α ≤ 2
√

3

3
,

according to the Corollary.
From here we state (

x2 + y2 − ax
)2 ≤ 4

3
a2
(
x2 + y2

)
.

That means that our geometric place belongs to C 2
√
3

3

(
zk0
2 ,
|zk0 |
2

)
, which confirms

this assertion.

Acknowledgements

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

References

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