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Research article

Geometric properties of harmonic functions associated with the symmetric conjecture points and exponential function

  • Received: 07 June 2020 Accepted: 24 August 2020 Published: 02 September 2020
  • MSC : 30C45, 30C65

  • In this paper, some classes of univalent harmonic functions are introduced by subordination, where the analytic parts of which are exponential starlike (or convex) functions with respect to the symmetric conjecture points. According to the relationships of the analytic part and the co-analytic part, the geometric properties, such as coefficient estimates, distortion theorems, integral expressions, estimates and growth conditions and covering theorem, of the classes are obtained.

    Citation: Lina Ma, Shuhai Li, Huo Tang. Geometric properties of harmonic functions associated with the symmetric conjecture points and exponential function[J]. AIMS Mathematics, 2020, 5(6): 6800-6816. doi: 10.3934/math.2020437

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  • In this paper, some classes of univalent harmonic functions are introduced by subordination, where the analytic parts of which are exponential starlike (or convex) functions with respect to the symmetric conjecture points. According to the relationships of the analytic part and the co-analytic part, the geometric properties, such as coefficient estimates, distortion theorems, integral expressions, estimates and growth conditions and covering theorem, of the classes are obtained.


    Let A be the class of functions h of the form

    h(z)=z+k=2akzk, (1.1)

    where h is analytic in U={zC:|z|<1}.

    We denote S,S and K the subclasses of A consisting of univalent, starlike and convex functions respectively ([1,2]) and denote P={p:p(0)=1,Rep(z)>0,zU}.

    An analytic function s:U={z:|z|<1}C is subordinate to an analytic function t:UC, if there is a function ν satisfying ν(0)=0and|ν(z)|<1(zU), such that s(z)=t(ν(z))(zU). Note that s(z)t(z). Especially, if t is univalent in U, then the following conclusion is true (see [1]):

    s(z)t(z)s(0)=t(0)ands(U)t(U).

    In 1933, a classical Fekete-Szegö problem for h(z)=z+n=2anznS was introduced by Fekete and Szegö [3] as follows,

    |a3μa22|{34μ,μ0,1+2exp(2μ1μ),0μ1,4μ3,μ1.

    The result is sharp.

    Using the subordination, the classes S(ϕ) and K(ϕ) of starlike and convex functions were defined by Ma and Minda [4] in 1994. The function h(z)S(ϕ) iff zh(z)h(z)ϕ(z) and zh(z)h(z)ϕ(z), where hA and ϕP. Moreover, Fekete-Szegö problems of the classes were obtained by Ma and Minda [4]. The problem of Fekete-Szegö has always been a hot topic in geometry function theory. Many authors studied and obtained many results (see [5,6,7]).

    Let ϕ(z)=1+Az1+Bz and 1B<A1. The classes S(ϕ) and K(ϕ) reduce to S(1+Az1+Bz) and K(1+Az1+Bz), which are the classes of Janowski starlike and convex functions respectively (see [8]).

    Without loss of generality, both S(1+z1z)=S and K(1+z1z)=K represent the well-known classes of starlike and convex function respectively.

    In 2015, Mediratta et al. [9] introduced the family of exponential starlike functions S(ez), that is

    S(ez)={hA:zh(z)h(z)ez,zU}

    or, equivalently

    S(ez)={hA:|logzh(z)h(z)|<1,zU}.

    According to the properties of the exponential function ez and the subordination relationship, the class S(ez) maps the unit disc U onto a region, which is symmetric with respect to the real axis and 1.

    In 1959, the class Ss of starlike functions with respect to symmetric points was introduced by Sakaguchi [10]. The function hSs if and only if

    Rezh(z)h(z)h(z)>0.

    In 1987, the classes Sc and Scs of starlike functions with respect to conjugate points and symmetric conjugate points were introduced by El-Ashwa and Thomas [11] as follows,

    hScRezh(z)h(z)+¯h(¯z)>0andhSscRezh(z)h(z)¯h(¯z)>0.

    For analytic functions h(z) and g(z)(zU). Let SH define the class of harmonic mappings with the following form (see [12,13])

    f(z)=h(z)+¯g(z),zU, (1.2)

    where

    h(z)=z+k=2akzkandg(z)=k=1bkzk,|b1|=α[0,1). (1.3)

    In particular, h is called the analytic part and g is called the co-analytic part of f.

    It is well known that the function f=h+¯g is locally univalent and sense preserving in U if and only if |h(z)|>|g(z)|(zU)([14]).

    According to the above conclusion, the coefficient estimations, distortion theorems, integral expressions, Jacobi estimates and growth condition in geometric properties of covering theorem of the co-analytic part can be obtained by using the analytic part of harmonic functions. In the recent years, various subclasses of SH were researched by many authors as follows.

    In 2007, the subclass of SH with hK was studied by Klimek and Michalski [15].

    In 2014, the subclass of SH with hS was studied by Hotta and Michalski [16].

    In 2015, the subclasses of SH with hS(1+(12β)z1z) and hK(1+(12β)z1z) were studied by Zhu and Huang [17].

    In this paper, by using subordination relationship, we studied the subclasses of SH with h(z)¯h(¯z)2S(ez) and h(z)¯h(¯z)2K(ez).

    Definition 1. Suppose f=h+¯gSH of the form (1.3). Let HS,αsc(e) denote the class of harmonic univalent exponential starlike functions with symmetric conjecture point consisting of f with hSsc(e). That is, the function f=h+¯gHS,αsc(e) if and only if

    2zh(z)h(z)¯h(¯z)ez.

    Also, let HKαsc(e) denote the class of harmonic univalent convex exponential functions with symmetric conjecture point consisting of f with hKsc(e), that is, f=h+¯gHKαsc(e) if and only if

    2(zh(z))(h(z)¯h(¯z))ez.

    We know that h(z)Ksc(e)zh(z)Ssc(e).

    In order to obtain our results, we need Lemmas as follows.

    Lemma 1. ([18]). Let ω(z)=c0+c1z++cnzn+ be analytic satisfying |ω(z)|1 in U. Then

    |cn|1|c0|2,n=1,2,, (2.1)

    and

    |c2γc21|max{1,|γ|}. (2.2)

    Lemma 2. Let

    2zh(z)h(z)¯h(¯z)=p(z),

    we have

    h(z)=z0p(η)expη0p(t)+¯p(¯t)22tdtdη.

    Lemma 3. If h(z)=z+n=2anznSsc(e), then

    |a2n|(2n1)!!(2n)!!and|a2n+1|(2n1)!!(2n)!!. (2.3)

    The estimate is sharp for h given by

    h(z)=1+z1z21.

    If h(z)=z+n=2anznKsc(e), then

    |a2n|(2n1)!!2n(2n)!!and|a2n+1|(2n1)!!(2n+1)(2n)!!. (2.4)

    The estimate is sharp for h given by

    h(z)=arcsinzlog(1+1z2)+log2.

    Proof. Let h(z)=z+n=2anznSsc(e). According to Definition 1, we have

    2zh(z)h(z)¯h(¯z)ez.

    There exists a function p(z)=1+k=1pkzk satisfying

    2zh(z)h(z)¯h(¯z)=p(z), (2.5)

    that is,

    1+k=1pkzkez=1+z+z22!+.

    Using the results of Rogosinski [19], we have |pk|1 for k1.

    By means of comparing the coefficients of the both sides of (2.5), we get

    2na2n=p2n1+a3p2n3++a2n1p1,

    and

    2na2n+1=p2n+a3p2n2++a2n1p2.

    Let ϕ(n)=1+|a3|++|a2n1|. It is easy to verify that

    |a2n|12nϕ(n) (2.6)

    and

    |a2n+1|12nϕ(n). (2.7)

    From (2.7), we have

    ϕ(n+1)(2n+1)!!(2n)!!. (2.8)

    According to (2.6), (2.7) and (2.8), we can obtain (2.3).

    If h(z)=z+n=2anznKsc(e), then zh(z)Ssc(e). Using the results in (2.3), we can obtain (2.4) easily.

    Lemma 4. Let μC.

    (1) If h(z)=z+n=2anznSsc(e), then

    |a3μa22|12max{1,12|μ1|}. (2.9)

    (2) If h(z)=z+n=2anznKsc(e), then

    |a3μa22|16max{1,18|3μ4|}. (2.10)

    The estimates are sharp.

    Proof. Let h(z)=z+n=2anznSsc(e). According to the subordination relationship and Definition 1, we get

    2zh(z)h(z)¯h(¯z)=eν(z),

    that is,

    log2zh(z)h(z)¯h(¯z)=ν(z), (2.11)

    where ν(z)=c1z+c2z2+ is an analytic function with ν(0)=0 and |ν(z)|<1(zU).

    By means of comparing the coefficients of two sides of (2.11), we get

    a2=12c1,a3=12c2+14c21anda4=14c3+38c2c1+548c31.

    Therefore, we have

    a3μa22=12{c212(μ1)c21}.

    Using the fact that (2.2) in Lemma 1, we obtain

    |a3μa22|12max{1,12|μ1|}.

    The bound is sharp for h given as follows,

    h(z)=z0exp(ξ+ξ0et+et22tdt)dξorh(z)=z0exp(ξ2+ξ0et21tdt)dξ.

    Taking h(z)=z+n=2anznKsc(e)zh(z)=z+n=2nanznSsc(e) into consideration, it is easy to obtain (2.10). The bound is sharp for h given as follows,

    h(z)=z01ηη0exp(ξ+ξ0et+et22tdt)dξdηorh(z)=z01ηη0exp(ξ2+ξ0et21tdt)dξdη.

    Lemma 5. Suppose h(z)A and |z|=r[0,1).

    (1) Let h(z)S(e). Then

    exp(r+r0eη1ηdη)<|h(z)|<exp(r+r0eη1ηdη) (2.12)

    and

    rexpr0eη1ηdη<|h(z)|<rexpr0eη1ηdη. (2.13)

    (2) Let h(z)K(e). Then

    expr0eη1ηdη<|h(z)|<expr0eη1ηdη (2.14)

    and

    rexpr0eη12ηdη<|h(z)|<rexpr0eη12ηdη. (2.15)

    Proof. Let h(z)S(e) and |z|=r[0,1). According to the subordination relationship and Definition 1, there exists an analytic function ν(z)=c1z+c2z2+ satisfying ν(0)=0 and |ν(z)|<|z|, such that

    zh(z)h(z)=eν(z).

    Thus, it is concluded that

    er<|zh(z)h(z)|<er, (2.16)

    that is,

    er<Rezh(z)h(z)<er. (2.17)

    Since

    Rezh(z)h(z)=rrlog|h(z)|. (2.18)

    By (2.17) and (2.18), we get

    rexpr0eη1ηdη<|h(z)|<rexpr0eη1ηdη. (2.19)

    From (2.16) and (2.19), we have

    rexp(r+r0eη1ηdη)<|zh(z)|<rexp(r+r0eη1ηdη).

    Similar to the previous proof. We let h(z)K(e) and |z|=r[0,1), then

    er<|1+zh(z)h(z)|<er.

    After simple calculation, we have

    er1<Rezh(z)h(z)<er1. (2.20)

    By (2.18) and (2.20), we get

    expr0eη1ηdη<|h(z)|<expr0eη1ηdη.

    Using the conclusion in [19], for h(z)K(e), we have 2zh(z)h(z)1ez. According to the subordination relationship, we have

    er<|2zh(z)h(z)1|<er,

    After simple calculation, we have

    er12<Rezh(z)h(z)1<er12. (2.21)

    By (2.18) and (2.21), we get

    rexpr0eη12ηdη<|h(z)|<rexpr0eη12ηdη.

    Therefore, we complete the proof of Lemma 5.

    Lemma 6. ([20]). If h(z)Ssc(e), then h(z)¯h(¯z)2S(e).

    Lemma 7. If h(z)Ksc(e), then h(z)¯h(¯z)2K(e).

    Lemma 8. Suppose h(z)A and |z|=r[0.1).

    (1) Let hSsc(e). Then

    ϕ1(r)<|h(z)|<ϕ2(r),

    where

    ϕ1(r)=exp(r+r0eη1ηdη),ϕ2(r)=exp(r+r0eη1ηdη). (2.22)

    (2) Let hKsc(e). Then

    ψ1(r)<|h(z)|<ψ2(r),

    where

    ψ1(r)=1rr0exp(t+t0eη1ηdη)dt,ψ2(r)=1rr0exp(t+t0eη1ηdη)dt. (2.23)

    Proof. Suppose h(z)Ssc(e). It is quite similar to the proof of Lemma 5, we have

    er|h(z)¯h(¯z)2||zh(z)|er|h(z)¯h(¯z)2|. (2.24)

    According to Lemma 6 and (2.13) of Lemma 5, we have

    rexpr0eη1ηdη<|h(z)¯h(¯z)2|<rexpr0eη1ηdη. (2.25)

    By (2.24) and (2.25), we can obtain (2.22).

    If h(z)Ksc(e), then

    er|(h(z)¯h(¯z))2||(zh(z))|er|(h(z)¯h(¯z))2|. (2.26)

    According to Lemmas 7 and (2.14) of Lemma 5, we have

    expr0eη1ηdη<|(h(z)¯h(¯z))2|<expr0eη1ηdη. (2.27)

    By (2.26) and (2.27), we get

    exp(r+r0eη1ηdη)|(zh(z))|exp(r+r0eη1ηdη). (2.28)

    By (2.28), integrating along a radial line ξ=teiθ, we obtained immediately,

    |zh(z)|r0exp(t+t0eη1ηdη)dt

    The verification for the remainder of (2.23) is given as follows. Let H(z):=zh(z), which is univalent. Suppose that ξ1Γ=H({z:|z|=r}) is the nearest point to the origin. By means of rotation, we suppose that ξ1>0 and z1=H1(ξ1). Let γ={ξ:0ξξ1} and L=H1(γ). If ς=H1(ξ), then dξ=H(ς)dς. Hence

    ξ1=ξ10dξ=z10H(ς)dςr0|H(teiθ)|dtr0exp(t+t0eη1ηdη)dt.

    Thus the proof of Lemma 8 is completed.

    Next, the integral expressions for functions of the classes defined in Definition 1 are obtained.

    Theorem 1. Let ω and ν be analytic in U with |ω(0)|=α,ν(0)=0,|ω(z)|<1 and |ν(z)|<1. If f=h+¯gHS,αsc(e). Then

    f(z)=z0φ(ξ)dξ+¯z0ω(ξ)φ(ξ)dξ, (3.1)

    where

    φ(ξ)=eν(ξ)expξ0eν(t)+e¯ν(¯t)22tdt. (3.2)

    Proof. Let f=h+¯gHS,αsc(e). Using Definition 1 and the subordination relationship, there exist analytic functions ω and ν satisfying ω(0)=b1,ν(0)=0,|ω(z)|<1 and |ν(z)|<1(zU), such that

    2zh(z)h(z)¯h(¯z)=eν(z), (3.3)

    and

    g(z)=ω(z)h(z). (3.4)

    If we substitute z by ¯z in (3.4), we obtain

    2¯zh(¯z)h(¯z)¯h(z)=eν(¯z). (3.5)

    It follows from (3.4) and (3.5) that

    2z(h(z)¯h(¯z))h(z)¯h(¯z)=eν(z)+e¯ν(¯z). (3.6)

    A routine computation for the equality (3.6) gives rise to the following equation,

    ¯h(z)h(¯z)2=zexpz0eν(t)+e¯ν(¯t)22tdt. (3.7)

    Plugging (3.7) back into (3.3), we have

    h(z)=eν(z)expz0eν(t)+e¯ν(¯t)22tdt. (3.8)

    If the equality (3.8) is integrated from the both sides of it, then

    h(z)=z0eν(ξ)expξ0eν(t)+e¯ν(¯t)22tdtdξ.

    Inserting (3.8) into (3.4), it is easy to show that

    g(z)=z0ω(ξ)eν(ξ)expξ0eν(t)+e¯ν(¯t)22tdtdξ.

    Thus, the proof of Theorem 1 is completed.

    Taking Theorem 1 and hKsc(e)zh(z)Ssc(e) into consideration, we get the following result.

    Theorem 2. Let ω and ν be analytic in U satisfying |ω(0)|=α,ν(0)=0,|ω(z)|<1 and |ν(z)|<1. If fHKαsc(e), then

    f(z)=z01ηη0φ(ξ)dξdη+¯z0ω(η)ηη0φ(ξ)dξdη.

    where φ(ξ) is defined by (3.2).

    In the following, the coefficient estimates of the class HS,αsc(e) will be obtained.

    Theorem 3. Let h and g be given by (1.3). If f=h+¯gHS,αsc(e), then

    |b2n|{1α22+α2,n=1,(1α2)2n(1+nk=2(4k3)(2k3)!!(2k2)!!)+α(2n1)!!(2n)!!,n2, (3.9)

    and

    |b2n+1|{2(1α2)3+α2,n=1,(1α2)2n+1(1+nk=2(4k3)(2k3)!!(2k2)!!+(2n1)!!(2n2)!!)+α(2n1)!!(2n)!!,n2. (3.10)

    The estimates are sharp and the extremal function is

    fα0(z)=1+z1z21+¯z0(α+(1α2α)t)(1t)2(1t2)dt.

    Specially, if fHS,0sc(e), then

    |b2n|{12,n=1,12n(1+nk=2(4k3)(2k3)!!(2k2)!!),n2,

    and

    |b2n+1|{23,n=1,12n+1(1+nk=2(4k3)(2k3)!!(2k2)!!+(2n1)!!(2n2)!!),n2.

    The estimates are sharp and the extremal function is

    f01(z)=1+z1z21+¯2z3+3z213(1z2)32+13.

    Proof. Let h and g be given by (1.3). Using the fact that g=ωh satisfying ω(z)=c0+c1z+c2z2+ analytic in U, we obtain

    2nb2n=2np=1papc2np(a1=1,n1) (3.11)

    and

    (2n+1)b2n+1=2n+1p=1papc2n+1p(a1=1,n1). (3.12)

    It is easy to show that

    2n|b2n|2np=1p|ap||c2np|

    and

    (2n+1)|b2n+1|2n+1p=1p|ap||c2n+1p|.

    Since g=ωh, it follows that c0=b1. By (2.1), it is obvious that |ck|1α2 for kN. Therefore,

    |b2n|{1α22+|a2|α,n=1,(1α2)2n(1+2n1k=2k|ak|)+α|a2n|,n2, (3.13)

    and

    |b2n+1|{1α23(1+2|a2|)+|a3|α,n=1,(1α2)2n+1(1+2nk=2k|ak|)+α|a2n+1|,n2. (3.14)

    According to Lemma 3, (3.13) and (3.14), after the simple calculation, (3.9) and (3.10) can be obtained easily. We also obtain the extreme function.

    By using the same methods in Theorem 3, the following results are obtained.

    Theorem 4. Let h and g of the form (1.3). If f=h+¯gHKαsc(e), then

    |b2n|{1α24+α4,n=1,(1α2)(2n)2(1+nk=2(4k3)(2k3)!!(2k2)!!)+α(2n1)!!2n(2n)!!,n2,

    and

    |b2n+1|{2(1α2)9+α6,n=1,(1α2)(2n+1)2(1+nk=2(4k3)(2k3)!!(2k2)!!+(2n1)!!(2n2)!!)+α(2n1)!!(2n+1)(2n)!!,n2.

    For functions of the classes defined in the paper, Fekete-Szegö inequality of which are listed below.

    Theorem 5. Let f=h+¯g with h and g given by (1.3) and μC.

    (1) If fHS,αsc(e), then

    |b3μb22|(1α2)3{1+3|μ|(1α2)4+|23μb1|2}+α2max{1,|μb11|2},
    |b2nb2n1|{12(1α2)+3α2,n=1,(1α2)((12n+12n1)(1+nk=2(4k3)(2k3)!!(2k2)!!)(2n3)!!(2n1)(2n4)!!)+α((2n1)!!(2n)!!+(2n3)!!(2n2)!!),n2,

    and

    |b2n+1b2n|(1α2)((12n+1+12n)(1+nk=2(4k3)(2k3)!!(2k2)!!)+(2n1)!!(2n+1)(2n2)!!)+2α(2n1)!!(2n)!!,n1.

    (2) If fHKαsc(e), then

    |b3μb22|(1α2)3{1+3|μ|(1α2)4+|23μb1|4}+α6max{1,|43μb1|8},
    |b2nb2n1|{12(1α2)+5α4,n=1,(1α2)((12n+12n1)(1+2nk=2(2k3)!!(2k2)!!)(2n3)!!(2n1)(2n2)!!)+α((2n1)!!(2n)(2n)!!+(2n3)!!(2n1)(2n2)!!),n2,

    and

    |b2n+1b2n|(1α2)((12n+1+12n)(1+2nk=2(2k3)!!(2k2)!!)+(2n1)!!(2n+1)(2n)!!)+α(12n+1+12n)(2n1)!!(2n)!!,n1.

    Proof. From the relation (3.11) and (3.12), we have

    2b2=c1+2a2c0,3b3=c2+2a2c1+3a3c0,

    and

    2nb2n=2np=1papc2np,(2n+1)b2n+1=2n+1p=1papc2n+1p(a1=1,n1).

    By (2.1), we have

    |b3μb22|1α23{1+3|μ|(1α2)4+|a2||23μb1|}+α|a3μb1a22|,
    |b2nb2n1|{12(1α2)+α(1+|a2|),n=1,(1α2)(12n2n1p=1p|ap|+12n12n2p=1p|ap|)+α(|a2n|+|a2n1|),n2,

    and

    |b2n+1b2n|(1α2)(12n+12np=1p|ap|+12n2n1p=1p|ap|)+α(|a2n+1|+|a2n|),n1.

    According to Lemma 3 and Lemma 4, we can compete the proof of Theorem 5. The estimates above are sharp.

    Paralleling the results of Zhu et al.[17], the corresponding results for functions of the classes defined in the paper can be obtained. For example, the estimates of distortion, growth of g and Jacobian of f and so on.

    Theorem 6. Let |z|=r[0,1).

    (1) If f=h+¯gHS,αsc(e), then

    max{αr,0}(1αr)ϕ1(r)|g(z)|α+r(1+αr)ϕ2(r), (3.15)

    where ϕ1(r) and ϕ2(r) are given by (2.22).

    Especially, let α=0, we have

    |g(z)|rexp(r+r0eη1ηdη).

    (2) If f=h+¯gHKαsc(e), then

    max{αr,0}(1αr)ψ1(r)|g(z)|(α+r)(1+αr)ψ2(r), (3.16)

    where ψ1(r) and ψ2(r) are given by (2.23).

    Especially, let α=0, we have

    |g(z)|r0exp(t+t0eη1ηdη)dt.

    Proof. According to the relation g=ωh, it is easy to see ω(z) satisfying |ω(0)|=|g(0)|=|b1|=α such that ([21]):

    |ω(z)ω(0)1¯ω(0)ω(z)||z|.

    It is easy to show

    |ω(z)ω(0)(1r2)1|ω(0)|2r2|r(1|ω(0)|2)1|ω(0)|2r2.

    A tedious calculation gives

    max{αr,0}1αr|ω(z)|α+r1+αr,zU. (3.17)

    Applying (3.17) and (2.22), we get (3.15). Similarly, applying (3.17) and (2.23), we get (3.16). Thus we complete the proof of Theorem 6.

    Using the method analogous to that in proof of Lemma 8, we can obtain the following results.

    Theorem 7. Let |z|=r[0,1).

    (1) If f=h+¯gHS,αsc(e), then

    r0max{αξ,0}(1αξ)ϕ1(ξ)dξ|g(z)|r0α+ξ(1+αξ)ϕ2(ξ)dξ,

    where ϕ1(ξ) and ϕ2(ξ) are given by (2.22).

    Especially, let α=0, we have

    |g(z)|r0ξexp(ξ+ξ0eη1ηdη)dξ.

    (2) If f=h+¯gHKαsc(e), then

    r0max{αξ,0}(1αξ)ψ1(ξ)dξ|g(z)|r0(α+ξ)(1+αξ)ψ2(ξ)dξ,

    where ψ1(ξ) and ψ2(ξ) are given by (2.23).

    Especially, let α=0, we have

    |g(z)|r0ξ0exp(t+t0eη1ηdη)dtdξ.

    Next, the Jacobian estimates and growth estimates of f are obtained.

    Theorem 8. Let |z|=r[0,1).

    (1) If f=h+¯gHS,αsc(e), then

    (1α2)(1r2)(1+αr)2ϕ21(r)Jf(z){(1α2)(1r2)(1αr)2ϕ22(r),r<α,ϕ22(r),rα.,

    where ϕ1(r) and ϕ2(r) are given by (2.22).

    (2) If f=h+¯gHKαsc(e), then

    (1α2)(1r2)(1+αr)2ψ21(r)Jf(z){(1α2)(1r2)(1αr)2ψ22(r),r<α,ψ22(r),rα,

    where ψ1(r) and ψ2(r) are given by (2.23).

    Proof. It is well known that Jacobian of f=h+¯g is

    Jf(z)=|h(z)|2|g(z)|2=|h(z)|2(1|ω(z)|2), (3.18)

    where ω satisfying g=ωh with ω(0)=0 and |ω(z)|<1 for zU.

    Let fHS,αsc(e), plugging (3.17) and (2.22) back into (3.18), we get

    Jf(z)(1α2)(1r2)(1+αr)2exp(2r+2r0eη1ηdη),

    and

    Jf(z)exp(2r+2r0eη1ηdη)(1(max{(αr),0})2(1αr)2)={exp(2r+2r0eη1ηdη)(1α2)(1r2)(1αr)2,r<α,exp(2r+2r0eη1ηdη),rα.

    Thus this completes the proof of (1). Plugging (3.17) and (2.23) back into (3.18), (2) of Theorem 8 can be proved by the same method as employed before.

    Theorem 9. Let |z|=r,0r<1.

    (1) If f=h+¯gHS,αsc(e), then

    r0(1α)(1ξ)(1+αξ)ϕ1(ξ)dξ|f(z)|r0(1+α)(1+ξ)(1+αξ)ϕ2(ξ)dξ, (3.19)

    where ϕ1(ξ) and ϕ2(ξ) are given by (2.22).

    (2) If f=h+¯gHKαsc(e), then

    r0(1α)(1ξ)(1+αξ)ψ1(ξ)dξ|f(z)|r0(1+α)(1+ξ)(1+αξ)ψ2(ξ)dξ, (3.20)

    where ψ1(ξ) and ψ2(ξ) are given by (2.23).

    Proof. For any point z=reiθU, let Ur={zU:|z|<r} and denote

    d=minzUr|f(Ur)|.

    It is easy to see that U(0,d)f(Ur)f(U). Thus, there is zrUr satisfying d=|f(zr)|. Let L(t)=tf(zr) for t[0,1], then (t)=f1(L(t)) is a well-defined Jordan arc. For f=h+¯gHS,αsc(β), by (2.22) and (3.17), we get

    d=|f(zr)|=L|dω|=|df|=|h(ρ)dρ+¯g(ρ)dˉρ||h(ρ)|(1|ω(ρ)|)|dρ|(1α)(1|ρ|)(1+α|ρ|)exp(|ρ|+|ρ|0eη1ηdη)|dρ|,=10(1α)(1|(t)|)(1+α|(t)|)exp(|(t)|+|(t)|0eη1ηdη)dt,r0(1α)(1ξ)(1+αξ)exp(ξ+ξ0eη1ηdη)dξ.

    Using (2.22) and (3.17), the right side of (3.19) is obtained. The remainder of proofs is similar to that in (3.20) and so we omit.

    According to (3.19) and (3.20), it follows that the covering theorems of f.

    Theorem 10. Let f=h+¯gSH.

    (1) If fHS,αsc(e), then UR1f(U), where

    R1=10(1α)(1ξ)(1+αξ)exp(ξ+ξ0eη1ηdη)dξ.

    (2) If fHKαsc(e), then UR2f(U), where

    R2=10(1α)(1ξ)ξ(1+αξ)ξ0exp(t+t0eη1ηdη)dtdξ.

    In this paper, with the help of the analytic part h satisfying certain conditions, we obtain the coefficients estimates of the co-analytic part g and the geometric properties of harmonic functions. Applying the methods in the paper, the geometric properties of the co-analytic part and harmonic function with the analytic part satisfying other conditions can be obtained, which can enrich the research field of univalent harmonic mapping.

    This work was supported by Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region(Grant No. NJZY20198; Grant No. NJZZ19209), Natural Science Foundation of Inner Mongolia Autonomous Region of China (Grant No. 2020MS01011; Grant No. 2019MS01023; Grant No. 2018MS01026) and the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region Under Grant (Grant No. NJYT-18-A14).

    The authors agree with the contents of the manuscript, and there is no conflict of interest among the authors.



    [1] P. L. Duren, Univalent Functions (Grundlehren der Mathematischen Wissenschaften 259), Springer-Verlag, New York, 1983.
    [2] H. M. Srivastava, S. Owa, Current Topics in Analytic Function Theory, World Scientific, London, 1992.
    [3] M. Fekete, G. Szegö, Eine bemerkung uber ungerade schlichte funktionen, J. Lond. Math. Soc., s1-8 (1933), 85-89.
    [4] W. Ma, D. Minda, A unified treatment of some special classes of univalent functions, In: Proceeding of the Conference on Complex Analysis, International Press, Boston, USA, 1994, 157-169.
    [5] R. M. EL-Ashwah, A. H. Hassan, Fekete-Szegö inequalities for certain subclass of analytic functions defined by using Sălăgean operator, Miskolc Math. Notes, 17 (2017), 827-836. doi: 10.18514/MMN.2017.1495
    [6] W. Koepf, On the Fekete-Szegö problem for close-to-convex functions, P. Am. Math. Soc., 101 (1987), 89-95.
    [7] F. M. Sakar, S. Aytaş, H. Ö. Güney, On The Fekete-Szegö problem for generalized class Mα,γ(β) defined by differential operator, J. Nat. Appl. Sci., Süleyman Demirel University, 20 (2016), 456-459.
    [8] W. Janowski, Some extremal problems for certain families of analytic functions, Ann. Pol. Math., 28 (1973), 297-326. doi: 10.4064/ap-28-3-297-326
    [9] R. Mendiratta, S. Nagpal, V. Ravichandran, On a subclass of strongly starlike functions associated with exponential function, B. Malays. Math. Sci. Soc., 38 (2015), 365-386. doi: 10.1007/s40840-014-0026-8
    [10] K. Sakaguchi, On a certain univalent mapping, J. Math. Soc. JPN, 11 (1959), 72-75. doi: 10.2969/jmsj/01110072
    [11] R. El-Ashwah, D. Thomas, Some subclasses of closed-to-convex functions, J. Ramanujan Math. Soc., 2 (1987), 85-100.
    [12] J. Clunie, T. Sheil-Small, Harmonic univalent functions, Ann. Acad. Sci. Fenn. Math., 9 (1984), 3-25. doi: 10.5186/aasfm.1984.0905
    [13] P. L. Duren, Harmonic mappings in the plane, Cambridge University Press, England, 2004.
    [14] H. Lewy, On the non-vanishing of the Jacobian in certain one-to-one mappings, B. Am. Math. Soc., 42 (1936), 689-692. doi: 10.1090/S0002-9904-1936-06397-4
    [15] D. Klimek, A. Michalski, Univalent anti-analytic perturbations of convex analytic mappings in the unit disc, Ann. Univ. Mariae Curie-Skłodowska Sect. A, 61 (2007), 39-49.
    [16] I. Hotta, A. Michalski, Locally one-to-one harmonic functions with starlike analytic part, arXiv: 1404.1826, 2014.
    [17] M. Zhu, X. Huang, The distortion theorems for harmonic mappings with analytic parts convex or starlike functions of order β, J. Math., 2015 (2015), 460191.
    [18] G. Kohr, I. Graham, Geometric Function Theory in One and Higher Dimensions, Marcel Dekker, New York, 2003.
    [19] W. Rogosinski, On the coefficients of subordinate functions, P. Lond. Math. Soc., 2 (1945), 48-82.
    [20] Y. Polatoğlu, M. Bolcal, A. Sen, Two-point distortion theorems for certain families of analytic functions in the unit disc, Int. J. Math. Math. Sci., 66 (2003) 4183-4193.
    [21] G. M. Goluzin, Geometric Theory of Functions of a Complex Variable, American Mathematical Society, Rhode Island, 1969.
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