Circularly Symmetric Locally Univalent Functions

Bulletin of the Malaysian Mathematical Sciences Society, Feb 2016

Let \(D\subset \mathbb {C}\) and \(0\in D\). A set D is circularly symmetric if, for each \(\varrho \in \mathbb {R}^+\), a set \(D\cap \{\zeta \in \mathbb {C}:|\zeta |=\varrho \}\) is one of three forms: an empty set, a whole circle, a curve symmetric with respect to the real axis containing \(\varrho \). A function f analytic in the unit disk \(\Delta \equiv \{\zeta \in \mathbb {C}:|\zeta |<1\}\) and satisfying the normalization condition \(f(0)=f^{\prime }(0)-1=0\) is circularly symmetric, if \(f(\Delta )\) is a circularly symmetric set. The class of all such functions is denoted by X. In this paper, we focus on the subclass \(X^{\prime }\) consisting of functions in X which are locally univalent. We obtain the results concerned with omitted values of \(f\in X^{\prime }\) and some covering and distortion theorems. For functions in \(X^{\prime }\) we also find the upper estimate of the n-th coefficient, as well as the region of variability of the second and the third coefficients. Furthermore, we derive the radii of starlikeness, convexity and univalence for \(X^{\prime }\).

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Circularly Symmetric Locally Univalent Functions

Circularly Symmetric Locally Univalent Functions Leopold Koczan 0 1 Paweł Zaprawa 0 1 Mathematics Subject Classification 0 1 B Paweł Zaprawa 0 1 0 Department of Mathematics, Lublin University of Technology , Nadbystrzycka 38D, 20-618 Lublin , Poland 1 Communicated by V. Ravichandran Let D ⊂ C and 0 ∈ D. A set D is circularly symmetric if, for each ∈ R+, a set D ∩ {ζ ∈ C : |ζ | = } is one of three forms: an empty set, a whole circle, a curve symmetric with respect to the real axis containing . A function f analytic in the unit disk ≡ {ζ ∈ C : |ζ | < 1} and satisfying the normalization condition f (0) = f (0) − 1 = 0 is circularly symmetric, if f ( ) is a circularly symmetric set. The class of all such functions is denoted by X . In this paper, we focus on the subclass X consisting of functions in X which are locally univalent. We obtain the results concerned with omitted values of f ∈ X and some covering and distortion theorems. For functions in X we also find the upper estimate of the n-th coefficient, as well as the region of variability of the second and the third coefficients. Furthermore, we derive the radii of starlikeness, convexity and univalence for X . Locally univalent functions; Radius of univalence; Radius of starlikeness; Coefficients' estimates 1 Introduction Let A denote the class of all functions analytic in the unit disk ≡ {ζ ∈ C : |ζ | < 1} which satify the condition f (0) = f (0) − 1 = 0. A function f is said to be typically real if the inequality (Im z)(Im f (z)) ≥ 0 holds for all z ∈ . The class of functions which are typically real is denoted by T˜ and the class of typically real functions which belong to A is denoted by T . For a typically real function f , z ∈ + ⇔ f (z) ∈ C+ and z ∈ − ⇔ f (z) ∈ C−. The symbols +, −, C+, C− stand for the following sets: the upper and the lower halves of the disk , the upper halfplane and the lower halfplane, respectively. Jenkins [3] established the following definitions. Definition 1 Let D ⊂ C, 0 ∈ D. A set D is circularly symmetric if, for each ∈ R+, a set D ∩ {ζ ∈ C : |ζ | = } is one of three forms: an empty set, a whole circle, a curve symmetric with respect to the real axis containing . Definition 2 A function f ∈ A is circularly symmetric if f ( ) is a circularly symmetric set. The class of all such functions is denoted by X . In fact Jenkins considered only those circularly symmetric functions which are univalent. This assumption is rather restrictive. A number of interesting problems appear while discussing non-univalent circularly symmetric functions. For these reasons, we decided to define a circularly symmetric function as in Definition 2. In this paper, we focus on the set X consisting of locally univalent circularly symmetric functions. According to Jenkins (see, [3]), if f ∈ X is univalent then z ff ((zz)) is a typically real function. Additionally, he observed that the property z ff ((zz)) ∈ T˜ does not guarantee the univalence of f . In fact, we have ( 1 ) ( 2 ) ( 3 ) f ∈ X ⇔ z f (z) f (z) ∈ T˜ . Jenkins also gave a nice geometric property of f in X . He proved that f ∈ X if and only if, for a fixed r ∈ ( 0, 1 ), a function | f (r eiϕ )| is nonincreasing for ϕ ∈ (0, π ) and nondecreasing for ϕ ∈ (π, 2π ). From ( 1 ), all coefficients of the Taylor series expansion of f ∈ X are real. In [9] the following relation between X and T was proved: f ∈ X ⇔ z ff ((zz)) = (1 + z)2 h(zz) , h ∈ T . It is known that each function of the class T can be represented by the formula h(z) = 1 z −1 1 − 2t z + z2 where μ is a probability measure on [−1, 1] (see, [7]). Applying ( 3 ) in ( 2 ) and integrating it, one can write a function f ∈ X in the form we obtain the relation between X and the set PR of functions with positive real part which have real coefficients: z f (z) 1 + z f ∈ X ⇔ f (z) = 1 − z p(z), p ∈ PR. It is known (Robertson, [6]) that the set of extreme points for T has the form 1−2zzt+z2 : t ∈ [−1, 1] . Putting these functions into formula ( 2 ) as h, we get It is easy to check that the functions f , which satisfy ( 8 ), are of the form ft (z) = z exp i cot log , t ∈ [−1, 1), 1 − zeiψ 1 − ze−iψ f (z) = z exp 0 1 2(1 + t ) −1 1 − 2t ζ + ζ 2 dμ(t ) dζ . Putting cos ψ instead of t in ( 4 ) and integrating it with respect to ζ , we get the integral representation of a function in X : z f (z) (1 + z)2 f (z) = 1 − 2zt + z2 . f1(z) = z exp 4z 1 − z 1 − zeiψ log 1 − ze−iψ dμ(ψ ) , where μ is a probability measure on [0, π ]. Applying the well-known equivalence ( 4 ) ( 5 ) ( 6 ) ( 7 ) ( 8 ) ( 9 ) ( 10 ) (11) where t = cos ψ . Obviously, t ∈ [−1, 1) ⇔ ψ ∈ (0, π ]. Furthermore, lim ψ→0+ i cot log 1 − zeiψ 1 − ze−iψ 4z , = 1 − z so we can write Besides ft , we need another family of functions belonging to X . Since the set T is convex, every linear combination of any two functions from T also belongs to T . Hence, taking 1+2t (1−z)2 + 2 (1+z)2 , t ∈ [−1, 1] as h in ( 2 ), we obtain z 1−t z Let us denote by gt the solutions of Eq. (11). From this equation gt (z) = z exp , t ∈ [−1, 1]. 2 Properties of ft and gt g−1(z) = f−1(z) = z and g1(z) = f1(z) = z exp 4z 1 − z . , Hence and Firstly, we shall describe the sets ft ( ) and gt ( ), where ft , gt are defined by ( 9 ) and (12), respectively. For ft , t ∈ ( −1, 1 ) (i.e. for ψ ∈ (0, π )), from ( 9 ) we obtain ft (eiϕ ) = exp cot arg 1 − ei(ϕ−ψ) 1 − ei(ϕ+ψ) We shall derive the argument which appears in the above expression. To do this, we need the following identity: arg 1 + eiα α = 2 − π · α + π 2π for α = (1 + 2k)π, k ∈ Z. ft (eiϕ ) = exp −ψ cot for ϕ ∈ (ψ, 2π − ψ ) ft (eiϕ ) = exp (π − ψ ) cot for ϕ ∈ [0, ψ ) ∪ (2π − ψ, 2π ]. From the above expressions, it follows that a function | ft (eiϕ )|, with fixed t ∈ ( −1, 1 ), does not depend on the variable ϕ. Moreover, | ft (eiϕ )| in (13) is less than 1 and | ft (eiϕ )| in (14) is greater than 1. Additionally, arg ft (eiϕ ) = ϕ + cot log sin ϕ+ψ 2 sin ϕ−ψ 2 which means that the curves { ft (eiϕ ), ϕ ∈ [0, φ)} and { ft (eiϕ ), ϕ ∈ (φ, π ]} wind around circles given by (13) and (14) infinitely many times. Hence, for t ∈ ( −1, 1 ), ft ( ) ⊂ w ∈ C : |w| < exp (π − ψ ) cot It is easily seen that f−1( ) = . Now, we shall show that f1 omits only one point on the real axis. Theorem 1 The condition f1(z) = −e−2 holds for all z ∈ . Proof On the contrary, suppose that there exists z ∈ such that f1(z) = −e−2. In fact, we can assume that arg z ∈ [0, π ] because the coefficients of f1 are real. For this reason, or equivalently, Let z = r eiϕ , ϕ ∈ [0, π ]. Comparing the arguments of both sides of this equality, we get (17) (18) (19) (20) (21) For a fixed r ∈ ( 0, 1 ), let us consider a function 2r 1 h(ϕ) = 1 − r 2 sin ϕ + log r (π − ϕ), ϕ ∈ [0, π ]. For ϕ ∈ [0, π ] the function h (ϕ) decreases and where z exp 2(1 + z) 1 − z = −1, 21r −sinr 2ϕ = −π l−ogϕr . 2r h (ϕ) ≥ − (1 − r 2) log r · g(r ), g(r ) = log r + Since g (r ) = −(1 − r )2/2r 2, the function g(r ) decreases for r ∈ ( 0, 1 ) and Both factors on the right-hand side of (20) are positive. Consequently, h(ϕ) increases for ϕ ∈ (0, π ), so h(ϕ) ≤ h(π ) = 0. Taking into account the last inequality, we can see that ϕ = π is the only solution of (19). It means that Eq. (18) is satisfied only for z = r eiπ = −r ; so Let us denote the left-hand side of (21) by k(r ). Since k is increasing for r ∈ ( 0, 1 ), sup{k(r ) : r ∈ ( 0, 1 )} = k( 1 ) = 0. Therefore, (21) has no solutions in the open set ( 0, 1 ), which contradicts (18), and, consequently we obtain the desired result. Applying a similar argument, one can prove the following more general theorem. Theorem 2 For gt , t ∈ (−1, 1] given by (12), (i) gt (z) = −e−1−t for z ∈ , (ii) the equation gt (z) = − e−1−t has a solution for any > 1, (iii) the equation gt (z) = −e−1−t eiθ has a solution for any θ ∈ (−π, π ). Proof ad (i) Suppose that there exists z ∈ such that gt (z) = −e−1−t . Then z exp (1 + t ) = −1, 1 + z 1 − z (22) (23) (25) (26) (27) and putting z = r eiϕ , we have 1 + r eiϕ 1 1 − r eiϕ = 1 + t (− log r + i (π − ϕ)) . Comparing the arguments on both sides, we obtain the same function h as in the proof for Theorem 4; consequently h(ϕ) ≤ 0. Therefore, Eq. (23) holds only if z = −r , but in this case 1 − r 1 log r = 0. (24) 1 + r + 1 + t Let the left-hand side of (24) be denoted by k(r ), which is an increasing and nonpositive function of r ∈ ( 0, 1 ). It yields that (24) has no solutions in the open set ( 0, 1 ), which contradicts the assumption. ad (ii) Consider an equation gt (z) = − e−1−t with fixed > 1. It takes the following form Putting z = r eiϕ into (25), we have z exp (1 + t ) = − . 1 + z 1 − z 1 + r eiϕ 1 1 − r eiϕ = 1 + t log r + i (π − ϕ) . 2r sin ϕ π − ϕ 1 − r 2 − log r Let h(ϕ) denote the left-hand side of this equality. The function h (ϕ) is decreasing; h (0) = 1 −2rr2 + log1 r > 0 and h (π ) = 1−−2rr2 + log1 r . One can easily prove that there exists only one number r0 ∈ ( 0, 1 ) such that h (π ) > 0 for r ∈ (0, r0) and h (π ) < 0 for r ∈ (r0, 1). Hence, for suitably taken r , the function h first increases and then decreases. Furthermore, h(0) = −π/ log r < 0 and h(π ) = 0. Consequently, there exists ϕ0 ∈ (0, π ) such that h(ϕ0) = 0. It means that (26) is satisfied by z0 = r eiϕ0 ; so (25) holds for z = z0. ad (iii) The proof of this part is similar to the proof of (ii). Corollary 1 gt ( ) = C\{−e−1−t } for all t ∈ (−1, 1]. Proof Let t ∈ (−1, 1] be fixed. For r ∈ ( 0, 1 ), we have gt (−r ) = −r exp(−2(1 + t )r/(1 + r )). Observe that |gt (−r )| is a continuous and increasing function of r ∈ [0, 1). For this reason, gt (−r ) achieves all values in (−e−1−t , 0]. From the definition of circularly symmetric function it follows that if c belongs to the negative real axis and c ∈ gt ( ), then the whole circle with radius |c| centered at the origin is also contained in this set. Hence, for each in [0, 1) we have {w ∈ C : |w| = e−1−t } ⊂ gt ( ). By Theorem 2, part (ii), (−∞, −e−1−t ) is contained in gt ( ). Let − e−1−t , > 1 be an arbitrary point of this ray. Applying the same argument as above, we conclude that for any > 1, {w ∈ C : |w| = e−1−t } ⊂ gt ( ). Combining these facts with points (i) and (iii) of Theorem 2 completes the proof. Let us consider a function b(ϕ) = arg gt (r eiϕ ), ϕ ∈ (0, π ) where t ∈ [−1, 1] and r ∈ ( 0, 1 ) are fixed. Analyzing the derivative of this function it can be observed that if t − 3 + 4σ 2 ≤ 0, where σ = 1 +2rr2 , then b(ϕ) increases in (0, π ). On the other hand, if t − 3 + 4σ 2 > 0 then, for ϕ ∈ (0, π ), the function b(ϕ) increases at the beginning, then it decreases, only to increase again at the end. From this observation we conclude that for small r , a set gt ({z ∈ √C : |z| < r, Im z ≥ 0}) is contained in the upper halfplane. If r is greater than rt = 2+√3−1 +tt , then this set is not contained in the upper halfplane; its boundary is wound around the origin. If r = 1, then gt (eiϕ ) = exp −1 − t + i ϕ + (1 + t ) cot ϕ 2 . gt (eiϕ ) = exp (−1 − t ) (28) (29) Hence and ϕ arg gt (eiϕ ) = ϕ + (1 + t ) cot . (30) 2 Therefore, gt ( ) is wound around the origin infinitely many times, or, more precisely, it is wound around the circle with radius e−1−t . 3 Coefficients of Functions in X To start with, let us look into the coefficients of f1 given by ( 10 ). Although it is complicated to find an explicit formula for the n-th coefficient of this function, the formula of the logarithmic coefficients γn of f1 can be easily derived. Indeed, thus where γn = 2 for all n ∈ N. The Taylor series expansion of f1 is given by = z + Denoting the n-th coefficient of f1 by An, we can write n−2 n−2 4n−1−s n − 2 , An = s=0 bs,n−1−s = s=0 (n − 1 − s)! n − 2 − s and consequently, The first four values of An are An = On the other hand, for An the formula holds for n ≥ 2. Indeed, expanding both sides of the equality 92, A5 = 2132. A2 = 4, A3 = 12, A4 = 3 nAn+1 = (2n + 2)An − (n − 2)An−1 zf1(z) = f1(z) 11 −+ zz 2 , ∞ n=1 nAnzn = ∞ Anzn · 1 + ∞ 2zk 2 . n=1 k=1 we get Comparing the coefficients at zn, we obtain (32). We shall now prove that the upper bound of n-th coefficient of a function f ∈ X is achieved when f is equal to f1. To do this, we apply the relation ( 7 ). Suppose that functions f ∈ X and p ∈ PR are of the form f (z) = and p(z) = n∞=0 pn zn with a1 = 1, p0 = 1. Equation ( 7 ) yields n∞=1 an zn z + ∞ n=2 nan zn = z + an zn · 1 + ∞ n=2 ∞ n=1 cn zn , where Comparing the coefficients at zn, n ≥ 2, we obtain cn = pn + 2 pk . n−1 j=1 n−1 k=0 n−1 j=1 a j ⎝ ⎛ an ≤ An. n−1 j=1 (n − 1)an = a j cn− j = pn− j + 2 n− j−1 ⎞ pk ⎠ . k=0 Taking into account (33) and the coefficient estimates of a function in PR, we conclude (n − 1)an ≤ 4 |a j |(n − j ), n ≥ 2. 1+z . Equality in (34) holds only if all pi in (33) are equal to 2, which means that p(z) = 1−z From (34), when n = 2, there is a2 ≤ 4|a1| = 4. Equality in this estimate holds for f1 only. Now, it is sufficient to apply mathematical induction in order to prove that successive coefficients an of any f ∈ X are bounded by corresponding coefficients An of f1. Hence Theorem 3 Let f ∈ X have the form f (z) = z + (31). Then, for n ≥ 2, n∞=2 an zn and let An be given by (33) (34) Our next problem is to find the set of variability of (a2, a3) for a function in X . For a given class of analytic functions A, let Ai, j ( A) denote a set {(ai ( f ), a j ( f )) : f ∈ A}. For the class PR of functions with positive real part and having real coefficients, the following result is known: A1,2( PR) = {(x , y) : −2 ≤ x ≤ 2, x 2 − 2 ≤ y ≤ 2}. (35) Based on this result, we can prove A2,3(X ) = (x , y) : 0 ≤ x ≤ 4, x 2 − x ≤ y ≤ 21 x 2 + x Corollary 2 Let f ∈ X have the form f (z) = z + 1 n∞=2 an zn. Then a3 ≥ − 4 . Proof of Theorem 4 Let f (z) = z + PR. It follows from (33) that n∞=2 an zn ∈ X and p(z) = 1 + n=1 pn zn ∈ ∞ Theorem 4 and a2 = p1 + 2, 2a3 = p2 + 2 p1 + 2 + a2( p1 + 2), p1 = a2 − 2, Combining these relations with the estimates given in (35) completes the proof. and ( 4, 12 ) correspond to the functions f−1(z) = z and f1(z) = z exp The points of intersection of two parabolas described in Theorem 4, i.e.: (0, 0) 4z 1−z = z + 4z2 + 12z3 + · · · , respectively. Observe that the class X is not convex. Indeed, if X is a convex set, then, for every fixed α ∈ ( 0, 1 ), a function α f1(z) + (1 − α) f−1(z) = αz exp 14−zz + (1 − α)z = z + 4αz2 + 12αz3 + · · · , would be in X . This will imply that (4α, 12α) ∈ A2,3(X ), a contradiction with Theorem 4. 4 Distortion Theorems Directly from the definition of a circularly symmetric function, it follows that | f (−r )| ≤ | f (r eiϕ )| ≤ | f (r )| (36) for every function f ∈ X and for all ϕ ∈ [0, 2π ] and r ∈ ( 0, 1 ). From ( 4 ), for any function f ∈ X and any number r ∈ ( 0, 1 ), f (r ) = r exp ≤ r exp r 0 r Similarly, | f (−r )| = r exp = r exp − , Equalities in the above estimates hold only if μ is a measure concentrated in point 1; z it means that h(z) = (1−z)2 . We have proved Theorem 5 For any f ∈ X and r = |z| ∈ ( 0, 1 ), r exp −4r 1 + r ≤ | f (z)| ≤ r exp (39) Equalities in the above estimates hold only for f1 and points z = −r and z = r . Corollary 3 For any f ∈ X , we have f ( ) ⊃ The estimates of | f (z)| for f ∈ X can be obtained from ( 2 ) and Theorem 5. Theorem 6 For any f ∈ X and |z| = r ∈ ( 0, 1 ), Equalities in the above estimates hold only for f1 and points z = −r and z = r . Proof Let f ∈ X . From ( 2 ), f (z) = (1 + z)2 h(zz) f (zz) , where h ∈ T . Therefore, if |z| = r ∈ ( 0, 1 ) then and | f (z)| ≤ (1 + r )2 1 (1 − r )2 f1(r ) r | f (z)| ≥ (1 − r )2 1 (1 + r )2 | f1(−r )| r , which is equivalent to (40). Moreover, equalities in both estimates appear when h is z equal to (1−z)2 and z is equal to r and −r , respectively. It means that f1 is the extremal function for (40). Finally, we shall prove two lemmas which will be useful in our research on the convexity of functions in X . Lemma 1 For a fixed point z ∈ +, the set z ff ((zz)) , while f varies in X , is of the form (z) of variability of the expression (z) = conv γ (z), where γ (z) is an upper halfplane located arc of a circle containing three nonlinear points: z0 = 0, z1 = 1, z2 = ( 11+−zz )2, with endpoints z1 and z2. Lemma 2 For any f ∈ X and z ∈ , Re (41) (42) where μ is a probability measure on [−1, 1]. With a fixed z ∈ , we denote by qz (t ) an integrand in (42). The image set {qz (t ) : t ∈ R} coincides with a circle going through the origin. Furthermore, qz (−1) = 1 and qz ( 1 ) = ( 1+z )2. For z su1−chz that Im z > 0, 1 − z = Im where w = z + 1/z. Hence, the set {qz (t ) : t ∈ [−1, 1]} is an arc of the circle with endpoints qz (−1) and qz ( 1 ), which does not contain the origin. For this reason, this set coincides with γ (z) and one endpoint of this arc is always 1, independent of z. Finally, (z) is a section of the disk bounded by γ (z) and the line segment with endpoints qz (−1) and qz ( 1 ). Proof of Lemma 2 Every function f in X has real coefficients, so f ( ) is symmetric with respect to the real axis. Hence, it is sufficient to prove (41) only for z ∈ +. But Lemma 1 leads to z f (z) inf Re f (z) : f ∈ X = Re qz ( 1 ) Re qz (−1) for Re(z + 1/z) ≤ 2 for Re(z + 1/z) ≥ 2. (43) It is easy to check that for z ∈ , Re 1 − z 5 Starlikeness and Convexity The relation ( 2 ) and the estimates of the argument for typically real functions imply that for z ∈ +, arg z f (z) f (z) = 2 arg(1 + z) + arg g(z) z Furthermore, The condition for starlikeness | arg z ff ((zz)) | ≤ π2 and the bounds given above result in . rS∗ (X ) = z Equality in (45) holds for g(z) = (1−z)2 , and, consequently, for f = f1. This result will be generalized in two ways. First, we estimate Re z ff ((zz)) for z in H = {z ∈ : |1 + z2| > 2|z|}. This set appears in the research on typically real functions. It is the domain of univalence and local univalence in T (see, [2]). The set H , called the Golusin lens, is the common part of two disks with radii √2 which have the centers in points i and −i . Moreover, H = z ∈ C : Re 1 − z > 0 . From Lemma 2, we obtain Theorem 7 For each f ∈ X and z ∈ H , z f (z) Re f (z) ≥ 0. (46) (47) It is worth noticing that this theorem is still true even if X is replaced by T . This property is very interesting because the classes X and T have a non-empty intersection, but one is not included in the other. As a corollary, from Theorem 7 we obtain (47). Another generalization of (47) refers to the radius of starlikeness of order α and the radius of strong starlikeness of order α (for definitions and other details the reader is referred to [1,5,8]). Theorem 8 The radius of starlikeness of order α, α ∈ [0, 1), in X is equal to 2 1−2α − 1−√α ⎩ 1+√α 1+2α 1−2α Theorem 9 The radius of strong starlikeness of order α, α ∈ (0, 1], in X is equal to rSS∗(α)(X ) = tan π 8 α . Corollary 5 rSS∗(2/3)(X ) = 2 − √3. Proof of Theorem 8 By Lemma 2, z f (z) Re f (z) ≥ z 1 + 4 Re (1−z)2 1 Let r = |z| be a fixed number, 0 < r ≤ 2 − √3. It is known that h(z) = (1−zz)2 is convex for |z| ≤ 2 − √3. Hence Re z −r (1 − z)2 ≥ (1 + r )2 with equality for z = −r . From (48) and (49) it follows that and so 4r 1 − (1+r)2 1 for Re(z + 1/z) ≥ 2, z f (z) Re f (z) ≥ 1 + r , with equality for z = −r . For this z, the condition Re(z + 1/z) ≤ 2 is satisfied. (48) (49) (50) as a function h(cos ϕ), ϕ ∈ [0, 2π ], where h(x ) = r((11−+2rr2x)+x−r22)r22 . If r ∈ (2 − one can check that (51) (52) where Thus Consequently with equalities for points z0 = r eiϕ0 and z0, where ϕ0 = arccos x0. Furthermore, Re(z0 + 1/z0) − 2 = (1/r + r ) cos ϕ0 − 2 = (1/r + r )x0 − 2 = − The condition Re(z0 + 1/z0) ≤ 2 is satisfied in this case also. Combining (50) and (51), we get . Re In the first case, substituting 11+−rr 2 by α, we obtain r = 11+−√√αα . The condition r ∈ (0, 2 − √3] is equivalent to α ∈ [1/3, 1). While discussing the second possibility in (52), we should remember that the radius of starlikeness in X is equal to √2 − 1. For this reason, we substitute 12−(16−r2r+2)r24 = α only for r ∈ [2 − √3, √2 − 1]. This results in r = The proof of Theorem 9 is easier. In fact, we need the condition for strong starlikeness and inequality (46). Thus we obtain 2 arctan Proof From ( 2 ), if f ∈ X then z ff ((zz)) = (1 + z)2 h(zz) , where h ∈ T . Hence . In further calculation, we shall apply Lemma 2. Let r ≤ 2 − √3. For z such that Re(z + 1/z) ≥ 2, Estimate (53) and the inequality Re 12+zz ≥ − 12−rr result in This estimate is not sharp because equalities in the two previous inequalities appear only if z = −r , but in this case Re(z + 1/z) < 2. From the above, we conclude that if Re(z + 1/z) ≥ 2 and r ∈ [0, √5 − 2) then Re 1 + zff ((zz)) > 0. (53) (54) Assume now that Re(z + 1/z) ≤ 2. In this case . The first two components can be estimated as above. Based on (49), the third one is greater than or equal to (1−+4rr)2 . Since each estimate is sharp (with equality for z = −r ), z f (z) is also sharp. Consequently, the estimate of the expression Re 1 + f (z) . The function in the numerator of the right-hand side of this inequality is decreasing for t ∈ R. For this reason, it has in ( 0, 1 ) the only solution r0. We have proven that if Re(z + 1/z) ≤ 2 and r ∈ [0, r0] then Re 1 + zff ((zz)) ≥ 0. But r0 < √5 − 2. Taking into account both parts of the proof, we obtain the assertion. Equality in z (53) holds for h(z) = (1−z)2 and z = −r . It means that (55) is sharp, with equality for f1 and z = −r . 6 Univalence The problems of the univalence of functions in X are more complicated. Based on the already proved results, we know that the radius of univalence rS(X ) is greater than or equal to √2 − 1. On the other hand, one can easily find the upper estimate of rS(X ). Namely, discuss a function F (z) = r1∗ f1(r∗z), where f1 is given by ( 10 ) and r∗ is equal to rS(X ) which we want to derive. The function F is univalent in and it has normalization F (0) = F (0)−1 = 0. From (31) it follows that F (z) = z +4r∗z2 +. . .. The estimate of the second coefficient of functions in S results in r∗ ≤ 1/2. The main theorem of this section is as follows. Theorem 11 The radius of univalence in X is equal to rS(X ) = r1, where r1 = 0.454 . . . is the only solution of equation arcsin 1 − r 2 2r 2(1 − r 2) + 1 + r 2 − √−1 + 6r 2 − r 4 = π (56) in (√2 − 1, 1). The extremal function is f1. In the proof of this theorem we need two lemmas. Lemma 3 For each ft , t ∈ [−1, 1] given by ( 9 ) and for each r ∈ (0, √3/3) and ϕ ∈ [0, π ] the following inequality is true: Lemma 4 The function f1 is univalent in the disk |z| < r1, where r1 is the only solution of (56). Proof of Lemma 3 Let t ∈ [−1, 1] and r ∈ (0, √3/3) be fixed. Let us denote by g(ψ ) the argument of ft (r eiϕ ) with a fixed ϕ ∈ [0, π ], where ψ and t are connected by t = cos ψ . Applying ( 9 ), g can be written as (57) for ϕ and ψ in [0, π ], we conclude that k(ψ ) ≥ 0 for all ψ ∈ [0, π ]. Now, we shall show that g(ψ ) is a decreasing function of the variable ψ . We have g(ψ ) = ϕ + cot ψ 2 k(ψ ), k(ψ ) = log 1 − r ei(ϕ+ψ) 1 − r ei(ϕ−ψ) . 1 − r ei(ϕ+ψ) 1 − r ei(ϕ−ψ) ≥ 1, −1 g (ψ ) = 2 sin2 ψ · h(ψ ), 2 h(ψ ) = k(ψ ) − k (ψ ) sin ψ. W (0) = (q − cos ϕ)(3q2 − 4 + cos2 ϕ) and W (π ) = −(q + cos ϕ)3. where Since where Hence A long and tedious calculation shows that h (ψ ) = sin ϕ(1 − cos ψ ) [(q cos ψ − cos ϕ)2 + (q2 − 1) sin2 ψ ]2 · W (ψ ), with q = (1 + r 2)/2r, q > 1 and − cos3 ϕ − q cos2 ϕ − (q2 − 1) cos ϕ + q(q2 − 1). W (ψ ) = [(1 − 2q2) cos ϕ − q] cos2 ψ + 2[q cos2 ϕ + cos ϕ + q(q2 − 1)] cos ψ It is obvious that W (π ) < 0. On the other hand, W (0) > 0, providing that r ∈ (0, √3/3), or equivalently, q2 > 4/3. From these observations, taking into account that W is a quadratic function of cos ψ , we can see that W (ψ ) has exactly one solution in [0, π ]; let us denote it by ψ0. Hence, h (ψ ) has only one solution ψ0 in (0, π ). Thus, h (ψ ) for ψ ∈ (0, ψ0) increases, and for ψ ∈ (ψ0, π ) decreases. Combining it with h(0) = h(π ) = 0, we obtain h(ψ ) ≥ 0 for ψ ∈ [0, π ]. This implies g (ψ ) ≤ 0 for ψ ∈ (0, π ); so g(ψ ) is decreasing in (0, π ). Finally, g(ψ ) ≤ g(0+) for all ψ ∈ [0, π ], where k(ψ ) 2 sin ϕ g(0+) = ψl→im0+ g(ψ ) = ϕ + ψl→im0+ tan ψ2 = ϕ + 2k (0) = ϕ + q − cos ϕ . Moreover, if ϕ = 0 then g(ψ ) = 0, and, if ϕ = π then g(ψ ) = π for all ψ ∈ [0, π ]. Consequently, (57) holds also for ϕ = 0 and ϕ = π . Proof of Lemma 4 Consider a level curve f1({z ∈ C : |z| = r }) with a fixed r ∈ ( 0, 1 ). Since f1 is a circularly symmetric function, f1(r eiϕ ) decreases for ϕ from 0 to π . Hence, f1 is univalent when the level curves has no self-intersection points. It happens at small r , ie. when r < √2 − 1, because f1 is starlike in this case. So it is enough to discuss for which r ∈ [√2 − 1, 1/2] the level curve f1({z ∈ C : |z| = r }) is tangent to the real axis. Denoting the point of tangency by w0, and denoting by z0 = r eiϕ0 the corresponding point on circle |z| = r for which f (z0) = w0, we obtain arg f (z0) = π . The tangency of the level curve to the real axis in w0 ensures that arg f1(r eiϕ ) is increasing for ϕ ∈ (0, ϕ0), decreasing for ϕ ∈ (ϕ0, ϕ1), and once again increasing for ϕ ∈ (ϕ1, π ), where ϕ1 is a number from the interval (ϕ0, π ). Hence, Re z0f1f1(z(0z0)) = 0. For this reason, we need to solve the system (58) (59) ϕ0 = arcsin . f1(r eiϕ0 ) = r eiϕ0 exp 4r eiϕ0 − 4r 2 1 − 2r cos ϕ0 + r 2 , obtain But The first equation can be written, using ( 8 ), as Re 11+−zz00 2 = 0. Since z0 = r eiϕ , we so . This means that b(r ) increases from π/2 + √2 to infinity, while r ∈ (√2 − 1, 1). For this reason, (56) has only one solution. Proof of Theorem 11 Let L = {log f (zz) , f ∈ X }. In the paper [9], the authors showed that the extreme points of the class L are as follows: lψ (z) = i cot ψ 2 max {Im l(z) : l ∈ L} = max Im lψ (z) : ψ ∈ [0, π ] . But Im l(z) = arg f (zz) for l ∈ L. Therefore max arg f (z) z : f ∈ X = max arg ftz(z) : t ∈ [−1, 1] , where ft is given by ( 9 ). Hence, for z ∈ \{0}, max arg f (z) : f ∈ X = max {arg ft (z) : t ∈ [−1, 1]} . Applying Lemma 3, we conclude that for every f ∈ X , r ∈ (0, √3/3) and ϕ ∈ [0, π ], the following inequality holds: arg f (r eiϕ ) ≤ arg f1(r eiϕ ). Consequently, for every function f ∈ X , from arg f1(r eiϕ ) ≤ π , it yields that arg f (r eiϕ ) ≤ π , which combined with Lemma 4 gives the assertion. In the paper [4], the class T of semi-typically real functions was defined. Namely, f ∈ T if z ∈ ( 0, 1 ) if and only if f (z) > 0. (60) (61) (62) (63) This equivalence means that the values of f belonging to T are positive real numbers if and only if z ∈ is positive and real. According to this definition, T ⊂ T . Based on the proof of Theorem 11, one can anticipate that functions f ∈ X are semi-typically real at most in the disk with radius r . The number r is chosen such that the level curves f ({z ∈ C : |z| = r }) for r < rTTand f ∈ X mayTwind around the origin, yet they do not touch the positive real halfaxis. Moreover, one can anticipate that the extremal function is still f1. Conjecture. The radius of semi-typical reality in X is equal to r ( X ) = r2, where T r2 = 0.718 . . . is the only solution of equation arcsin + Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. 1. 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Leopold Koczan, Paweł Zaprawa. Circularly Symmetric Locally Univalent Functions, Bulletin of the Malaysian Mathematical Sciences Society, 2016, 1615-1635, DOI: 10.1007/s40840-016-0329-z