Covering Problems for Functions \(n\) -Fold Symmetric and Convex in the Direction of the Real Axis II

Bulletin of the Malaysian Mathematical Sciences Society, Oct 2015

Let \({\mathcal {F}}\) denote the class of all functions univalent in the unit disk \(\Delta \equiv \{\zeta \in {\mathbb {C}}\,:\,\left| \zeta \right| <1\}\) and convex in the direction of the real axis. The paper deals with the subclass \({\mathcal {F}}^{(n)}\) of these functions \(f\) which satisfy the property \(f(\varepsilon z)=\varepsilon f(z)\) for all \(z\in \Delta \), where \(\varepsilon =e^{2\pi i/n}\). The functions of this subclass are called \(n\)-fold symmetric. For \({\mathcal {F}}^{(n)}\), where \(n\) is odd positive integer, the following sets, \(\bigcap _{f\in {\mathcal {F}}^{(n)}} f(\Delta )\)—the Koebe set and \(\bigcup _{f\in {\mathcal {F}}^{(n)}} f(\Delta )\)—the covering set, are discussed. As corollaries, we derive the Koebe and the covering constants for \({\mathcal {F}}^{(n)}\).

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Covering Problems for Functions \(n\) -Fold Symmetric and Convex in the Direction of the Real Axis II

Bull. Malays. Math. Sci. Soc. Covering Problems for Functions n-Fold Symmetric and Convex in the Direction of the Real Axis II Leopold Koczan 0 Paweł Zaprawa 0 Mathematics Subject Classification 0 0 Primary 30C45 Let F denote the class of all functions univalent in the unit disk ≡ {ζ ∈ C : |ζ | < 1} and convex in the direction of the real axis. The paper deals with the subclass F (n) of these functions f which satisfy the property f (εz) = ε f (z) for all z ∈ , where ε = e2πi/n . The functions of this subclass are called n-fold symmetric. For F (n), where n is odd positive integer, the following sets, f ∈F (n) f ( )-the Koebe set and f ∈F (n) f ( )-the covering set, are discussed. As corollaries, we derive the Koebe and the covering constants for F (n). Communicated by Saminathan Ponusammy. Covering domain; Koebe domain; Convexity in one direction; n-fold symmetry 1 Introduction Let F denote the class of all functions f which are univalent in ≡ {ζ ∈ C : |ζ | < 1}, convex in the direction of the real axis and normalized by f (0) = f (0) − 1 = 0. Recall that an analytic function f is said to be convex in the direction of the real axis if the intersection of f ( ) with each horizontal line is either a connected set or empty. For a given subclass A of F , the following sets: f ∈A f ( ) and f ∈A f ( ) are called the Koebe set for A and the covering set for A. We denote them by K A and L A, respectively. The radius of the largest disk with center at the origin contained in K A is called the Koebe constant for A. Analogously, the radius of the smallest disk with center at the origin that contains L A is called the covering constant for A. In the class F , we consider functions which satisfy the property of n-fold symmetry: f (εz) = ε f (z) for all z ∈ , where ε = e2πi/n . The subclass of F consisting of n-fold symmetric functions is denoted by F (n). By the definition, for every f ∈ F (n) a set f ( ) is n-fold symmetric, which means that f ( ) = ε f ( ). In other words, f ( ) may be obtained as the union of rotations about a multiple of 2π/n from a set f ( ) ∩ {w : arg w ∈ [0, 2π/n]}. From this reason, the following notation is useful: 2π n 0 = , j = ε j 0 , j = 1, 2, . . . , n − 1 ∗ = . The main aim of the paper is to find the Koebe set and the covering set for the class F (n) when n is an odd positive integer. Similar problems in related classes were discussed, for instance, in [1,2,5] and in the papers of the authors [3,4]. At the beginning, let us consider the general properties of the Koebe sets and the covering sets for F (n). In [4], we proved that Theorem 1 The sets K (n) and LF(n) , for n ∈ N, are symmetric with respect to both F axes of the coordinate system. Theorem 2 The sets KF(n) and LF(n) , for n ∈ N, are n-fold symmetric. To prove both the theorems, it is enough to consider functions and and Obviously, g(z) = f (z) h(z) = − f (−z). f ∈ F (n) ⇔ g, h ∈ F (n). ( 1 ) ( 2 ) ( 3 ) Moreover, if D = f ( ) then g( ) = D, h( ) = −D. Taking ( 3 ) into account, it is clear that the coordinate axes are the lines of symmetry for both the sets KF(n) and LF(n) for all n ∈ N. Furthermore, Lemma 1 Each straight line ε j/4 · {ζ = t , t ∈ R}, j = 0, 1, . . . , 4n − 1 is the line of symmetry of KF(n) and LF(n) for every positive odd integer n. Proof Let n be a positive odd integer and let D be one of the two sets: K (n) or L (n) . F F Since D is symmetric with respect to the real axis and the positive real half-axis contains one side of the sector 0, each rotation of the real axis about a multiple of 2π/n is the line of symmetry of D. Because of the equality {ζ = t , t ∈ R} · ε1/2 = {ζ = t , t ∈ R} · ε(n+1)/2 , our claim is true for all even j , j = 0, 1, . . . , 4n − 1. Let n = 4k + 1, k ≥ 1. The bisector of (n−1)/4 divides this sector into two subsectors: {w : arg w ∈ [π/2 − π/2n, π/2 + π/2n]} and {w : arg w ∈ [π/2 + π/2n, π/2 + 3π/2n]}. Hence the imaginary axis is the bisector of the former. For this reason, each rotation of the imaginary axis about a multiple of 2π/n is the line of symmetry of D. Moreover, {ζ = i t , t ∈ R} · ε1/2 = {ζ = i t , t ∈ R} · ε(n+1)/2. Hence our claim is valid also for all odd j , j = 0, 1, . . . , 4n − 1. If n = 4k + 3, k ≥ 0, then the bisector of (n−3)/4 divides this sector into two subsectors: {w : arg w ∈ [π/2 − 3π/2n, π/2 − π/2n]} and {w : arg w ∈ [π/2 − π/2n, π/2 + π/2n]}. The imaginary axis is the bisector of the latter. Similar argument to the one for n = 4k + 1 completes the proof for this choice of n. Theorem 3 The sets KF(n) and LF(n) , for positive odd integers n, are 2n-fold symmetric. pPoroinotfbLeleotnDginbge otonethoef bthoeuntwdaorsyeotsf: DKFsu(nc) horthLaFta(nr)g. ϕL0et∈w[00,=π|/w20n|]e.iϕ0 be an arbitrary It is sufficient to apply Lemma 1. Firstly, the symmetric point to w0 with respect to the straight line ε1/4 · {ζ = t , t ∈ R} is w1 = |w0|ei(π/n−ϕ0). Secondly, the symmetric point to w0 with respect to the real axis is w2 = |w0|e−iϕ0 . Both points w1, w2 also belong to the boundary of D. Consequently, which results in w1 = ε1/2 · w2 , D = ε1/2 · D. Remark 1 On the basis of this lemma, we can observe that in order to find sets K and L, it is enough to determine their boundaries only in the sector of the measure π/2n. 2 Extremal Polygons and Functions Let n be a fixed positive odd integer, n ≥ 5, and let K denote the Koebe set for F (n). We denote by W, the family of n-fold symmetric polygons W convex in the direction of the real axis and which have 2n sides and interior angles π + π/n and π − 3π/n alternately. Suppose that w∗ ∈ ∂ K ∩ ∗, where ∂ K stands for the boundary of K . According to Theorem 1, −w∗ ∈ ∂ K . Consider the straight horizontal line containing a segment I = {−λw∗ + (1 − λ)w∗ : λ ∈ ( 0, 1 )}. There are two possibilities: the intersection of this line with K is either I or the empty set. Assume now that I ⊂ K . We shall see that the second case holds only if Re w∗ = 0. Since K is n-fold symmetric, all points w∗ · ε j , j = 0, 1, . . . , n − 1 belong to ∂ K . On the one hand, w∗ ∈ ∗ means that w∗ has the greatest imaginary part among points w∗ · ε j . On the other hand, w∗ ∈ ∂ K means that there exists f∗ ∈ F (n) such that w∗ ∈ ∂ f∗( ). From the convexity of f∗ in the direction of the real axis, at least one of the two horizontal rays emanating from w∗ is disjoint from f∗( ). Since {w∗ − t : t ≥ 0} ∩ K = I , it is a ray l = {w∗ + t : t ≥ 0} is disjoint from f∗( ). Taking into account the n-fold symmetry of f∗, all rays l · ε j , j = 0, 1, . . . , n − 1 are disjoint from f∗( ). Observe that the point w∗ · ε(n+1)/2 has the lowest imaginary part among points w∗ · ε j , j = 0, 1, . . . , n − 1. Only if arg w∗ = π/2, this point is one of two points with the same imaginary part. The convexity in the direction of the real axis of f∗ implies tIhfathtiosnreayofistwoofthhoerfizoormntakl1ra=ys{wem∗ a·εn(ant+in1g)/f2r+omt :wt∗ ≥·ε(0n+},1t)h/e2nis(akl1s·oεdji)s∩joifn∗t(fro)m= f∅∗(and),. conseqeuntly, f∗( ) is included in a polygon of the family W. Indeed, the rays l and k1 · ε(n−1)/2 form a sector with the vertex in w∗ and the opening angle π + π/n. From the n-fold symmetry of f∗, we obtain the polygon mentioned above. The conjugate angle to this opening angle is the vertex angle of the polygon at w∗. It is easy to check that the angle of the polygon between l · ε and k1 · ε(n+1)/2 has the measure π − 3π/n. But there is another possibility, i.e., k2 ∩ f∗( ) = ∅, where k2 = {w∗ · ε(n+1)/2 − t : t ≥ 0}. If arg w∗ = π/2, the set C \ {k2 · ε j , j = 0, 1, . . . , n − 1} consists of two parts: an unbounded part and a bounded one which is a regular n-gon, see Fig. 1. A regular polygon is convex and it can be treated as the generalization of a set of the family W. Every second side of this generalized polygon has the length 0. If arg w∗ ∈ [π/2 − π/n, π/2), then the set C \ {l · ε j , k2 · ε j , j = 0, 1, . . . , n − 1} is not bounded and it is not convex in the direction of the real axis. Since w∗ ·ε ∈ ∂ f∗( ), one of two horizontal rays emanating from this point is also disjoint from f∗( ). If m1 = {w∗ · ε + t : t ≥ 0} has no common points with f∗( ), then w∗ ∈/ ∂ f∗( ), bme2ca∩usfe∗(Im)(w=∗ε∅), w<heImre wm∗2 a=nd{wR∗e(·wε∗−ε)t <:tR≥e w0}∗.,Haecnocnetr(amd2ic·tiεojn).∩Fofr∗(this) r=ea∅so.n, In this way, we obtain 3n rays emanating from n points: w∗ ·ε j , j = 0, 1, . . . , n −1. Let us take three rays starting from w∗. These rays are: l ∗=− t{ w·ε∗−+1 t t :≥t 0}≥. T0he, } k2ε−(n+1)/2 = {w∗ − t · ε−(n+1)/2 : t ≥ 0} and m2ε−1 = {w : angles between them and the positive real half-axis are equal to: 0, −π/n, π − 2π/n. It means that l lies in the sector with the vertex in w∗ and with the sides k2ε−(n+1)/2 and m2ε−1. The opening angle of this sector is equal to π − 3π/n. Consequently, f∗( ) is included in a polygon generated by k2ε j and m2ε j , j = 0, 1, . . . , n − 1. This polygon belongs to the family W. Two examples of members of W are shown in Fig. 2. From now on, we assume that all members of W are open sets. Let fα(z) = gα(z) = 0 0 z z (1 − ζ ne−inα) n1 (1 + ζ ne−inα/3)− n dζ , α ∈ 3 (1 − ζ ne−inα) n1 (1 − ζ ne−i(nα/3+5π/3))− n dζ , α ∈ 3 We choose the principal branch of n-th root. Since the exponential function is periodic, in the above definitions we restrict the range of variability of α to the intervals of length 6π/n. The choice of these intervals depends on the properties of fα and gα. Some additional information will be given in Remark 2. The definition of the family W may be extended for n = 3. In this case, the sets belonging to W may be treated as generalized polygons. The measure of the angles is equal to 4π/3 and 0 alternately. These sets have the shape of an unbounded threepointed star, see Fig. 3. Moreover, for n = 3, the functions gα map onto these generalized polygons. Lemma 2 All functions fα, α ∈ − 3nπ , 3nπ , belong to F (n) for n = 4k + 1, k ≥ 1. Lemma 3 All functions gα, α ∈ − 2nπ , 4nπ , belong to F (n) for n = 4k + 3, k ≥ 0. Proof of Lemma 2 At the beginning, we shall show fα, α ∈ [−3π/n, 3π/n] are univalent. Observe that that the functions where and with a = nα, b = nα/3. A Möbius function p1(z) = (1 − ze−ia )/(1 + ze−ib) , a, b ∈ R satisfies the condition Re eiβ p1(z) > 0 with some β ∈ R; hence, for all n ∈ N, the inequality , p(z) = 1 − zne−ia 1/n 1 + zne−ib z h(z) = (1 + zne−ib)2/n ( 6 ) Re eiβ p(z) = Re eiβ p1(zn)1/n > 0 holds with the same β. A function h1(z) = z/(1 + ze−ib)2, b ∈ R is starlike, so is h(z) = √nh1(zn). Combining these two facts with ( 6 ), we conclude that which means that fα is close-to-convex and consequently univalent. Next, we claim that each polygon fα( ) is a set which is convex in the direction of the real axis. It is sufficient to discuss the argument of the tangent line to ∂ fα( ). Observe that arg ∂ ∂ϕ fα(eiϕ ) = arg fα(eiϕ )i eiϕ 1 = n arg 1 + ei(π+nϕ−nα) arg 1+ei(nϕ−nα/3) Let ϕ ∈ [α/3 − π/n, α). Then, π + nϕ − nα as well as nϕ − nα/3 are in [−π, π ) and from (7) we get arg ∂ ∂ϕ fα(eiϕ ) 1 3 = 2n (π + nϕ − nα)− 2n nϕ − n α 3 π π π + 2 + ϕ = 2 + 2n . The above means that the tangent for ϕ in (α/3−π/n, α) has the constant argument π/2 + π/2n. Since fα( ) is a polygon with angles measuring π + π/n and π − 3π/n alternately, the argument of the tangent line takes values π/2 + π/2n + 2 j π/n and π/2 − π/2n + 2 j π/n, j = 1, 2, . . . , n alternately. What is more, putting j = k in π/2 + π/2n + 2 j π/n, we obtain the argument equal to π and putting j = 3k + 1 in π/2 − π/2n + 2 j π/n, we obtain the argument equal to 2π . Hence two of the sides of fα( ) are horizontal; consequently fα ∈ F (n). The proof of Lemma 3 is similar. Remark 2 A polygon fα( ) has vertices in points fα(eiα)·ε j and fα(ei(α/3+π/n))·ε j , j = 0, 1, . . . , n − 1. These vertices correspond to angles measuring π + π/n and π − 3π/n, respectively. It is worth pointing out some particular cases of polygons belonging to W. For α = −3π/2n and α = 3π/2n, they become regular n-gons and for α = −3π/n, α = 0, and α = 3π/n these sets are n-pointed stars symmetric with respect to the real axes. The functions f−3π/n, f0, and f3π/n have real coefficients. In all other cases, coefficients are nonreal. These particular functions are as follows: f−3π/2n(z) = f0(z) = 0 z 0 z (1 − ζ n) n1 (1 + ζ n)− n dζ, 3 2 (1 + i ζ n)− n dζ , f3π/2n(z) = 2 (1 − i ζ n)− n dζ , 0 z and and which means that Consequently, f−3π/n (z) = f3π/n(z) = 0 z Likewise, for α = −π/2n and α = 5π/2n the sets gα( ) are regular n-gons, and for α = −2π/n, α = π/n and α = 4π/n these sets are n-pointed stars. These functions gα which map on n-pointed stars have real coefficients. Let n = 4k + 1, k ≥ 1 be fixed. Let us denote by Wα, a set fα( ) for a fixed α ∈ [−3π/n, 3π/n]. Observe that the following equalities hold: f−α(z) = fα(z) for α ∈ 3π 3π − n , n f 32πn −γ (−z) = f 32πn +γ (z) for γ ∈ 0, 3π 3π − n , n − f 32πn −γ (−z) = f 32πn +γ (z) for γ ∈ 0, 3π 2n For this reason, a polygon W as well as W and −W belong to the family W. From the geometric construction of polygons in W, it follows that the ratio of lengths of any two adjacent sides of a polygon varies from 0 to infinity as α is changing in [−3π/n, 3π/n]; in each case a polygon is convex in the direction of the real axis. Multiplying sets Wα, α ∈ [−3π/n, 3π/n] by λ > 0 we obtain all members of the set W. We have proved one part of the following lemma (the second one can be proved analogously) Lemma 4 1. W = {λ · fα( ), λ > 0, α ∈ [−3π/n, 3π/n]} for n = 4k + 1, k ≥ 1 , 2. W = {λ · gα( ), λ > 0, α ∈ [−2π/n, 4π/n]} for n = 4k + 3, k ≥ 0. 3 Koebe Sets for F (n) Let us define and and From ( 4 ) and ( 5 ) F (α) ≡ fα(eiα), α ∈ G(α) ≡ gα(eiα), α ∈ Theorem 4 The Koebe set K (n) , for a fixed n = 4k + 1, k ∈ N, is a bounded and F 2n-fold symmetric domain such that where αF is the only solution of the equation π arg F (α) = 2n in [0, 3π/2n]. Proof Let K denote the Koebe set for F (n). Let us consider a polygon Vα = fα( ) belonging to W, such that one of its vertices, let say v∗, lies in ∗ (its argument is in [π/2 − π/n, π/2]) and the interior angle at v∗ has the measure π(1 + 1/n). Suppose additionaly that w∗ is a point of the boundary of K , such that arg w∗ = arg v∗ and |w∗| < |v∗|. We denote the quotient w∗/v∗ = |w∗|/|v∗| by λ. Hence λ < 1. (9) (10) (11) (12) Since w∗ ∈ ∂ K , there exists f∗ ∈ F (n) such that f∗( ) ⊂ λVα Vα = fα( ). (13) cTohnetrreafdoircet,s (f1∗3≺). Itfαmaenadns1th=at vf∗∗(=0) w≤∗, foαr(0in) o=the1r. wCoorndsse,qwue∗nctloyincf∗ide=s wfαit,hwshoimche vertex of fα( ). Hence w∗ is equal to fα(eiα) rotated about a multiple of 2π/n, namely about 2π/n · (n − 1)/4. However, it is true only for those α, for which w∗ = F (α) · ε(n−1)/4 is in ∗. Observe that for α ∈ [−3π/n, 3π/n], we have that is, From this and (9), (12) it follows that F (−α) = F (α) , arg F (−α) = − arg F (α) . F (α) · ε(n−1)/4 : α ∈ [−αF , αF ] is the boundary of the Koebe set for F (n) in ∗. Combining this with Theorem 3, the equality (11) follows. Finally, we claim that αF is the only solution of (12) in [0, 3π/2n]. On the contrary, assume that there exist two different numbers α1, α2 ∈ [0, 3π/2n] such that or equivalently, arg F (α1) = arg F (α2) , arg fα1 (eiα1 ) = arg fα2 (eiα2 ). Wα1 ⊂ Wα2 or Wα2 ⊂ Wα1 . The sets Wα1 = fα1 ( ), Wα2 = fα2 ( ) are polygons of the family W. This and the definition of W (or Lemma 4) result in The normalization of fα1 and fα2 leads to Wα1 = Wα2 . Hence α1 = α2, a contradiction. This means that (12) has only one solution in the set [0, 3π/2n]. The above proof gives more. Namely, F is starlike for α ∈ [−3π/2n, 3π/2n]. Moreover, arg F 3π α + n 3π = arg F (α) + n . It implies that F is starlike for α ∈ [−π, π ]. Furthermore, it is not difficult to see that there do not exist two different points innotththeesectas∂eK, Fi.e(n.), t∩here∗ ewxiitsht vth1easnadmve2 ismuacghinthaartyvp1ar=t. Fv2o,rvc1o,nvt2ra∈ry∂suKpFp(on)se∩tha∗t iatnids Im v1 = Im v2. With use of an argument similar to those in the proof of previous theorem, we can see that there exist two polygons V1, V2 ∈ W with vertices v1, v2, respectively. The angles at these vertices have the same measure. Hence the sides of these polygons are pairwise parallel and if Re v1 < Re v2, then V1 ⊂ V2. This means that there exist f1, f2 ∈ F (n) that f1( ) = V1, f2( ) = V2, and f1 ≺ f2. But the normalization of f1 and f2 is the same, hence f1 = f2; a contradiction. Theorem 5 The Koebe set KF(n) , for a fixed n = 4k + 3, k ≥ 0 is a bounded and 2n-fold symmetric domain such that , (14) where αG is the only solution of the equation π arg G (α) = 2n (15) in [−π/2n, π/n]. Proof A consideration similar to the above shows that αG is the only solution of the Eq. (15) in [−π/2n, π/n]. Suppose that w∗ ∈ ∂ K ∩ ∗. The analogous argument to this in the proof of Theorem 4 yields that K is contained in some polygon W of the family W. Let g∗ be a function from F (n) for which w∗ ∈ ∂g∗( ). We have g∗( ) ⊂ W = gα( ) for some α ∈ [−π/2n, 5π/n]. For this reason g∗ ≺ gα, but taking into account the normalization of both functions, we obtain g∗ = gα. Hence w∗ = gα(eiα)·ε(n−3)/4, but only if w∗ ∈ ∗. For α ∈ [−π/2n, π/n], and so From this, (10) and (15), we conclude that G 2π n − α = εG (α) , arg G 2π n − α 2π = n − arg G (α) . G(α) · ε(n−3)/4 : α ∈ αG , is the boundary of the Koebe set for F (n) in our theorem. Now, we can derive the Koebe constant for F (n). Theorem 6 For a fixed positive odd integer n, n ≥ 3 and for every function f ∈ F (n) the disk rn , where rn = B(1/n, 1/2n + 1/2)/n √n4, is included in f ( ). The number rn cannot be increased. The symbol B stands for the Beta and r , r > 0 means r = {ζ ∈ C : |ζ | < r }. Proof According to Theorems 4 and 5, the Koebe constant is equal to ∗. Theorem 3 concludes the proof of or But Likewise, where min {|F (α)| : α ∈ [−αF , αF ]} for n = 4k + 1 , and the integrand in this expression is nonnegative; thus |F (α)| ≥ (1 − t n) n1 Re qF (α, t )dt, qF (α, t ) = (1 + t ne2inα/3)− n . 3 |G(α)| ≥ (1 − t n) n1 Re qG (α, t )dt, qG (α, t ) = (1 − t nei(2nα/3−5π/3))− n . 3 It is easy to check that for n ≥ 3 the functions p(z) = (1 ± t n z)−3/n are convex in and they have real coefficients. This means that Applying (16) for both qF and qG , we get 3 3 Re(1 ± t n z)− n ≥ (1 + t n)− n . |F (α)| ≥ q0 and |G(α)| ≥ q0 , 0 1 q0 = (1 − t n) n1 (1 + t n)− n dt. 3 (16) (17) and min {|F (α)| : α ∈ [−αF , αF ]} = |F (0)| Moreover, substituting t n = tan2(x /2) in (17), we get . Corollary 1 For a fixed positive odd integer n, n ≥ 3, the Koebe constant for F (n) is equal to rn = B(1/n, 1/2n + 1/2)/n √n4. The Koebe sets and the Koebe disks for n = 3 and n = 5 are shown in Fig. 4. Remark 3 The results established in Theorem 4 and in Corollary 1 are actually valid also for n = 1. They were obtained by Złotkiewicz and Reade in [6]. One can check that for n = 1, the function F takes the form 1 1 − t e2iα/3 F (α) = eiα 0 (1 + t e2iα/3)3 dt = 4 cos(α/3) . The Eq. (12) gives αF = 3π/4; thus the boundary of the Koebe set in the upper half-plane can be written as follows cos(2α/3) 1 u = 4 cos(α/3) , v = 2 sin(α/3) , α ∈ and This fact, we can rewrite in a different way The extremal functions fα given by ( 4 ) are of the form KF = {w ∈ C : 8|w| (|w| + | Re w|) < 1} . gα(z) = − fα(−z) , where α ∈ [−3π/4, 3π/4]. The image set fα( ) for a fixed α ∈ (−3π/4, 3π/4) coincides with the plane with a horizontal ray excluded. For α = −3π/4, 3π/4, the sets fα( ) are half-planes. Moreover, r1 = B( 1, 1 )/4 = 1/4. 4 Covering Domains for F (n) Theorem 7 The covering set LF(n) for odd n ≥ 5 is a bounded and 2n-fold symmetric domain such that ∂ LF(n) ∩ π w : arg w ∈ 0, 2n = H 0, π 2n . (18) In the proof of this theorem, we need the following lemma. Lemma 5 Let n ≥ 5 be a fixed odd integer. If f ∈ F (n) and w ∈ f ( ) ∩ ∗, then f ( ) contains a polygon W ∈ W such that W has one of its vertices at w and the interior angle at w has the measure π − 3π/n. Proof Let f ∈ F (n) and w ∈ f ( ) ∩ ∗, i.e., arg w ∈ [π/2 − π/n, π/2]. Because of the n-fold symmetry of f every point w · ε j , j = 0, 1, . . . , n − 1 belongs to f ( ). It can be easily checked that max Im w · ε j , j = 0, 1, . . . , n − 1 = Im (w) and min Im w · ε j , j = 0, 1, . . . , n − 1 = Im w · ε2k+1 . Let w1 = w · ε and w2 = w · ε2k . The point w1 has the second biggest imaginary part among points w, w · ε, . . . , w · εn−1. Likewise, w2 has the second lowest imaginary part among those points. Let, moreover, l1 and l2 stand for two horizontal rays emanating from w1 and w2: l1 = {w1 + t : t ≥ 0}, l2 = {w2 + t : t ≥ 0}, respectively. From the inequality Im w2 > Im w · ε2k+1, we conclude that the point w · ε2k+1 lies on the opposite side of the straight line which contains l2 with respect to the origin. As a consequence, w1 lies on the other side of the straight line including l2 · ε−2k with respect to the origin. Hence, two rays l1 and l2 · ε−2k have a common point, let say w0. We shall show that w0 also belongs to f ( ). Suppose, contrary to our claim, that w0 ∈/ f ( ). The points w0, w1 lie on the ray l1 and w1 ∈ f ( ). Therefore, taking into account the convexity in the direction of the real axis of f , a ray m1 = {w0 + t : t ≥ 0} is disjoint from f ( ). Since w0, w belong to l2 · ε−2k , the points w0 · ε2k , w · ε2k belong to l2. Moreover, w0 · ε2k ∈/ f ( ) and w2 ∈ f ( ). Consequently, m2 = {w0 · ε2k + t : t ≥ 0} is disjoint from f ( ), and, generally, m2ε j ∩ f ( ) = ∅, j = 0, 1, . . . , n − 1. We have proved that the rays m1 and m2ε−2k with the common vertex w0 are disjoint from f ( ). It means that the reflex sector with the vertex in w0 and these two rays as the sides has no common points with f ( ). But w1 lies in this reflex sector; hence w1 ∈/ f ( ), a contradiction. From the argument given above all points wε j , w0ε j , j = 0, 1, . . . , n − 1 belong to f ( ). Applying n-fold symmetry and the convexity of f in the direction of the real axis, we can see that a polygon W with succeeding vertices at points w, w0, wε, w0ε, . . . , wεn−1, w0εn−1 is contained in f ( ). It is easy to check that this polygon has the interior angles π − 3π/n and π + π/n alternately. For this reason, W is in W. According to Lemmas 2 and 3, every function in F (n) mapping onto a polygon of the family W has the form ( 4 ) and ( 5 ) with appropriately taken α. These functions may be written in the form fβ (z) = gβ (z) = 0 0 z z (1 + ζ ne−3inβ ) n1 (1 − ζ ne−inβ )− n dζ, β ∈ 3 (1 + ζ ne−3inβ ) n1 (1 − ζ ne−inβ )− n dζ, β ∈ 3 , , (19) (20) equivalent to ( 4 ) and ( 5 ). In fact, the functions defined by ( 4 ) and (19) are connected by the relation β = α/3 + π/n and the functions in ( 5 ) and (20) are connected by β = α/3 + 5π/3n. Let us define and H (β) = fβ (eiβ ) for β ∈ 0, H (β) = gβ (eiβ ) for β ∈ 2π n π 3π , n n , . (21) (22) (23) Hence Observe that Furthermore, 0 1 H (β) ≡ eiβ Proof of Theorem 7 Let L denote the covering set for F (n). We additionaly assume that n = 4k + 1, k ≥ 1. The proof for the case n = 4k + 3, k ≥ 0 is almost similar. Let us consider a polygon Wβ = fβ ( ) belonging to W, such that one of its vertices, let say w∗, lies in ∗ and the interior angle at w∗ has the measure π(1 − 3/n). Suppose additionaly that v∗ is a point of the boundary of L such that arg v∗ = arg w∗ and |v∗| > |w∗|. We denote the quotient v∗/w∗ = |v∗|/|w∗| by μ. Hence μ > 1. Since v∗ ∈ ∂ L, there exists f ∗ ∈ F (n) such that v∗ is a boundary point of f ∗( ). From Lemma 5 f ∗( ) ⊃ μWβ Wβ = fβ ( ). Therefore, fβ ≺ f ∗ and 1 = fβ (0) ≤ f ∗ (0) = 1. Consequently fβ = f ∗, which contradicts (23). It means that w∗ = v∗, or in other words, v∗ coincides with some vertex of fβ ( ). Hence v∗ is equal to fβ (eiβ ) rotated about a multiple of 2π/n, namely about 2π/n · (n − 1)/4. It is enough to take such β that v∗ = H (β) · ε(n−1)/4 is in ∗. From this, we conclude that β ∈ [0, π/2n]. Theorem 8 For a fixed odd integer n ≥ 5 and for every function f ∈ F (n), the set f ( ) is included in Rn , where Rn = B(1/n, 1/2 − 3/2n)/n √n4. The number Rn cannot be decreased. Proof We have 0 1 |H (β)| ≤ (1 + t ne−2inβ ) n1 (1 − t n)− n dt ≤ 3 1 (1 + t n) n1 0 (1 − t n) n3 dt = |H (0)| . It can be shown that H (0) = B(1/n, 1/2 − 3/2n)/n √n4. Corollary 2 For a fixed odd integer n ≥ 5, the covering constant for F (n) is equal to Rn = B(1/n, 1/2 − 3/2n)/n √n4. The results presented above are valid for positive odd integers greater than or equal to 5. In the last part of this section, we turn to the case n = 3. As it was said in Section 2 (see also Fig. 3) for n = 3 and β ∈ [π/3, π ]\{π/2, 5π/6}, the functions given by (20) map onto the polygons with the interior angles 4π/3 and and Let 0 alternately, and the vertices in points a · ε j , ∞ · a · ε j , j = 0, 1, 2 alternately, where a = gβ (eiβ ) = H (β). Both sides adjacent to every vertex in infinity are parallel. Hence gβ ( ) are star-shaped sets with three unbounded strips. The strips have the direction π/3, π , 5π/3 if β ∈ [π/3, π/2) ∪ (5π/6, π ] and 0, 2π/3, 4π/3 if β ∈ (π/2, 5π/6). The thickness of the strips is changing as β varies in β ∈ [π/3, π ] \ {π/2, 5π/6}, but when β tends to π/2 or 5π/6, the thickness of the strips tends to 0. For β = π/2 and β = 5π/6, the functions map onto the equilateral triangles symmetric with respect to the imaginary axis. The first triangle has one of its vertices in the point i c, the second one - in the point −i c, where g π (z) = 2 Theorem 9 The covering domain LF( 3 ) is an unbounded and 6-fold symmetric domain 5 j=0 LF( 3 ) = j π i e 3 · 0. Proof Let L denote the covering set for F (n) and let L∗ stand for 5j=0 e jπi/3 · 0. At the beginning, we can see that L includes six-pointed star obtained as a union of gπ/2( ) and g5π/6( ). We know that for β ∈ (π/2, 5π/6) each set gβ ( ) contains a part of a horizontal strip between two rays emanating from a/ε and a, where a = H (β). From (21), it follows that the arguments of these points vary continuously from −π/6 to π/6 for the point a/ε and from π/2 to 5π/6 for the point a. This and the symmetry of L with respect to the imaginary axis result in L∗ ⊂ L. Now, we shall prove that L ⊂ L∗. On the contrary, assume that w0 ∈/ L∗ but w0 ∈ L. It means that there exists a function f0 ∈ F ( 3 ) such that w0 ∈ f0( ). Without loss of generality, we can assume that arg w0 ∈ (0, π/6) because of Lemma 1 and Remark 1. From the 3-fold symmetry of f0, we know that w0ε, w0ε2 ∈ f0( ). Moreover, Im w0 = |w0| sin ϕ0 < |w0| sin ϕ0 + because ϕ0 = arg w0 ∈ (0, π/6). Observe that the point w1 = {w0 − t : t ≥ 0} ∩ (ε · {w0 − t : t ≥ 0}) also belongs to f0( ). If it were not the case, the points w1ε, w1ε2 would not be in f0( ) either. But w1, w1ε2 ∈ {w0 − t : t ≥ 0}. Combining w1, w1ε2 ∈/ f0( ) with w0 ∈ f0( ) yields that the segment connecting w1 and w1ε2 has no common points with f0( ). From this and the 3-fold symmetry, all three segments connecting w1, w1ε, w1ε2 and, as a consequence, the equilateral triangle T with vertices in these points, would be disjoint with f0( ), a contradiction. This means that w1, w1ε, w1ε2 ∈ f0( ), which results in T ⊂ f0( ). g 5π ( ) ⊂ T 6 and g 5π ( ) = T . 6 (24) (25) But From (24) and (25), g5π/6 is subordinated to f , but g5π/6 and f have the same normalization, a contradiction. It means that if w0 ∈ L , then w0 ∈ L ∗, which completes the proof. The covering domains for F ( 3 ) and F ( 5 ) are shown in Fig. 5. Open Access This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited. 1. Goodman , A.W. , Saff , E.B. : On univalent functions convex in one direction . Proc. Am. Math. Soc . 73 , 183 - 187 ( 1979 ) 2. Koczan , L. : Typically real functions convex in the direction of the real axis . Ann. Univ. Mariae CurieSkłodowska Sect A 43 , 23 - 29 ( 1989 ) 3. Koczan , L. , Zaprawa , P. : On functions convex in the direction of the real axis with real coefficients . Bull. Belg. Math. Soc. Simon Stevin 18 ( 2 ), 321 - 335 ( 2011 ) 4. Koczan , L. , Zaprawa , P. : Covering problems for functions n-fold symmetric and convex in the direction of the real axis . Appl. Math. Comput . 219 ( 3 ), 947 - 958 ( 2012 ) 5. Krzyz , J. , Reade , M.O. : Koebe domains for certain classes of analytic functions . J. Anal. Math . 18 , 185 - 195 ( 1967 ) 6. Reade , M.O. , Złotkiewicz , E.: On univalent functions with two preassigned values . Proc. Am. Math. Soc . 30 , 539 - 544 ( 1971 )


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Leopold Koczan, Paweł Zaprawa. Covering Problems for Functions \(n\) -Fold Symmetric and Convex in the Direction of the Real Axis II, Bulletin of the Malaysian Mathematical Sciences Society, 2015, 1637-1655, DOI: 10.1007/s40840-014-0107-8