International Journal of Analysis and Applications Volume 18, Number 6 (2020), 1037-1047 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-1037 MEROMORPHIC STARLIKE FUNCTIONS WITH RESPECT TO SYMMETRIC POINTS MUHAMMAD GHAFFAR KHAN1,∗, MASLINA DARUS2, BAKHTIAR AHMAD3, GANGADHARAN MURUGUSUNDARAMOORTHY4, RAEES KHAN5, NASIR KHAN5 1Department of Mathematics, Abdul Wali Khan University Mardan, Pakistan 2Department of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600, Bangi, Selangor, Malaysia 3Govt. Degree College Mardan, 23200 Mardan, Pakistan 4Department of Mathematics, SAS, Vellore Institute of Technology, Deemed to be University, Vellore 632014, India 5Department of Mathematics, Fata university, Dara Adam Khel, N.M.D. Kohat, KP, Pakistan ∗Corresponding author: ghaffarkhan020@gmail.com Abstract. The main purpose of this article is to introduce a class of meromorphic functions associated with the symmetric points in circular domain. We investigate the necessary and sufficient conditions, distortions theorem for this class. Furthermore, we obtain closure and convolutions properties, radii of starlikeness and partial sum results for these functions. 1. Introduction Denoted by M, the class of functions f which are analytic in the D∗ = D�{0} , where D = {z ∈ C : |z| < 1} and having the following series expansion form (1.1) f(z) = 1 z + ∞∑ n=1 an z n, z ∈ D∗. Received August 5th, 2020; accepted September 4th, 2020; published October 14th, 2020. 2010 Mathematics Subject Classification. 30C45, 30C50. Key words and phrases. meromorphic functions; subordinations; convolutions; Janowski functions. ©2020 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 1037 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-1037 Int. J. Anal. Appl. 18 (6) (2020) 1038 We say that an analytic function f1 (z) is subordinate to f2 (z) in D, symbolically represented as f1 (z) ≺ f2 (z), if there exists an analytic function w(z) with conditions |w(z)| < 1 and w(0) = 1 such that f1 (z) = f2 (w(z)). Moreover, if f2 (z) is univalent, then we have the following equivalency from [1] and [2], f1 (0) = f2 (0) and f1 (D) ⊆ f2 (D) . For two functions f1 (z) = 1 z + ∑∞ n=1 an,1 z n and f2 (z) = 1 z + ∑∞ n=1 an,2 z n in D∗ the convolution or Hadamard product is defined by (f1 ∗f2) (z) = 1 z + ∞∑ n=1 an,1an,2z n. A function f ∈ M is said to be in the class MS∗ (α) of meromorphic starlike functions of order α if it satisfies the inequality (1.2) < ( zf′(z) f(z) ) < −α, z ∈ D∗ 0 ≤ α < 1. For some recent investigation of meromorphic functions see [3–13]. Motivated by aforementioned and recent work of [14], we define the functions’ class as below: Let −1 ≤ B < A ≤ 1. Then the function f is in the class MS∗∗ [A,B] if it satisfies (1.3) − 2zf′(z) f(z) −f(z) ≺ 1 + Az 1 + Bz , (z ∈ D∗), or equivalently (1.4) ∣∣∣∣∣∣ 2zf′(z) f(z)−f(−z) + 1 B 2zf′(z) f(z)−f(−z) + A ∣∣∣∣∣∣ < 1 (z ∈ D∗) . 2. Coefficient Inequalities Theorem 2.1. Let f ∈M and assumed as in (1.1) then f ∈MS∗∗ [A,B], if and only if (2.1) ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an| ≤ A−B. This inequality is sharp. Proof. Let us assume that condition (2.1) holds. To show f ∈ MS∗∗ [A,B] , we only need to show the inequality (1.4) holds. For this consider∣∣∣∣∣∣ 2zf′(z) f(z)−f(−z) + 1 B 2zf′(z) f(z)−f(−z) + A ∣∣∣∣∣∣ = ∣∣∣∣∣ zf ′(z) + f(z)−f(−z) 2 Bzf′(z) + A f(z)−f(−z) 2 ∣∣∣∣∣ = ∣∣∣∣ ∑∞n=1 ( n+ 1−(−1)n 2 ) an (B−A)+ ∑∞ n=1(Bn+A 1−(−1)n 2 )an ∣∣∣∣ ≤ ∑∞ n=1 ( n+ 1−(−1)n 2 ) an (B−A)− ∑∞ n=1(Bn+A 1−(−1)n 2 )an < 1. Int. J. Anal. Appl. 18 (6) (2020) 1039 Now for other part let suppose f ∈ MS∗∗ [A,B] . We are to show that the inequality (2.1) , holds true. Consider ∣∣∣∣∣∣ 2zf′(z) f(z)−f(−z) + 1 B 2zf′(z) f(z)−f(−z) + A ∣∣∣∣∣∣ = ∣∣∣∣∣ zf ′ (z) + f(z)−f(−z) 2 Bzf′ (z) + A f(z)−f(−z) 2 ∣∣∣∣∣ = ∣∣∣∣ ∑∞n=1 ( n+ 1−(−1)n 2 ) an (A−B)+ ∑∞ n=1(Bn+A 1−(−1)n 2 )an ∣∣∣∣ , since the <(z) ≤ |z| , we have (2.2) < { ∑∞ n=1 ( n+ 1−(−1)n 2 ) anz n−1 (A−B)+ ∑∞ n=1(Bn+A 1−(−1)n 2 )anzn−1 } < 1 , now if we choose the value of z on real axis then 2zf ′ (z) f(z)−f(−z) is real. Letting z → 1 − on real axis and some simple calculation in (2.2) , lead us to (2.1) . � Theorem 2.2. Let f ∈M and assumed as in (1.1) then f ∈MS∗∗ [A,B], if and only if (2.3) z [ f(z) ∗ ( (1 − 2z) ( 1 + Beiθ ) z (1 −z)2 − z ( 1 + Aeiθ ) 1 −z2 )] 6= 0, (z ∈ D) . Proof. It is easy to verify the relations (2.4) f(z) ∗ z 1 −z2 = f(z) −f (−z) 2 and f(z) ∗ [ 1 z (1 −z)2 − 2 (1 −z)2 ] = −zf ′ (z). To prove (2.3) , if f ∈MS∗∗ [A,B] then we write (1.3), by using definition of subordination as (2.5) − 2zf ′ (z) f (z) −f (−z) = 1 + Aw (z) 1 + Bw (z) , which is equivalent to − 2zf ′ (z) f(z) −f (−z) 6= 1 + Aeiθ 1 + Beiθ , z ∈ D, θ ∈ [0, 2π] , which implies that (2.6) −zf ′ (z) ( 1 + Beiθ ) − f(z) −f (−z) 2 ( 1 + Aeiθ ) 6= 0. Using the relation (2.4), (2.6) become z [ f(z) ∗ ( (1 − 2z) ( 1 + Beiθ ) z (1 −z)2 − z ( 1 + Aeiθ ) 1 −z2 )] 6= 0, for z ∈ D. Conversly, suppose that the condition (2.3) hold, it follows that zf (z) 6= 0 for all z ∈ D. Hence Φ (z) = − 2zf ′ (z) f(z)−f(−z) is analytic in D with Φ (0) = 1. Since (2.7) − 2zf ′ (z) f(z) −f (−z) 6= 1 + Aeiθ 1 + Beiθ . If we denote Ψ (z) = 1 + Az 1 + Bz , Int. J. Anal. Appl. 18 (6) (2020) 1040 the relation (2.7) , show that Φ (D)∩Ψ (D) = ∅. Therefore the simply connected domain Φ (D) is contained in connected component of C�Ψ (∂D) . The univalence of ”Φ” togather with the fact Ψ (0) = Φ (0) = 1, this show that Φ ≺ Ψ which shows that f ∈MS∗∗ [A,B] . � Theorem 2.3. The class MS∗∗ [A,B] is closed under convex combination. Proof. Let fi (z) ∈MS∗∗ [A,B] , such that fi (z) = 1 z + ∞∑ n=1 an,iz n, i ∈ N. Then by equation (2.1) , we have ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an,i| ≤ A−B. For ∑∞ i=1 δi = 1, 0 ≤ δ ≤ 1, we have ∞∑ i=1 δifi (z) = 1 z + ∞∑ n=1 ( ∞∑ i=1 δian,i ) zn. Using 2.1, we have ∞∑ n=1 ( ∞∑ i=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) δi |an,i| ) ≤ ∞∑ i=1 δi { ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an,i| } ≤ (A−B) ∞∑ i=1 δi = A−B. Hence MS∗∗ [A,B] is convex. � Theorem 2.4. Let f ∈MS∗∗ [A,B] , |z| = r. Then (2.8) 1 r − A−B A + B + 2 r ≤ |f (z)| ≤ 1 r + A−B A + B + 2 r. Proof. As |f (z)| = ∣∣∣∣∣1z + ∞∑ n=1 anz n ∣∣∣∣∣ ≤ 1 r + ∞∑ n=1 |an| |r| n ≤ 1 r + A−B A + B + 2 r. Where we have used Theorem 2.1, on similar argument we have Int. J. Anal. Appl. 18 (6) (2020) 1041 |f (z)| = ∣∣∣∣∣1z + ∞∑ n=1 anz n ∣∣∣∣∣ ≥ 1 r − ∞∑ n=1 |an| |r| n ≥ 1 r − A−B A + B + 2 r. Thus prove the result. � Theorem 2.5. Let f (z) ∈MS∗∗ [A,B] , |z| = r. Then (2.9) 1 r2 − 2 (A−B) A + B + 2 r ≤ |f′(z)| ≤ 1 r2 + 2 (A−B) A + B + 2 r. Proof. As |f′(z)| = ∣∣∣∣∣− 1z2 + ∞∑ n=1 nanz n−1 ∣∣∣∣∣ ≤ 1 r2 + ∞∑ n=1 |an| |r| n−1 ≤ 1 r2 + 2 (A−B) A + B + 2 r. Where we have used Theorem 2.1, and |f′(z)| = ∣∣∣∣∣− 1z2 + ∞∑ n=1 nanz n−1 ∣∣∣∣∣ ≥ 1 r2 − ∞∑ n=1 |an| |r| n ≥ 1 r2 − 2 (A−B) A + B + 2 r. Thus prove the result. � Theorem 2.6. Let f (z) ∈MS∗∗ [A,B] of the form (1.1) , and h (z) = 1 z + ∑∞ n=1 bnz n ∈MS∗∗ (A,B) with |bn| ≤ 1, then f (z) ∗h (z) ∈MS∗∗ [A,B] . Proof. Since by Theorem 2.1, we have ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an| ≤ A−B. Int. J. Anal. Appl. 18 (6) (2020) 1042 Since ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |anbn| = ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an| |bn| ≤ ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an| ≤ A−B. Thus f (z) ∗h (z) ∈MS∗∗[A,B]. � Theorem 2.7. If f ∈MS∗∗[A,B]. Then f ∈MS∗ (α) for |z| < r1, where (2.10) r1 = ( (1−α) ( (1+B)n+(1+A) 1−(−1)n 2 ) (n+α)(A−B) ) 1 n+1 . Proof. Let f ∈MS∗∗[A,B]. To prove f ∈MS∗ (α) , we only need to show∣∣∣∣ zf′(z) + f(z)zf′(z) − (1 − 2α) f(z) ∣∣∣∣ < 1. Using (1.1) along with some simple computation yields (2.11) ∞∑ n=1 n + α 1 −α |an| |z| n+1 ≤ 1. As f is in the class MS∗∗[A,B] so we have from (2.1) , ∞∑ n=1 (1 + B) n + (1 + A) 1−(−1)n 2 A−B |an| ≤ 1. Now inequality (2.11) will be true, if the following holds ∑∞ n=1 n+α 1−α |an| |z| n+1 < ∑∞ n=1 (1+B)n+(1+A) 1−(−1)n 2 A−B |an| , which implies that |z|n+1 < (1−α) ( (1+B)n+(1+A) 1−(−1)n 2 ) (n+α)(A−B) , and so |z| < ( (1−α) ( (1+B)n+(1+A) 1−(−1)n 2 ) (n+α)(A−B) ) 1 n+1 = r1, we get the required condition. � Int. J. Anal. Appl. 18 (6) (2020) 1043 Theorem 2.8. If f0 (z) = 1 z and for n ≥ 1 fn (z) = 1 z + A−B (1 + B) n + (1 + A) 1−(−1)n 2 zn. Then f ∈MS∗∗[A,B] if and only if (2.12) f (z) = ∞∑ n=0 δnfn (z) , where δn ≥ 0 and ∞∑ n=1 δn = 1. Proof. Let f (z) be expressed in the form (2.12) , then f (z) = 1 z + ∞∑ n=1 δn A−B (1 + B) n + (1 + A) 1−(−1)n 2 zn, and for above function, we have ∞∑ n=1 [ (1 + B) n + (1 + A) 1 − (−1)n 2 ] ×δn A−B (1 + B) n + (1 + A) 1−(−1)n 2 = (A−B) (1 − δ0) ≤ A−B. Thus by Theorem 2.1, f (z) ∈MS∗∗[A,B]. Conversly, let f (z) ∈MS∗∗[A,B], since by Theorem 2.1, we have |an| ≤ A−B (1 + B) n + (1 + A) 1−(−1)n 2 , n ≥ 1, we set δn = (1 + B) n + (1 + A) 1−(−1)n 2 A−B |an| , n ≥ 1, and δ0 = 1 − ∞∑ n=1 δn, so, it follows that f (z) = ∞∑ n=0 δnfn (z) . Hence proof is complete. � Int. J. Anal. Appl. 18 (6) (2020) 1044 3. Partial Sums Silverman [17] determined sharp lower bounds on the real part of the quotients between the normalized starlike or convex functions and their sequences of partial sums. As a natural extension, one is interested to search results analogous to those of Silverman for meromorphic univalent functions. In this section, motivated essentially by the work of Silverman [17] and Cho and Owa [15]( also see [16, 18]) we will investigate the ratio of a function of the form (3.1) f(z) = 1 z + ∞∑ n=1 anz n, to its sequence of partial sums (3.2) f1(z) = 1 z and fk(z) = 1 z + k∑ n=1 anz n when the coefficients are sufficiently small to satisfy the condition analogous to ∞∑ n=1 ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) |an| ≤ A−B. For the sake of brevity we rewrite it as (3.3) ∞∑ n=1 dn|an| ≤ A−B, where (3.4) dn(A,B) := ( (1 + B) n + (1 + A) 1 − (−1)n 2 ) More precisely we will determine sharp lower bounds for <{f(z)/fk(z)} and <{fk(z)/f(z)}. In this con- nection we make use of the well known results that< { 1+w(z) 1−w(z) } > 0 (z ∈ D∗) if and only if ω(z) = ∞∑ n=1 cnz n satisfies the inequality |ω(z)| ≤ |z|. Unless otherwise stated, we will assume that f is of the form (1.1) and its sequence of partial sums is denoted by fk(z) = 1 z + ∑k n=1 anz n. Theorem 3.1. Let f ∈MS∗∗ [A,B] be given by (1.1)satisfies condition (2.1),then (3.5) Re { f(z) fk(z) } ≥ dk+1(A,B) + B −A dk+1(A,B) (z ∈ D∗) where (3.6) dn(A,B) ≥   A−B, if n = 1, 2, 3, . . . ,kdk+1(A,B), if n = k + 1,k + 2, . . . . The result (3.5) is sharp with the function given by (3.7) f(z) = 1 z + A−B dk+1(A,B) zk+1. Int. J. Anal. Appl. 18 (6) (2020) 1045 Proof. Define the function w(z) by 1 + w(z) 1 −w(z) = dk+1(A,B) A−B [ f(z) fk(z) − dk+1(A,B) + B −A dk+1(A,B) ] (3.8) = 1 + k∑ n=1 anz n+1 + ( dk+1(A,B) A−B ) ∞∑ n=k+1 anz n+1 1 + k∑ n=1 anzn+1 . It suffices to show that |w(z)| ≤ 1. Now, from (3.8) we can write w(z) = ( dk+1(A,B) A−B ) ∞∑ n=k+1 anz n+1 2 + 2 k∑ n=1 anzn+1 + ( dk+1(A,B) A−B ) ∞∑ k=n+1 anzn+1 . Hence we obtain |w(z)| ≤ ( dk+1(A,B) A−B ) ∞∑ k=n+1 |an| 2 − 2 k∑ n=1 |an|− ( dk+1(A,B) A−B ) ∞∑ n=k+1 |an| . Now |w(z)| ≤ 1 if 2 ( dk+1(A,B) A−B ) ∞∑ n=k+1 |an| ≤ 2 − 2 k∑ n=1 |an| or, equivalently, k∑ n=1 |an| + dk+1(A,B) A−B ∞∑ n=k+1 |an| ≤ 1. From the condition (2.1), it is sufficient to show that k∑ n=1 |an| + dk+1(A,B) A−B ∞∑ n=k+1 |an| ≤ ∞∑ n=1 dn(A,B) A−B |an| which is equivalent to k∑ n=1 ( dn(A,B) + B −A A−B ) |an| + ∞∑ n=k+1 ( dn(A,B) −dk+1(A,B) A−B ) |an| ≥ 0.(3.9) Int. J. Anal. Appl. 18 (6) (2020) 1046 To see that the function given by (3.7) gives the sharp result, we observe that for z = reiπ/k f(z) fk(z) = 1 + A−B dk+1(A,B) zn → 1 − A−B dk+1(A,B) = dk+1(A,B) + B −A dk+1(A,B) when r → 1−. which shows the bound (3.5) is the best possible for each k ∈ N. � We next determine bounds for fk(z)/f(z). Theorem 3.2. If f of the form (3.2) satisfies the condition (2.1), then (3.10) Re { fk(z) f(z) } ≥ dk+1(A,B) dk+1(A,B) + B −A (z ∈ D∗), where (3.11) dk(A,B) ≥   A−B, if k = 1, 2, 3, . . . ,ndk+1(A,B), if k = n + 1,n + 2, . . . . . The result (3.10) is sharp with the function given by (3.7). Proof. We write 1 + w(z) 1 −w(z) = dk+1(A,B) + B −A A−B [ fk(z) f(z) − dk+1(A,B) dk+1(A,B) + B −A ] = 1 + k∑ n=1 anz n+1 − ( dk+1(A,B) A−B ) ∞∑ n=k+1 anz n+1 1 + ∞∑ n=1 anzn+1 , where |w(z)| ≤ ( dk+1(A,B)+B−A A−B ) ∞∑ n=k+1 |an| 2 − 2 k∑ n=1 |an|− ( dk+1(A,B)+B−A A−B ) ∞∑ n=k+1 |an| ≤ 1. This last inequality is equivalent to k∑ n=1 |an| + dk+1(A,B) A−B ∞∑ n=k+1 |an| ≤ 1. Make use of (2.1) to get (3.9). Finally, equality holds in (3.10) for the extremal function f(z) given by (3.7). � Acknowledgment: The second author is supported by UKM Grant: GUP-2019-032. Authors contributions: All authors jointly worked on the results and they read and approved the final manuscript. Int. J. Anal. 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