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New Criteria for Convex-Exponent Product of Log-Harmonic Functions

New Criteria for Convex-Exponent Product of Log-Harmonic Functions axioms Article New Criteria for Convex-Exponent Product of Log-Harmonic Functions 1 1 2, 3 Rasoul Aghalary , Ali Ebadian , Nak Eun Cho * and Mehri Alizadeh Department of Mathematics, Faculty of Science, Urmia University, Urmia 57561-51818, Iran; r.aghalary@urmia.ac.ir (R.A.); a.ebadian@urmia.ac.ir (A.E.) Department of Applied Mathematics, College of Natural Sciences, Pukyong National University, Busan 48513, Republic of Korea Department of Mathematics, Faculty of Science, Payame Noor University, Tehran 19556-43183, Iran; a.alizadeh@pnu.ac.ir * Correspondence: necho@pknu.ac.kr Abstract: In this study, we consider different types of convex-exponent products of elements of a certain class of log-harmonic mapping and then find sufficient conditions for them to be starlike log-harmonic functions. For instance, we show that, if f is a spirallike function, then choosing a 2g suitable value of g, the log-harmonic mapping F(z) = f (z)j f (z)j is a-s piralike of order r. Our results generalize earlier work in the literature. Keywords: product; log-harmonic function; convex-exponent combination; starlike and spirallike functions MSC: 30C45; 30C80 1. Introduction Let E be the open unit disk E = fz 2 C : jzj < 1g and H(E) denote the linear space of all analytic functions defined on E. Additionally, letA be a subclass consisting of f 2 H(E) such that f (0) = f (0) 1 = 0. A C -function defined in E is said to be harmonic if D f = 0, and a log-harmonic Citation: Aghalary, R.; Ebadian, A.; function f is a solution of the nonlinear elliptic partial differential equation Cho, N.E.; Alizadeh, M. New Criteria for Convex-Exponent Product of f f Log-Harmonic Functions. Axioms = a , (1) 2023, 12, 409. https://doi.org/ 10.3390/axioms12050409 where the second dilation function a 2 H(E) is such that ja(z)j < 1 for all z 2 E. In the Academic Editor: Georgia Irina Oros above formula, f means ( f ). Observe that f is log-harmonic if log f is harmonic. The au- thors in [1] have proven that, if f is a non-constant log-harmonic mapping that vanishes Received: 3 April 2023 only at z = 0, then f should be in the form Revised: 20 April 2023 Accepted: 20 April 2023 m 2mb f (z) = z jzj h(z)g(z), (2) Published: 22 April 2023 where m is a nonnegative integer, Reb > , while h and g are analytic functions in H(E) satisfying g(0) = 1 and h(0) 6= 0. The exponent b in (2) depends only on a(0) and is given by Copyright: © 2023 by the authors. 1 + a(0) Licensee MDPI, Basel, Switzerland. b = a(0) . (3) 1ja(0)j This article is an open access article distributed under the terms and We remark that f (0) 6= 0 if and only if m = 0 and that a univalent log-harmonic conditions of the Creative Commons mapping in E vanishes at the origin if and only if m = 1, that is, f has the form Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 2b f (z) = zjzj h(z)g(z), 4.0/). Axioms 2023, 12, 409. https://doi.org/10.3390/axioms12050409 https://www.mdpi.com/journal/axioms Axioms 2023, 12, 409 2 of 13 where Reb > and 0 2 / hg(E). Recently, the class of log-harmonic functions has been extensively studied by many authors; for instance, see [1–10]. The Jacobian of log-harmonic function f is given by 2 2 J (z) = j f j (1ja(z)j ) (4) f z and is positive. Therefore, all non-constant log-harmonic mappings are sense-preserving in the unit disk E. Let B denote the class of functions a 2 H(E) with ja(z)j < 1 and B denote a 2 B such that a(0) = 0. It is easy to see that, if f (z) = zh(z)g(z), then the functions h and g, and the dilation a satisfy 0 0 zg (z) zh (z) = a(z) 1 + . (5) g(z) h(z) 2b Definition 1. (See [2].) Let f = zjzj h(z)g(z) be a univalent log-harmonic mapping. We say that f is a starlike log-harmonic mapping of order a if iq ¶ arg f (re ) z f z f = Re > a, 0  a < 1 ¶q f for all z 2 E. Denote by ST (a) the class of all starlike log-harmonic mappings. L H By taking b = 0 and g(z) = 1 in Definition 1, we obtain the class of starlike analytic functions in A, which we denote by S (a). The following lemma shows the relationship of the classes ST (a) and S (a). L H 2b Lemma 1. (See [2].) Let f (z) = zjzj h(z)g(z) be a log-harmonic mapping on E, 0 2 / hg(E). zh(z) Then, f 2 ST (a) if and only if j(z) = 2 S (a). L H g(z) In [2], the authors studied the class of a s pirallike functions and proved that, if 2b f (z) = zjzj h(z)g(z) is a log-harmonic mapping on E, 0 2 / hg(E), then f is a s pirallike if z f z f z z ia Re e > 0, 0  a < 1 for all z 2 E. We remark that a simply connected domain W in C containing the origin is p p ia said to be a s pirallike, < a < if w exp(te ) 2 W for all t  0 whenever w 2 W 2 2 and that f is an a s pirallike function, if f (E) is an a-s piralike domain. Motivated by this, we define the class of a s pirallike log-harmonic mappings of order r as follows: 2b Definition 2. Let f (z) = zjzj h(z)g(z) be a univalent log-harmonic mapping on E, with 0 2 / hg(E). Then, we say that f is an a s pirallike log-harmonic mapping of order r (0  r < 1) if z f z f z z ia Re e > r cos a (z 2 E) f (z) p a for some real a(jaj < ). The class of these functions is denoted by S (r). Furthermore, we 2 L H a a define S (1) = S (r). 0r<1 L H L H Additionally, we denote by S (r) the subclass of all f 2 A such that f is a-s piralike of a a order r and S (1) = S (r). 0r<1 Axioms 2023, 12, 409 3 of 13 2b 1 Lemma 2. ([2]) If f (z) = zjzj h(z)g(z) is log-harmonic on E and 0 2 / hg(E), with Reb > , zh(z) a a then f 2 S (r) if and only if y(z) = 2 S (r). 2ia L H g(z) In the celebrated paper [11], the authors introduce a new way of studying harmonic functions in Geometric Function Theory. Additionally, many authors investigated the linear combinations of harmonic functions in a plane; see, for example, [12–14]. In Section 2 of this paper, taking the convex-exponent product combination of two elements, a specified class of new log-harmonic functions is constructed. Indeed, we show that, if f (z) = zh(z)g(z) is spirallike log-harmonic of order r, then by choosing suitable parameters of a and g, the 2g function F(z) = f (z)j f (zj is log-harmonic spirallike of order a. Additionally, in Section 3, we provide some examples that are constructed from Section 2. 2. Main Results Theorem 1. Let f (z) = zh(z)g(z) 2 ST (r), (0  r < 1) with respect to a 2 B , L H 0 a b f 2 S (g), (0  g < 1) and a, b be real numbers with a + b = 1. Then, F(z) = f (z) K(z) is starlike log-harmonic mapping of order ar + bg with respect to a, where a(s) f (s) K(z) = f(z) exp 2Re ds . 1 a(s) f(s) Proof. By definition of F, we have F f K F f K z z z z z z = a + b and = a + b . (6) F f K F f K Additionally direct computations show that 0 0 K 1 f (z) K a(z) f (z) z z = , and = . (7) K 1 a(z) f(z) 1 a(z) f(z) Now, in view of Equations (6) and (7), z z f F z K a + b a + b f f K a ˆ(z) = = = a(z) = a(z). F f f z z K z K z z a + b a + b f K f K On the other hand, zF zF z f zK z f zK z z z z z z Re = Re a + b Re a + b F f K f K z f z f zK zK z z z z = aRe + bRe f K f K > ar + bg. The above relation shows that F is a log-harmonic starlike function of order ar + bg, and the proof is complete. Theorem 2. Let f (z) = zh(z)g(z) 2 S (r) with respect to a 2 B and g be a constant with L H 2g Reg > . Then, F(z) = f (z)j f (z)j is an a s pirallike log-harmonic mapping of order r with respect to (1 + g ¯)a(z) + g ¯ a ˆ(z) = , 1 + g + ga(z) tan b+2Img where jbj < and a = tan . 2 1+2Reg Axioms 2023, 12, 409 4 of 13 Proof. By definition of F, we have 2g 1+g g F(z) = f (z)j f (z)j = z z H(z)G(z), where 1+g g g ¯ 1+g ¯ H(z) = h (z)g (z) and G(z) = h (z)g (z). With a straightforward calculation and using Equation (5), 0 0 0 zF zh (z) zg (z) zh (z) = (1 + g) 1 + + g = 1 + ((1 + g) + ga(z)), F h(z) g(z) h(z) and ! ! 0 0 0 z ¯F zh (z) zg (z) zh (z) z ¯ = g 1 + + (1 + g) = 1 + (g + (1 + g)a(z)). h(z) g(z) h(z) If we consider z ¯F (z) z ¯ F(z) a ˆ(z) = , zF (z) F(z) then ¯ ¯ g + (1 + g)a(z) a ˆ(z) = . (1 + g) + ga(z) Now, in view of ja(z)j < 1, it easy to see that ja ˆ(z)j < 1 provided that < 1, 1+g 2 2 which evidently holds jgj < j1 + gj since Reg > , and this means that F is a log- harmonic function. Additionally, by putting z H(z) y(z) = , 2ia G(z) we have 1+g g z H(z) zh(z) g(z) y(z) = = . 2ia 2ia e g ¯ 1+g ¯ e G(z) (h (z)g (z)) Then, we obtain 0 0 0 zy (z) zh (z) zg (z) ia ia ia ia ia ia e = e + [(1 + g)e ge ] [(1 + g)e ge ] y(z) h(z) g(z) zh (z) ia ia ia ia = (ge + g ¯ e ) + [(1 + g)e ge ] 1 + h(z) zg (z) ia ia [(1 + g)e ge ] . g(z) The condition on a ensures that cos a cos a ia ia ib ia ia ib (1 + g)e ge = e and (1 + g)e ge = e , cos b cos b because by letting g = g + ig , the first equality holds true if and only if 1 2 cos b cos a i(1 + 2g ) sin a cos b + i2g cos b cos a = cos a cos b i cos a sin b 1 2 or, equivalently, after simplification 2g cot b (1 + 2g ) tan a cot b = 1 2 1 Axioms 2023, 12, 409 5 of 13 or tan b + 2Img a = tan . 1 + 2Reg Thus, by hypothesis, 0 0 0 zy (z) cos a zh (z) zg (z) ia ib ib Refe g = Re e (1 + ) e > r cos a y(z) cos b h(z) g(z) and it follows that F is an a-spirallike log-harmonic mapping of order r in which the dilation is a ˆ(z). Theorem 3. Let f (z) = zh (z)g (z) 2 S (r) with k = 1, 2 and with respect to the same k k k L H a 2 B and g be a constant with Reg > . Moreover, let 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 1l Then, F(z) = F (z)F (z) is an a-spirallike log-harmonic mapping of order r with 1 2 respect to (1 + g ¯)a(z) + g ¯ a ˆ(z) = , 1 + g + ga(z) tan b+2Img where jbj < and a = tan . 2 1+2Reg Proof. According to the definitions of F and F , we have 1 2 l 2g l F (z) = ( f (z)j f (z)j ) 1 1 1+g g g 1+g 2g = (zjzj h (z)g (z)h (z)g (z)) 1 1 1 1 and 1l 2g 1l F (z) = ( f (z)j f (z)j ) 2 2 1l 1+g g g 1+g 2g = (zjzj h (z)g (z)h (z)g (z)) . 2 2 2 2 l 1l Putting the values of F and F on F, we obtain 1 2 l 1l 1+g g g 1+g 1+g g g 1+g 2g 2g F(z) = (zjzj h (z)g (z)h (z)g (z)) (zjzj h (z)g (z)h (z)g (z)) 1 1 1 1 2 2 2 2 2g = zjzj H(z)G(z), where l(1+g) lg (1l)(1+g) (1l)g H(z) = h (z) g (z) h (z) g (z) (8) 1 1 2 2 and lg l(1+g) (1l)g (1l)(1+g) G(z) = h (z) g (z) h (z) g (z) . (9) 1 1 2 2 Now, we show that the second dilation of F i.e., m(z) satisfies the condition jm(z)j < 1. For this, since F (z) F(z) m(z) = , F (z) F(z) Axioms 2023, 12, 409 6 of 13 we have F (z) F (z) 1z 2z l + (1 l) F (z) F (z) 1 2 m(z) = F (z) F (z) 1z 2z l + (1 l) F (z) F (z) 1 2 0 0 0 0 zh zg zh zg 1 1 2 2 l[g(1 + ) + (1 + g) ] + (1 l)[g(1 + ) + (1 + g) ] h g h g 1 1 2 2 0 0 0 0 zh zg zh zg 1 1 2 2 l[(1 + g)(1 + ) + g ] + (1 l)[(1 + g)(1 + ) + g ] h g h g 1 2 1 2 zh zh 1 2 l(1 + )[g + (1 + g)a(z)] + (1 l)(1 + )[g + (1 + g)a(z)] h h 1 2 0 0 zh zh 1 2 l(1 + )[(1 + g) + ga(z)] + (1 l)(1 + )[(1 + g) + ga(z)] (10) h h 1 2 0 0 zh zh 1 2 [l(1 + ) + (1 l)(1 + )][g + (1 + g)a(z)] h h 1 2 0 0 zh zh 1 2 [l(1 + ) + (1 l)(1 + )][(1 + g) + ga(z)] h h 1 2 [g + (1 + g)a(z)] [(1 + g) + ga(z)] a(z) + (1 + g) 1+g = , a(z)g (1 + g) 1 + 1+g and the condition Reg > ensures that jm(z)j < 1 in E, which implies that F is a locally univalent log-harmonic mapping. Now, to prove l 1l a F(z) = F (z)F (z) 2 S (r), 1 L H z H(z) we have to show that y(z) = 2 S (r). However, a direct calculation shows that 2ia G(z) l(1+g) lg (1l)(1+g) (1l)g [zh (z)g (z)h (z)g (z)] z H(z) 2 2 1 1 y(z) = = . 2ia l(1+g) (1l)g (1l)(1+g) e lg 2ia G(z) [h (z)g (z)h (z)g (z)] 1 1 2 2 Now, zy (z) ia y(z) 0 0 zh (z) zg (z) ia 2ia 1 2ia 1 = e 1 + l(((1 + g) e g) ((1 + g)e g) ) h (z) g (z) 1 1 0 0 zh (z) zg (z) ia 2ia 2 2ia 2 + e (1 l)(((1 + g) e g) ((1 + g)e g) ) h (z) g (z) 2 2 ia ia = ge + e g ¯ 0 0 zh (z) zg (z) ia ia 1 ia ia 1 + l ((1 + g)e e g)(1 + ) ((1 + g)e ge ) h (z) g (z) 1 1 0 0 zh (z) zg (z) ia ia ia ia 2 2 + (1 l) ((1 + g)e e g)(1 + ) ((1 + g)e ge ) . h (z) g (z) 2 2 By hypothesis, we know that cos a cos a ia ia ib ia ia ib (1 + g)e ge = e and (1 + g)e ge = e , cos b cos b Axioms 2023, 12, 409 7 of 13 so zy (z) ia Refe g y(z) 0 0 zh (z) zg (z) cos a ib 1 ib 1 = l Re e (1 + ) e cos b h (z) g (z) 1 1 0 0 zh (z) zg (z) cos a ib 2 ib 1 + (1 l) Re e (1 + ) e cos b h (z) g (z) 2 1 > r cos a and the proof is completed. Theorem 4. Let f (z) = zh (z)g (z) 2 S (r) with respect to a 2 B (k = 1, 2). Moreover, k k k k L H suppose that Reg > , 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 If " # 0 0 zh (z) zh (z) 1 2 Re (1 a (z)a (z)) 1 + 1 +  0 ( f or any z 2 E), 1 2 h (z) h (z) 1 2 then l 1l a F(z) = F (z)F (z) 2 S (r), L H 1 2 tan b+2I mg p 1 where jbj < , 0  l  1 and a = tan . 2 1+2Reg Proof. Using the same argument as in Theorem 3, we have 2g F(z) = zjzj H(z)G(z), where H(z) and G(z) are defined by Equations (8) and (9). Now, we show that the second dilation of F, i.e., m(z), satisfies the condition jm(z)j < 1. For this, since F (z) F(z) m(z) = , F (z) F(z) using a similar argument to the relation Equation (10) of Theorem 3, we have 0 0 zh zh 1 2 l(1 + )[g + (1 + g)a (z)] + (1 l)(1 + )[g + (1 + g)a (z)] 1 2 h h 1 2 jm(z)j = . 0 0 zh zh 1 2 l(1 + )[(1 + g) + ga (z)] + (1 l)(1 + )[(1 + g) + ga (z)] 1 2 h h 1 2 However, by hypothesis, we obtain zh zh 1 2 l(1 + )[(1 + g) + ga (z)] + (1 l)(1 + )[(1 + g) + ga (z)] 1 2 h h 1 2 0 0 zh zh 1 2 l(1 + )[g + (1 + g)a (z)] + (1 l)(1 + )[g + (1 + g)a (z)] 1 2 h h 1 2 2 2 0 0 zh zh 2 1 2 2 2 2 = (2Reg + 1) l 1 + (1ja j ) + (1 l) 1 + (1ja j ) 1 2 h h 1 2 0 0 zh zh 1 2 + (2Reg + 1) 2l(1 l)Re[(1 a a )(1 + )(1 + )] > 0. 1 2 h h 1 Axioms 2023, 12, 409 8 of 13 Therefore, jm(z)j < 1 in E, which implies that F is a locally univalent mapping. Moreover, by following a similar proof to that in Theorem 3, we observe that l 1l a F(z) = F (z)F (z) 2 S (r), 1 2 L H and the proof is completed. Theorem 5. Let f (z) = zh (z)g (z) be univalent log-harmonic functions with respect to k k a 2 B (k = 1, 2) and Reg > . Moreover, suppose that zh g = f (z), where k 0 k k k a (t) f (z) = zex p 2 dt t(1 a (t)) and 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 Then, l 1l a F(z) = F (z)F (z) 2 S (1) 1 L H 2Img where 0  l  1 and a = tan . 1+2Reg Proof. Since zh g = f (z), by definition of a (z) and f (z), we obtain k k k k k zh (z) 1 + = (k = 1, 2). h (z) 1 a (z) k k Let F (z) F(z) m(z) = . F (z) F(z) Using a similar argument to the relation in Equation (10) of Theorem 3, we obtain l(1 a (z))[g + (1 + g)a (z)] + (1 l)((1 a (z))[g + (1 + g)a (z)] 2 1 1 2 jm(z)j = . l(1 a (z))[(1 + g) + ga (z)] + (1 l)(1 a (z))[(1 + g) + ga (z)] 2 1 1 2 Now, jm(z)j < 1 is equivalent to y(l) := jl(1 a (z))[(1 + g) + ga (z)] + (1 l)(1 a (z))[(1 + g) + ga (z)]j 2 2 1 1 jl(1 a (z))[g + (1 + g)a (z)] + (1 l)((1 a (z))[g + (1 + g)a (z)]j 2 1 1 2 2 2 2 = (2Reg + 1)[l j1 a (z)j (1ja (z)j ) 2 1 + 2l(1 l)Re[(1 a (z))(1 a (z))(1 a (z)a (z))] 2 1 1 2 2 2 2 + (1 l) j1 a (z)j (1ja (z)j )] > 0. 1 2 However, by taking the derivative of y(l), we have y (l) = 2(2Reg + 1) h i 2 2 Re[(1 a (z))(1 a (z))(1 a (z)a (z))]j1 a (z)j (1ja (z)j ) , 2 1 1 2 1 2 which shows that y is a continuous monotonic function of l in the interval [0, 1]. Since y(0) = (2Reg + 1)j1 a (z)j (1ja (z)j ) > 0 2 1 Axioms 2023, 12, 409 9 of 13 and y(1) = (2Reg + 1)j1 a (z)j (1ja (z)j ) > 0, 1 2 we deduce that y(l) > 0 for all l 2 [0, 1], which implies that F is a locally univalent mapping. Now, to prove l 1l a F = F F 2 S (11) 1 2 L H z H(z) we have to show that y(z) = 2 S (1), where H(z) and G(z) are defined by 2ia G(z) Equations (8) and (9). A direct computation such as that in Theorem 3 shows that ia ia ia ia (1 + g)e ge (1 + g)e ge = = 1. cos a cos a Additionally, we note that 0 0 0 0 zh zg zh zg 1 1 2 2 1 + = 1 + = 1. h g h g 1 1 2 2 Using these relation and the same argument as that made in Theorem 3, we obtain z H(z) y(z) = 2 S (1), and the proof is complete. 2ia G(z) Theorem 6. Let f (z) = zh (z)g (z)(k = 1, 2) be log-harmonic functions with respect to k k a 2 B . Moreover, suppose that zh g = z and k 0 k k 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 Then, l 1l a F(z) = F (z)F (z) 2 S (1), 1 2 L H 2Img where 0  l  1 and a = tan . (1+2Reg) Proof. Since zh g = z, by definition of a (z), we obtain k k k zh (z) 1 + = (k = 1, 2). h (z) 1 + a (z) k k Using the same argument as that in Theorem 5, we obtain our result, but we omit the details. 3. Examples We provide several examples in this section. Example 1. Let Reg > and ib [cos b(1r)e 1] ib 2ib ib (1 + z) [(1r) cos be e ] (1r) cos be f (z) = z (1 + z) (1 z) . ib (1r) cos be (1 z) Then, it is easy to see that f is a b-spirallike log-harmonic mapping of order r with respect to 2ib 2g a(z) = ze . Now, Theorem 2 implies that the function F(z) = f (z)j f (z)j is a a-spirallike log-harmonic mapping of order r with respect to 2ib (1 + g ¯)ze + g ¯ a ˆ(z) = , 2ib (1 + g) ge z Axioms 2023, 12, 409 10 of 13 where tan b + 2Img a = tan . 1 + 2Reg The image in Example 1 is shown in Figure 1. Figure 1. Image of F(z) for b = 0.5, r = 1, and g = 0.25 in Example 1. Example 2. Let Reg > , 0 < a < 1, f be the function defined in Example 1 and (1+a2r) ib [cos b e 1] (1+a2r) 1+a (1+a2r) ib ib 2ib (1 + z) cos be [ cos be e ] a(1+a) 1+a f (z) = z (1 + z) (1 az) . (1+a2r) ib cos be 1+a (1 az) Then, it is easy to see that f and f are b-spirallike log-harmonic mappings of order r with 1 2 2ib respect to a (z) = a (z) = ze . Additionally, suppose that 2 1 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 Then, Theorem 3 shows that l 1l a F(z) = F (z)F (z) 2 S (r), 1 2 L H tan b+2Img where 0  l  1 and a = tan . (1+2Reg) Example 3. Let Reg > , z 1 z ¯ f (z) = j1 + zj 1 z and z 1 Re 1z f (z) = e . 1 z Firstly, we show that f and f are log-harmonic starlike functions of order 1/2 with respect 1 2 to a (z) = z and a (z) = , respectively. A direct computation shows that 2z z( f ) 1 z ¯( f ) z 1 z 1 z = , = 2 2 f 1 z f 1 z 1 1 z( f ) 2 z z( f ) z 2 z 2 z ¯ = , = . 2 2 f 2(1 z ) f 2(1 z ) 2 2 Axioms 2023, 12, 409 11 of 13 Therefore, we obtain z ¯( f ) z( f ) z ¯( f ) z( f ) 1 z ¯ 1 z 2 z ¯ 2 z = a (z) and = a (z) , 1 2 f f f f 1 1 2 2 and this means that f and f are locally univalent log-harmonic functions. Additionally, 1 2 z( f ) z ¯( f ) 1 z 1 1 1 z 1 z ¯ Re = Re + = Re > , 2 2 f 1 z 1 z 1 z 2 and z( f ) z ¯( f ) 2 z z 1 1 2 z 2 z ¯ Re = Re = Re > . 2 2 f 2(1 z ) 2(1 z ) 1 + z 2 Hence, f and f are starlike log-harmonic functions of order 1/2. Additionally, let 1 2 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 iq Since for z = re , 0 0 zh zh 1 2 Re(1 a a )(1 + )(1 + ) 1 2 h h 1 2 1 1 1jzj 1 = (1jzj )Re = Re 2 2 2 (1 z ¯) 1 z j1 zj (1 z)(1 + z) 1 r = (1 r ) > 0. i 2 j1 re qj Theorem 4 implies that l 1l a F(z) = F (z)F (z) 2 S ( ), L H 1 2 2Img where 0  l  1 and a = tan . 1+2Reg The images in Example 2–4 are shown in Figures 2–4. 1 1 Example 4. Let Reg > , a (z) = z, and h (z) = g (z) = . Moreover, let a (z) = z 1 1 1 2 2 1z and h (z) = g (z) = . Then, it is easy to verify that all conditions of Theorem 5 are satisfied. 2 2 1+z Hence, according to Theorem 5, by taking 2g zjzj F (z) = 1+2g 1+2g (1 z) (1 z ¯) and 2g zjzj F (z) = , 1+2g 1+2g (1 + z) (1 + z ¯) we have l 1l a F(z) = F (z)F (z) 2 S (1), 1 L H 2Img where 0  l  1 and a = tan . 1r+2Reg 1 1 Example 5. Let Reg > , a (z) = z and h (z) = , g(z) = 1 z. Moreover, let 1 1 2 1z a (z) = z and h (z) = , g (z) = 1 + z. Then, it is easy to verify that all conditions of 2 2 2 1+z Theorem 6 are satisfied. Hence, according to Theorem 6, by taking Axioms 2023, 12, 409 12 of 13 2g 2g zjzj (1 z ¯) zjzj (1 + z ¯) F (z) = and F (z) = , 1 2 (1 z) (1 + z) we have l 1l a F(z) = F (z)F (z) 2 S (1), 1 L H 2Img where 0  l  1 and a = tan . 1r+2Reg Figure 2. Images of f (z) and f (z) in Example 3. 1 2 Figure 3. Images of F (z) and F (z) for g = 1 + i in Example 3. 1 2 Figure 4. Image of F(z) for g = 1 + i and l = 0.5 in Example 3. Axioms 2023, 12, 409 13 of 13 4. Conclusions In this paper, we have shown that, if f (z) = zh(z)g ¯(z) is spirallike log-harmonic of 2g order r, then by choosing suitable parameters of a and g, the function F(z) = f (z)j f (zj is log-harmonic spirallike of order a. Moreover, we provide some examples for the obtained results. Author Contributions: Conceptualization: R.A. and A.E.; original draft preparation: R.A.; writing—review and editing: A.E. and N.E.C.; investigation: M.A. All authors read and approved the final manuscript. Funding: The third author was supported by the Basic Science Research Program through the National Research Foundation of the Republic of Korea (NRF) funded by the Ministry of Education, Science and Technology (grant No. 2019R1I1A3A01050861). Data Availability Statement: Not applicable. Acknowledgments: The authors thank the anonymous referees for their invaluable comments in improving the first draft of this paper. Conflicts of Interest: The authors declare no conflict of interest. References 1. Abdulhadi, Z.; Bshouty, D. Univalent functions in H H(D). Trans. Am. Math. Soc. 1988, 305, 841–849. 2. Abdulhadi, Z.; Hengartner, W. Spirallike log-harmonic mappings. Complex Var. Theory Appl. 1987, 9, 121–130. 3. Abdulhadi, Z. Close-to-starlike logharmonic mappings. Internat. J. Math. Math. Sci. 1996, 19, 563–574. [CrossRef] 4. Aydogan, M.; Polatolu, Y. A certain class of starlike log-harmonic mappings. J. Comput. Appl. Math. 2014, 270, 506–509. [CrossRef] 5. Aydogan, M. Some results on a starlike log-harmonic mapping of order alpha. J. Comput. Appl. Math. 2014, 256, 77–82. [CrossRef] 6. Abdulhadi, Z.; Ali, R.M. Univalent log-harmonic mapping in the plane. J. Abstr. Appl. 2012, 2012, 721943. 7. Abdulhadi, Z.; Alareefi, N.M.; Ali, R.M. On the convex-exponent product of log-harmonic mappings. J. Inequal. Appl. 2014, 2014, 485. [CrossRef] 8. Li, P.; Ponnusamy, S; Wang, X. Some properties of planar p-harmonic and log-p-harmonic mappings. Bull. Malays. Math. Sci. Soc. 2013, 36, 595–609. 9. Liu, Z.; Ponnusamy, S. Some properties of univalent log-harmonic mappings. Filomat 2018, 32, 5275–5288. [CrossRef] 10. Seoudy, T.; Aouf, M.K. Fekete-Szeg problem for certain subclass of analytic functions with complex order defined by q-analogue of Ruscheweyh operator. Constr. Math. Anal. 2020, 3, 36–44. 11. Clunie, J.; Sheil-Small, T. Harmonic univalent functions. Ann. Acad. Sci. Fenn. Ser. A. I Math. 1984, 9, 3–25. [CrossRef] 12. Sun, Y.; Jiang, Y.; Wang, Z. On the convex combinations of slanted half-plane harmonic mappings. Houst. J. Math. Anal. Appl. 2015, 6, 46–50. 13. Sun, Y.; Rasila, A.; Jiang, Y. Linear combinations of harmonic quasiconformal mappings convex in one direction. J. Kodai Math. Appl. 2016, 39, 1323–1334. [CrossRef] 14. Wang, Z.G.; Liu, Z.H.; Li, Y.C. On the linear combinations of harmonic univalent mappings. J. Math. Anal. Appl. 2013, 400, 452–459. [CrossRef] Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Axioms Multidisciplinary Digital Publishing Institute

New Criteria for Convex-Exponent Product of Log-Harmonic Functions

Aghalary, Rasoul; Ebadian, Ali; Cho, Nak Eun; Alizadeh, Mehri
Axioms , Volume 12 (5) – Apr 22, 2023
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axioms Article New Criteria for Convex-Exponent Product of Log-Harmonic Functions 1 1 2, 3 Rasoul Aghalary , Ali Ebadian , Nak Eun Cho * and Mehri Alizadeh Department of Mathematics, Faculty of Science, Urmia University, Urmia 57561-51818, Iran; r.aghalary@urmia.ac.ir (R.A.); a.ebadian@urmia.ac.ir (A.E.) Department of Applied Mathematics, College of Natural Sciences, Pukyong National University, Busan 48513, Republic of Korea Department of Mathematics, Faculty of Science, Payame Noor University, Tehran 19556-43183, Iran; a.alizadeh@pnu.ac.ir * Correspondence: necho@pknu.ac.kr Abstract: In this study, we consider different types of convex-exponent products of elements of a certain class of log-harmonic mapping and then find sufficient conditions for them to be starlike log-harmonic functions. For instance, we show that, if f is a spirallike function, then choosing a 2g suitable value of g, the log-harmonic mapping F(z) = f (z)j f (z)j is a-s piralike of order r. Our results generalize earlier work in the literature. Keywords: product; log-harmonic function; convex-exponent combination; starlike and spirallike functions MSC: 30C45; 30C80 1. Introduction Let E be the open unit disk E = fz 2 C : jzj < 1g and H(E) denote the linear space of all analytic functions defined on E. Additionally, letA be a subclass consisting of f 2 H(E) such that f (0) = f (0) 1 = 0. A C -function defined in E is said to be harmonic if D f = 0, and a log-harmonic Citation: Aghalary, R.; Ebadian, A.; function f is a solution of the nonlinear elliptic partial differential equation Cho, N.E.; Alizadeh, M. New Criteria for Convex-Exponent Product of f f Log-Harmonic Functions. Axioms = a , (1) 2023, 12, 409. https://doi.org/ 10.3390/axioms12050409 where the second dilation function a 2 H(E) is such that ja(z)j < 1 for all z 2 E. In the Academic Editor: Georgia Irina Oros above formula, f means ( f ). Observe that f is log-harmonic if log f is harmonic. The au- thors in [1] have proven that, if f is a non-constant log-harmonic mapping that vanishes Received: 3 April 2023 only at z = 0, then f should be in the form Revised: 20 April 2023 Accepted: 20 April 2023 m 2mb f (z) = z jzj h(z)g(z), (2) Published: 22 April 2023 where m is a nonnegative integer, Reb > , while h and g are analytic functions in H(E) satisfying g(0) = 1 and h(0) 6= 0. The exponent b in (2) depends only on a(0) and is given by Copyright: © 2023 by the authors. 1 + a(0) Licensee MDPI, Basel, Switzerland. b = a(0) . (3) 1ja(0)j This article is an open access article distributed under the terms and We remark that f (0) 6= 0 if and only if m = 0 and that a univalent log-harmonic conditions of the Creative Commons mapping in E vanishes at the origin if and only if m = 1, that is, f has the form Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 2b f (z) = zjzj h(z)g(z), 4.0/). Axioms 2023, 12, 409. https://doi.org/10.3390/axioms12050409 https://www.mdpi.com/journal/axioms Axioms 2023, 12, 409 2 of 13 where Reb > and 0 2 / hg(E). Recently, the class of log-harmonic functions has been extensively studied by many authors; for instance, see [1–10]. The Jacobian of log-harmonic function f is given by 2 2 J (z) = j f j (1ja(z)j ) (4) f z and is positive. Therefore, all non-constant log-harmonic mappings are sense-preserving in the unit disk E. Let B denote the class of functions a 2 H(E) with ja(z)j < 1 and B denote a 2 B such that a(0) = 0. It is easy to see that, if f (z) = zh(z)g(z), then the functions h and g, and the dilation a satisfy 0 0 zg (z) zh (z) = a(z) 1 + . (5) g(z) h(z) 2b Definition 1. (See [2].) Let f = zjzj h(z)g(z) be a univalent log-harmonic mapping. We say that f is a starlike log-harmonic mapping of order a if iq ¶ arg f (re ) z f z f = Re > a, 0  a < 1 ¶q f for all z 2 E. Denote by ST (a) the class of all starlike log-harmonic mappings. L H By taking b = 0 and g(z) = 1 in Definition 1, we obtain the class of starlike analytic functions in A, which we denote by S (a). The following lemma shows the relationship of the classes ST (a) and S (a). L H 2b Lemma 1. (See [2].) Let f (z) = zjzj h(z)g(z) be a log-harmonic mapping on E, 0 2 / hg(E). zh(z) Then, f 2 ST (a) if and only if j(z) = 2 S (a). L H g(z) In [2], the authors studied the class of a s pirallike functions and proved that, if 2b f (z) = zjzj h(z)g(z) is a log-harmonic mapping on E, 0 2 / hg(E), then f is a s pirallike if z f z f z z ia Re e > 0, 0  a < 1 for all z 2 E. We remark that a simply connected domain W in C containing the origin is p p ia said to be a s pirallike, < a < if w exp(te ) 2 W for all t  0 whenever w 2 W 2 2 and that f is an a s pirallike function, if f (E) is an a-s piralike domain. Motivated by this, we define the class of a s pirallike log-harmonic mappings of order r as follows: 2b Definition 2. Let f (z) = zjzj h(z)g(z) be a univalent log-harmonic mapping on E, with 0 2 / hg(E). Then, we say that f is an a s pirallike log-harmonic mapping of order r (0  r < 1) if z f z f z z ia Re e > r cos a (z 2 E) f (z) p a for some real a(jaj < ). The class of these functions is denoted by S (r). Furthermore, we 2 L H a a define S (1) = S (r). 0r<1 L H L H Additionally, we denote by S (r) the subclass of all f 2 A such that f is a-s piralike of a a order r and S (1) = S (r). 0r<1 Axioms 2023, 12, 409 3 of 13 2b 1 Lemma 2. ([2]) If f (z) = zjzj h(z)g(z) is log-harmonic on E and 0 2 / hg(E), with Reb > , zh(z) a a then f 2 S (r) if and only if y(z) = 2 S (r). 2ia L H g(z) In the celebrated paper [11], the authors introduce a new way of studying harmonic functions in Geometric Function Theory. Additionally, many authors investigated the linear combinations of harmonic functions in a plane; see, for example, [12–14]. In Section 2 of this paper, taking the convex-exponent product combination of two elements, a specified class of new log-harmonic functions is constructed. Indeed, we show that, if f (z) = zh(z)g(z) is spirallike log-harmonic of order r, then by choosing suitable parameters of a and g, the 2g function F(z) = f (z)j f (zj is log-harmonic spirallike of order a. Additionally, in Section 3, we provide some examples that are constructed from Section 2. 2. Main Results Theorem 1. Let f (z) = zh(z)g(z) 2 ST (r), (0  r < 1) with respect to a 2 B , L H 0 a b f 2 S (g), (0  g < 1) and a, b be real numbers with a + b = 1. Then, F(z) = f (z) K(z) is starlike log-harmonic mapping of order ar + bg with respect to a, where a(s) f (s) K(z) = f(z) exp 2Re ds . 1 a(s) f(s) Proof. By definition of F, we have F f K F f K z z z z z z = a + b and = a + b . (6) F f K F f K Additionally direct computations show that 0 0 K 1 f (z) K a(z) f (z) z z = , and = . (7) K 1 a(z) f(z) 1 a(z) f(z) Now, in view of Equations (6) and (7), z z f F z K a + b a + b f f K a ˆ(z) = = = a(z) = a(z). F f f z z K z K z z a + b a + b f K f K On the other hand, zF zF z f zK z f zK z z z z z z Re = Re a + b Re a + b F f K f K z f z f zK zK z z z z = aRe + bRe f K f K > ar + bg. The above relation shows that F is a log-harmonic starlike function of order ar + bg, and the proof is complete. Theorem 2. Let f (z) = zh(z)g(z) 2 S (r) with respect to a 2 B and g be a constant with L H 2g Reg > . Then, F(z) = f (z)j f (z)j is an a s pirallike log-harmonic mapping of order r with respect to (1 + g ¯)a(z) + g ¯ a ˆ(z) = , 1 + g + ga(z) tan b+2Img where jbj < and a = tan . 2 1+2Reg Axioms 2023, 12, 409 4 of 13 Proof. By definition of F, we have 2g 1+g g F(z) = f (z)j f (z)j = z z H(z)G(z), where 1+g g g ¯ 1+g ¯ H(z) = h (z)g (z) and G(z) = h (z)g (z). With a straightforward calculation and using Equation (5), 0 0 0 zF zh (z) zg (z) zh (z) = (1 + g) 1 + + g = 1 + ((1 + g) + ga(z)), F h(z) g(z) h(z) and ! ! 0 0 0 z ¯F zh (z) zg (z) zh (z) z ¯ = g 1 + + (1 + g) = 1 + (g + (1 + g)a(z)). h(z) g(z) h(z) If we consider z ¯F (z) z ¯ F(z) a ˆ(z) = , zF (z) F(z) then ¯ ¯ g + (1 + g)a(z) a ˆ(z) = . (1 + g) + ga(z) Now, in view of ja(z)j < 1, it easy to see that ja ˆ(z)j < 1 provided that < 1, 1+g 2 2 which evidently holds jgj < j1 + gj since Reg > , and this means that F is a log- harmonic function. Additionally, by putting z H(z) y(z) = , 2ia G(z) we have 1+g g z H(z) zh(z) g(z) y(z) = = . 2ia 2ia e g ¯ 1+g ¯ e G(z) (h (z)g (z)) Then, we obtain 0 0 0 zy (z) zh (z) zg (z) ia ia ia ia ia ia e = e + [(1 + g)e ge ] [(1 + g)e ge ] y(z) h(z) g(z) zh (z) ia ia ia ia = (ge + g ¯ e ) + [(1 + g)e ge ] 1 + h(z) zg (z) ia ia [(1 + g)e ge ] . g(z) The condition on a ensures that cos a cos a ia ia ib ia ia ib (1 + g)e ge = e and (1 + g)e ge = e , cos b cos b because by letting g = g + ig , the first equality holds true if and only if 1 2 cos b cos a i(1 + 2g ) sin a cos b + i2g cos b cos a = cos a cos b i cos a sin b 1 2 or, equivalently, after simplification 2g cot b (1 + 2g ) tan a cot b = 1 2 1 Axioms 2023, 12, 409 5 of 13 or tan b + 2Img a = tan . 1 + 2Reg Thus, by hypothesis, 0 0 0 zy (z) cos a zh (z) zg (z) ia ib ib Refe g = Re e (1 + ) e > r cos a y(z) cos b h(z) g(z) and it follows that F is an a-spirallike log-harmonic mapping of order r in which the dilation is a ˆ(z). Theorem 3. Let f (z) = zh (z)g (z) 2 S (r) with k = 1, 2 and with respect to the same k k k L H a 2 B and g be a constant with Reg > . Moreover, let 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 1l Then, F(z) = F (z)F (z) is an a-spirallike log-harmonic mapping of order r with 1 2 respect to (1 + g ¯)a(z) + g ¯ a ˆ(z) = , 1 + g + ga(z) tan b+2Img where jbj < and a = tan . 2 1+2Reg Proof. According to the definitions of F and F , we have 1 2 l 2g l F (z) = ( f (z)j f (z)j ) 1 1 1+g g g 1+g 2g = (zjzj h (z)g (z)h (z)g (z)) 1 1 1 1 and 1l 2g 1l F (z) = ( f (z)j f (z)j ) 2 2 1l 1+g g g 1+g 2g = (zjzj h (z)g (z)h (z)g (z)) . 2 2 2 2 l 1l Putting the values of F and F on F, we obtain 1 2 l 1l 1+g g g 1+g 1+g g g 1+g 2g 2g F(z) = (zjzj h (z)g (z)h (z)g (z)) (zjzj h (z)g (z)h (z)g (z)) 1 1 1 1 2 2 2 2 2g = zjzj H(z)G(z), where l(1+g) lg (1l)(1+g) (1l)g H(z) = h (z) g (z) h (z) g (z) (8) 1 1 2 2 and lg l(1+g) (1l)g (1l)(1+g) G(z) = h (z) g (z) h (z) g (z) . (9) 1 1 2 2 Now, we show that the second dilation of F i.e., m(z) satisfies the condition jm(z)j < 1. For this, since F (z) F(z) m(z) = , F (z) F(z) Axioms 2023, 12, 409 6 of 13 we have F (z) F (z) 1z 2z l + (1 l) F (z) F (z) 1 2 m(z) = F (z) F (z) 1z 2z l + (1 l) F (z) F (z) 1 2 0 0 0 0 zh zg zh zg 1 1 2 2 l[g(1 + ) + (1 + g) ] + (1 l)[g(1 + ) + (1 + g) ] h g h g 1 1 2 2 0 0 0 0 zh zg zh zg 1 1 2 2 l[(1 + g)(1 + ) + g ] + (1 l)[(1 + g)(1 + ) + g ] h g h g 1 2 1 2 zh zh 1 2 l(1 + )[g + (1 + g)a(z)] + (1 l)(1 + )[g + (1 + g)a(z)] h h 1 2 0 0 zh zh 1 2 l(1 + )[(1 + g) + ga(z)] + (1 l)(1 + )[(1 + g) + ga(z)] (10) h h 1 2 0 0 zh zh 1 2 [l(1 + ) + (1 l)(1 + )][g + (1 + g)a(z)] h h 1 2 0 0 zh zh 1 2 [l(1 + ) + (1 l)(1 + )][(1 + g) + ga(z)] h h 1 2 [g + (1 + g)a(z)] [(1 + g) + ga(z)] a(z) + (1 + g) 1+g = , a(z)g (1 + g) 1 + 1+g and the condition Reg > ensures that jm(z)j < 1 in E, which implies that F is a locally univalent log-harmonic mapping. Now, to prove l 1l a F(z) = F (z)F (z) 2 S (r), 1 L H z H(z) we have to show that y(z) = 2 S (r). However, a direct calculation shows that 2ia G(z) l(1+g) lg (1l)(1+g) (1l)g [zh (z)g (z)h (z)g (z)] z H(z) 2 2 1 1 y(z) = = . 2ia l(1+g) (1l)g (1l)(1+g) e lg 2ia G(z) [h (z)g (z)h (z)g (z)] 1 1 2 2 Now, zy (z) ia y(z) 0 0 zh (z) zg (z) ia 2ia 1 2ia 1 = e 1 + l(((1 + g) e g) ((1 + g)e g) ) h (z) g (z) 1 1 0 0 zh (z) zg (z) ia 2ia 2 2ia 2 + e (1 l)(((1 + g) e g) ((1 + g)e g) ) h (z) g (z) 2 2 ia ia = ge + e g ¯ 0 0 zh (z) zg (z) ia ia 1 ia ia 1 + l ((1 + g)e e g)(1 + ) ((1 + g)e ge ) h (z) g (z) 1 1 0 0 zh (z) zg (z) ia ia ia ia 2 2 + (1 l) ((1 + g)e e g)(1 + ) ((1 + g)e ge ) . h (z) g (z) 2 2 By hypothesis, we know that cos a cos a ia ia ib ia ia ib (1 + g)e ge = e and (1 + g)e ge = e , cos b cos b Axioms 2023, 12, 409 7 of 13 so zy (z) ia Refe g y(z) 0 0 zh (z) zg (z) cos a ib 1 ib 1 = l Re e (1 + ) e cos b h (z) g (z) 1 1 0 0 zh (z) zg (z) cos a ib 2 ib 1 + (1 l) Re e (1 + ) e cos b h (z) g (z) 2 1 > r cos a and the proof is completed. Theorem 4. Let f (z) = zh (z)g (z) 2 S (r) with respect to a 2 B (k = 1, 2). Moreover, k k k k L H suppose that Reg > , 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 If " # 0 0 zh (z) zh (z) 1 2 Re (1 a (z)a (z)) 1 + 1 +  0 ( f or any z 2 E), 1 2 h (z) h (z) 1 2 then l 1l a F(z) = F (z)F (z) 2 S (r), L H 1 2 tan b+2I mg p 1 where jbj < , 0  l  1 and a = tan . 2 1+2Reg Proof. Using the same argument as in Theorem 3, we have 2g F(z) = zjzj H(z)G(z), where H(z) and G(z) are defined by Equations (8) and (9). Now, we show that the second dilation of F, i.e., m(z), satisfies the condition jm(z)j < 1. For this, since F (z) F(z) m(z) = , F (z) F(z) using a similar argument to the relation Equation (10) of Theorem 3, we have 0 0 zh zh 1 2 l(1 + )[g + (1 + g)a (z)] + (1 l)(1 + )[g + (1 + g)a (z)] 1 2 h h 1 2 jm(z)j = . 0 0 zh zh 1 2 l(1 + )[(1 + g) + ga (z)] + (1 l)(1 + )[(1 + g) + ga (z)] 1 2 h h 1 2 However, by hypothesis, we obtain zh zh 1 2 l(1 + )[(1 + g) + ga (z)] + (1 l)(1 + )[(1 + g) + ga (z)] 1 2 h h 1 2 0 0 zh zh 1 2 l(1 + )[g + (1 + g)a (z)] + (1 l)(1 + )[g + (1 + g)a (z)] 1 2 h h 1 2 2 2 0 0 zh zh 2 1 2 2 2 2 = (2Reg + 1) l 1 + (1ja j ) + (1 l) 1 + (1ja j ) 1 2 h h 1 2 0 0 zh zh 1 2 + (2Reg + 1) 2l(1 l)Re[(1 a a )(1 + )(1 + )] > 0. 1 2 h h 1 Axioms 2023, 12, 409 8 of 13 Therefore, jm(z)j < 1 in E, which implies that F is a locally univalent mapping. Moreover, by following a similar proof to that in Theorem 3, we observe that l 1l a F(z) = F (z)F (z) 2 S (r), 1 2 L H and the proof is completed. Theorem 5. Let f (z) = zh (z)g (z) be univalent log-harmonic functions with respect to k k a 2 B (k = 1, 2) and Reg > . Moreover, suppose that zh g = f (z), where k 0 k k k a (t) f (z) = zex p 2 dt t(1 a (t)) and 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 Then, l 1l a F(z) = F (z)F (z) 2 S (1) 1 L H 2Img where 0  l  1 and a = tan . 1+2Reg Proof. Since zh g = f (z), by definition of a (z) and f (z), we obtain k k k k k zh (z) 1 + = (k = 1, 2). h (z) 1 a (z) k k Let F (z) F(z) m(z) = . F (z) F(z) Using a similar argument to the relation in Equation (10) of Theorem 3, we obtain l(1 a (z))[g + (1 + g)a (z)] + (1 l)((1 a (z))[g + (1 + g)a (z)] 2 1 1 2 jm(z)j = . l(1 a (z))[(1 + g) + ga (z)] + (1 l)(1 a (z))[(1 + g) + ga (z)] 2 1 1 2 Now, jm(z)j < 1 is equivalent to y(l) := jl(1 a (z))[(1 + g) + ga (z)] + (1 l)(1 a (z))[(1 + g) + ga (z)]j 2 2 1 1 jl(1 a (z))[g + (1 + g)a (z)] + (1 l)((1 a (z))[g + (1 + g)a (z)]j 2 1 1 2 2 2 2 = (2Reg + 1)[l j1 a (z)j (1ja (z)j ) 2 1 + 2l(1 l)Re[(1 a (z))(1 a (z))(1 a (z)a (z))] 2 1 1 2 2 2 2 + (1 l) j1 a (z)j (1ja (z)j )] > 0. 1 2 However, by taking the derivative of y(l), we have y (l) = 2(2Reg + 1) h i 2 2 Re[(1 a (z))(1 a (z))(1 a (z)a (z))]j1 a (z)j (1ja (z)j ) , 2 1 1 2 1 2 which shows that y is a continuous monotonic function of l in the interval [0, 1]. Since y(0) = (2Reg + 1)j1 a (z)j (1ja (z)j ) > 0 2 1 Axioms 2023, 12, 409 9 of 13 and y(1) = (2Reg + 1)j1 a (z)j (1ja (z)j ) > 0, 1 2 we deduce that y(l) > 0 for all l 2 [0, 1], which implies that F is a locally univalent mapping. Now, to prove l 1l a F = F F 2 S (11) 1 2 L H z H(z) we have to show that y(z) = 2 S (1), where H(z) and G(z) are defined by 2ia G(z) Equations (8) and (9). A direct computation such as that in Theorem 3 shows that ia ia ia ia (1 + g)e ge (1 + g)e ge = = 1. cos a cos a Additionally, we note that 0 0 0 0 zh zg zh zg 1 1 2 2 1 + = 1 + = 1. h g h g 1 1 2 2 Using these relation and the same argument as that made in Theorem 3, we obtain z H(z) y(z) = 2 S (1), and the proof is complete. 2ia G(z) Theorem 6. Let f (z) = zh (z)g (z)(k = 1, 2) be log-harmonic functions with respect to k k a 2 B . Moreover, suppose that zh g = z and k 0 k k 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 Then, l 1l a F(z) = F (z)F (z) 2 S (1), 1 2 L H 2Img where 0  l  1 and a = tan . (1+2Reg) Proof. Since zh g = z, by definition of a (z), we obtain k k k zh (z) 1 + = (k = 1, 2). h (z) 1 + a (z) k k Using the same argument as that in Theorem 5, we obtain our result, but we omit the details. 3. Examples We provide several examples in this section. Example 1. Let Reg > and ib [cos b(1r)e 1] ib 2ib ib (1 + z) [(1r) cos be e ] (1r) cos be f (z) = z (1 + z) (1 z) . ib (1r) cos be (1 z) Then, it is easy to see that f is a b-spirallike log-harmonic mapping of order r with respect to 2ib 2g a(z) = ze . Now, Theorem 2 implies that the function F(z) = f (z)j f (z)j is a a-spirallike log-harmonic mapping of order r with respect to 2ib (1 + g ¯)ze + g ¯ a ˆ(z) = , 2ib (1 + g) ge z Axioms 2023, 12, 409 10 of 13 where tan b + 2Img a = tan . 1 + 2Reg The image in Example 1 is shown in Figure 1. Figure 1. Image of F(z) for b = 0.5, r = 1, and g = 0.25 in Example 1. Example 2. Let Reg > , 0 < a < 1, f be the function defined in Example 1 and (1+a2r) ib [cos b e 1] (1+a2r) 1+a (1+a2r) ib ib 2ib (1 + z) cos be [ cos be e ] a(1+a) 1+a f (z) = z (1 + z) (1 az) . (1+a2r) ib cos be 1+a (1 az) Then, it is easy to see that f and f are b-spirallike log-harmonic mappings of order r with 1 2 2ib respect to a (z) = a (z) = ze . Additionally, suppose that 2 1 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 Then, Theorem 3 shows that l 1l a F(z) = F (z)F (z) 2 S (r), 1 2 L H tan b+2Img where 0  l  1 and a = tan . (1+2Reg) Example 3. Let Reg > , z 1 z ¯ f (z) = j1 + zj 1 z and z 1 Re 1z f (z) = e . 1 z Firstly, we show that f and f are log-harmonic starlike functions of order 1/2 with respect 1 2 to a (z) = z and a (z) = , respectively. A direct computation shows that 2z z( f ) 1 z ¯( f ) z 1 z 1 z = , = 2 2 f 1 z f 1 z 1 1 z( f ) 2 z z( f ) z 2 z 2 z ¯ = , = . 2 2 f 2(1 z ) f 2(1 z ) 2 2 Axioms 2023, 12, 409 11 of 13 Therefore, we obtain z ¯( f ) z( f ) z ¯( f ) z( f ) 1 z ¯ 1 z 2 z ¯ 2 z = a (z) and = a (z) , 1 2 f f f f 1 1 2 2 and this means that f and f are locally univalent log-harmonic functions. Additionally, 1 2 z( f ) z ¯( f ) 1 z 1 1 1 z 1 z ¯ Re = Re + = Re > , 2 2 f 1 z 1 z 1 z 2 and z( f ) z ¯( f ) 2 z z 1 1 2 z 2 z ¯ Re = Re = Re > . 2 2 f 2(1 z ) 2(1 z ) 1 + z 2 Hence, f and f are starlike log-harmonic functions of order 1/2. Additionally, let 1 2 2g 2g F (z) = f (z)j f (z)j and F (z) = f (z)j f (z)j . 1 1 1 2 2 2 iq Since for z = re , 0 0 zh zh 1 2 Re(1 a a )(1 + )(1 + ) 1 2 h h 1 2 1 1 1jzj 1 = (1jzj )Re = Re 2 2 2 (1 z ¯) 1 z j1 zj (1 z)(1 + z) 1 r = (1 r ) > 0. i 2 j1 re qj Theorem 4 implies that l 1l a F(z) = F (z)F (z) 2 S ( ), L H 1 2 2Img where 0  l  1 and a = tan . 1+2Reg The images in Example 2–4 are shown in Figures 2–4. 1 1 Example 4. Let Reg > , a (z) = z, and h (z) = g (z) = . Moreover, let a (z) = z 1 1 1 2 2 1z and h (z) = g (z) = . Then, it is easy to verify that all conditions of Theorem 5 are satisfied. 2 2 1+z Hence, according to Theorem 5, by taking 2g zjzj F (z) = 1+2g 1+2g (1 z) (1 z ¯) and 2g zjzj F (z) = , 1+2g 1+2g (1 + z) (1 + z ¯) we have l 1l a F(z) = F (z)F (z) 2 S (1), 1 L H 2Img where 0  l  1 and a = tan . 1r+2Reg 1 1 Example 5. Let Reg > , a (z) = z and h (z) = , g(z) = 1 z. Moreover, let 1 1 2 1z a (z) = z and h (z) = , g (z) = 1 + z. Then, it is easy to verify that all conditions of 2 2 2 1+z Theorem 6 are satisfied. Hence, according to Theorem 6, by taking Axioms 2023, 12, 409 12 of 13 2g 2g zjzj (1 z ¯) zjzj (1 + z ¯) F (z) = and F (z) = , 1 2 (1 z) (1 + z) we have l 1l a F(z) = F (z)F (z) 2 S (1), 1 L H 2Img where 0  l  1 and a = tan . 1r+2Reg Figure 2. Images of f (z) and f (z) in Example 3. 1 2 Figure 3. Images of F (z) and F (z) for g = 1 + i in Example 3. 1 2 Figure 4. Image of F(z) for g = 1 + i and l = 0.5 in Example 3. Axioms 2023, 12, 409 13 of 13 4. Conclusions In this paper, we have shown that, if f (z) = zh(z)g ¯(z) is spirallike log-harmonic of 2g order r, then by choosing suitable parameters of a and g, the function F(z) = f (z)j f (zj is log-harmonic spirallike of order a. Moreover, we provide some examples for the obtained results. Author Contributions: Conceptualization: R.A. and A.E.; original draft preparation: R.A.; writing—review and editing: A.E. and N.E.C.; investigation: M.A. All authors read and approved the final manuscript. Funding: The third author was supported by the Basic Science Research Program through the National Research Foundation of the Republic of Korea (NRF) funded by the Ministry of Education, Science and Technology (grant No. 2019R1I1A3A01050861). Data Availability Statement: Not applicable. Acknowledgments: The authors thank the anonymous referees for their invaluable comments in improving the first draft of this paper. Conflicts of Interest: The authors declare no conflict of interest. References 1. Abdulhadi, Z.; Bshouty, D. Univalent functions in H H(D). Trans. Am. Math. Soc. 1988, 305, 841–849. 2. Abdulhadi, Z.; Hengartner, W. Spirallike log-harmonic mappings. Complex Var. Theory Appl. 1987, 9, 121–130. 3. Abdulhadi, Z. Close-to-starlike logharmonic mappings. Internat. J. Math. 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Axioms – Multidisciplinary Digital Publishing Institute

Published: Apr 22, 2023

Keywords: product; log-harmonic function; convex-exponent combination; starlike and spirallike functions

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