#### 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
+
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