EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 4, 2017, 871-876
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Sufficient conditions for starlikeness of reciprocal order

B.A. Frasin1, M. Ab. Sabri2,∗

1 Faculty of Science, Department of Mathematics, Al al-Bayt University, Mafraq, Jordan
2 College of Basic Education, Department of Mathematics, University of Mustansiriya,
Baghdad, Iraq

Abstract. The object of the present paper is to derive certain sufficient conditions for starlikeness
of reciprocal order of analytic functions in the open unit disk.

2010 Mathematics Subject Classifications: 30C45

Key Words and Phrases: Analytic functions, starlike and convex functions, starlike function of
reciprocal order, sufficient conditions

1. Introduction and definitions

Let A denote the class of functions f(z) defined by

f(z) = z +

∞∑
n=2

anz
n (1)

which are analytic and univalent in the open unit disk U = {z : |z| < 1}. A function f
∈ A is said to be starlike of order α if it satisfies

R

(
zf ′(z)

f(z)

)
> α (z ∈ U) (2)

for some α(0 ≤ α < 1). We denote by S∗(α) the subclass of A consisting of functions
which are starlike of order α in U . Clearly S∗(α) ⊆ S∗(0) = S∗, where S∗is the class of
functions that are starlike in U .

A function f ∈ A is said to be starlike of reciprocal order α if

R

{
f(z)

zf ′(z)

}
> α (z ∈ U) (3)

for some α(0 ≤ α < 1). We denote the class of such functions by S−1∗(α) (see, [1, 4, 8]).

∗Corresponding author.

Email addresses: bafrasin@yahoo.com (B.A. Frasin), mustafasabri.edbs@uomustansiriyah.edu.iq (M.
Ab. Sabri )

http://www.ejpam.com 871 c© 2017 EJPAM All rights reserved.



B. A. Frasin, M. Ab. Sabri / Eur. J. Pure Appl. Math, 10 (4) (2017), 871-876 872

In view of the fact that

Rp(z) > 0⇒ R
1

p(z)
= R

p(z)

|p(z)|2
> 0,

it follows that a starlike function of reciprocal order 0 is same as a starlike function. In
particular, every starlike function of reciprocal order α ≥ 0 is starlike and hence univalent
(cf. [10, Example 1]).

Example 1. The function f(z) = ze(1−α)z is a starlike function of reciprocal order 1/(2−
α) [10, Example 2].

Sufficient conditions were studied by various authors for starlikeness [e.g., see [2–7, 9–
12]). The object of the present paper is to derive certain sufficient conditions for starlike-
ness of reciprocal order α by using the same techniques as in [9].

In order to establish our main results, we require the following lemma due to
Nunokawa et al. [9].

Lemma 1. Let p(z) = 1 +
∞∑
n=1

cnz
n be analytic in U and suppose that there exists a point

z0 ∈ U such that
R {p(z)} > 0 for |z| < |z0| (4)

and
R {p(z0)} = 0. (5)

Then we have

z0p
′
(z0) ≤ −

1

2
(1 + |p(z0)|2), (6)

where z0p
′
(z0) is a negative real number.

2. Sufficient conditions for starlikeness of reciprocal order

Our first result is contained in the following.

Theorem 2. Let f(z) ∈ A satisfies f(z) f
′
(z) 6= 0 in 0 < |z| < 1 and

R

{
f(z)

zf ′(z)

(
1− αzf

′′
(z)

f ′(z)

)}
> −α

2

(
3 +

∣∣∣∣ f(z)

zf ′(z)

∣∣∣∣2
)

(z ∈ U ;α > 0). (7)

Then f(z) is starlike of reciprocal order 0 in U and thus, f(z) is starlike in U .



B. A. Frasin, M. Ab. Sabri / Eur. J. Pure Appl. Math, 10 (4) (2017), 871-876 873

Proof. Let us define the function p(z) by

p(z) =
f(z)

zf ′(z)
. (8)

Then p(z) is analytic in U and p(0) = 1. Differentiating (8) logarithmically we obtain

f(z)

zf ′(z)

(
1− αzf

′′
(z)

f ′(z)

)
= αzp

′
(z) + (α+ 1)p(z)− α. (9)

Suppose that there exists a point z0 ∈ U such that

R {p(z)} > 0 for |z| < |z0|

and
R {p(z0)} = 0,

then from Lemma 1, we have,

z0p
′
(z0) ≤ −

1

2
(1 + |p(z0)|2).

Therefore from (9),we have

R

{
f(z0)

z0f
′(z0)

(
1− αz0f

′′
(z0)

f ′(z0)

)}
= R

{
αz0p

′
(z0) + (α+ 1)p(z0)− α

}
.

≤ −α
2

(
1 + |p(z0)|2

)
− α

≤ −α
2

(
3 +

∣∣∣∣ f(z0)

z0f
′(z0)

∣∣∣∣2
)
.

which contradicts our condition (6) of Theorem 2. Thus we complete the proof of
Theorem 2.

Next, we derive the following.

Theorem 3. Let f(z) ∈ A satisfies f(z) f
′
(z) 6= 0 in 0 < |z| < 1 and

R

{
f(z)

zf ′(z)

(
−1− zf

′′
(z)

f ′(z)

)}
> −5

4
− 1

4

∣∣∣∣ 2f(z)

zf ′(z)
− 1

∣∣∣∣2 (z ∈ U).

Then f(z) is starlike of reciprocal order 1
2 in U .

Proof. Putting

p(z) = 2

(
f(z)

zf ′(z)
− 1

2

)
, (10)



B. A. Frasin, M. Ab. Sabri / Eur. J. Pure Appl. Math, 10 (4) (2017), 871-876 874

then we have p(0) = 1. Suppose that there exists a point z0 ∈ U satisfies the conditions
(4) and (5) of Lemma 1, from (10) we have

R

{
f(z0)

z0f
′(z0)

(
−1− zf

′′
(z0)

f ′(z0)

)}
= R

{
1

2
z0p

′
(z0)− 1

}
. (11)

Using (6) of Lemma 1 in (11), it follows that

R

{
f(z0)

z0f
′(z0)

(
−1− z0f

′′
(z0)

f ′(z0)

)}
≤ −1

4

(
1 + |p(z0)|2

)
− 1

≤ −5

4
− 1

4
|p(z0)|2

≤ −5

4
− 1

4

∣∣∣∣ 2f(z0)

z0f
′(z0)

− 1

∣∣∣∣2 .
which contradicts the hypothesis of Theorem 3 and therefore, we have

R {p(z)} > 0 (z ∈ U)

or

R

{
f(z)

zf ′(z)

}
>

1

2
(z ∈ U).

Finally, we discuss the following theorem.

Theorem 4. Let f(z) ∈ A satisfies

R

{
f(z)

zf ′(z)

(
1− αzf

′′
(z)

f ′(z)

)}
> − α

(2− α)

∣∣∣∣ f(z)

zf ′(z)
− α

2

∣∣∣∣2 +
α

4
(3α−4) (z ∈ U ; 0 ≤ α < 2).

(12)
Then f(z) is starlike of reciprocal order α

2 in U .

Proof. Let the function p(z) be defined by

f(z)

zf ′(z)
=
(

1− α

2

)
p(z) +

α

2
, p(0) = 1. (13)

Suppose that there exists a point z0 ∈ U satisfies the conditions (4) and (5) of Lemma 1,
from (13) we have

R

{
f(z0)

z0f
′(z0)

(
1− αz0f

′′
(z0)

f ′(z0)

)}
= R

{
α
(

1− α

2

)
z0p

′
(z0) + (1 + α)

(
1− α

2

)
p(z0) +

α

2
(α− 1)

}
. (14)



REFERENCES 875

Thus, by using (5) and (6) of Lemma 1 in (14), it follows that

R

{
f(z0)

z0f
′(z0)

(
1− αz0f

′′
(z0)

f ′(z0)

)}
≤ −α

2

(
1− α

2

)(
1 + |p(z0)|2

)
+
α

2
(α− 1)

≤ −α
2

(
1− α

2

)
|p(z0)|2 +

α

4
(3α− 4)

≤ − α

(2− α)

∣∣∣∣ f(z0)

z0f
′(z0)

− α

2

∣∣∣∣2 +
α

4
(3α− 4)

which contradicts the hypothesis (12). It follows that

R

{
f(z)

zf ′(z)

}
>
α

2
(z ∈ U).

Thus proof of the Theorem 4 is completed.

References

[1] Muhammad Arif, Maslina Darus, Mohsan Raza and Qaiser Khan, Coefficient bounds
for some families of starlike and convex functions of reciprocal order, The scientific
world journal, Volume 2014, Article ID 989640, 6 pages.

[2] B.A. Frasin, New sufficient conditions for analytic and univalent functions, Acta Univ
Apul. No. 17 (2009), 1-7.

[3] B.A. Frasin, On sufficient conditions for strongly starlikeness and strongly convex
functions, Kyungpook Math.J. 46(2006), 131-137.

[4] B.A. Frasin,Y. Talafha and Tariq Al-Hawary, Subordination results for classes of
functions of reciprocal order, Tamsui Oxford Journal of Mathematical Sciences, 30
(2014) 81-89.

[5] Z. Lewandowski, S.S. Miller and E. Zlotkiewicz, Generating functions for some classes
of univalent functions, Proc. Amer. Math. Soc. 56 (1976), 111–117.

[6] Jian-Lin Li and S. Owa, Sufficient conditions for starlikeness, Indian J. Pure Appl.
Math. 33, 9 (2002), 1385–1390.

[7] Jin-Lin Liu, Some argument inequalities for certain analytic functions, Math. Slovaca
62 (2012), No. 1, 25–28.

[8] J. Nishiwaki and S. Owa, Coefficient inequalities for starlike and convex functions of
reciprocal order, Electronic Journal of Mathematical Analysis and Applications, vol.
1, no. 2, pp. 212–216, 2013.

[9] Mamoru Nunokawa, S.P. Goyal and Rakesh Kumar, Sufficient conditions for starlike-
ness, Journal of Classical Analysis, Volume 1, Number 1 (2012), 85–90.



REFERENCES 876

[10] M. Nunokawa, S. Owa, J. Nishiwaki, K. Kuroki and T. Hayami, Differential subordi-
nation and argumental property, Comput. Math. Appl. 56 (10) (2008) 2733–2736.

[11] M. Nunokawa, S.Owa, S.K. Lee, M. Obradovic, M.K. Aouf, H. Saitoh, H. Ikada and
N. Koika, Sufficient conditions for starlikeness, Chinese Journal of Mathematics 24
(1996), 265–270.

[12] C. Ramesha, S. Kumar and K.S. Padmanbham, A sufficient condition for starlikeness,
Chinese J. Math. 23 (1995), 167–171.