Circularly Symmetric Locally Univalent Functions

Let $$D\subset \mathbb {C}$$D⊂C and $$0\in D$$0∈D. A set D is circularly symmetric if, for each $$\varrho \in \mathbb {R}^+$$ϱ∈R+, a set $$D\cap \{\zeta \in \mathbb {C}:|\zeta |=\varrho \}$$D∩{ζ∈C:|ζ|=ϱ} is one of three forms: an empty set, a whole circle, a curve symmetric with respect to the real axis containing $$\varrho $$ϱ. A function f analytic in the unit disk $$\Delta \equiv \{\zeta \in \mathbb {C}:|\zeta |<1\}$$Δ≡{ζ∈C:|ζ|<1} and satisfying the normalization condition $$f(0)=f^{\prime }(0)-1=0$$f(0)=f′(0)-1=0 is circularly symmetric, if $$f(\Delta )$$f(Δ) is a circularly symmetric set. The class of all such functions is denoted by X. In this paper, we focus on the subclass $$X^{\prime }$$X′ consisting of functions in X which are locally univalent. We obtain the results concerned with omitted values of $$f\in X^{\prime }$$f∈X′ and some covering and distortion theorems. For functions in $$X^{\prime }$$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 $$X^{\prime }$$X′.


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 byT 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.

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 f (z) f (z) is a typically real function. Additionally, he observed that the property z f (z) f (z) ∈T does not guarantee the univalence of f . In fact, we have 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 (re iϕ )| 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: It is known that each function of the class T can be represented by the formula 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 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 : where μ is a probability measure on [0, π].
Applying the well-known equivalence we obtain the relation between X and the set P R of functions with positive real part which have real coefficients: It is known (Robertson, [6]) that the set of extreme points for T has the form 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 where t = cos ψ.
so we can write Besides f t , 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+t (2), we obtain Let us denote by g t the solutions of Eq. (11). From this equation In particular, we have 2 Properties of f t and g t Firstly, we shall describe the sets f t ( ) and g t ( ), where f t , g t are defined by (9) and (12), respectively. For f t , t ∈ (−1, 1) (i.e. for ψ ∈ (0, π)), from (9) we obtain We shall derive the argument which appears in the above expression. To do this, we need the following identity: and From the above expressions, it follows that a function | f t (e iϕ )|, with fixed t ∈ (−1, 1), does not depend on the variable ϕ. Moreover, | f t (e iϕ )| in (13) is less than 1 and | f t (e iϕ )| in (14) is greater than 1. Additionally, which means that the curves { f t (e iϕ ), ϕ ∈ [0, φ)} and { f t (e iϕ ), ϕ ∈ (φ, π ]} wind around circles given by (13) and (14) infinitely many times. Hence, for t ∈ (−1, 1), It is easily seen that f −1 ( ) = . Now, we shall show that f 1 omits only one point on the real axis.
Proof On the contrary, suppose that there exists z ∈ such that f 1 (z) = −e −2 . In fact, we can assume that arg z ∈ [0, π] because the coefficients of f 1 are real. For this reason, or equivalently, Let z = re iϕ , ϕ ∈ [0, π]. Comparing the arguments of both sides of this equality, we get 2r sin ϕ For a fixed r ∈ (0, 1), let us consider a function For ϕ ∈ [0, π] the function h (ϕ) decreases and where g(r ) = log r + 1 − r 2 2r .
Applying a similar argument, one can prove the following more general theorem.
Theorem 2 For g t , t ∈ (−1, 1] given by (12), Proof ad (i) Suppose that there exists z ∈ such that g t (z) = −e −1−t . Then and putting z = re iϕ , we have 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 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 g t (z) = − e −1−t with fixed > 1. It takes the following form Putting z = re iϕ into (25), we have Hence Let h(ϕ) denote the left-hand side of this equality. The function h (ϕ) is decreasing; One can easily prove that there exists only one number r 0 ∈ (0, 1) such that h (π ) > 0 for r ∈ (0, r 0 ) and h (π ) < 0 for r ∈ (r 0 , 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 ϕ 0 ∈ (0, π) such that h(ϕ 0 ) = 0. It means that (26) is satisfied by z 0 = re iϕ 0 ; so (25) holds for z = z 0 . ad (iii) The proof of this part is similar to the proof of (ii).
Observe that |g t (−r )| is a continuous and increasing function of r ∈ [0, 1). For this reason, g t (−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 ∈ g t ( ), then the whole circle with radius |c| centered at the origin is also contained in this set. Hence, for each in 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 } ⊂ g t ( ).
Combining these facts with points (i) and (iii) of Theorem 2 completes the proof.
Let us consider a function b(ϕ) = arg g t (re iϕ ), ϕ ∈ (0, π) where t ∈ [−1, 1] and r ∈ (0, 1) are fixed. Analyzing the derivative of this function it can be observed that 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 g t ({z ∈ C : |z| < r, Im z ≥ 0}) is contained in the upper halfplane. If r is greater than 1+t , then this set is not contained in the upper halfplane; its boundary is wound around the origin.
If r = 1, then Hence and Therefore, g t ( ) is wound around the origin infinitely many times, or, more precisely, it is wound around the circle with radius e −1−t .

Coefficients of Functions in X
To start with, let us look into the coefficients of f 1 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 γ n of f 1 can be easily derived. Indeed, The Taylor series expansion of f 1 is given by Denoting the n-th coefficient of f 1 by A n , we can write and consequently, The first four values of A n are On the other hand, for A n the formula holds for n ≥ 2. Indeed, expanding both sides of the equality Comparing the coefficients at z n , 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 f 1 . To do this, we apply the relation (7).
Suppose that functions f ∈ X and p ∈ P R are of the form f (z) = ∞ n=1 a n z n and p(z) = ∞ n=0 p n z n with a 1 = 1, p 0 = 1. Equation (7)  Comparing the coefficients at z n , n ≥ 2, we obtain Taking into account (33) and the coefficient estimates of a function in P R , we conclude (n − 1)a n ≤ 4 Equality in (34) holds only if all p i in (33) are equal to 2, which means that p(z) = 1+z 1−z . From (34), when n = 2, there is a 2 ≤ 4|a 1 | = 4. Equality in this estimate holds for f 1 only. Now, it is sufficient to apply mathematical induction in order to prove that successive coefficients a n of any f ∈ X are bounded by corresponding coefficients A n of f 1 . Hence Theorem 3 Let f ∈ X have the form f (z) = z + ∞ n=2 a n z n and let A n be given by (31). Then, for n ≥ 2, a n ≤ A n .
Our next problem is to find the set of variability of (a 2 , a 3 ) for a function in X . For a given class of analytic functions A, For the class P R of functions with positive real part and having real coefficients, the following result is known: Based on this result, we can prove Theorem 4 and Corollary 2 Let f ∈ X have the form f (z) = z + ∞ n=2 a n z n . Then a 3 ≥ − 1 4 .
Proof of Theorem 4 Let f (z) = z + ∞ n=2 a n z n ∈ X and p(z) = 1 + ∞ n=1 p n z n ∈ P R . It follows from (33) that or equivalently, Combining these relations with the estimates given in (35) completes the proof.

Distortion Theorems
Directly from the definition of a circularly symmetric function, it follows that 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), Similarly, Equalities in the above estimates hold only if μ is a measure concentrated in point 1; it means that h(z) = z (1−z) 2 . We have proved Theorem 5 For any f ∈ X and r = |z| ∈ (0, 1), Equalities in the above estimates hold only for f 1 and points z = −r and z = r.

Corollary 3 For any 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 f 1 and points z = −r and z = r.
Proof Let f ∈ X . From (2), where h ∈ T . Therefore, if |z| = r ∈ (0, 1) then which is equivalent to (40). Moreover, equalities in both estimates appear when h is equal to z (1−z) 2 and z is equal to r and −r , respectively. It means that f 1 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) of variability of the expression z f (z) f (z) , while f varies in X , is of the form
where γ (z) is an upper halfplane located arc of a circle containing three nonlinear points: z 0 = 0, z 1 = 1, z 2 = ( 1+z 1−z ) 2 , with endpoints z 1 and z 2 . Lemma 2 For any f ∈ X and z ∈ , (41) Proof of Lemma 1 Let z ∈ + . Applying (2) and the representation formula for a function in T , we have where μ is a probability measure on [−1, 1]. With a fixed z ∈ , we denote by q z (t) an integrand in (42). The image set {q z (t) : t ∈ R} coincides with a circle going through the origin. Furthermore, q z (−1) = 1 and q z (1) = ( 1+z 1−z ) 2 . For z such that Im z > 0, where w = z + 1/z. Hence, the set {q z (t) : t ∈ [−1, 1]} is an arc of the circle with endpoints q z (−1) and q z (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 q z (−1) and q z (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 It is easy to check that for z ∈ , Consequently, (41) can be written as follows:

Starlikeness and Convexity
The relation (2) and the estimates of the argument for typically real functions imply that for z ∈ + , (45) Furthermore, The condition for starlikeness | arg z f (z) f (z) | ≤ π 2 and the bounds given above result in Equality in (45) holds for g(z) = z (1−z) 2 , and, consequently, for f = f 1 . This result will be generalized in two ways.
First, we estimate Re z f (z) f (z) for z in H = {z ∈ : |1 + z 2 | > 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,

From Lemma 2, we obtain
Theorem 7 For each f ∈ X and z ∈ H, 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]). Now, suppose that r ∈ (2− √ 3, 1). The real part of z (1−z) 2 for z = re iϕ can be written as a function h(cos ϕ), ϕ ∈ [0, 2π ], where h(x) = r (1+r 2 )x−2r 2 (1−2r x+r 2 ) 2 . If r ∈ (2 − √ 3, 1), one can check that . Consequently with equalities for points z 0 = re iϕ 0 and z 0 , where ϕ 0 = arccos x 0 . Furthermore, The condition Re(z 0 + 1/z 0 ) ≤ 2 is satisfied in this case also. Combining (50) and (51), we get In the first case, substituting 1−r 1+r 2 by α, we obtain r = 1− √ α 1+ √ α . 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 1−6r 2 +r 4 2(1−r 2 ) 2 = α only for r ∈ [2 − √ 3, 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 2r 1 − r 2 ≤ π 2 α, and hence Solving this inequality with respect to r , the assertion of Theorem 9 follows. The next theorem is concerned with the problem of convexity of a function in X .

Theorem 10 The radius of convexity for X is equal to r CV
The extremal function is f 1 .
In the proof of this theorem, we need the following result of Todorov for h ∈ T (see [10]): 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 2z 1+z ≥ − 2r 1−r 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 + z f (z) Assume now that Re(z + 1/z) ≤ 2. In this case (55) The first two components can be estimated as above. Based on (49), the third one is greater than or equal to −4r (1+r ) 2 . Since each estimate is sharp (with equality for z = −r ), the estimate of the expression Re 1 + z f (z) f (z) is also sharp. Consequently, 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 r 0 . We have proven that if Re(z + 1/z) ≤ 2 and r ∈ [0, r 0 ] then Re 1 Taking into account both parts of the proof, we obtain the assertion. Equality in (53) holds for h(z) = z (1−z) 2 and z = −r . It means that (55) is sharp, with equality for f 1 and z = −r .

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 r S (X ) is greater than or equal to √ 2 − 1. On the other hand, one can easily find the upper estimate of r S (X ). Namely, discuss a function F(z) = 1 r * f 1 (r * z), where f 1 is given by (10) and r * is equal to r S (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 * z 2 +. . .. 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 r S (X ) = r 1 , where r 1 = 0.454 . . . is the only solution of equation The extremal function is f 1 .
In the proof of this theorem we need two lemmas.

Lemma 4
The function f 1 is univalent in the disk |z| < r 1 , where r 1 is the only solution of (56).
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 T . The number r T is chosen such that the level curves f ({z ∈ C : |z| = r }) for r < r T and f ∈ X may wind around the origin, yet they do not touch the positive real halfaxis. Moreover, one can anticipate that the extremal function is still f 1 . Conjecture. The radius of semi-typical reality in X is equal to r T (X ) = r 2 , where r 2 = 0.718 . . . is the only solution of equation It is worth emphasizing that in the proof of Lemma 3 we did apply the assumption r ∈ (0, √ 3/3), which is equivalent to q 2 > 4/3. The number √ 3/3 in this expression is not necesserily sharp. Hence, the argument given in the proof of Theorem 11 is not sufficient to prove this conjecture.