ON THE PARTIAL SUMS OF CONVEX HARMONIC UNIVALENT FUNCTIONS

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A. ZIREH1_{, M. M. SHABANI}1_, _§

Abstract. Partial sums of analytic univalent functions and partial sums of starlike have

been investigated extensively by several researchers. In this paper, we investigate a partial sums of convex harmonic functions that are univalent and sense preserving in the open unit disk.

Keywords: Harmonic, Univalent, Convex, Partial sums.

AMS Subject Classification: 30C45; 30C50.

1. Introduction

A continuous function f = u+iv is a complex valued harmonic function in a complex domain Ω⊂_Cif both uand v are real harmonic in Ω.

In any simply connected domain Ω⊂C, we may writef =h+g, whereh andg are analytic in Ω. We call h the analytic part andg the co-analytic part off. A necessary and sufficient condition forf to be locally univalent and sense-preserving in Ω is that|h0(z)|>|g0(z)|in Ω. (See [2]).

Denote by SH the class of functions f = h+g that are harmonic univalent and

sense-preserving inD={z∈C:|z|<1}for which f(0) =fz(0)−1 = 0. Then forf =h+g∈ SH,

the analytic functionsh and gcan be expressed as

h(z) =z+

∞ X

k=2

akzk, g(z) =

∞ X

k=1

bkzk, |b1|<1. (1) A function f of the form (1) is harmonic convex of order α,0≤α <1, denoted byKH(α), if it satisfies

∂ ∂θ

n

arg ∂ ∂θf(re

iθo

=Renz zh

0_(z)0₊_{z zg}0_(z)0

zh0(z)−zg0(z)

o ≥α,

where 0≤θ≤2π,|z|=r <1.

As shown recently by Jahangiri [6] a sufficient condition for a function of the form (1) to be

1

Department of Mathematics, Shahrood University of Technology, Shahrood, P.O.Box: 316-36155, Iran. e-mail: [email protected]; ORCID: https://orcid.org/0000-0002-3405-853X.

e-mail: [email protected]; ORCID: https://orcid.org/0000-0003-1416-7795. § Manuscript received: April 5, 2017; accepted: June 20, 2017.

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inKH(α) is that

∞ X

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

≤2. (2)

In 1985, Silvia[11] studied the partial sums of convex functions of orderα. Later, Silverman [10], Abubaker and Darus[1], Dixit and Porwal[3], Frasin[4, 5], Raina and Bansal[8], Rosy et al.[9] and Porwal and Dixit[7] exhibited some results on partial sums for various classes of analytic functions. Here, we investigate a partial sums of convex harmonic functions.

Now, we let the sequences of partial sums of functions of the form (1) withb1= 0, have forms

fm(z) =z+ m

X

k=2

akzk+

∞ X

k=2 bkzk,

fn(z) =z+

∞ X

k=2

akzk+ n

X

k=2 bkzk_,

fm,n(z) =z+ m

X

k=2

akzk+ n

X

k=2 bkzk.

In the present paper, we determine sharp lower bounds for Ren f fm

o

, Renfm f

o

, Renf

0

f0

m

o

,

Renf

0

m f0

o

,Renf fn

o

,Renfn f

o

,Ren f fm,n

o

,Renfm,n f

o

,Ren f

0

f0

m,n

o

and Renf

0

m,n f0

o

,

wheref0(z) = ∂ ∂θf(re

iθ_{) =}_{i zh}0_(z)₋_zg0_(z)

.

2. Main Results

Theorem 2.1. If f(z) of the form (1) with b1= 0 satisfies condition (2), then

Re

n f(z)

fm(z)

o

≥ m(m+ 2−α)

(m+ 1)(m+ 1−α), (z∈D) (3)

The result (3) is sharp with the function

f(z) =z+ 1−α

(m+ 1)(m+ 1−α)z

m+1_. ₍₄₎

Proof. To obtain sharp lower bound given by (3), let us put

(m+ 1)(m+ 1−α) 1−α

h f(z) fm(z)

− m(m+ 2−α) (m+ 1)(m+ 1−α)

i =

1 + m

P

k=2

rk−1_ei(k−1)θ_a k+

∞

P

k=2

rk−1_e−i(k+1)θ_b k+

(m+ 1)(m+ 1−α) 1−α

h ∞ P

k=m+1

rk−1_ei(k−1)θ_a k

i

1 + m

P

k=2

∞

P

k=2

rk−1_e−i(k+1)θ_b k

(3)

Set 1 +A(z) 1 +B(z) =

1 +ω(z)

1−ω(z), so thatω(z) =

A(z)−B(z)

2 +A(z) +B(z). Then

ω(z) =

(m+ 1)(m+ 1−α) 1−α

h ∞ P

k=m+1

rk−1ei(k−1)θak

i

2 + 2 m

P

k=2

∞

P

k=2

+(m+ 1)(m+ 1−α) 1−α

∞ P

k=m+1

Hence

|ω(z)| ≤

(m+ 1)(m+ 1−α) 1−α

h ∞ P

k=m+1

|ak|

i

2−2

m P

k=2

|ak|+

∞ P

k=2

|bk|

−(m+ 1)(m+ 1−α)

1−α

∞

P

k=m+1

|ak|

The last expression is bounded above by 1, if and only if m

X

k=2

|ak|+

∞ X

k=2

|bk|+

(m+ 1)(m+ 1−α) 1−α

X∞

k=m+1

|ak|

≤1 (5)

It suffices to show that the L. H. S. of (5) is bounded above by

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

,

which is equivalent to m

X

k=2

k(k−α)

1−α −1

|ak|+

∞ X

k=2

k(k+α)

1−α −1

|bk|

+

∞ X

k=m+1

k(k−α)

1−α −

(m+ 1)(m+ 1−α) 1−α

|ak| ≥0.

To seef(z) =z+ 1−α

(m+ 1)(m+ 1−α)z

m+1_{gives the sharp result, we observe that for}_z₌_rei_mπ

we have f(z)

fm(z) = 1 +

1−α

(m+ 1)(m+ 1−α)z

m_−→₁₋ 1−α

(m+ 1)(m+ 1−α) =

m(m+ 2−α) (m+ 1)(m+ 1−α)

when r→1−. This completes the proof of Theorem 2.1.

Theorem 2.2. If f(z) of the form (1) with b1= 0 satisfies condition (2), then Re

nfm(z)

f(z)

o

≥ (m+ 1)(m+ 1−α)

m(m+ 2−α) + 2(1−α), (z∈D) (6)

The result (6) is sharp with the function given by (4).

Proof. We may write

1 +ω(z) 1−ω(z) =

m(m+ 2−α) + 2(1−α) 1−α

hf_m(z) f(z) −

(m+ 1)(m+ 1−α) m(m+ 2−α) + 2(1−α)

i =

1 + m

P

k=2

∞

P

k=2

rk−1_e−i(k+1)θ_b k−

(m+ 1)(m+ 1−α) 1−α

∞ P

k=m+1

1 +

∞

P

k=2

∞

P

k=2

(4)

where

|ω(z)| ≤

m(m+ 2−α) + 2(1−α) 1−α

h ∞ P

k=m+1

|ak|i

2−2

m P

k=2

|ak|+

∞ P

k=2

|bk|−m(m+ 2−α)

1−α

∞

P

k=m+1

|ak| ≤1.

Equivalently m

X

k=2

|ak|+

∞ X

k=2

|bk|+m(m+ 2−α) + (1−α) 1−α

_X∞

k=m+1

|ak|≤1. (7)

since the L. H. S. of (7) is bounded above by

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

, the proof is

complete.

Theorem 2.3. If f(z) of the form (1) with b1= 0 satisfies condition (2), then Renf

0_(z)

f0

m(z)

o

≥ m

m+ 1−α, (z∈D) (8)

Proof. We have

1 +ω(z) 1−ω(z) =

m+ 1−α 1−α

hf0(z)

f0

m(z)

− m

m+ 1−α

i

= 1 +

m

P

k=2

krk−1ei(k−1)θak−

∞ P

k=2

krk−1e−i(k+1)θbk+

m+ 1−α 1−α

h ∞ P

k=m+1

krk−1ei(k−1)θak

i

1 + m

P

k=2

krk−1_ei(k−1)θ_a k−

∞ P

k=2

krk−1_e−i(k+1)θ_b k

.

Then

ω(z) =

m+ 1−α 1−α

h ∞ P

k=m+1

krk−1ei(k−1)θak

i

2 + 2 m

P

k=2

∞ P

k=2

+m+ 1−α 1−α

∞

P

k=m+1

krk−1_ei(k−1)θ_a k

In a similar fashion as in Theorem 2.1. the proof is complete.

Renf

0

m(z) f0(z)

o

≥ m+ 1−α

m+ 2(1−α), (z∈D) (9)

Proof. Since

1 +ω(z) 1−ω(z) =

m+ 2(1−α) 1−α

hf0

m(z) f0(z) −

m+ 1−α m+ 2(1−α)

i

,

proceeding exactly as in the proof of Theorem 2.3, we evidently have the required result.

Renf(z) fn(z)

o

≥ n(n+ 2 +α)

(5)

The result (10) is sharp with the function

f(z) =z+ 1−α

(n+ 1)(n+ 1 +α)z

n+1_. ₍₁₁₎

Proof. Write

1 +ω(z) 1−ω(z) =

(n+ 1)(n+ 1 +α) 1−α

hf(z) fn(z)

− n(n+ 2 +α) (n+ 1)(n+ 1 +α)

i =

1 +

∞

P

k=2

n

P

k=2

rk−1_e−i(k+1)θ_b k+

(n+ 1)(n+ 1 +α) 1−α

h ∞ P

k=n+1

i

1 +

∞

P

k=2

n

P

k=2

.

where

ω(z) =

(n+ 1)(n+ 1 +α) 1−α

h ∞ P

k=n+1

i

2 + 2

∞

P

k=2

n

P

k=2

+(n+ 1)(n+ 1 +α) 1−α

∞ P

k=n+1

.

Then

|ω(z)| ≤

(n+ 1)(n+ 1 +α) 1−α

h ∞ P

k=n+1

|bk|i

2−2

∞ P

k=2

|ak|+ n

P

k=2

|bk|− (n+ 1)(n+ 1 +α)

1−α

∞

P

k=n+1

|bk|

.

This last expression is bounded above by 1, if and only if

∞ X

k=2

|ak|+ n

X

k=2

|bk|+(n+ 1)(n+ 1 +α) 1−α

_X∞

k=n+1

|bk|≤1. (12)

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

,

which is equivalent to

∞ X

k=2

k(k−α)

1−α −1

|ak|+

n

X

k=2

k(k+α)

1−α −1

|bk|

+

∞ X

k=n+1

k(k+α)

1−α −

(n+ 1)(n+ 1 +α) 1−α

|bk| ≥0.

To see that f(z) = z+ 1−α

(n+ 1)(n+ 1 +α)z

n+1 _{gives the sharp result, we observe that for}

z=reinπ+2 _{one obtains}

f(z) fn(z)

= 1 + 1−α

(n+ 1)(n+ 1 +α)r n_e− iπ

n+2(n+2)−→1− 1−α

(n+ 1)(n+ 1 +α)=

n(n+ 2 +α) (n+ 1)(n+ 1 +α)

when r→1−. This completes the proof.

Theorem 2.6. If f(z) of the form (1) with b1= 0 satisfies condition (2), then Renfn(z)

f(z)

o

≥ (n+ 1)(n+ 1 +α)

n(n+ 2 +α) + 2 , (z∈D) (13)

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Proof. It is easy to see that

1 +ω(z) 1−ω(z) = n(n+ 2 +α) + 2

1−α

hf_n(z) f(z) −

(n+ 1)(n+ 1 +α) n(n+ 2 +α) + 2

i =

1 +

∞

P

k=2

n

P

k=2

rk−1_e−i(k+1)θ_b k−

(n+ 1)(n+ 1 +α) 1−α

h ∞ P

k=n+1

i

1 +

∞

P

k=2

∞

P

k=2

.

Rest of the proof is omitted since it runs parallel to that from Theorem 2.2.

Theorem 2.7. If f(z) of the form (1) with b1= 0 satisfies condition (2), then (i) Ren f(z)

fm,n(z)

o

≥ m(m+ 2−α)

(m+ 1)(m+ 1−α),(z ∈ D) if n(n+ 2 +α) + 2α ≥ m(m+ 2−α) or bk= 0 ∀k≥2.

(ii) Ren f(z) fm,n(z)

o

≥ n(n+ 2 +α)

(n+ 1)(n+ 1 +α),(z ∈ D) if n(n+ 2 +α) + 2α ≤ m(m+ 2−α) or ak = 0 ∀k≥2.

Proof. To prove (i), we may write

1 +ω(z) 1−ω(z) =

(m+ 1)(m+ 1−α) 1−α

h f(z)

fm,n(z)

− m(m+ 2−α)

(m+ 1)(m+ 1−α)

i

= P

1 + m

P

k=2

n

P

k=2

,

where

P = 1 + m

X

k=2

rk−1ei(k−1)θak+ n

X

k=2

rk−1e−i(k+1)θbk

+(m+ 1)(m+ 1−α) 1−α

h X∞

k=m+1

rk−1ei(k−1)θak+

∞ X

k=n+1

i

.

So that

ω(z) =

(m+ 1)(m+ 1−α) 1−α

h ∞ P

k=m+1

∞ P

k=n+1

i

Q ,

where

Q= 2+2

_Xm

k=2

X

k=2

+(m+ 1)(m+ 1−α) 1−α

X∞

k=m+1

∞ X

k=n+1

.

Then

|ω(z)| ≤

(m+ 1)(m+ 1−α) 1−α

h ∞ P

k=m+1

|ak|+

∞ P

k=n+1

|bk|

i

2−2

m P

k=2

|ak|+ n

P

k=2

|bk|−(m+ 1)(m+ 1−α)

1−α

∞

P

k=m+1

|ak|+

∞ P

k=n+1

|bk|

(7)

m

X

k=2

|ak|+ n

X

k=2

|bk|+

(m+ 1)(m+ 1−α) 1−α

X∞

k=m+1

|ak|+

∞ X

k=n+1

|bk|

≤1. (14)

Since the L. H. S. of (14) is bounded above by

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

, it yields

the following inequality

m

X

k=2

k(k−α)

1−α −1

|ak|+

∞ X

k=m+1

k(k−α)

1−α −

(m+ 1)(m+ 1−α) 1−α

|ak|

+ n

X

k=2

k(k+α)

1−α −1

|bk|+

∞ X

k=n+1

k(k+α)

1−α −

(m+ 1)(m+ 1−α) 1−α

|bk| ≥0.

(m+ 1)(m+ 1−α)z

m+1_{gives the sharp result, we observe that for}_z₌_rei_mπ

that

f(z)

fm,n(z) = 1 +

1−α

(m+ 1)(m+ 1−α)z

m_−→₁₋ 1−α

(m+ 1)(m+ 1−α) =

m(m+ 2−α) (m+ 1)(m+ 1−α)

when r→1−.

To prove (ii), note that

1 +ω(z) 1−ω(z) =

(n+ 1)(n+ 1 +α) 1−α

h f(z)

fm,n(z) −

n(n+ 2 +α) (n+ 1)(n+ 1 +α)

i

= P

1 + m

P

k=2

rk−1_ei(k−1)θ_ak₊ Pn

k=2

rk−1_e−i(k+1)θ_bk .

where

P = 1 + m

X

k=2

X

k=2

+(n+ 1)(n+ 1 +α) 1−α

h _X∞

k=m+1

∞ X

k=n+1

i

.

So that

ω(z) =

(n+ 1)(n+ 1 +α) 1−α

h ∞ P

k=m+1

∞ P

k=n+1

i

Q ,

where

Q= 2 + 2

_Xm

k=2

X

k=2

+(n+ 1)(n+ 1 +α) 1−α

_X∞

k=m+1

∞ X

k=n+1

(8)

Consequently

|ω(z)| ≤

(n+ 1)(n+ 1 +α) 1−α

h ∞ P

k=m+1

|ak|+

∞ P

k=n+1

|bk|i

2−2

m P

k=2

|ak|+ n

P

k=2

|bk|−(n+ 1)(n+ 1 +α)

1−α

∞

P

k=m+1

|ak|+

∞ P

k=n+1

|bk|

.

m

X

k=2

|ak|+ n

X

k=2

|bk|+

(n+ 1)(n+ 1 +α) 1−α

X∞

k=m+1

|ak|+

∞ X

k=n+1

|bk|

≤1. (15)

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

,

which is equivalent to

m

X

k=2

k(k−α)

1−α −1

|ak|+

∞ X

k=m+1

k(k−α)

1−α −

(n+ 1)(n+ 1 +α) 1−α

|ak|

+ n

X

k=2

k(k+α)

1−α −1

|bk|+

∞ X

k=n+1

k(k+α)

1−α −

(n+ 1)(n+ 1 +α) 1−α

|bk| ≥0.

(n+ 1)(n+ 1 +α)z

n+1_{gives the sharp result, we observe that for}_z₌_rei_nπ₊₂

we get

f(z) fm,n(z)

= 1 + 1−α

(n+ 1)(n+ 1 +α)r n_e− iπ

n+2(n+2)−→1− 1−α

(n+ 1)(n+ 1 +α)=

n(n+ 2 +α) (n+ 1)(n+ 1 +α)

when r→1−. The result follows.

(i) Re

nf_m,n(z)

f(z)

o

≥ (m+ 1)(m+ 1−α)

m(m+ 2−α) + 2(1−α),(z ∈D) if n(n+ 2 +α) + 2α≥ m(m+ 2−α) or bk= 0 ∀k≥2.

(ii) Renfm,n(z) f(z)

o

≥ (n+ 1)(n+ 1 +α)

n(n+ 2 +α) + 2 ,(z ∈ D) if n(n+ 2 +α) + 2α ≤ m(m+ 2−α) or ak = 0 ∀k≥2.

Proof. To prove (i), we may write

1 +ω(z) 1−ω(z) =

m(m+ 2−α) + 2(1−α) 1−α

hfm,n(z)

f(z) −

(m+ 1)(m+ 1−α) m(m+ 2−α) + 2(1−α)

i

= P

1 +

∞ P

k=2

rk−1_ei(k−1)θ_ak₊

∞ P

k=2

rk−1_e−i(k+1)θ_bk ,

where

P = 1 + m

X

k=2

X

k=2

−(m+ 1)(m+ 1−α)

1−α

h _X∞

k=m+1

∞ X

k=n+1

i

(9)

Then

|ω(z)| ≤

m(m+ 2−α) + 2(1−α) 1−α

h ∞ P

k=m+1

|ak|+

∞ P

k=n+1

|bk|i

2−2

m P

k=2

|ak|+ n

P

k=2

|bk|−m(m+ 2−α)

1−α

∞

P

k=m+1

|ak|+

∞ P

k=n+1

|bk| ≤1.

This last inequality is equivalent to

m

X

k=2

|ak|+ n

X

k=2

|bk|+

m(m+ 2−α) + (1−α) 1−α

_X∞

k=m+1

|ak|+

∞ X

k=n+1

|bk|

≤1. (16)

Sufficiently, the L. H. S. of (16) is bounded above by

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

, the

proof is complete.

To prove (ii), we consider that

1 +ω(z) 1−ω(z) =

n(n+ 2 +α) + 2 1−α

hf_m,n(z)

f(z) −

(n+ 1)(n+ 1 +α) n(n+ 2 +α) + 2

i

= P

1 +

∞ P

k=2

∞ P

k=2

,

where

P = 1 + m

X

k=2

X

k=2

−(n+ 1)(n+ 1 +α)

1−α

h _X∞

k=m+1

∞ X

k=n+1

i

.

Then

|ω(z)| ≤

n(n+ 2 +α) + 2 1−α

h ∞ P

k=m+1

|ak|+

∞ P

k=n+1

|bk|

i

2−2 m

P

k=2

|ak|+ n

P

k=2

|bk|

−n(n+ 2 +α) + 2α

1−α

∞

P

k=m+1

|ak|+

∞ P

k=n+1

|bk|

≤1.

This last inequality is equivalent to

m

X

k=2

|ak|+ n

X

k=2

|bk|+

n(n+ 2 +α) + (1 +α) 1−α

X∞

k=m+1

|ak|+

∞ X

k=n+1

|bk|

≤1. (17)

Sufficiently, the L. H. S. of (17) is bounded above by

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

, which

completes the proof.

Theorem 2.9. If f(z) of the form (1) with b1= 0 satisfies condition (2), then Re

n f0(z)

f0

m,n(z)

o

≥ m

m+ 1−α, (z∈D) if n > m (18)

(10)

Proof. Consider

1 +ω(z) 1−ω(z) =

m+ 1−α 1−α

h f0(z)

f_m,n0 (z) − m m+ 1−α

i

= P

1 + m

P

k=2

n

P

k=2

,

where

P = 1 + m

X

k=2

krk−1ei(k−1)θak− n

X

k=2

krk−1e−i(k+1)θbk

+m+ 1−α 1−α

h _X∞

k=m+1

∞ X

k=n+1

i

.

Then

ω(z) =

m+ 1−α 1−α

h ∞ P

k=m+1

krk−1ei(k−1)θak− ∞ P

k=n+1

i

Q ,

where

Q= 2 + 2

_Xm

k=2

n

X

k=2

+m+ 1−α 1−α

_X∞

k=m+1

∞ X

k=n+1

.

Consequently, we get

|ω(z)| ≤

m+ 1−α 1−α

h ∞ P

k=m+1

k|ak| −

∞ P

k=n+1 k|bk|

i

2−2 m

P

k=2

k|ak|+ n

P

k=2 k|bk|

−m+ 1−α

1−α

∞

P

k=m+1

k|ak|+

∞ P

k=n+1 k|bk|

≤1.

This last inequality is equivalent to m

X

k=2

k|ak|+ n

X

k=2

k|bk|+ m+ 1−α 1−α

_X∞

k=m+1

k|ak|+

∞ X

k=n+1

k|bk|≤1. (19)

Since the L. H. S. of (19) is bounded above by

∞ P

k=1

k(k−α)

1−α |ak|+

k(k+α) 1−α |bk|

, the proof

is complete.

Theorem 2.10. If f of the form (1) withb1 = 0 satisfies condition (2), then

Re

nf_m,n0 (z)

f0(z)

o

≥ m+ 1−α

m+ 2(1−α), (z∈D) (20)

The result (20) is sharp with the function f(z) =z+ 1−α

(m+ 1)(m+ 1−α)z m+1_.

Proof. Proceeding exactly as in the proof of Theorem 2.9, we evidently have the required

result.

(11)

References

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[3] Dixit, K. K., and Porwal, S., (2009), A convolution approach on partial sums of certain analytic and univalent functions, J. Inequal. Pure Appl. Math., 10(4), Art. 101, pp 1-17.

[4] Frasin, B. A., (2005), Partial sums of certain analytic and univalent functions, Acta Math. Acad. Paed. Nyir., 21, pp 135-145.

[5] Frasin, B. A., (2008), Generalization of partial sums of certain analytic and univalent functions, Applied Mathematics Letters, 21, no. 7, pp 735-741.

[6] Jahangiri, J. M., (1998), Coefficient bounds and univalence criteria for harmonic functionswith negative coefficients, Ann. Univ. Mariae Cruie-Sklodowska Sect. A 52, no. 2, pp 57-66.

[7] Porwal, S., and Dixit, K. K., (2010), Partial sums of starlike harmonic univalent functions, Kyungpook Mathematical Journal, 50, no. 3, pp 433-445.

[8] Raina, R. K., and Bansal, D., (2008), Some properties of a new class of analytic functions defined in terms of a Hadamard product, J. Inequal. Pure. Appl. Math. 9, No. 1, Art 22, pp 1-9.

[9] Rosy, T., Subramanian, K. G., and Murugusundaramoorthy, G., (2003), Neighbourhoods and partial sums of starlike functions based on Ruscheweyh derivatives, J.Ineq. Pure and Appl. Math., 4(64).

[10] Silverman, H., (1997), Partial sums of starlike and convex functions, Journal of Mathematical Analysis and Applications, 209, no. 1, pp 221-227.

[11] Silvia, E.M., (1985), Partial sums of convex functions of orderα, Houston.J.Math., Math.Soc., 11(3), pp 397-404.

Ahmad Zirehis an associated professor of Shahrood University of Technology and his research interests are Complex Analysis and Complex Dynamics. He has published more than 33 research papers in various international journals. He has supervised 39 master theses and 6 doctoral theses.