# New Subclasses of Analytic Functions with Respect to Symmetric and Conjugate Points

Authors: Huo Tang; Guan-Tie Deng
Journal: Journal of Complex Analysis
 (2013)

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## Abstract

We introduce new subclasses of close-to-convex and quasiconvex functions with respect to symmetric and conjugate points. The coefficient estimates for functions belonging to these classes are obtained.


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## Body

## 1. Introduction

LetU be the class of functions which are analytic and univalent in the open unit disk E={z:|z|<1} given by
(1)ω(z)=∑k=1∞ckzk
and satisfying the conditions ω(0)=0, |ω(z)|≤1, z∈E.LetS denote the class of functions f which are analytic and univalent in E of the form
(2)f(z)=z+∑n=2∞anzn,z∈E.LetSs* be the subclass of functions f(z)∈S and satisfying the condition
(3)Re(zf′(z)f(z)-f(-z))>0,z∈E.
These functions are called starlike with respect to symmetric points and were introduced by Sakaguchi [1].Also, letSc* be the subclass of functions f(z)∈S and satisfying the condition
(4)Re(zf′(z)f(z)+f(z-)¯)>0,z∈E.
These functions are called starlike with respect to conjugate points and were introduced by El-Ashwah and Thomas [2]. Further results on starlike functions with respect to symmetric points or conjugate points can be found in [3–5].Then, Das and Singh [6] introduced another class Cs, namely, convex functions with respect to symmetric points and satisfying the condition
(5)Re((zf′(z))′(f(z)-f(-z))′)>0,z∈E.Suppose thatf and g are two analytic functions in E. Then, we say that the function g is subordinate to the function f, and we write g(z)≺f(z), z∈E, if there exists a Schwarz function ϖ(z) with ϖ(0)=0 and |ϖ(z)|<1 such that g(z)=f(ϖ(z)),z∈E.In view of subordination definition, Goel and Mehrok [7] introduced a subclass of Ss* denoted by Ss*(A,B).LetSs*(A,B) be the class of functions of the form (2) and satisfying the condition
(6)2zf′(z)f(z)-f(-z)≺1+Az1+Bz,-1≤B<A≤1,z∈E.Following them, many authors introduced the analogue definitions by extension as follows (see [8, 9]).Definition 1.
LetSc*(A,B) be the subclass of S consisting of functions given by (2) satisfying the condition
(7)2zf′(z)f(z)+f(z-)¯≺1+Az1+Bz,-1≤B<A≤1,z∈E.
LetCs(A,B) be the subclass of S consisting of functions given by (2) satisfying the condition
(8)2(zf′(z))′(f(z)-f(-z))′≺1+Az1+Bz,-1≤B<A≤1,z∈E.
LetCc(A,B) be the subclass of S consisting of functions given by (2) satisfying the condition
(9)2(zf′(z))′(f(z)+f(z-)¯)′≺1+Az1+Bz,-1≤B<A≤1,z∈E.Motivated by the pervious classes, Selvaraj and Vasanthi [10] defined the following classes of functions with respect to symmetric and conjugate points.Definition 2.
LetMs(α,A,B) be the subclass of S consisting of functions given by (2) satisfying the condition
(10)2zf′(z)+2αz2f′′(z)(1-α)(f(z)-f(-z))+αz(f(z)-f(-z))′≺1+Az1+Bz,-1≤B<A≤1,0≤α≤1,z∈E.
LetMc(α,A,B) be the subclass of S consisting of functions given by (2) satisfying the condition
(11)2zf′(z)+2αz2f′′(z)(1-α)(f(z)+f(z-)¯)+αz(f(z)+f(z-)¯)′≺1+Az1+Bz,-1≤B<A≤1,0≤α≤1,z∈E.In this paper, we introduce the classKs*(α,A,B;C,D) consisting of analytic functions f of the form (2) and satisfying
(12)2zf′(z)+2αz2f′′(z)(1-α)(g(z)-g(-z))+αz(g(z)-g(-z))′≺1+Cz1+Dz,-1≤D≤B<A≤C≤1,0≤α≤1,z∈E,
where g(z)=z+∑n=2∞bnzn∈Ms(α,A,B).In addition, we introduce the classKc*(α,A,B;C,D) consisting of analytic functions f of the form (2) and satisfying
(13)2zf′(z)+2αz2f′′(z)(1-α)(g(z)+g(z-)¯)+αz(g(z)+g(z-)¯)′≺1+Cz1+Dz,-1≤D≤B<A≤C≤1,0≤α≤1,z∈E,
where g(z)=z+∑n=2∞bnzn∈Mc(α,A,B).We note that(i)
forα=0, Ks*(0,A,B;C,D) = Ks(A,B;C,D) (see Mehrok et al. [11]) and Kc*(0,A,B;C,D) = Kc(A,B;C,D);(ii)
forα=0,C=1 and D=-1, Ks*(0,A,B;1,-1)=Ks(A,B) (see Janteng and Halim [12]) and Kc*(0,A,B;1,-1)=Kc(A,B);(iii)
forα=0,A=C=1 and B=D=-1, Ks*(0,1,-1;1,-1)≡Ks and Kc*(0,1,-1;1,-1)≡Kc;(iv)
forα=1, Ks*(1,A,B;C,D)=Ks*(A,B;C,D) and Kc*(1,A,B;C,D)=Kc*(A,B;C,D);(v)
forα=1,C=1 and D=-1, Ks*(1,A,B;1,-1)=Ks*(A,B) (see Janteng and Halim [13]) and Kc*(1,A,B;1,-1)=Kc*(A,B);(vi)
forα=1,A=C=1 and B=D=-1, Ks*(1,1,-1;1,-1)≡Ks* and Kc*(1,1,-1;1,-1)≡Kc*.By the definition of subordination, it follows thatf∈Ks*(α,A,B;C,D) if and only if
(14)2zf′(z)+2αz2f′′(z)(1-α)(g(z)-g(-z))+αz(g(z)-g(-z))′=1+Cω(z)1+Dω(z)=P(z),ω(z)∈U,
and that f∈Kc*(α,A,B;C,D) if and only if
(15)2zf′(z)+2αz2f′′(z)(1-α)(g(z)+g(z-)¯)+αz(g(z)+g(z-)¯)′=1+Cω(z)1+Dω(z)=P(z),ω(z)∈U,
where
(16)P(z)=1+∑n=1∞pnzn.In the next section, we discuss the coefficient estimates for functions belonging to the classesKs*(α,A,B;C,D) and Kc*(α,A,B;C,D).


## 2. Some Preliminary Lemmas

We will require the following lemmas for proving our main results.Lemma 3 (see [7]).
IfP(z) is given by (14), (15) and (16), then for  -1≤D<C≤1,
(17)|pn|≤(C-D),n=1,2,….Lemma 4 (see [10]).
Letg(z)=z+∑n=2∞bnzn∈Ms(α,A,B). Then, for n≥1, 0≤α≤1,
(18)|b2n|≤(A-B)2n·n!(1+(2n-1)α)∏j=1n-1‍(A-B+2j),|b2n+1|≤(A-B)2n·n!(1+2nα)∏j=1n-1‍(A-B+2j).Lemma 5 (see [10]).
Letg(z)=z+∑n=2∞bnzn∈Mc(α,A,B). Then, for n≥1, 0≤α≤1,
(19)|b2n|≤(A-B)(2n-1)!(1+(2n-1)α)∏j=12n-2‍(A-B+j),|b2n+1|≤(A-B)(2n)!(1+2nα)∏j=12n-1‍(A-B+j).


## 3. Main Results

Unless otherwise mentioned, we will assume in the reminder of this paper that-1 ≤ D ≤ B < A ≤ C ≤ 1, 0 ≤ α ≤ 1, and z∈E.Theorem 6.
Letf∈Ks*(α,A,B;C,D), then for n≥1,
(20)|a2n|≤(C-D)2n·n!(1+(2n-1)α)∏j=1n-1‍(A-B+2j),(21)|a2n+1|≤1(2n+1)(1+2nα)×{∏j=1n-1‍[(C-D)+(A-B)2n]×[12n-1·(n-1)!∏j=1n-1‍(A-B+2j)]}.Proof.
Sinceg(z)=z+∑n=2∞bnzn∈Ms(α,A,B), it follows that
(22)2zg′(z)+2αz2g′′(z)=[(1-α)(g(z)-g(-z))+αz(g(z)-g(-z))′]K(z),
for z∈E, with Re(K(z))>0, where K(z)=1+d1z+d2z2+d3z3+⋯.
On equating the coefficients of like powers ofz in (22), we get
(23)2(1+α)b2=d1,2(1+2α)b3=d2,4(1+3α)b4=d3+(1+2α)b3d1,4(1+4α)b5=d4+(1+2α)b3d2,
and continuing in this way, we obtain
(24)2n(1+(2n-1)α)b2n=d2n-1+(1+2α)b3d2n-3+⋯+(1+(2n-2)α)b2n-1d1,(25)2n(1+2nα)b2n+1=d2n+(1+2α)b3d2n-2+⋯+(1+(2n-2)α)b2n-1d2.
From (14) and (16), we have
(26)(z+2a2z2+3a3z3+4a4z4+5a5z5+⋯+2na2nz2n+⋯)+α(2a2z2+6a3z3+12a4z4+20a5z5+⋯+(2n-1)2na2nz2n+⋯)=[(1-α)(z+b3z3+b5z5+⋯+b2n-1z2n-1+b2n+1z2n+1+⋯)+α(z+3b3z3+5b5z5+⋯+(2n-1)b2n-1z2n-1+(2n+1)b2n+1z2n+1+⋯)]×(1+p1z+p2z2+p3z3+p4z4+p5z5+⋯+p2n-1z2n-1+p2nz2n+⋯).
On equating the coefficients, we obtain(27)2(1+α)a2=p1,3(1+2α)a3=p2+(1+2α)b3,(28)4(1+3α)a4=p3+(1+2α)b3p1,5(1+4α)a5=p4+(1+2α)b3p2+(1+4α)b5,
and so
(29)2n(1+(2n-1)α)a2n=p2n-1+(1+2α)b3p2n-3+⋯+(1+(2n-2)α)b2n-1p1,(2n+1)(1+2nα)a2n+1=p2n+(1+2α)b3p2n-2+⋯+(1+(2n-2)α)b2n-1p2+(1+2nα)b2n+1.
By using Lemma 3 and (27), we have
(30)|a2|≤(C-D)2·1·(1+α),|a3|≤(A-B)+2(C-D)3·2·(1+2α).
Again, by applying Lemma 3 and using (23), we obtain from (28)
(31)|a4|≤(C-D)(A-B+2)4·2·(1+3α),|a5|≤(A-B+2)[(A-B)+4(C-D)]5·8·(1+4α).
It follows that (20) and (21) hold for n=1,2. We now prove (20) and (21) by induction.
Equation (29) together with Lemma 3 yield
(32)|a2n|≤(C-D)2n(1+(2n-1)α)[1+∑k=1n-1(1+2kα)|b2k+1|],(33)|a2n+1|≤1(2n+1)(1+2nα)×{(C-D)[1+∑k=1n-1(1+2kα)|b2k+1|]+(1+2nα)|b2n+1|∑k=1n-1‍}.
Again, using Lemma3 in (25), we have
(34)|b2n+1|≤(A-B)2n(1+2nα)[1+∑k=1n-1(1+2kα)|b2k+1|].
Using (34) in (33), we obtain
(35)|a2n+1|≤1(2n+1)(1+2nα)×{[(C-D)+(A-B)2n][1+∑k=1n-1(1+2kα)|b2k+1|]}.
We suppose that (20) and (21) hold for k=3,4,…,(n-1).
Using Lemma4 in (32) and (35), we get
(36)|a2n|≤(C-D)2n(1+(2n-1)α)×[1+∑k=1n-1(A-B)2k·k!∏j=1k-1‍(A-B+2j)],(37)|a2n+1|≤1(2n+1)(1+2nα)×{∏j=1k-1‍[(C-D)+(A-B)2n]×[1+∑k=1n-1(A-B)2k·k!∏j=1k-1‍(A-B+2j)]}.
In order to prove (20), it is sufficient to show that
(38)(C-D)2m(1+(2m-1)α)[1+∑k=1m-1(A-B)2k·k!∏j=1k-1‍(A-B+2j)]=(C-D)2m·m!(1+(2m-1)α)∏j=1m-1‍(A-B+2j)(m=3,4,…,n).
Thus, (38) is valid for m=3.
Let us assume that (38) is true for all m,3<m≤(n-1). Then, from (36), we have
(39)(C-D)2n(1+(2n-1)α)[1+∑k=1n-1(A-B)2k·k!∏j=1k-1‍(A-B+2j)]=(n-1)n{∏j=1k-1‍(C-D)2(n-1)(1+(2n-1)α)×[1+∑k=1n-2‍(A-B)2k·k!∏j=1k-1‍(A-B+2j)]}+(C-D)2n(1+(2n-1)α)·(A-B)2n-1·(n-1)!×∏j=1n-2‍(A-B+2j)=(n-1)n·(C-D)2n-1·(n-1)!(1+(2n-1)α)×∏j=1n-2‍(A-B+2j)+(C-D)2n(1+(2n-1)α)·(A-B)2n-1·(n-1)!×∏j=1n-2‍(A-B+2j)=(C-D)2n·n!(1+(2n-1)α)∏j=1n-2‍(A-B+2j)×(A-B+2(n-1))=(C-D)2n·n!(1+(2n-1)α)∏j=1n-1‍(A-B+2j).
Thus, (38) holds for m=n, and, hence, (20) follows. Next, we prove (21).
From (38), we have
(40)1+∑k=1n-1(A-B)2k·k!∏j=1k-1‍(A-B+2j)=12n-1·(n-1)!∏j=1n-1‍(A-B+2j).
By using (40) in (37), we obtain
(41)|a2n+1|≤1(2n+1)(1+2nα)×{∏j=1n-1‍[(C-D)+(A-B)2n]×[12n-1·(n-1)!∏j=1n-1‍(A-B+2j)]}
which proves (21).By takingα=0 and α=1 in Theorem 6, respectively, we can readily deduce the following corollaries.Corollary 7 (see [11]).
Letf∈Ks(A,B;C,D), then, for n≥1,
(42)|a2n|≤(C-D)2n·n!∏j=1n-1‍(A-B+2j),|a2n+1|≤12n+1{∏j=1n-1‍[(C-D)+(A-B)2n]×[12n-1·(n-1)!∏j=1n-1‍(A-B+2j)]}.Corollary 8.
Letf∈Ks*(A,B;C,D), then, for n≥1,
(43)|a2n|≤(C-D)2n+1·n·n!∏j=1n-1‍(A-B+2j),|a2n+1|≤1(2n+1)2{∏j=1n-1‍[(C-D)+(A-B)2n]×[12n-1·(n-1)!∏j=1n-1‍(A-B+2j)]}.
Further, putting C=1 and D=-1 in Corollary 8, one obtains the following.Corollary 9.
Letf∈Ks*(A,B), then, for n≥1,
(44)|a2n|≤12n·n·n!∏j=1n-1‍(A-B+2j),|a2n+1|≤1(2n+1)2×{∏j=1n-1‍[2+(A-B)2n]×[12n-1·(n-1)!∏j=1n-1‍(A-B+2j)]}.Remark 10.
Corollary9 improves the result obtained by Janteng and Halim [13, Theorem 3.1].Theorem 11.
Letf∈Kc*(α,A,B;C,D), then, for n≥1,
(45)|a2n|≤12n(1+(2n-1)α)×{[(C-D)+(A-B)2n-1][1(2n-2)!∏j=12n-2‍(A-B+j)]},(46)|a2n+1|≤1(2n+1)(1+2nα)×{[(C-D)+(A-B)2n][1(2n-1)!∏j=12n-1‍(A-B+j)]}.Proof.
Sinceg(z)=z+∑n=2∞bnzn∈Mc(α,A,B), it follows that
(47)2zg′(z)+2αz2g′′(z)=[(1-α)(g(z)+g(z-)¯)+αz(g(z)+g(z-)¯)′]K(z),
where K(z)=1+d1z+d2z2+d3z3+⋯.
On equating the coefficients of like powers ofz in (47), we get
(48)(1+α)b2=d1,2(1+2α)b3=d2+(1+α)b2d1,(49)3(1+3α)b4=d3+(1+α)b2d2+(1+2α)b3d1,(50)4(1+4α)b5=d4+(1+α)b2d3+(1+2α)b3d2+(1+3α)b4d1,
and continuing in this way, we obtain
(51)(2n-1)(1+(2n-1)α)b2n=d2n-1+(1+α)b2d2n-2+⋯+(1+(2n-2)α)b2n-1d1,(52)2n(1+2nα)b2n+1=d2n+(1+α)b2d2n-1+⋯+(1+(2n-1)α)b2nd1.
From (15) and (16), we have
(53)(z+2a2z2+3a3z3+4a4z4+5a5z5+⋯+2na2nz2n+⋯)+α(2a2z2+6a3z3+12a4z4+20a5z5+⋯+(2n-1)2na2nz2n+⋯)=[(1-α)(z+b2z2+b3z3+b4z4+b5z5+⋯+b2nz2n+⋯)+α(z+2b2z2+3b3z3+4b4z4+5b5z5+⋯+2nb2nz2n+⋯)]×(1+p1z+p2z2+p3z3+p4z4+p5z5+⋯+p2nz2n+⋯).
On equating the coefficients, we obtain(54)2(1+α)a2=p1+(1+α)b2,3(1+2α)a3=p2+(1+α)b2p1+(1+2α)b3,(55)4(1+3α)a4=p3+(1+α)b2p2+(1+2α)b3p1+(1+3α)b4,(56)5(1+4α)a5=p4+(1+α)b2p3+(1+2α)b3p2+(1+3α)b4p1+(1+4α)b5,
and so
(57)2n(1+(2n-1)α)a2n=p2n-1+(1+α)b2p2n-2+(1+2α)b3p2n-3+⋯+(1+(2n-2)α)b2n-1p1+(1+(2n-1)α)b2n,(58)(2n+1)(1+2nα)a2n+1=p2n+(1+α)b2p2n-1+(1+2α)b3p2n-2+⋯+(1+(2n-1)α)b2np1+(1+2nα)b2n+1.
By using Lemma 3, (48), and (54), we have
(59)|a2|≤(C-D)+(A-B)2·1·(1+α),|a3|≤(A-B+1)[(A-B)+2(C-D)]3·2·(1+2α).
Again, by applying Lemma 3 and using (48)–(50), we obtain from (55) and (56)
(60)|a4|≤(A-B+1)(A-B+2)[(A-B)+3(C-D)]4·6·(1+3α),|a5|≤(A-B+1)(A-B+2)(A-B+3)×[(A-B)+4(C-D)](5·24·(1+4α))-1.
It follows that (45) and (46) hold for n=1,2. We now prove (45) by induction.
Equation (57) together with Lemma 3 yield
(61)|a2n|≤12n(1+(2n-1)α)×{∑k=1n-1‍(C-D)[1+∑k=1n-1‍(1+(2k-1)α)|b2k|+∑k=1n-1‍(1+2kα)|b2k+1|]+(1+(2n-1)α)|b2n|∑k=1n-1‍}.
Again, by using Lemma 3 in (51), we have
(62)|b2n|≤(A-B)(2n-1)(1+(2n-1)α)×[1+∑k=1n-1(1+(2k-1)α)|b2k|+∑k=1n-1(1+2kα)|b2k+1|].
Using (62) in (61), we obtain
(63)|a2n|≤12n(1+(2n-1)α)×{∑k=1n-1(1+2kα)|b2k+1|[(C-D)+(A-B)2n-1]×[1+∑k=1n-1(1+(2k-1)α)|b2k|+∑k=1n-1(1+2kα)|b2k+1|]}.
We suppose that (45) holds for k=3,4,…,(n-1).
Using Lemma5 in (63), we get
(64)|a2n|≤12n(1+(2n-1)α)×{∏j=12k-1‍[(C-D)+(A-B)2n-1]×[1+∑k=1n-1‍(A-B)(2k-1)!∏j=12k-2‍(A-B+j)+∑k=1n-1‍(A-B)(2k)!∏j=12k-1‍(A-B+j)]}.
In order to prove (45), it is sufficient to show that
(65)12m(1+(2m-1)α)×{∏j=12k-1‍[(C-D)+(A-B)2m-1]×[1+∑k=1m-1(A-B)(2k-1)!∏j=12k-2‍(A-B+j)+∑k=1m-1(A-B)(2k)!∏j=12k-1‍(A-B+j)]}=12m(1+(2m-1)α)×{∏j=12m-2‍[(C-D)+(A-B)2m-1]×[1(2m-2)!∏j=12m-2‍(A-B+j)]}(m=3,4,…,n).
Thus, (65) is valid for m=3.
Let us assume that (65) is true for all m, 3<m≤(n-1). Then, from (64), we have
(66)12n(1+(2n-1)α)×{+∑k=1n-1‍(A-B)(2k)!∏j=12k-1‍(A-B+j)[(C-D)+(A-B)2n-1]×[1+∑k=1n-1‍(A-B)(2k-1)!∏j=12k-2‍(A-B+j)+∑k=1n-1‍(A-B)(2k)!∏j=12k-1‍(A-B+j)]}=(n-1n){+∑k=1n-1‍(A-B)(2k)!∏j=12k-1‍(A-B+j)12(n-1)(1+(2n-1)α)×[(C-D)+(A-B)2n-1]×[1+∑k=1n-2‍(A-B)(2k-1)!∏j=12k-2‍(A-B+j)+∑k=1n-2‍(A-B)(2k)!∏j=12k-1‍(A-B+j)]}+12n(1+(2n-1)α)[(C-D)+(A-B)2n-1]×[(A-B)(2(n-1)-1)!∏j=12n-4‍(A-B+j)+(A-B)(2(n-1))!∏j=12n-3‍(A-B+j)]=(n-1n)·12(n-1)(1+(2n-1)α)×[(C-D)+(A-B)2n-1]×[1(2(n-1)-2)!∏j=12n-4‍(A-B+j)]+12n(1+(2n-1)α)[(C-D)+(A-B)2n-1]×[(A-B)(2(n-1)-1)!∏j=12n-4‍(A-B+j)+(A-B)(2(n-1))!∏j=12n-3‍(A-B+j)]=12n(1+(2n-1)α)[(C-D)+(A-B)2n-1]×[1(2(n-1)-1)!∏j=12n-4‍(A-B+j)×(A-B+(2n-3))+(A-B)(2(n-1))!∏j=12n-3‍(A-B+j)]=12n(1+(2n-1)α)[(C-D)+(A-B)2n-1]×[1(2(n-1)-1)!∏j=12n-3‍(A-B+j)+(A-B)(2(n-1))!∏j=12n-3‍(A-B+j)]=12n(1+(2n-1)α)[(C-D)+(A-B)2n-1]×[1(2(n-1))!∏j=12n-3‍(A-B+j)×(A-B+(2n-2))∏j=12n-3‍]=12n(1+(2n-1)α)[(C-D)+(A-B)2n-1]×[1(2n-2)!∏j=12n-2‍(A-B+j)].
Thus, (65) holds for m=n, and, hence, (45) follows. Similarly, we can prove (46).By takingα=0 and α=1 in Theorem 11, respectively, we can easily derive the following corollaries.Corollary 12.
Letf∈Kc(A,B;C,D), then, for n≥1,
(67)|a2n|≤12n{∏j=12n-2‍[(C-D)+(A-B)2n-1]×[1(2n-2)!∏j=12n-2‍(A-B+j)]},|a2n+1|≤12n+1{∏j=12n-2‍[(C-D)+(A-B)2n]×[1(2n-1)!∏j=12n-1‍(A-B+j)]}.Corollary 13.
Letf∈Kc*(A,B;C,D), then, for n≥1,
(68)|a2n|≤1(2n)2{∏j=12n-2‍[(C-D)+(A-B)2n-1]×[1(2n-2)!∏j=12n-2‍(A-B+j)]},|a2n+1|≤1(2n+1)2{∏j=12n-1‍[(C-D)+(A-B)2n]×[1(2n-1)!∏j=12n-1‍(A-B+j)]}.



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*Source: 578036-2013-02-04.xml*