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. 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