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Some results of analytic functions related to concave domains
Applied Mathematics Letters 25 (2012) 1505–1509
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Some results of analytic functions related to concave domains R. Aghalary a,∗ , Zhi-Gang Wang b a
Department of Mathematics, Faculty of Science, Urmia University, Urmia, Iran
b
School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan 455002, People’s Republic of China
article
abstract
info
Article history: Received 10 August 2011 Accepted 1 January 2012
The main purpose of the present paper is to derive some results for analytic functions whose takes values on concave domains. As special cases of these results, several new sufficient conditions for univalency and starlikeness of analytic functions are obtained. © 2012 Elsevier Ltd. All rights reserved.
Keywords: Concave domains Starlike function Univalent functions
1. Introduction and preliminaries Let An denote the class of functions of the form f (z ) = z +
∞
ak+n z k+n
(n ∈ N = {1, 2, . . .})
k=1
which are analytic in the open unit disc U = {z : |z | < 1}. In particular, we set A1 = A. As usual, we denote by S the subclass of An , consisting of functions which are univalent in U. Also we define the subclass S ∗ of starlike functions whenever f (U ) is a domain that is starlike with respect to the origin. We denote the subclass of An consisting of functions f (z ) satisfying f (z ) Re( z ) > 0 by Sn0 . Set S10 = S0 . Let β < 1 and k be a real number such that k > 1, we define Dk (β) = {ω ∈ C; |ω − kβ| > |Reω − β|}, and
Ω (β) = {ω ∈ C; |ω − 2β + 1| > |Reω − β|}. It is easy to see that Dk (β) is a concave region laying on the left side of the parabola
γ : y2 = 2β x(k − 1) − (k2 − 1)β 2 . Also Ω (β) is a concave set on the right side of the parabola
γ : y2 = 2(β − 1)x − 3β 2 + 4β − 1. We shall use the following lemmas to prove our main result.
∗
Corresponding author. E-mail addresses: [email protected] (R. Aghalary), [email protected] (Z.-G. Wang).
0893-9659/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2012.01.005
1506
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509
Lemma 1.1 ([1]). Let ω(z ) = ωn z n + ωn+1 z n+1 + · · · be analytic for |z | ≤ |z0 | < 1 with ω(0) = 0. Then if
|ω(z0 )| = max{|ω(z )| : |z | ≤ |z0 |}, then
z0 ω′ (z0 ) ω(z0 )
= m, where m ≥ n.
Lemma 1.2 ([2]). If f (z ) = z + a2 z 2 + a3 z 3 + · · · ∈ S0 , then the partial sums s n ( z ) = z + a2 z 2 + a3 z 3 + · · · + an z n are univalent in |z | <
1 4
(n = 2, 3, . . .)
.
Lemma 1.3 ([3]). If f (z ) ∈ S0 , then f (z ) is univalent in |z | <
√ 2 − 1.
2. Main results γn
Theorem 2.1. Let n ∈ N = {1, 2, . . .}, k > 1, γ > 0, k+2 1 < β < 1 and 0 < α ≤ (k−1)β . If q (be a nonvanishing function in zq′ (z ) the (U − {0})) belongs to An and the function h(z ) = (1 − γ )( q(zz ) )α + γ q(z ) ( q(zz ) )α satisfies h(U ) ⊂ Dk (β), then
Re
α
z
> β.
q(z )
Proof. Suppose p(z ) = ( q(zz ) )α . It is easy to check that
(1 − γ )
z q(z )
α
+γ
zq′ (z ) q(z )
z q(z )
α
= p(z ) −
γ ′ zp (z ). α
(2.1)
We define the function ω(z ) by p(z ) =
1 + (1 − 2β)ω(z ) 1 − ω(z )
.
(2.2)
Then ω(z ) is analytic in U and ω(0) = 0. It follows from (2.2) that p(z ) −
1 + (1 − 2β)ω(z ) γ 2(1 − β)z ω′ (z ) γ ′ zp (z ) = − . α 1 − ω(z ) α (1 − ω(z ))2
Suppose that there exists a point z0 ∈ U such that max |ω(z )| = |ω(z0 )| = 1.
|z |≤|z0 |
Then, using Lemma 1.1, we have z0 ω′ (z0 ) = mω(z0 )
(m is real and m ≥ n),
and ω(z0 ) = e where 0 ≤ x ≤ 2π . Thus from (2.3) we obtain that ix
p(z0 ) −
γ (1 − β) sin x γ m(1 − β) z0 p′ (z0 ) = β + i + , α 1 − cos x α 1 − cos x
so
2 2 γ γ p(z0 ) − z0 p′ (z0 ) − kβ = p(z0 ) − z0 p′ (z0 ) − β − (k − 1)β α α 2 γ m(1 − β) (1 − β)2 sin2 x = − (k − 1)β + α 1 − cos x (1 − cos x)2 2 γ = Re p(z0 ) − z0 p′ (z0 ) − β α
(2.3)
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509
1507
γ 1)β α (1
(k − 1)2 β 2 (1 − cos x) + (1 − β)2 (1 + cos x) − 2(k − − β)m + 1 − cos x 2 γ ≤ Re p(z0 ) − z0 p′ (z0 ) − β α (k − 1)2 β 2 (1 − cos x) + (1 − β)2 (1 + cos x) − 2(k − 1)β γα (1 − β)n + . 1 − cos x γ
If we let f (t ) = t [(1 − β)2 − (k − 1)2 β 2 ] + [(k − 1)2 β 2 + (1 − β)2 − 2(k − 1)β α n], where t = cos x, then under the hypothesis of the theorem we obtain
γ n ≤ 0. α γ γ Hence we have |p(z0 ) − α z0 p′ (z0 ) − kβ| ≤ |Re(p(z0 ) − α z0 p′ (z0 )) − β| and this implies that h(z0 ) ̸∈ Dk (β) which contradicts z α the hypothesis of the theorem so Re( q(z ) ) > β (z ∈ U ) and the proof is complete.
f (t ) ≤ 2(k − 1)β (k − 1)β −
By putting γ = α = 1 and q(z ) =
f (z ) f ′ (z )
in the Theorem 2.1 we have.
Corollary 2.1. Let f be univalent and belongs to An , h(z ) =
zf ′ (z )
f (z )
zf ′ (z ) f (z )
−
zf ′′ (z )
f ′ (z )
2 3
< β < 1 and the function
,
satisfies the condition h(U ) ⊆ D2 (β), zf ′ (z )
then Re f (z ) > β . Also by choosing γ = α = 1 and q(z ) =
z 2 f ′ (z ) f (z )
in the Theorem 2.1 we obtain.
Corollary 2.2. Let f be univalent and belongs to An , h(z ) =
f (z )
2+
zf ′ (z )
zf (z ) ′′
f ′ (z )
2 3
< β < 1 and the function
− 1,
satisfies the condition h(U ) ⊆ D2 (β), f (z )
then Re zf ′ (z ) > β . γn
Corollary 2.3. Let n ∈ N = {1, 2, . . .}, k > 1, γ > 0, k+2 1 < β < 1 and 0 < α ≤ (k−1)β . If q (be a nonvanishing function in the (U − {0})) belongs to An and
Re (1 − γ )
z q(z )
α
+γ
zq′ (z ) q(z )
z q(z )
α
<
(k + 1)β 2
then
Re
z
α
q(z )
> β.
Proof. It is easy to see that the region Γ = {z ∈ C; Rez < result.
(k+1)β 2
} is contained in Dk (β) and so by Theorem 2.1 we get our γn
Corollary 2.4. Let n ∈ N = {1, 2, . . .}, k > 1, γ > 0, k+2 1 < β < 1 and 0 < α ≤ (k−1)β . If q (be a nonvanishing function on (U − {0})) belongs to An and
α α z zq′ (z ) z k−1 arg (1 − γ ) +γ > arctan k + 1 q(z ) q(z ) q(z ) then
Re
z q(z )
α
> β.
1508
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509 1+(1−(k+1)β) 1 −z
It is known that the function g (z ) =
α
α
(k+1)β 2
for (z ∈ U ). Set p(z ) = ( q(zz ) )α and
γ ′ 1 + (1 − (k + 1)β) zp (z ) = . α 1−z ∞ n Solving this differential equation we find that p(z ) = 1 − α(2 − (k + 1)β) n=1 γ nz−α . Now using Corollary 2.3 we obtain (1 − γ )
z
q(z )
+γ
zq′ (z )
satisfies the condition Reg (z ) <
q(z )
z
q(z )
= p(z ) −
the following example. γ
Example 2.1. If k > 1, γ > 0, k+2 1 < β < 1 and 0 < α < min{ (k−1)β , γ } then we have
Re 1 − α(2 − (k + 1)β)
∞
zn
n =1
γn − α
> β.
Theorem 2.2. Let n ∈ N = {1, 2, . . .}, γ > 0, β < 1 and 0 < α ≤ γ n. If f ∈ An and h(z ) = (1 − γ )( satisfies the condition
f (z ) α z
) + γ zff ((zz)) ( f (zz ) )α ′
h(U ) ⊆ Ω (β), then
Re
f (z )
α
z
> β.
Proof. Defining the function p(z ) by p(z ) =
f (z )
α
z
,
we have
(1 − γ )
f (z )
α
z
+γ
zf ′ (z ) f (z )
f (z ) z
α
= p(z ) +
γ ′ zp (z ). α
(2.4)
If we define the function ω(z ) by p(z ) =
1 + (1 − 2β)ω(z ) 1 − ω(z )
,
(2.5)
then ω(z ) is analytic in U and ω(0) = 0. Assuming that there exists a point z0 ∈ U such that max |ω(z )| = |ω(z0 )| = 1.
|z |≤|z0 |
Lemma 1.1, implies that p(z0 ) +
γ (1 − β) sin θ γ m(1 − β) z0 p′ (z0 ) = β + i − , α 1 − cos θ α 1 − cos θ
where ω(z0 ) = eiθ . So
2 γ 2(1 − β)2 m2 γ 2 (1 − β)2 (1 − β)2 γ m + 2 − 2 p(z0 ) + z0 p′ (z0 ) − β + (1 − β) = α (1 − cos θ ) α (1 − cos θ )2 α(1 − cos θ ) γ 2 2 1 − α m (1 − β)2 γ ′ = Re p(z0 ) + z0 p (z0 ) − β + α 1 − cos θ γ 2 2 1 − n (1 − β)2 γ α ≤ Re p(z0 ) + z0 p′ (z0 ) − β + α 1 − cos θ 2 γ ′ ≤ Re p(z0 ) + z0 p (z0 ) − β . α Thus h(z0 ) ̸∈ Ω (β) which contradicts the hypothesis of the Theorem 2.2. Hence, we conclude that |ω(z )| < 1 for all z ∈ U and the proof is complete.
By putting γ = α = 1 and n = 1 on the Theorem 2.2 we obtain the result in [4] (see Theorem 1).
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509
Corollary 2.5. Let n ∈ N = {1, 2, . . .} and γ ≥ zf (z ) ′
f (z )
zf (z ) ′
1+γ −
zf (z ) ′′
+
f (z )
1 . n
1509
If f ∈ An , satisfies the condition
∈ Ω (β) z ∈ U ,
f ′ (z )
zf ′ (z )
then Re f (z ) > β . For β < 1, γ > 0 and α = γ the function h(z ) = z
1 + (1 − 2β)z 1−z
satisfies
(1 − γ ) with
α1
h(z )
α
z
zh′ (z )
+γ
h(z )
h(z )
α
∈ Ω (β) z ∈ U ,
z
α α h(z ) zh′ (z ) h(z ) ∂ (1 − γ ) +γ = Ω (β), z h(z ) z h(eiθ )
h(z )
and Re( z )α > β (z ∈ U ), (Re( eiθ )α = β for eiβ ̸= 1). Therefore in the case α = γ > 0 the last β in Theorem 2.2 cannot be replaced by a smaller number. Corollary 2.6. Let n ∈ N = {1, 2, . . .}, γ ≥ 0 and 0 < α ≤ γ n. If f ∈ An , satisfies the condition
(1 − γ ) then Re(
f (z )
α
z
+γ
zf ′ (z )
f (z )
f (z )
α
z
∈Ω
1
2
z ∈ U,
f (z ) α z
) > 12 .
It is know [5] that if Re(
f (z ) z
)>
1 2
for |z | < 1, then f (z ) is starlike for |z | < 2
Corollary 2.7. Let n ∈ N = {1, 2, . . .}, γ ≥
(1 − γ )
f (z )
z
+ γ f ′ (z ) ∈ Ω
then f is starlike for |z | <
1 . n
1
−1 2
. So from Corollary 2.6 we obtain.
If f ∈ An , and z ∈ U,
2
√ 2 . 2
Also by using Lemma 1.3 we have. Corollary 2.8. Let n ∈ N = {1, 2, . . .}, γ ≥
(1 − γ )
f (z )
z
1 . n
If f ∈ An , and
+ γ f ′ (z ) ∈ Ω (0) z ∈ U ,
then f is univalent for |z | <
√ 2 − 1.
Finally, using Lemma 1.2 and Theorem 2.2 we obtain. Corollary 2.9. Let n ∈ N = {1, 2, . . .}, γ ≥
(1 − γ )
f (z )
z
1 . n
If f (z ) = z + an+1 z n+1 + an+2 z n+2 + · · · + an+m z n+m + · · · ∈ An , and
+ γ f ′ (z ) ∈ Ω (0) z ∈ U ,
then the partial sums Sm (z ) = z + an+1 z n+1 + an+2 z n+2 + · · · + an+m z n+m are univalent for |z | <
1 . 4
References [1] [2] [3] [4] [5]
I.S. Jack, Functions starlike and convex of order α , J. Lond. Math. Soc. 3 (1971) 469–474. f (z ) K. Yamagushi, On functions satisfying Re[ z ] > 0, Proc. Amer. Math. Soc. 17 (1966) 588–591. K. Noshiro, On the univalency of certain analytic functions, J. Fac. Sci. Hokkaido. Imp. Univ. 2 (1934) 89–101. J. Sokół, On functions with derivative satisfying a geometric condition, Appl. Math. Comput. 204 (2008) 116–119. T.H. Macgregor, The radius of convexity for starlike functions of order 12 , Proc. Amer. Math. Soc. 14 (1963) 71–76.
Contents lists available at SciVerse ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Some results of analytic functions related to concave domains R. Aghalary a,∗ , Zhi-Gang Wang b a
Department of Mathematics, Faculty of Science, Urmia University, Urmia, Iran
b
School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan 455002, People’s Republic of China
article
abstract
info
Article history: Received 10 August 2011 Accepted 1 January 2012
The main purpose of the present paper is to derive some results for analytic functions whose takes values on concave domains. As special cases of these results, several new sufficient conditions for univalency and starlikeness of analytic functions are obtained. © 2012 Elsevier Ltd. All rights reserved.
Keywords: Concave domains Starlike function Univalent functions
1. Introduction and preliminaries Let An denote the class of functions of the form f (z ) = z +
∞
ak+n z k+n
(n ∈ N = {1, 2, . . .})
k=1
which are analytic in the open unit disc U = {z : |z | < 1}. In particular, we set A1 = A. As usual, we denote by S the subclass of An , consisting of functions which are univalent in U. Also we define the subclass S ∗ of starlike functions whenever f (U ) is a domain that is starlike with respect to the origin. We denote the subclass of An consisting of functions f (z ) satisfying f (z ) Re( z ) > 0 by Sn0 . Set S10 = S0 . Let β < 1 and k be a real number such that k > 1, we define Dk (β) = {ω ∈ C; |ω − kβ| > |Reω − β|}, and
Ω (β) = {ω ∈ C; |ω − 2β + 1| > |Reω − β|}. It is easy to see that Dk (β) is a concave region laying on the left side of the parabola
γ : y2 = 2β x(k − 1) − (k2 − 1)β 2 . Also Ω (β) is a concave set on the right side of the parabola
γ : y2 = 2(β − 1)x − 3β 2 + 4β − 1. We shall use the following lemmas to prove our main result.
∗
Corresponding author. E-mail addresses: [email protected] (R. Aghalary), [email protected] (Z.-G. Wang).
0893-9659/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2012.01.005
1506
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509
Lemma 1.1 ([1]). Let ω(z ) = ωn z n + ωn+1 z n+1 + · · · be analytic for |z | ≤ |z0 | < 1 with ω(0) = 0. Then if
|ω(z0 )| = max{|ω(z )| : |z | ≤ |z0 |}, then
z0 ω′ (z0 ) ω(z0 )
= m, where m ≥ n.
Lemma 1.2 ([2]). If f (z ) = z + a2 z 2 + a3 z 3 + · · · ∈ S0 , then the partial sums s n ( z ) = z + a2 z 2 + a3 z 3 + · · · + an z n are univalent in |z | <
1 4
(n = 2, 3, . . .)
.
Lemma 1.3 ([3]). If f (z ) ∈ S0 , then f (z ) is univalent in |z | <
√ 2 − 1.
2. Main results γn
Theorem 2.1. Let n ∈ N = {1, 2, . . .}, k > 1, γ > 0, k+2 1 < β < 1 and 0 < α ≤ (k−1)β . If q (be a nonvanishing function in zq′ (z ) the (U − {0})) belongs to An and the function h(z ) = (1 − γ )( q(zz ) )α + γ q(z ) ( q(zz ) )α satisfies h(U ) ⊂ Dk (β), then
Re
α
z
> β.
q(z )
Proof. Suppose p(z ) = ( q(zz ) )α . It is easy to check that
(1 − γ )
z q(z )
α
+γ
zq′ (z ) q(z )
z q(z )
α
= p(z ) −
γ ′ zp (z ). α
(2.1)
We define the function ω(z ) by p(z ) =
1 + (1 − 2β)ω(z ) 1 − ω(z )
.
(2.2)
Then ω(z ) is analytic in U and ω(0) = 0. It follows from (2.2) that p(z ) −
1 + (1 − 2β)ω(z ) γ 2(1 − β)z ω′ (z ) γ ′ zp (z ) = − . α 1 − ω(z ) α (1 − ω(z ))2
Suppose that there exists a point z0 ∈ U such that max |ω(z )| = |ω(z0 )| = 1.
|z |≤|z0 |
Then, using Lemma 1.1, we have z0 ω′ (z0 ) = mω(z0 )
(m is real and m ≥ n),
and ω(z0 ) = e where 0 ≤ x ≤ 2π . Thus from (2.3) we obtain that ix
p(z0 ) −
γ (1 − β) sin x γ m(1 − β) z0 p′ (z0 ) = β + i + , α 1 − cos x α 1 − cos x
so
2 2 γ γ p(z0 ) − z0 p′ (z0 ) − kβ = p(z0 ) − z0 p′ (z0 ) − β − (k − 1)β α α 2 γ m(1 − β) (1 − β)2 sin2 x = − (k − 1)β + α 1 − cos x (1 − cos x)2 2 γ = Re p(z0 ) − z0 p′ (z0 ) − β α
(2.3)
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509
1507
γ 1)β α (1
(k − 1)2 β 2 (1 − cos x) + (1 − β)2 (1 + cos x) − 2(k − − β)m + 1 − cos x 2 γ ≤ Re p(z0 ) − z0 p′ (z0 ) − β α (k − 1)2 β 2 (1 − cos x) + (1 − β)2 (1 + cos x) − 2(k − 1)β γα (1 − β)n + . 1 − cos x γ
If we let f (t ) = t [(1 − β)2 − (k − 1)2 β 2 ] + [(k − 1)2 β 2 + (1 − β)2 − 2(k − 1)β α n], where t = cos x, then under the hypothesis of the theorem we obtain
γ n ≤ 0. α γ γ Hence we have |p(z0 ) − α z0 p′ (z0 ) − kβ| ≤ |Re(p(z0 ) − α z0 p′ (z0 )) − β| and this implies that h(z0 ) ̸∈ Dk (β) which contradicts z α the hypothesis of the theorem so Re( q(z ) ) > β (z ∈ U ) and the proof is complete.
f (t ) ≤ 2(k − 1)β (k − 1)β −
By putting γ = α = 1 and q(z ) =
f (z ) f ′ (z )
in the Theorem 2.1 we have.
Corollary 2.1. Let f be univalent and belongs to An , h(z ) =
zf ′ (z )
f (z )
zf ′ (z ) f (z )
−
zf ′′ (z )
f ′ (z )
2 3
< β < 1 and the function
,
satisfies the condition h(U ) ⊆ D2 (β), zf ′ (z )
then Re f (z ) > β . Also by choosing γ = α = 1 and q(z ) =
z 2 f ′ (z ) f (z )
in the Theorem 2.1 we obtain.
Corollary 2.2. Let f be univalent and belongs to An , h(z ) =
f (z )
2+
zf ′ (z )
zf (z ) ′′
f ′ (z )
2 3
< β < 1 and the function
− 1,
satisfies the condition h(U ) ⊆ D2 (β), f (z )
then Re zf ′ (z ) > β . γn
Corollary 2.3. Let n ∈ N = {1, 2, . . .}, k > 1, γ > 0, k+2 1 < β < 1 and 0 < α ≤ (k−1)β . If q (be a nonvanishing function in the (U − {0})) belongs to An and
Re (1 − γ )
z q(z )
α
+γ
zq′ (z ) q(z )
z q(z )
α
<
(k + 1)β 2
then
Re
z
α
q(z )
> β.
Proof. It is easy to see that the region Γ = {z ∈ C; Rez < result.
(k+1)β 2
} is contained in Dk (β) and so by Theorem 2.1 we get our γn
Corollary 2.4. Let n ∈ N = {1, 2, . . .}, k > 1, γ > 0, k+2 1 < β < 1 and 0 < α ≤ (k−1)β . If q (be a nonvanishing function on (U − {0})) belongs to An and
α α z zq′ (z ) z k−1 arg (1 − γ ) +γ > arctan k + 1 q(z ) q(z ) q(z ) then
Re
z q(z )
α
> β.
1508
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509 1+(1−(k+1)β) 1 −z
It is known that the function g (z ) =
α
α
(k+1)β 2
for (z ∈ U ). Set p(z ) = ( q(zz ) )α and
γ ′ 1 + (1 − (k + 1)β) zp (z ) = . α 1−z ∞ n Solving this differential equation we find that p(z ) = 1 − α(2 − (k + 1)β) n=1 γ nz−α . Now using Corollary 2.3 we obtain (1 − γ )
z
q(z )
+γ
zq′ (z )
satisfies the condition Reg (z ) <
q(z )
z
q(z )
= p(z ) −
the following example. γ
Example 2.1. If k > 1, γ > 0, k+2 1 < β < 1 and 0 < α < min{ (k−1)β , γ } then we have
Re 1 − α(2 − (k + 1)β)
∞
zn
n =1
γn − α
> β.
Theorem 2.2. Let n ∈ N = {1, 2, . . .}, γ > 0, β < 1 and 0 < α ≤ γ n. If f ∈ An and h(z ) = (1 − γ )( satisfies the condition
f (z ) α z
) + γ zff ((zz)) ( f (zz ) )α ′
h(U ) ⊆ Ω (β), then
Re
f (z )
α
z
> β.
Proof. Defining the function p(z ) by p(z ) =
f (z )
α
z
,
we have
(1 − γ )
f (z )
α
z
+γ
zf ′ (z ) f (z )
f (z ) z
α
= p(z ) +
γ ′ zp (z ). α
(2.4)
If we define the function ω(z ) by p(z ) =
1 + (1 − 2β)ω(z ) 1 − ω(z )
,
(2.5)
then ω(z ) is analytic in U and ω(0) = 0. Assuming that there exists a point z0 ∈ U such that max |ω(z )| = |ω(z0 )| = 1.
|z |≤|z0 |
Lemma 1.1, implies that p(z0 ) +
γ (1 − β) sin θ γ m(1 − β) z0 p′ (z0 ) = β + i − , α 1 − cos θ α 1 − cos θ
where ω(z0 ) = eiθ . So
2 γ 2(1 − β)2 m2 γ 2 (1 − β)2 (1 − β)2 γ m + 2 − 2 p(z0 ) + z0 p′ (z0 ) − β + (1 − β) = α (1 − cos θ ) α (1 − cos θ )2 α(1 − cos θ ) γ 2 2 1 − α m (1 − β)2 γ ′ = Re p(z0 ) + z0 p (z0 ) − β + α 1 − cos θ γ 2 2 1 − n (1 − β)2 γ α ≤ Re p(z0 ) + z0 p′ (z0 ) − β + α 1 − cos θ 2 γ ′ ≤ Re p(z0 ) + z0 p (z0 ) − β . α Thus h(z0 ) ̸∈ Ω (β) which contradicts the hypothesis of the Theorem 2.2. Hence, we conclude that |ω(z )| < 1 for all z ∈ U and the proof is complete.
By putting γ = α = 1 and n = 1 on the Theorem 2.2 we obtain the result in [4] (see Theorem 1).
R. Aghalary, Z.-G. Wang / Applied Mathematics Letters 25 (2012) 1505–1509
Corollary 2.5. Let n ∈ N = {1, 2, . . .} and γ ≥ zf (z ) ′
f (z )
zf (z ) ′
1+γ −
zf (z ) ′′
+
f (z )
1 . n
1509
If f ∈ An , satisfies the condition
∈ Ω (β) z ∈ U ,
f ′ (z )
zf ′ (z )
then Re f (z ) > β . For β < 1, γ > 0 and α = γ the function h(z ) = z
1 + (1 − 2β)z 1−z
satisfies
(1 − γ ) with
α1
h(z )
α
z
zh′ (z )
+γ
h(z )
h(z )
α
∈ Ω (β) z ∈ U ,
z
α α h(z ) zh′ (z ) h(z ) ∂ (1 − γ ) +γ = Ω (β), z h(z ) z h(eiθ )
h(z )
and Re( z )α > β (z ∈ U ), (Re( eiθ )α = β for eiβ ̸= 1). Therefore in the case α = γ > 0 the last β in Theorem 2.2 cannot be replaced by a smaller number. Corollary 2.6. Let n ∈ N = {1, 2, . . .}, γ ≥ 0 and 0 < α ≤ γ n. If f ∈ An , satisfies the condition
(1 − γ ) then Re(
f (z )
α
z
+γ
zf ′ (z )
f (z )
f (z )
α
z
∈Ω
1
2
z ∈ U,
f (z ) α z
) > 12 .
It is know [5] that if Re(
f (z ) z
)>
1 2
for |z | < 1, then f (z ) is starlike for |z | < 2
Corollary 2.7. Let n ∈ N = {1, 2, . . .}, γ ≥
(1 − γ )
f (z )
z
+ γ f ′ (z ) ∈ Ω
then f is starlike for |z | <
1 . n
1
−1 2
. So from Corollary 2.6 we obtain.
If f ∈ An , and z ∈ U,
2
√ 2 . 2
Also by using Lemma 1.3 we have. Corollary 2.8. Let n ∈ N = {1, 2, . . .}, γ ≥
(1 − γ )
f (z )
z
1 . n
If f ∈ An , and
+ γ f ′ (z ) ∈ Ω (0) z ∈ U ,
then f is univalent for |z | <
√ 2 − 1.
Finally, using Lemma 1.2 and Theorem 2.2 we obtain. Corollary 2.9. Let n ∈ N = {1, 2, . . .}, γ ≥
(1 − γ )
f (z )
z
1 . n
If f (z ) = z + an+1 z n+1 + an+2 z n+2 + · · · + an+m z n+m + · · · ∈ An , and
+ γ f ′ (z ) ∈ Ω (0) z ∈ U ,
then the partial sums Sm (z ) = z + an+1 z n+1 + an+2 z n+2 + · · · + an+m z n+m are univalent for |z | <
1 . 4
References [1] [2] [3] [4] [5]
I.S. Jack, Functions starlike and convex of order α , J. Lond. Math. Soc. 3 (1971) 469–474. f (z ) K. Yamagushi, On functions satisfying Re[ z ] > 0, Proc. Amer. Math. Soc. 17 (1966) 588–591. K. Noshiro, On the univalency of certain analytic functions, J. Fac. Sci. Hokkaido. Imp. Univ. 2 (1934) 89–101. J. Sokół, On functions with derivative satisfying a geometric condition, Appl. Math. Comput. 204 (2008) 116–119. T.H. Macgregor, The radius of convexity for starlike functions of order 12 , Proc. Amer. Math. Soc. 14 (1963) 71–76.