# Linear Combinations of Univalent Mappings with Complex Coefficients

Authors: Zhihong Liu; Yulin Cai; Ziwei Wang; Lei Shi; Yingchun Li
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we study the linear combinations of two univalent mappings with complex coefficients, and we derive several sufficient conditions that the linear combinations of these mappings are locally univalent and convex in one direction. Several examples are constructed to demonstrate the main results.


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## Body

## 1. Introduction

Call a complex-valued functionf=u+iv is harmonic in the unit disk D=z∈ℂ:z<1, where u and v are real-valued harmonic functions in D. Note that any harmonic function f has decomposition f=h+g¯, where h and g are analytic in D. The functions h and g are called the analytic and coanalytic parts of f, respectively. The Jacobian of f=h+g¯ is given by Jfz=h′z2−g′z2. Lewy [1] proved that every harmonic function f=h+g¯ is locally univalent and sense-preserving in D if and only if Jfz>0 in D which is equivalent to the existence of an analytic function ωz=g′z/h′z in D such that ωz<1 for all z∈D. The function ωz is called the dilatation of f.Letℋ denote the class of all locally univalent and sense-preserving harmonic mappings f=h+g¯ defined on D. Also, let SH denote the subclass of ℋ normalized by the conditions f0=fz0−1=0. Moreover, let SH0 denote the subclass of SH with fz¯0=0.A domainΩ is convex in the direction μ∈0,π if for every a∈ℂ, the set Ω∩a+teiμ:t∈ℝ is either connected or empty. In particular, a domain is convex in the horizontal direction (CHD) if every line parallel to the real axis has a connected intersection with Ω. A function f∈ℋ is convex in the horizontal (or imaginary) direction if it maps D onto a domain Ω convex in the horizontal (or imaginary) direction.For two analytic functionsf1 and f2 and a real number λ0≤λ≤1, the function f=λf1+1−λf2 is called the linear combination of f1 and f2. Generally, the function f may not possess the univalency or other geometry properties by f1 and f2; one can refer to the survey paper by Campbell [2].In 2001, Morgan [3] investigated the conditions for the linear combination of an analytic function and a harmonic mapping to be locally univalent and sense-preserving. The following theorem gives a criterion for when the mapping will be locally univalent and sense-preserving.Theorem 1 (see [3], Theorem 3.1.7).
Letf=λf1z+1−λf2z, where f1z is an analytic mapping and f2z is a sense-preserving harmonic mapping, for fixed λ, 0≤λ≤1. Then, f is a locally univalent and sense-preserving harmonic mapping if 21−λReh2′z/f1′z¯>−λ.The linear combination of two harmonic mappingsf1=h1+g1¯ and f2=h2+g2¯ can be written as(1)f=λf1+1−λf2=λh1+1−λh2+λg1+1−λg2¯,0≤λ≤1.For this case, Dorff [4] presented some sufficient conditions for the linear combination f=λf1+1−λf2 of two harmonic mappings f1 and f2 to be univalent and convex in the direction of the imaginary axis provided the dilatation of f1 and f2 is the same. Several other works on this topic have been studied in [5–10] and the references cited therein.So far, there are only few results on the complex coefficients of the linear combination. Recently, Khurana et al. [11] investigated the linear combination of two harmonic mappings by taking the constant λ as a complex number instead of the real one. They determined the radius of close-to-convexity of f and established some sufficient conditions for f to be locally univalent and sense-preserving. Moreover, they presented the sufficient condition for the linear combination of some univalent harmonic mappings with complex number λ to be convex in the horizontal direction.Theorem 2 (see [11], Theorem 2.4).
Forj=1,2, let fj=hj+gj¯∈SH such that(2)h1−e−2iαg1=∫0ze−iα1+ζeiθ1+ζe−iθdζ,θ∈ℝ,h2−e−2iβg2=∫0ze−iβ1+ζeiθ1+ζe−iθdζ,θ∈ℝ,where α=argλ and β=arg1−λ. Then, F=λf1+1−λf2 is convex in the horizontal direction provided it is locally univalent.In this paper, motivated by the work carried out in [11–13], we study the linear combination f=λf1+1−λf2 in the cases when λ takes some complex number.In particular, we show that the combinationf=λf1+1−λf2 of the mappings fj=hj+gj¯∈SH,j=1,2, satisfying hj′−e2iμ−αjgj′=e−iαjψμ,νzpjz, is univalent and convex in thedirection μ for some specific dilatations of f1 and f2, where Repjz>0, and ψμ,νz is given by(3)ψμ,νz=11−2ze−iμcosν+z2e−2iμ,μ,ν∈0,2π.The following result by Royster and Ziegler [14] is used to check the convexity in a particular direction of analytic mappings.Lemma 1.
Letϕ be a nonconstant analytic mapping in D. Then, ϕ maps D onto a domain convex in the direction γ0≤γ<π if and only if there are real numbers μ and ν0≤ν<2π such that(4)Reeiμ−γ1−2ze−iμcosν+z2e−2iμϕ′z≥0,z∈D.The well-known shear construction by Clunie and Sheil-Small [15] is used to check the convexity in a particular direction of harmonic mappings.Lemma 2.
A locally univalent and sense-preserving harmonic mappingf=h+g¯ on D is univalent and maps D onto a domain convex in the direction γ0≤γ<π if and only if the analytic mapping h−e2iγg is univalent and maps D onto a domain convex in the direction γ.


## 2. Main Results

Firstly, we obtain the univalency criterion of the linear combination of an analytic function and a harmonic mapping for a complex numberλ, which is a generalization of Theorem 1.Theorem 3.
Letf=λf1+1−λf2, where f1 is an analytic mapping and f2=h2+g2¯ is a locally univalent and sense-preserving harmonic mapping, for fixed complex number λ. Then, f is a locally univalent and sense-preserving harmonic mapping if(5)λ2+2Reλ1−λh2′f1′¯>0.Proof.
Observe thatf=λf1+1−λh2+1−λg2¯≕h+g¯ is locally univalent and sense-preserving provided g′<h′. It is equivalent to(6)1−λg2′2<λf1′+1−λh2′2.
And thus,(7)1−λ2g2′2<λ2f1′2+1−λ2h2′2+2Reλf1′1−λh2′¯.
Sincef2 is locally univalent and sense-preserving, g2′<h2′, it is sufficient to have(8)λ2f1′2+2Reλf1′1−λh2′¯>0.
Dividing both sides byf1′2, we get inequality (5). This completes the theorem.The next two theorems deal with the linear combinations of two harmonic mappings with a complex numberλ.Theorem 4.
Forj=1,2, let fj=hj+gj¯∈SH such that(9)h1−e2iμ−αg1=∫0ze−iαψμ,νζp1ζdζ,h2−e2iμ−βg2=∫0ze−iβψμ,νζp2ζdζ,where α=argλ, β=arg1−λ, pj is an analytic mapping with Repj>0 on D, and ψμ,ν is given by (3). Then, f=λf1+1−λf2 is convex in the direction μ for a complex constant λ if it is locally univalent and sense-preserving.Proof.
Since(10)f=λf1+1−λf2=λh1+1−λh2+λ¯g1+1−λ¯g2¯≕h+g¯,thus(11)F=h−e2iμg=λh1+1−λh2−e2iμλ¯g1+1−λ¯g2=λh1−e2iμλ¯g1+1−λh2−e2iμ1−λ¯g2.
Settingλ=λeiα and 1−λ=1−λeiβ, we get(12)F=λeiαh1−e2iμ−βg1+1−λeiβh2−e2iμ−βg2.
Sinceψμ,νz=1/1−2ze−iμcosν+z2e−2iν, it follows that(13)ReF′zψμ,νza=λReeiαh1′−e2iμ−αg1′ψμ,νz+1−λReeiβh2′−e2iμ−βg2′ψμ,νz=λRep1+1−λRep2.
Then, by Lemma1, F=h−e2iμg is convex in the direction μ. Subsequently, in view of Lemma 2, we know that f=h+g¯ is univalent and convex in the direction μ.Remark 1.
In Theorem4, if we set μ=0 and p1z=p2z=1, it is reduced to Theorem 2. Also, Theorem 4 provides an extension of [12], Theorem 2.1, for the case of a real number λ.Theorem 5.
Forj=1,2, let fj=hj+gj¯∈SH and satisfy(14)h1−e2iμ−αg1=∫0ze−iαψμ,νζp1ζdζ,(15)h2−e2iμ−βg2=∫0ze−iβψμ,νζp2ζdζ,where α=argλ,β=arg1−λ, λ is a complex number, pj is an analytic mapping with Repj>0 on D, and ψμ,ν is given by (3). Let ωfj be the dilatation of fj; then, the mapping f=λf1+1−λf2 is univalent and is convex in the direction μ if ωfj and pj satisfy(1)
ωf1=ωf2 and β=α+kπk∈ℤ(2)
p1/1−e2iμ−αωf1=p2/1−e2iμ−βωf2(3)
p1=p2 and β=α+kπk∈ℤ(4)
ωf2=−ωf1, β=α+kπk∈ℤ, and Re1−e2iμ−αωf1p1/1−e2iμ−αωf2p2>0Proof.
In view of Theorem4, it is enough to show that f is locally univalent and sense-preserving. Now, differentiating both sides of (14) and (15), we have(16)h1′−e2iμ−αg1′=e−iαψμ,νp1,h2′−e2iμ−βg2′=e−iβψμ,νp2.
Sincegj′=ωfjhj′j=1,2, the above equations give(17)h1′=e−iαψμ,νp11−e2iμ−αωf1,h2′=e−iβψμ,νp21−e2iμ−βωf2.
Then,(18)ωf=g′h′=λ¯g1′+1−λ¯g2′λh1′+1−λh2′=λ¯ωf11−e2iμ−βωf2e−iαp1+1−λ¯ωf21−e2iμ−αωf1e−iβp2λ1−e2iμ−βωf2e−iαp1+1−λ1−e2iμ−αωf1e−iβp2=λωf11−e2iμ−βωf2e−2iαp1+1−λωf21−e2iμ−αωf1e−2iβp2λ1−e2iμ−βωf2p1+1−λ1−e2iμ−αωf1p2.
Letωf1=ωf2 and β=α+kπk∈ℤ; then, (18) gives that ωf=e−2iαωf1=e−2iβωf2, and hence, ωf<1.
Also, letpj and ωfj be given by (2); then, (18) gives that(19)ωf=λωf1e−2iα+1−λωf2e−2iβλ+1−λ.
Observeωf1<1 and ωf2<1; then, we have λωf1e−2iα+1−λωf2e−2iβ<λ+1−λ. So, we can get ωf<1.
Moreover, letp1=p2 and β=α+kπk∈ℤ; then, (18) gives that(20)ωf=e−2iαλωf11−e2iμ−βωf2+1−λωf21−e2iμ−αωf1λ1−e2iμ−βωf2+1−λ1−e2iμ−αωf1=e−2iαλωf11−e2iμ−αωf2+1−λωf21−e2iμ−αωf1λ1−e2iμ−αωf2+1−λ1−e2iμ−αωf1≕e−2iαω^f,where(21)ω^f=λωf11−e2iμ−αωf2+1−λωf21−e2iμ−αωf1λ1−e2iμ−αωf2+1−λ1−e2iμ−αωf1.
Therefore,ω^f<1 implies that(22)Re1+e2iμ−αω^f1−e2iμ−αω^f=λλ+1−λRe1+e2iμ−αωf11−e2iμ−αωf1+1−λλ+1−λRe1+e2iμ−αωf21−e2iμ−αωf2>0.
Hence,ωf=ω^f<1.
Lastly, letωf2=−ωf1 and β=α+kπk∈ℤ; then, from (18), we have(23)ωf=e−2iαωf1λ1+e2iμ−αωf1p1−1−λ1−e2iμ−αωf1p2λ1+e2iμ−αωf1p1+|1−λ|1−e2iμ−αωf1p2≕e−2iαωf1φ,where(24)φ=λ1+e2iμ−αωf1p1−1−λ1−e2iμ−αωf1p2λ1+e2iμ−αωf1p1+1−λ1−e2iμ−αωf1p2.
To proveωf<1, it is enough to show φ<1. Now, by the assumption in (4), we obtain(25)Re1+φ1−φ=Reλ1+e2iμ−αp11−λ1−e2iμ−αp2>0.
Hence,φ<1. This completes the proof.Remark 2.
Whenα=β+kπk∈ℤ in Theorem 5, since α=argλ and β=arg1−λ, we obtain that λ is a real number in this case, and then it is reduced to [12], Theorem 2.3.
Now, we remove some restrictions and get the following two conclusions.Theorem 6.
Forj=1,2, let fj=hj+gj¯∈SH such that(26)h1+e2iμ−αg1=∫0ze−iαψμ,νζdζ,h2+e2iμ−βg2=∫0ze−iβψμ,νζdζ,where α=argλ,β=arg1−λ, and ψμ,ν is given by (3). Then, the mapping f=λf1+1−λf2 is univalent and is convex in the direction μ for a complex constant λ if it is locally univalent and sense-preserving.Proof.
In view of (11), it follows that(27)h−e2iμg=λh1+1−λh2−e2iμλ¯g1+1−λ¯g2=λh1−λ¯e2iμg1+1−λh2−1−λ¯e2iμg2=λeiαh1−e2iμ−αg1+1−λeiβh2−e2iμ−βg2.
On the contrary, we note that(28)h1′−e2iμ−αg1′=h1′+e2iμ−αg1′h1′−e2iμ−αg1′h1′+e2iμ−αg1′=h1′+e2iμ−αg1′1−e2iμ−αωf11+e2iμ−αωf1≕e−iαψμ,νzp1z.
Similarly, one can derive(29)h2′−e2iμ−βg2′=e−iβψμ,νzp2z,where(30)p1z=1−e2iμ−αωf11+e2iμ−αωf1,p2z=1−e2iμ−βωf21+e2iμ−βωf2,satisfy Repjz>0,∀z∈D,j=1,2. Thus, from (27), we find that(31)Reh′−e2iμg′ψμ,ν=λRep1z+1−λRep2z>0,on D for a complex number λ. Hence, by Lemma 1, h−e2iμg is convex in the direction μ. The results now follow by Lemma 2.Theorem 7.
Forj=1,2, let fj=hj+gj¯∈SH satisfy (26), where ψμ,ν is given by (3), α=argλ,andβ=arg1−λ. Let ωfj be the dilatation of fj; if ωf1=e2iα−βωf2, then the mapping f=λf1+1−λf2 is convex in the direction μ for a complex number λ.Proof.
In view of Theorem6, it is enough to show that ωf<1 on D, where ωf is the dilatation of f. Differentiating (26) gives(32)h1′+e2iμ−αg1′=e−iαψμ,νz,h2′+e2iμ−βg2′=e−iβψμ,νz.
The above equations along withgj′=ωjhj′j=1,2 yield(33)h1′=e−iαψμ,νz1+e2iμ−αωf1,h2′=e−iβψμ,νz1+e2iμ−βωf2.
In view of (11) and combining with (33), ωf is given by(34)ωf=g′h′=λ¯g1′+1−λ¯g2′λh1′+1−λh2′=λ¯ωf1h1′+1−λ¯ωf2h2′λh1′+1−λh2′=λ¯e−iαωf11+e2iμ−βωf2+1−λ¯e−iβωf21+e2iμ−αωf1λe−iα1+e2iμ−βωf2+1−λe−iβ1+e2iμ−αωf1=λe−2iαωf11+e2iμ−βωf2+1−λe−2iβωf21+e2iμ−αωf1λ1+e2iμ−βωf2+1−λ1+e2iμ−αωf1.(1)
Ifβ=α+kπk∈ℤ, then ωf1=ωf2. By (34), we have ωf=e−2iαωf1. Since ωf1<1, it follows that ωf<1.(2)
Ifβ≠α+kπk∈ℤ, subset ωf1=e2iα−βωf2 into (34) yields(35)ωf=λe−2iαωf1+1−λe−2iβωf2λ+1−λ=e−2iβωf2=e−2iαωf1.
Hence,ωf<1. This completes the proof.


## 3. Examples

In this section, we give several examples to illustrate our results.Example 1.
Consider the linear combinationf=λf1+1−λf2 with λ=i; let f1 be a univalent analytic function:(36)f1z=3z−1/2z21−z2.
Suppose thatf2=h2+g2¯, where h2−g2=z/1−z, and choose ωf2=g2′/h2′=z. So, the corresponding harmonic univalent function is(37)f2z=h2z+g2z¯=z−1/2z21−z2+1/2z21−z2¯.
It is obvious that1/3f1′z=h2′z=1/1−z3; then, we have(38)λ2+2Reλ1−λh2′zf1′z¯=1−23=13>0,which verifies that (5) holds.
By Theorem3, we know that f=λf1+1−λf2 is locally univalent and sense-preserving. The images of D under f1,f2, and f=if1+1−if2 are presented in Figures 1(a)–1(c), respectively.Figure 1
Images ofD under f1,f2, and f. (a) f1. (b) f2. (c) f=if1+1−if2.
(a)(b)(c)Example 2.
Consider the linear combinationf=λf1+1−λf2; if we set λ=i, then α=argλ=π/2 and β=arg1−λ=−π/4. Take μ=ν=π/2 in (3); we get(39)ψμ,νz=11−2ze−iμcosν+z2e−2iμ=11−z2.
Then, from (14), it follows that(40)h1′z−g1′z=−ip1z1−z2.
If we takep1z=p2z=1+z/1−z and ωf1=g1′z/h1′z=−z, applying the shear construction, we have(41)h1z=−i12z1−z+14log1+z1−z,g1z=−i−12z1−z+14log1+z1−z.
The image ofD under f1z=h1z+g1z¯ is presented in Figure 2(a).
According to condition (2) in Theorem 5, we have ωf2=iωf1=g2′z/h2′z=−iz. Derivatives on both sides of (15) can be obtained:(42)h2′z+ig2′z=eπ/4ip2z1−z2.
Computingh2 and g2 as in the previous yields(43)h2z=1+i4212z1−z+14log1+z1−z,g2z=1+i4212z1−z−14log1+z1−z.
The image ofD under f2z=h2z+g2z¯ and fz=λf1z+1−λf2z with λ=i is shown in Figures 2(b) and 2(c), respectively. It follows from Theorem 5 that the mapping fz=if1z+1−if2z is convex in the imaginary direction.Figure 2
Images ofD under f1,f2, and f. (a) f1. (b) f2. (c) f=if1+1−if2.
(a)(b)(c)Example 3.
Consider the linear combinationf=λf1+1−λf2; if we set λ=1+i/2, then α=argλ=π/4 and β=arg1−λ=−π/4.
Setμ=ν=π/2 in (3); we get(44)ψμ,νz=11−2ze−iμcosν+z2e−2iμ=11−z2.
Letf1=h1+g1¯ such that(45)h1+e2iμ−αg1=∫0ze−iαψμ,νζdζ,with ωf1=g1′/h1′=z2. After a simple calculation, it yields that(46)h1z=1+i4log1+1−i/2z1−1−i/2z−24ilog1+z1−z,g1z=−1−i4log1+1−i/2z1−1−i/2z−24ilog1+z1−z.
Next, letf2=h2+g2¯ such that(47)h2+e2iμ−βg2=∫0ze−iβψμ,νζdζ,with ωf2=g2′/h2′=e−2iα−βωf1=−z2. Then,(48)h2z=−1−i4log1+1−i/2z1−1−i/2z+24log1+z1−z,g2z=1+i4log1+1−i/2z1−1−i/2z−24log1+z1−z.
Thus, the linear combination of the univalent harmonic mapping with complex coefficients can be represented as(49)fz=1+i2f1z+1−i2f2z.
Sincef1z and f2z satisfy the conditions of (26), it follows from Theorem 7 that the mapping fz=λf1z+1−λf2z is convex in the imaginary direction. The images of D under f1z,f2z, and fz=λf1z+1−λf2z with λ=1+i/2 are shown in Figures 3(a)–3(c), respectively.Figure 3
Images ofD under f1,f2, and f. (a) f1. (b) f2. (c) f=1+i/2f1+1−i/2f2.
(a)(b)(c)



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