3_544_goyal.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 4, No. 3, 2011, 230-236

ISSN 1307-5543 – www.ejpam.com

Sufficient Conditions For Sakaguchi Type Functions of Order β

S. P. Goyal1,∗, Pramila Vijaywargiya1, Pranay Goswami2

1 Department of Mathematics, University of Rajasthan, Jaipur-302004, India
2 Department of Mathematics, Amity University Rajasthan, Jaipur-302002, India

Abstract. In this paper, we obtain some sufficient conditions for Sakaguchi type function of order β ,

defined on the open unit disk. Several interesting consequences of our results are also pointed out.

2000 Mathematics Subject Classifications: 30C45.

Key Words and Phrases: Sakaguchi type functions of order β , Univalent functions.

1. Introduction

Let An be the class of all functions f (z) = z + an+1zn+1 + . . . , which are analytic in the

open unit disk ∆= {z : z ∈ C ; |z| < 1} and let A1 = A.

A function f (z) ∈ An is said to be in class Sn(β , t), if it satisfies

Re

¨

(1− t)z f ′(z)

f (z)− f (tz)

«

> β , (|t| ≤ 1, t 6= 1) (1)

for some β(0≤ β < 1) and for all z ∈∆. For n= 1, this class is reduced to S(β , t) (see, [5]).

The class S(0,−1) was introduced by Sakaguchi [7]. Therefore, a function f (z) ∈ S(β ,−1) is

called Sakaguchi function of order β (see, [1]). Recently Owa et al. [5], Goyal and Goswami

[2] have discussed some properties for functions f (z) ∈ S(β , t).

In this paper, we obtain some sufficient conditions for functions f (z) ∈ Sn(β , t). To prove

our results, we need the following :

Lemma 1 ([4]). Let Ω be a set in the complex plane C and suppose that Φ is a mapping from

C
2 ×∆ to C which satisfies Φ(i x , y; z) 6∈ Ω for z ∈∆, and for all real x, y such that

y ≤ −n(1+x2)/2. If the function p(z) = 1+cnzn+. . . is analytic in∆ and Φ(p(z), zp′(z); z) ∈ Ω
for all z ∈∆, then Re(p(z)) > 0.

∗Corresponding author.

Email addresses: somprg�gmail.
om (S. Goyal), pramilavijay1979�gmail.
om (P. Vijaywargiya),pranaygoswami83�gmail.
om (P. Goswami)

http://www.ejpam.com 230 c© 2011 EJPAM All rights reserved.



S. Goyal, P. Vijaywargiya, P. Goswami / Eur. J. Pure Appl. Math, 4 (2011), 230-236 231

2. Main Results

Theorem 1. If f (z) ∈ An satisfies

Re

�

(1− t)2z f ′(z)

f (z)− f (tz)

¨

αz f ′′(z)

f ′(z)
+
αtz f ′(tz)

f (z)− f (tz)
+ 1

«�

>

αβ

§

β +
n

2
(1− t)− (1− t)

ª

+

§

β −
nα

2

ª

(1− t)

for

(z ∈∆, 0≤ α ≤ 1, 0≤ β < 1, |t| ≤ 1 and t 6= 1),

then f (z) ∈ Sn(β , t).

Proof. Define p(z) by
(1− t)z f ′(z)

f (z)− f (tz)
= (1− β)p(z) + β .

Then p(z) = 1+ cnzn + . . . and is an analytic in ∆.

A computation shows that

z f ′′(z)

f ′(z)
+

tz f ′(tz)

f (z)− f (tz)
=
(1− t)(1− β)zp′(z) + [(1− β)p(z) + β]2 − (1− t)[(1− β)p(z) +β]

(1− t)[(1− β)p(z) + β]

and hence

(1− t)2z f ′(z)

f (z)− f (tz)

�

αz f ′′(z)

f ′(z)
+
αtz f ′(tz)

f (z)− f (tz)
+ 1

�

= α(1− t)(1− β)zp′(z) +α(1− β)2p2(z) + (1− β)[2αβ + (1−α)(1− t)]p(z) + β[αβ + (1−α)(1− t)]

= Φ(p(z), zp′(z); z) (say),

where

Φ(r, s; z) = α(1−t)(1−β)s+α(1−β)2r2+(1−β)[2αβ+(1−α)(1−t)]r+β[αβ+(1−α)(1−t)].

For all real x and y satisfying y ≤ −n(1+ x2)/2, we have

Re[Φ(i x , y; z)] = α(1− t)(1− β)y −α(1− β)2 x2+ β[αβ + (1−α)(1− t)]

≤ α(1− t)(1− β)
¦

−n(1+ x2)/2
©

−α(1− β)2 x2+ β[αβ + (1−α)(1− t)]

=
−αn

2
(1− t)(1− β)−

§αn

2
(1− t)(1− β) +α(1− β)2

ª

x2+ β[αβ + (1−α)(1− t)]

≤
−αn

2
(1− t)(1− β) + β[αβ + (1−α)(1− t)]

= αβ

§

β +
n

2
(1− t)− (1− t)

ª

+

§

β −
nα

2

ª

(1− t).



S. Goyal, P. Vijaywargiya, P. Goswami / Eur. J. Pure Appl. Math, 4 (2011), 230-236 232

Let Ω =
¦

w; Re(w) > αβ
¦

β + n

2
(1− t)− (1− t)

©

+
¦

β − nα

2

©

(1− t)
©

.

Then Φ(p(z), zp′(z); z) ∈ Ω and Φ(i x , y; z) 6∈ Ω for all real x and y ≤ −n(1+ x2)/2, z ∈ ∆.

By an application of Lemma 1, the result fellows.

On taking t = −1, in the Theorem 1, we have following

Corollary 1. If f (z) ∈ An satisfies

Re

�

z f ′(z)

f (z)− f (−z)

¨

αz f ′′(z)

f ′(z)
−
αz f ′(−z)

f (z)− f (−z)
+ 1

«�

>
αβ

4

�

β + n− 2
	

+

�

2β − nα

4

�

for

(z ∈∆, 0≤ α ≤ 1, 0≤ β < 1),

then f (z) ∈ Sn(β ,−1).

By taking β = 0 in Corollary 1, we have

Corollary 2. If f (z) ∈ An satisfies

Re

�

z f ′(z)

f (z)− f (−z)

¨

αz f ′′(z)

f ′(z)
−
αz f ′(−z)

f (z)− f (−z)
+ 1

«�

>
−nα

4

where

(z ∈∆, 0≤ α ≤ 1),

then f (z) ∈ Sn(0,−1).

If we take t = 0 in the Theorem 1, we have the following

Corollary 3 ([6]). If f (z) ∈ An satisfies

Re

�

z f ′(z)

f (z)

¨

αz f ′′(z)

f ′(z)
+ 1

«�

> αβ

§

β +
n

2
− 1

ª

+

§

β −
nα

2

ª

for

(z ∈∆, 0≤ α ≤ 1, 0≤ β < 1),

then f (z) ∈ Sn(β , 0) = S∗n(β).

If we take β = 0 and n= 1 in Corollary 3, we have

Corollary 4 ([3]). If f (z) ∈ A satisfies

Re

�

z f ′(z)

f (z)

¨

αz f ′′(z)

f ′(z)
+ 1

«�

>
−α

2
(z ∈∆),

for some α(α≥ 0), then f (z) ∈ S1(0,0) = S∗.

If we take β = α
2

and n= 1, in Corollary 3, we get the following



S. Goyal, P. Vijaywargiya, P. Goswami / Eur. J. Pure Appl. Math, 4 (2011), 230-236 233

Corollary 5 ([3]). If f (z) ∈ A satisfies

Re

�

z f ′(z)

f (z)

¨

αz f ′′(z)

f ′(z)
+ 1

«�

>
−α2

4
(1−α) (z ∈∆),

for some α(0≤ α < 2), then f (z) ∈ S1(
α

2
, 0) = S∗(α

2
).

Theorem 2. Let 0≤ β < 1, |t| ≤ 1, t 6= 1 with −1≤ t + β < 1,

λ= (1− β)2
§

1− β + (1− t)
n

2

ª2

, µ=

§

(1−β)(1− t)
n

2
− (β2 − (1− t)β)

ª2

,

ν =
¦

(1− β)2+ (β2 − (1− t)β)
©2

and σ = (1− β)2(2β − 1+ t)2 (2)

satisfy (λ+µ− ν +σ)β2 < (1− 2β)µ.

Also suppose that u0 be the positive real root of the equation

2λ(1− β)2u3 +
¦

(1− β)2(2λ+µ− ν +σ) + 3λβ2
©

u2 + 2β2(2λ+µ− ν +σ)u

+(λ+ 2µ− ν +σ)β2 − (1− β)2µ = 0 (3)

and

ρ2 =
(1− β)2(1+ u0)

(1− t)2
�

(1− β)2uo +β
2
	[λu2

0 + (λ+µ− ν +σ)u0 +µ]. (4)

Now if f (z) ∈ An satisfies

�

�

�

�

�

�

(1− t)z f ′(z)

f (z)− f (tz)
− 1

��

z f ′′(z)

f ′(z)
+

tz f ′(tz)

f (z)− f (tz)

�

�

�

�

�

�

≤ ρ (z ∈∆),

then f (z) ∈ Sn(β , t).

Proof. Define p(z) by
(1− t)z f ′(z)

f (z)− f (tz)
= (1− β)p(z) + β .

Then p(z) = 1+ cnzn + . . . and is an analytic in ∆.

A computation shows that

z f ′′(z)

f ′(z)
+

tz f ′(tz)

f (z)− f (tz)
=
(1− t)(1− β)zp′(z) + [(1− β)p(z) + β]2 − (1− t)[(1− β)p(z) +β]

(1− t)[(1− β)p(z) + β]

and hence

�

(1− t)z f ′(z)

f (z)− f (tz)
− 1

��

z f ′′(z)

f ′(z)
+

tz f ′(tz)

f (z)− f (tz)

�

=
(1− β)(p(z)− 1)

(1− t)[(1− β)p(z) + β]

�

(1− t)(1− β)zp′(z) + [(1− β)p(z) + β]2 −(1− t)[(1−β)p(z) + β]
	



S. Goyal, P. Vijaywargiya, P. Goswami / Eur. J. Pure Appl. Math, 4 (2011), 230-236 234

= Φ(p(z), zp′(z); z).

Then, for all real x and y satisfying y ≤ −n(1+ x2)/2, we have

|Φ(i x , y; z)|2 =
(1− β)2(1+ x2)

(1− t)2[(1−β)2 x2+ β2]

×
h

¦

(1− t)(1−β)y − β(1− t −β)− (1− β)2 x2
©2
+ (1− β)2(2β − 1+ t)2 x2

i

=
(1− β)2(1+ u)

(1− t)2[(1−β)2u+ β2]

×
h

¦

(1− t)(1−β)y − β(1− t −β)− (1− β)2u
©2
+ (1−β)2(2β − 1+ t)2u

i

=g(u, y)

where u= x2 > 0 and y ≤ −n(1+ x2)/2.

Since

∂ g

∂ y
=

2(1− β)3(1+ u)

(1− t)[(1− β)2u+ β2]

¦

(1− t)(1− β)y − β(1− t − β)− (1−β)2u
©

< 0,

therefore we have

h(u) = g[u,−n(1+ u)/2]≤ g(u, y),

where

h(u) =
(1− β)2(1+ u)

(1− t)2
�

(1− β)2u+ β2
	[λu2 + (λ+µ− ν +σ)u+µ], (5)

where λ, µ, ν and σ are given in (2).

Now differentaiting (5) and using h′(u) = 0, we get

2λ(1− β)2u3 +
¦

(1− β)2(2λ+µ− ν +σ) + 3λβ2
©

u2+

2β2(2λ+µ− ν +σ)u+ (λ+ 2µ− ν +σ)β2 − (1− β)2µ = 0

which is a cubic equation in u. Since u0 is the positive real root of this equation we have

h(u)≥ h(u0) and hence

|Φ(i x , y; z)|2 ≥ h(u0) = ρ
2.

Define Ω =
�

w; |w| < ρ
	

, then Φ(p(z), zp′(z); z) ∈ Ω and Φ(i x , y; z) 6∈ Ω for all real x and

y ≤ −n(1+ x2)/2, z ∈∆. Therefore by an application of Lemma 1. the result follows.

By taking t = −1, in Theorem 2, we have the following

Corollary 6. Let 0 ≤ β < 1, λ1 = (1 − β)
2
�

1− β + n
	2

, µ1 =
¦

(1− β)n− (β2− 2β)
©2

,

ν1 =
¦

(1− β)2+ (β2 − 2β)
©2

and σ1 = 4(1−β)4, satisfy (λ1+µ1−ν1+σ1)β
2 < (1−2β)µ1.

Also suppose that u1 be the positive real root of the equation

2λ1(1− β)
2u3 +
¦

(1− β)2(2λ1+µ1− ν1 +σ1) + 3λ1β
2
©

u2



S. Goyal, P. Vijaywargiya, P. Goswami / Eur. J. Pure Appl. Math, 4 (2011), 230-236 235

+2β2(2λ1+µ1− ν1 +σ1)u+ (λ1+ 2µ1− ν1 +σ1)β
2 − (1− β)2µ1 = 0 (6)

and

ρ2
1 =

(1− β)2(1+ u1)

4
�

(1− β)2u1 + β
2
	[λ1u2

1 + (λ1 +µ1− ν1 +σ1)u1 +µ1]. (7)

Now if f (z) ∈ An satisfies

�

�

�

�

�

�

2z f ′(z)

f (z)− f (−z)
− 1

��

z f ′′(z)

f ′(z)
−

z f ′(−z)

f (z)− f (−z)

�

�

�

�

�

�

≤ ρ1 (z ∈∆),

then f (z) ∈ Sn(β ,−1).

By taking β = 0 in Corollary 6, we get the following

Corollary 7. Let u2 be the positive real root of the equation

2(n+ 1)2u3 + (3n2+ 4n+ 5)u2− n2 = 0 (8)

and

ρ2
2 =
(1+ u2)

4u2

[(n+ 1)2u2
2 + 2(n2+ n+ 2)u2+ n2]. (9)

Now if f (z) ∈ An satisfies

�

�

�

�

�

�

2z f ′(z)

f (z)− f (−z)
− 1

��

z f ′′(z)

f ′(z)
−

z f ′(−z)

f (z)− f (−z)

�

�

�

�

�

�

≤ ρ2 (z ∈∆),

then f (z) ∈ Sn(0,−1).

By taking n = 1 in Corollary 7, we have u3 = 0.266048 . . ., thus we have the following

result

Corollary 8. If f (z) ∈ A satisfies

�

�

�

�

�

�

2z f ′(z)

f (z)− f (−z)
− 1

��

z f ′′(z)

f ′(z)
−

z f ′(−z)

f (z)− f (−z)

�

�

�

�

�

�

≤ ρ3 (z ∈∆),

where ρ3 = 2.0145979 . . ., then f (z) ∈ S(0,−1).

By taking t = 0 in Theorem 2, we get a known result due to Ravichandran et al. [6, Thm.

2.5]. For n = 1, β = 0 = t, our Theorem 2 reduces to another known result due to Li and

Owa [3].

ACKNOWLEDGEMENTS The first author (S P G) is thankful to CSIR, New Delhi, India for

awarding Emeritius Scientist under scheme No. 21(084)/10/EMR-II and second author (P

V) is thankful to the CSIR, India, for providing Senior Research Fellowship under research

scheme No. 09/149(0431)/2006-EMR-I.



REFERENCES 236

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