ON q-HYPERGEOMETRIC FUNCTIONS K. A. Challab1 , M. Darus1,∗ and F. Ghanim2 1
School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia Bangi 43600 Selangor D. Ehsan Malaysia e-mail: khalid
[email protected] e-mail:
[email protected] 2
Department of Mathematics College of Sciences University of Sharjah Sharjah, United Arab Emirates e-mail:
[email protected] Abstract. In this article, we studied some results on meromorphic functions defined by qhypergeometric functions. In addition, certain sufficient conditions for these meromorphic functions to satisfy a subordination property are also pointed out. In fact, these results extend known results of starlikeness, convexity, and close to convexity. 2010 Mathematics Subject Classification: 32Wxx, 30-XX. Keywords and phrases: analytic functions, meromorphic functions, starlike functions, convex functions, Hadamard product, subordination property; q-hypergeometric function. ∗ Corresponding Author
1. Introduction In the present paper, we initiate the study of functions which are meromorphic in the punctured disk U∗ = {z : 0 < |z| < 1} with a Laurent expansion about the origin; see [9]. Let A be the class of analytic functions ∪ h(z) with h(0) = 1, which are convex and univalent in the open unit disk U = U∗ {0} and for which (z ∈ U∗ ).
ℜ{h(z)} > 0,
(1.1)
For functions f and g analytic in U, we say that f is subordinate to g and write f ≺ g in U or f (z) ≺ g(z), (z ∈ U)
(1.2)
if there exists an analytic function ω(z) in U such that |ω(z)| ≤ |z|,
f (z) = g(ω(z))
(z ∈ U).
(1.3)
Futhermore, if the function g is univalent in U, then f (z) ≺ g(z) ⇔ f (0) = g(0) 1
(1.4)
f (U) ⊆ g(U),
(z ∈ U).
2. Perliminaries Let Σ denote the class of meromorphic function f (z) normalized by 1 ∑ + dn z n , z n=1 ∞
f (z) =
(2.1)
which are analytic in the punctured unit disk U∗ . For 0 ≤ β, we denote by S ∗ (β) and K(β) the subclasses of Σ consisting of all meromorphic functions which are, respectively, starlike of order β and convex of order β in U∗ . For functions fj (z) (j=1, 2) defined by 1 ∑ + dn,j z n , z n=1 ∞
fj (z) =
(2.2)
we denote the Hadamard product (or convolution) of f1 (z) and f2 (z) by 1 ∑ (f1 ∗ f2 ) = + dn,1 dn,2 z n . z n=1 ∞
(2.3)
Cho et al. [4] and Ghanim and Darus [7] studied the following function: )µ ∞ ( λ 1 ∑ z n (λ > 0, µ ≥ 0). qλ,µ (z) = + z n=1 n + 1 + λ
(2.4)
Corresponding to the function qλ,µ (z) and using the Hadamard product for f (z) ∈ Σ, Ghanim and Darus [8] defined a linear operator L(λ, µ) on Σ by 1 ∑ Lλ,µ f (z) = (f (z) ∗ qλ,µ (z)) = + z n=1 ∞
(
λ n+1+λ
)µ |dn |z n .
(2.5)
As for the result of this paper on applications involving generalized hypergeometric functions, we need to utilize the well- known q-hypergeometric function. For complex parameters a1 , ..., al and b1 , ..., bm (bj ̸= 0, −1, ...; j = 1, 2, ..., m) the qhypergeometric function l Ψm (z) is defined by ∞ ∑
n (a1 , q)n ...(al , q)n × [(−1)n q ( 2 ) ]1+m−l z n , (2.6) (q, q)n (b1 , q)n ...(bm , q)n n=0 (n ) ∪ with 2 = n(n − 1)/2, where q ̸= 0 when l > m + 1 (l, m ∈ N0 = N {0}; z ∈ U).
l Ψm (a1 , ..., al ; b1 , ..., bm ; q, z)
:=
The q-shifted factorial is defined for a, q ∈ C as a product of n factors by { (a : q)n =
( ) (1 − a) (1 − aq) ... 1 − aq n−1
(n ∈ N)
1
(n = 0)
2
(2.7)
and in terms of basic analogue of the gamma function (q a ; q)n =
Γq (a + n)(1 − q)n , Γq (a)
n > 0.
(2.8)
It is of interest to note that limq→−1 ((q a ; q)n /(1 − q)n ) = (a)n = a(a + 1)...(a + n − 1) is the familier Pochhammer symbol and ∞ ∑ (a1 )n ...(al )n z n . l Ψm (a1 , ..., al ; b1 , ..., bm ; z) := (b1 )n ...(bm )n n! n=0
(2.9)
Now for z ∈ U, 0 < |q| < 1, and l = m + 1, the basic hypergeometric function defined in (2.9) takes the form l Ψm (a1 , ..., al ; b1 , ..., bm ; q, z) :=
∞ ∑ n=0
(a1 , q)n ...(al , q)n zn, (q, q)n (b1 , q)n ...(bm , q)n
(2.10)
which converges absolutely in the open unit disk U. Corresponding to the function l Ψm (a1 , ..., al ; b1 , ..., bm ; q, z) recently for meromorphic functions f ∈ Σ consisting functions of the form (2.1), Aldweby and Darus [1] introduce q-analogue of Liu-Srivastava operator as below: l Υm (a1 , ..., al ; b1 , ..., bm ; q, z)
=
where
∏s
∗ f (z)
1 l Ψm (a1 , ..., al ; b1 , ..., bm ; q, z) ∗ f (z) z
∏l ∞ (ai , q)n+1 1 ∑ i=1∏ = + dn z n z n=1 (q, q)n+1 m (b , q) i n+1 i=1
k=1 (ak , q)n+1
(z ∈ U∗ ),
(2.11)
= (a1 , q)n+1 ...(as , q)n+1 , and
l Υm (a1 , ..., al ; b1 , ..., bm ; q, z)
=
1 l Ψm (a1 , ..., al ; b1 , ..., bm ; q, z) z
∏l ∞ (ai , q)n+1 1 ∑ i=1∏ zn. = + m z n=1 (q, q)n+1 i=1 (bi , q)n+1 Corresponding to the functions l Υm (a1 , ..., al ; b1 , ..., bm ; q, z) and qλ,µ (z) given in (2.4) and using the Hadamard product for f (z) ∈ Σ, we will define a new linear operator as Lλµ (a1 , a2 , ...al ; b1 , b2 , ...bm ; q) on Σ by Lλµ (a1 , a2 , ...al ; b1 , b2 , ...bm ; q)f (z) = (f (z) ∗l Υm (a1 , ..., al ; b1 , ..., bm ; q, z) ∗ qλ,µ (z)) ∏l ( )µ ∞ (ai , q)n+1 λ 1 ∑ i=1∏ = + |dn |z n . m z n=1 (q, q)n+1 i=1 (bi , q)n+1 n + 1 + λ (2.12) and for convenience, we shall henceforth denote
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Lλµ (a1 , a2 , ...al ; b1 , b2 , ...bm ; q)f (z) = Lλµ (al , bm , q)f (z).
(2.13)
Remark 2.1. (i) For µ = 0, ai = q ai , bj = qjb , ai > 0, bj > 0, (i = 1, ..., l, j = 1, ...m, l l = m + 1), q → 1, the operator Lλµ (al , bm , q)f (z) = Hm [al ]f (z) which was investigated by Liu and srivastava [12]. (ii) For µ = 0, l = 2, m = 1, a2 = q, q → 1, the operator Lλµ (al , q, bm , q)f (z) = L(al , bm )f (z) was introduced and studied by Liu and srivastava [11]. (iii) For µ = 0, l = 1, m = 0, ai = γ + 1, q → 1, the operator Lλµ (γ + 1, bm , q)f (z) = 1 γ Dγ f (z) = z(1−z) is the differential operator which was γ+1 ∗ f (z) (γ > −1), where D introduced by Ganigi and Uralegadi [5] and then it was generalized by Yang [14]. For a function f ∈ Lλµ (al , bm , q)f (z), we define I 0 (Lλµ (al , bm , q)f (z)) = Lλµ (al , bm , q)f (z), and for k = 1, 2, 3, ..., 2 I k (Lλµ (al , bm , q)f (z)) = z(I k−1 Lλµ (al , bm , q)f (z))′ + , z I k (Lλµ (al , bm , q)f (z)) ∏l ( )µ ∞ (ai , q)n+1 λ 1 ∑ k i=1∏ = + n |dn |z n . z n=1 (q, q)n+1 m (b , q) n + 1 + λ i n+1 i=1
(2.14)
We note that I k in (2.14) was studied by Ghanim and Darus [6], Challab and Darus [3] and Challab and Darus [2]. Also, it follows from (2.12) that z(Lλµ (al , bm , q)f (z))′ = nLλµ (al , bm , q)f (z) −
n+1 , z
(2.15)
also, from (2.15), we get n+1 . (2.16) z In this present paper, we obtain certain sufficient conditions for a function f ∈ Σ to satisfy either of the following subordinations: z(I k Lλµ (al , bm , q)f (z))′ = nI k Lλµ (al , bm , q)f (z) −
I k Lλµ (al + 1, bm , q)f (z) γ(1 − z) ≺ , k λ I Lµ (al , bm , q)f (z) γ−z I k Lλµ (al , bm , q)f (z) 1 + Az ≺ , z 1−z 4
I k Lλµ (al , bm , q)f (z) γ(1 − z) ≺ . z γ−z To prove our main results, we need the following: Lemma 2.1. (cf. Miller and Mocanu [13], Theorem 3.4h, p. 132]). Let q(z) be univalent in the unit disk U and let ϑ and φ be analytic in the a domain q(U) ⊂ D, with φ(ω) ̸= 0 when q(U) ∈ ω. Set Q(z) := zq ′ (z)φ(q(z)), h(z) := ϑ(q(z)) + Q(z). Suppose that (1) Q(z) starlike univalent in U, and ( ′ is ) zh (z) (2)ℜ Q(z) > 0 for z ∈ U. If p(z) is analytic in U with p(0) = q(0), p(U) ⊂ D and ϑ(p(z)) + zp′ (z)φ(p(z)) ≺ ϑ(q(z)) + zq ′ (z)φ(q(z)),
(2.17)
then p(z) ≺ q(z) and q(z) is the best dominant. 3. MAIN RESULTS Theorem 3.1. Let a ∈ R satisfy −1 ≤ a ≤ 1 z(I k Lλµ (al , bm , q)f (z)) ̸= 0 in U∗ and (
I k Lλµ (al + 1, bm , q)f (z) I k Lλµ (al , bm , q)f (z)
)a (
and
γ > 1. If
n+1 z(I k Lλµ (al + 1, bm , q)f (z))
f ∈ Σ
satisfies
) ≺ h(z),
(3.1)
where ( h(z) =
γ(1 − z) γ−z
)a+1 (
n+1 (γ − 1)z − k λ z(I Lµ (al + 1, bm , q)f (z)) γ(1 − z)2
) ,
then I k Lλµ (al + 1, bm , q)f (z) γ(1 − z) ≺ . k λ I Lµ (al , bm , q)f (z) γ−z Proof. . The condition (3.1) and z(I k Lλµ (al , bm , q)f (z)) ̸= 0 in U∗ implies that z(I k Lλµ (al + 1, bm , q)f (z)) ̸= 0 in U∗ . Define the function p(z) by p(z) :=
I k Lλµ (al + 1, bm , q)f (z) . I k Lλµ (al , bm , q)f (z)
Clearly, p(z) is analytic in U∗ . A computation shows that z[I k Lλµ (al + 1, bm , q)f (z)]′ z[I k Lλµ (al , bm , q)f (z)]′ zp′ (z) − k λ . = p(z) I k Lλµ (al + 1, bm , q)f (z) I Lµ (al , bm , q)f (z) By using the identity (2.16) and (3.2), we get
5
(3.2)
n+1 (n + 1)p(z) zp′ (z) = − . z(I k Lλµ (al + 1, bm , q)f (z)) z(I k Lλµ (al + 1, bm , q)f (z)) p(z)
(3.3)
Using (3.3) in (3.1), we get (n + 1)(p(z))a+1 − zp′ (z)(p(z))a−1 ≺ h(z). z(I k Lλµ (al + 1, bm , q)f (z))
(3.4)
Let q(z) be the function defined by γ(1 − z) . γ−z It is clear that q is convex univalent in U∗ . Since q(z) :=
(n + 1)(q(z))a+1 h(z) = − zq ′ (z)(q(z))a−1 , k λ z(I Lµ (al + 1, bm , q)f (z)) we see that (3.4) can be written as (2.17) when ϑ and φ are given by ϑ(ω) =
(n + 1) ω a+1 , + 1, bm , q)f (z))
and,
z(I k Lλµ (al
φ(ω) = ω a−1 .
Clearly, φ and ϑ are analytic in C \ 0. Now Q(z) := zq ′ (z)φ(q(z)) = zq ′ (z)(q(z))a−1 =
(1 − γ)zγ a (1 − z)a−1 , (γ − z)1+a
h(z) := ϑ(q(z)) + Q(z) = (
γ(1 − z) γ−z
)1+a (
(n + 1) (γ − 1)z − k λ z(I Lµ (al + 1, bm , q)f (z)) γ(1 − z)2
By our assumptions on the parameters a and γ, we see that ( ) zQ′ (z) z(1 − a) z ℜ =ℜ 1+ + (1 + a) Q(z) 1−z γ−z 1 (1 + a)γ > 1 − (1 − a) − 2 1+γ =
(1 + a)(γ − 1) > 0, 2(1 + γ)
and therefore Q(z) is starlike. Also, we have ℜ
zh′ (z) Q(z)
6
) .
= (n + 1)ℜ
γ(z − 1) [z(γ − z)(1 + a) − (1 − z)((γ − z)(n + 1) − z(1 + a))] z 2 (γ − z)(1 − γ)(I k Lλµ (al + 1, bm , q)f (z)) +ℜ
zQ′ (z) ≥ 0. Q(z)
By an application of Lemma 2.1, we have p(z) ≺ q(z) or I k Lλµ (al + 1, bm , q)f (z) γ(1 − z) ≺ . k λ I Lµ (al , bm , q)f (z) γ−z
This completes the proof of Theorem 3.1.
Theorem 3.2. Let −1 < a < 0 and −1 < A < 1. If f ∈ Σ satisfy the condition I k Lλ µ (al ,bm ,q)f (z) ̸= 0 in U∗ and z (
I k Lλµ (al , bm , q)f (z) z
)a (
n+1 z2
) ≺ h(z),
(3.5)
where ( h(z) =
1 + Az 1−z
)a+1 ( (n − 1) +
(1 + A)z (1 − z)(1 + Az)
) ,
then I k Lλµ (al , bm , q)f (z) 1 + Az ≺ . z 1−z Proof. Define the function p(z) by I k Lλµ (al , bm , q)f (z) . (3.6) z It is clear that p is analytic in U∗ . By using the identity (2.16), we get from (3.6), p(z) :=
n+1 = (n − 1)p(z) − zp′ (z). 2 z Using (3.7) in (3.5), we see that the subordination becomes
(3.7)
(n − 1)p(z)1+a − z(p(z))a p′ (z) ≺ h(z). Define the function q(z) by 1 + Az . 1−z It is clear that q(z) is univalent in U and q(U) is the region ℜq(z) > (1 − A)/2. Define the functions ϑ and φ by q(z) :=
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ϑ(ω) = (n − 1)ω a+1
φ(ω) = ω a .
and
We observe that (3.5) can be written as (2.17). Note that φ and ϑ are analytic in C\0. Also, we see that Q(z) := zq ′ (z)φ(q(z)) =
z(1 + A)(1 + Az)a (1 − z)2+a
and ( h(z) := ϑ(q(z)) + Q(z) =
1 + Az 1−z
)a+1 ( (n − 1) +
(1 + A)z (1 + Az)(1 − z)
) .
By our assumptions, we have [ ] zQ′ (z) Az z ℜ =ℜ 1+a + (2 + a) Q(z) 1 + Az 1−z >1+
a|A| 2+a −a(1 − |A|) − = > 0, 1 + |A| 2 2(1 + |A)
and [ ′ ] ϑ (q(z)) zQ′ (z) zh′ (z) zQ′ (z) ℜ =ℜ + = (n − 1)(1 + a) + ℜ ≥ 0. Q(z) φ(q(z)) Q(z) Q(z)
The results now follows by an application of Lemma 2.1.
Theorem 3.3. Let a ≥ −1, γ > 1, f ∈ Σ and I k Lλµ (al , bm , q)f (z)/z ̸= 0 in U∗ . If f satisfies (
I k Lλµ (al , bm , q)f (z) z
)a (
n+1 z2
)
( ≺
γ(1 − z) (γ − z)
)1+a ( (n − 1) −
z(1 − γ) (1 − z)(γ − z)
) , (3.8)
then I k Lλµ (al , bm , q)f (z) γ(1 − z) ≺ , z γ−z
(z ∈ U ∗ ).
Proof. Define the function p(z) by I k Lλµ (al , bm , q)f (z) . z clearly, p(z) is analytic in U∗ , we can compute to show p(z) =
z(I k Lλµ (al , bm , q)f (z))′ − I k Lλµ (al , bm , q)f (z) . z by using the identity (2.16), we can get from (3.10), zp′ (z) =
8
(3.9)
(3.10)
n+1 = (n − 1)p(z) − zp′ (z). 2 z Using (3.11) in (3.8), we obtain
(3.11)
(n − 1)(p(z))a+1 − zp′ (z)(p(z))a ( ≺
γ(1 − z) (γ − z)
)1+a ( (n − 1) −
z(1 − γ) (1 − z)(γ − z)
) .
(3.12)
Define the function q(z) by q(z) =
γ(1 − z) γ−z
which is univalent in U∗ . We see that (3.12) can be written as (2.17) when ϑ and φ are given by ϑ(ω) = (n − 1)ω a+1 , φ(ω) = −ω a such that ϑ and φ are analytic in C \ 0. Now Q(z) = zq ′ (z)φ(q(z)) = ( h(z) = ϑ(q(z)) + Q(z) =
γ(1 − z) (γ − z)
−zγ a+1 (1 − γ)(1 − z)a , (γ − z)a+z )1+a ( (n − 1) −
z(1 − γ) (1 − z)(γ − z)
) .
By our assumptions, we have ( ) zQ′ (z) z(1 − a) z ℜ =ℜ 1+ + (1 + a) Q(z) 1−z γ−z 1 (1 + a)γ > 1 − (1 − a) − 2 1+γ =
(1 + a)(γ − 1) > 0, 2(1 + γ)
hence Q(z) is starlike. Now [ ′ ] ϑ (q(z)) zQ′ (z) zh′ (z) zQ′ (z) =ℜ + ≥ 0. ℜ = (1 − n)(1 + a) + ℜ Q(z) φ(q(z)) Q(z) Q(z) Now we can apply lemma 2.1 to get p(z) ≺ q(z). We have I k Lλµ (al + 1, bm , q)f (z) γ(1 − z) ≺ . z γ−z This completes the proof of Theorem 3.3.
4. Acknowledgement The authors thank the anonymous referees for their valuable suggestions and constructive criticism which improved the presentation of the paper.
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