1 Introduction
Let
$\mathcal H$
denote the class of analytic functions in the unit disk
$\mathbb D=\{z\in \mathbb C: |z|<1\}$
while
$\mathcal A\subset \mathcal H$
denote the set of all functions g that are analytic in
$\mathbb D$
and has the series representation
$$ \begin{align} g(z)=z+\sum_{n=2}^{\infty }c_{n}z^{n},\quad z\in \mathbb D. \end{align} $$
Let
$\mathcal {S}\subset \mathcal {A}$
be the class of normalized univalent functions. In [Reference Robertson15], Robertson introduced the class
${\mathcal S}^*_{\alpha } $
of starlike functions of order
$\alpha $
,
$\alpha <1$
$$ \begin{align*} {\mathcal S}^*_{\alpha} :=& \left\{ g\in {\mathcal A}:\ \mathrm{{Re}}\frac{zg'(z)}{g(z)}>\alpha ,\ z\in\mathbb D\right \}. \end{align*} $$
In particular, when
$\alpha =0$
, the class
${\mathcal S}^*_0:=\mathcal S^*$
is the well-known class of starlike functions which map
$\mathbb D$
onto a domain starlike with the origin (see [Reference Emch3, Reference Nevanlinna12]). Furthermore, the related class
${\mathcal K}_{\alpha }$
of convex functions of order
$\alpha $
may be defined in the following way:
$$ \begin{align*} {\mathcal K}_{\alpha } :=& \left\{g\in\mathcal A:\ zg'(z)\in {\mathcal S}^*_{\alpha }\right\} \\ =& \left\{ g\in {\mathcal A}:\ \mathrm{{Re}}\left\{ 1+\frac{zg"(z)}{g'(z)} \right\}>\alpha ,\ z\in\mathbb D\right \}, \quad \alpha <1. \end{align*} $$
In particular, when
$\alpha =0$
, the class
$\mathcal K_0:=\mathcal K$
is the well-known class of convex functions which map
$\mathbb D$
onto a convex domain. Strohhäcker in [Reference Strrohhäcker18] proved that
${\mathcal K}_{0}\subset {\mathcal S}^*_{1/2}\subset {\mathcal S}^*_{0}$
. Wilken and Feng in [Reference Wilken and Feng20] showed that if
$g\in \mathcal K_{\alpha }$
, then
$g\in {\mathcal S}^*_{\delta (\alpha )}$
, where
$$ \begin{align} \delta(\alpha)=\left\{ \begin{array}{@{}lll} 1/\log 4, & \alpha =1/2, \\ (1-2\alpha)/(4^{1-\alpha}-2), & \alpha \neq 1/2. \end{array} \right. \end{align} $$
Let
${\mathcal {SS}}^*(\beta )$
denote the class of strongly starlike functions of order
$\beta $
,
$0<\beta < 2$
,
$$ \begin{align} {\mathcal {SS}}^*(\beta):=\left\{g\in {\mathcal A}:\left|\arg\frac{zg'(z) }{g(z)}\right|<\frac{\beta \pi} {2},\ \ z\in {\mathbb D}\right\},\ \ \end{align} $$
which was introduced in [Reference Brannan and Kirwan2, Reference Stankiewicz17]. Furthermore,
${\mathcal {SK}}(\beta )=\left \{g\in {\mathcal A}:\ zg'(z)\in \mathcal {SS}^*(\beta )\right \}$
denote the class of strongly convex functions of order
$\beta $
. Analogously to (1.2), in the work [Reference Nunokawa14], it was proved that if
$\beta \in (0,1)$
and
$g\in {\mathcal {SK}}(\alpha (\beta ))$
, then
$g\in {\mathcal {SS}}(\beta )$
, where
$$ \begin{align} \alpha(\beta)=\beta+\frac{2}{\pi}\tan^{-1}\left(\frac{\beta n(\beta)\sin(\pi(1-\beta)/2)}{m(\beta)+\beta n(\beta)\cos(\pi(1-\beta)/2)}\right), \end{align} $$
and where
$$ \begin{align*} m(\beta)=(1+\beta)^{(1+\beta)/2}\ \ \ n(\beta)=(1-\beta)^{(\beta-1)/2}. \end{align*} $$
It is well known that if
$g\in \mathcal H$
satisfies
$g(0)=0$
,
$g'(0)\neq 0,$
and
$$ \begin{align*} \mathrm{{Re}} \left\{ \frac{zg'(z)}{g(z)} \right\}>0, \quad z\in\mathbb D, \end{align*} $$
then g is univalent and starlike in
${\mathbb D}$
. Moreover, it is known that for
$g\in \mathcal A,$
the condition
$$ \begin{align*} \mathrm{{Re}}\left\{ 1+\frac{zg"(z)}{g'(z)} \right\}>0, \quad z\in{\mathbb D} \end{align*} $$
is sufficient for starlikeness of order
$1/2$
of g in
${\mathbb D}$
. The condition
$$ \begin{align*} \left| zg"(z)\right|<2|g'(z)|, \quad z\in\mathbb D \end{align*} $$
is also sufficient for starlikeness of g in
${\mathbb D}$
.
Robertson in [Reference Robertson16] obtained that if
$g\in \mathcal A$
with
$g(z)/z \neq 0$
,
$z\in \mathbb D$
and if there exists k,
$0<k\leq 2$
, such that
$$ \begin{align*} \left|\frac{zg"(z)}{g'(z)}\right|\leq k\left| \frac{zg'(z)}{g(z)} \right|, \quad z\in\mathbb D, \end{align*} $$
then
$g \in {\mathcal S}^*_{2/(2+k)}$
. Mocanu in [Reference Mocanu10] proved that if
$g\in \mathcal A$
with
$g(z)g'(z)/z\neq 0$
,
$z\in \mathbb D,$
and
$$ \begin{align*} \left|1+\frac{g"(z)}{g'(z)}\right|\leq \sqrt{2}\left|\frac{zg'(z)}{g(z)}+1\right|,\quad z\in\mathbb D, \end{align*} $$
then
$g\in {\mathcal S}^*_{0}$
.
The simplest analytic criteria for univalence have limited application because they imply special properties, such as starlikenes or close-to-convexity. These conditions are sufficient for univalence but far from necessary. Another important condition sufficient for univalence of
$g\in \mathcal H$
is due to Nehari [Reference Nehari11] and involves the Schwarzian derivative
$\{g,z\}$
:
$$ \begin{align} \left|\{g,z\}\right|=\left|\left\{\frac{g"(z)}{g'(z)}\right\}' -\frac{1}{2}\left\{\frac{g"(z)}{g'(z)}\right\}^2\right|<\frac{2}{1-|z|^2},\quad z\in\mathbb D. \end{align} $$
In (1.5), the constant
$2$
cannot be replaced by a smaller number. On the other hand, inequality (1.5) becomes a necessary condition for univalence of g if the constant
$2$
is replaced by
$6$
. Instead of Nehari’s condition (1.5), it is often simpler to work directly with the ratio
$g"(z)/g'(z)$
. Becker [Reference Becker1] applied Loewner’s method to show that the condition
$$ \begin{align} \left|\frac{g"(z)}{g'(z)}\right|<\frac{1}{1-|z|^2},\quad z\in\mathbb D, \end{align} $$
is sufficient for the univalence of
$g\in \mathcal H$
in
$\mathbb D$
. Becker also proved that constant
$1$
in the nominator cannot be replaced by a greater number. The well-known Noshiro–Warshawski theorem [Reference Noshiro13, Reference Warschawski19] says that
$$ \begin{align} \mathrm{Re}\{e^{i\phi}g'(z)\}>0,\quad z\in\mathbb D, \end{align} $$
for some
$\phi \in \mathbb R$
, is a sufficient condition for the univalence of
$g\in \mathcal A$
. On the other hand, the Noshiro–Warshawski condition with
$\phi =0$
is not a sufficient condition for
$g\in \mathcal A$
to be starlike (see [Reference Goodman5, p. 123]). The radius of starlikess in the class
$\{g\in \mathcal A: \mathrm {Re}g'(z)>0,~|z|<1\}$
is unknown but it is at least
$4\sqrt {2}-5=0.65\ldots $
, [Reference Goodman6, p. 95], [Reference Lewandowski8].
Several more complicated sufficient conditions for starlikeness and for convexity are collected in Chapter
$5$
of [Reference Miller and Mocanu9] (see also [Reference Goodman6]). The current work revolves around determining the optimal angle opening of the sector containing the image of
$f'$
that guaranties certain geometric properties of f, i.e., f is a starlike, convex, or Bazilevic̆ function. We utilize the notion of the dimension of a subclass
$\mathcal R$
of the class
$\mathcal A$
of normalized analytic functions in
$\mathbb D$
. That is, the largest number
$\varepsilon \in (0,2]$
for which the condition
$|\arg f'(z)| < \varepsilon /2$
, for
$f\in \mathcal A$
, implies that
$f\in \mathcal R$
. The main objective of the article is to obtain a lower bound (sharper than the trivial one
$1/2$
) for the dimension of the class of starlike functions.
2 Sector for derivative sufficient for starlikeness
Definition 2.1 Let
$\mathcal R$
be a subclass of
$\mathcal A$
. Set E to be the set of all
$\varepsilon \in (0,2]$
such that condition
$|\arg f'(z)| < \varepsilon /2$
, where
$z\in \mathbb D$
and
$f\in \mathcal A$
, implies
$f\in \mathcal R$
. The dimension of
$\mathcal R$
is defined as the supremum of E and is denoted by
$D{\mathcal R}\{|\arg f'(z)|<\varepsilon \pi /2\}$
.
For
$\phi =0,$
the known Noshiro–Warshawski condition (1.7) has the form
$$ \begin{align*} |\arg f'(z)|<\pi/2,\quad z\in \mathbb D \end{align*} $$
and this is a sufficient condition for
$f\in \mathcal A$
to be in the class
$\mathcal S$
of normalized univalent functions. Goodman [Reference Goodman4] proved that for each
$\epsilon>0,$
there is a function
$f\in \mathcal A,$
which satisfies
$$ \begin{align*} |\arg f'(z)|<\pi/2+\epsilon,\quad z\in \mathbb D \end{align*} $$
and f is not univalent in
$\mathbb D$
. Therefore, we have
$$ \begin{align*} D{\mathcal S}\{|\arg f'(z)|<\varepsilon\pi/2\}=1. \end{align*} $$
If
$h(z)\in \mathcal A$
is such that
$$ \begin{align*} h'(z)=\left\{\frac{1+z}{1-z}\right\}^\alpha \quad\mathrm{or}\quad|\arg h'(z)|<\alpha\pi/2,\quad z\in \mathbb D, \end{align*} $$
then
$$ \begin{align*} 1+\frac{zh"(z)}{h'(z)}=1+\frac{2\alpha z}{1-z^2}. \end{align*} $$
This shows that there is no an
$\alpha>0$
such that the condition
$|\arg \{h'(z)\}|<\alpha \pi /2$
implies that
$h(z)$
is a convex function. Thus,
$D\mathcal K\{|\arg f'(z)|<\varepsilon \pi /2\}$
doesn’t exist.
For two functions f and g analytic in
$ \mathbb D$
, we say that f is subordinate to g, written by
$f\prec g$
, if there exists an analytic function
$w(z)$
with
$w\left ( 0\right ) =0$
and
$\left \vert w(z)\right \vert <1$
such that
$ f(z)=g\left (w(z)\right )$
.
It is known [Reference Hallenbeck and Ruscheweyh7] that for a convex function
$q(z)=1+q_1z+\cdots $
, we have
$$ \begin{align*} \left[f\in\mathcal A, \quad f'(z)\prec q(z)\right]\quad\Rightarrow\quad\left[f(z)/z\prec q(z)\right]. \end{align*} $$
Therefore, the condition
$$ \begin{align*} f'(z)\prec\sqrt{\frac{1+z}{1-z}}\quad\mathrm{or}\quad|\arg f'(z)|<\pi/4,\quad z\in \mathbb D \end{align*} $$
implies
$$ \begin{align*} \frac{f(z)}{z}\prec\sqrt{\frac{1+z}{1-z}}\quad\mathrm{or} \quad\left|\arg \frac{f(z)}{z}\right|<\pi/4,\quad z\in \mathbb D \end{align*} $$
and so if
$|\arg \{f'(z)\}|<\pi /4$
, then
$$ \begin{align*} \left|\arg\frac{zf'(z)}{f(z)}\right| &\leq\left|\arg f'(z)\right|+\left|\arg \frac{f(z)}{z}\right|\\ &<\pi/4+\pi/4=\pi/2,\quad z\in \mathbb D. \end{align*} $$
Then, we can see that
$$ \begin{align} |\arg f'(z)|<\pi/4,\quad z\in \mathbb D, \end{align} $$
is a sufficient condition for the starlikeness of
$f\in \mathcal A$
and we can see that
$D{\mathcal S^*}\{|\arg \{f'(z)\}|<\varepsilon \pi /2\}$
satisfies
$$ \begin{align} D{\mathcal S^*}\{|\arg f'(z)|<\varepsilon\pi/2\}\geq 1/2. \end{align} $$
In this article, we want to replace in (2.2) the number
$1/2$
by a greater number.
Lemma 2.2 [Reference Nunokawa14]
Let
$P(z)$
be analytic function in
$|z|<1$
with
$P(0)=1$
,
$P(z)\neq 0$
. If there exists a point
$z_0$
,
$|z_0|<1$
, such that
$|\arg P(z)|<\pi \gamma /2$
for
$|z|<|z_0|$
and
$|\arg P(z_0)|=\pi \gamma /2$
for some
$\gamma \in (0,2]$
, then we have
$$ \begin{align} \frac{z_0P'(z_0)}{P(z_0)}=ik\gamma \end{align} $$
for some
$k\geq (a+a^{-1})/2\geq 1$
when
$\arg P(z_0)=\pi \gamma /2$
or
$k\leq -(a+a^{-1})/2\leq -1$
when
$\arg P(z_0)=-\pi \gamma /2$
, where
$ \left \{P(z_0)\right \}^{1/\gamma }=\pm ia$
, and
$a>0$
.
3 Main results
Theorem 3.1 If
$q(z)\in \mathcal H$
,
$q(0)=0$
,
$q'(0)\neq 0,$
and
$$ \begin{align} \left|\arg \frac{q'(z)}{q'(0)}\right|<\frac{\alpha\pi}{2},\quad z\in\mathbb D, \end{align} $$
for some
$\alpha \in (0,2]$
, and
$$ \begin{align*} \frac{q(z)}{z}\neq 0,\quad z\in\mathbb D, \end{align*} $$
then
$$ \begin{align*} \left|\arg\frac{q(z)}{zq'(0)}\right|<\frac{\pi\gamma}{2} <\frac{\alpha\pi}{2},\quad z\in\mathbb D, \end{align*} $$
where
$\gamma \in (0,2]$
is the smallest positive root of the equation
$$ \begin{align} \frac{\pi\gamma}{2}+\tan^{-1}\gamma=\frac{\alpha\pi}{2}. \end{align} $$
Proof Because
$q(0)=0,$
the function
$$ \begin{align*} s(z)=\frac{q(z)}{zq'(0)}, \quad z\in\mathbb D, \end{align*} $$
is analytic in
$\mathbb D$
,
$s(0)=1$
and it follows that
$$ \begin{align*} \frac{q'(z)}{q'(0)} = s(z)+zs'(z). \end{align*} $$
If there exists a point
$z_1\in \mathbb D$
, such that
$$ \begin{align*} |\arg s(z)|<\frac{\pi\gamma}{2},\quad (|z|<|z_1|) \end{align*} $$
and at the point
$z_1$
is
$$ \begin{align*} |\arg s(z_1)|=\frac{\pi\gamma}{2}, \end{align*} $$
then from Lemma 2.2, we have
$$ \begin{align*} \frac{z_1 s'(z_1)}{s(z_1)}=ik\gamma, \end{align*} $$
where
$$ \begin{align*} k\geq\frac{1}{2}\left(a+\frac{1}{a}\right)\geq1 \ \ \ \text{ when }\ \ \arg p(z_1)=\frac{\pi\gamma}{2} \end{align*} $$
and
$$ \begin{align*} k\leq-\frac{1}{2}\left(a+\frac{1}{a}\right)\leq-1 \ \ \ \mathrm{when}\ \ \arg p(z_1)=-\frac{\pi\gamma}{2}, \end{align*} $$
where
$$ \begin{align*} \left\{p(z_1)\right\}^{1/\gamma}=\pm ia, \ \ \mathrm{and}\ \ a>0. \end{align*} $$
Therefore, we have
$$ \begin{align*} \left|\arg\frac{q'(z_1)}{q'(0)} \right| &= \left|\arg\left\{s(z_1)\left(1+\frac{z_1s'(z_1)}{s(z_1)}\right)\right\}\right| \\ &= \left|\arg\left\{s(z_1)\right\}+\arg\left\{1+\frac{z_1s'(z_1)}{s(z_1)}\right\}\right| \\ &= \left|\frac{\pi\gamma}{2}+\arg\left\{1+\frac{z_1s'(z_1)}{s(z_1)}\right\}\right| \\ &= \left|\frac{\pi\gamma}{2}+\arg\left\{1+i\gamma k\right\}\right| \\ &\geq\frac{\pi\gamma}{2}+\tan^{-1}\gamma \\ &= \frac{\alpha\pi}{2} \end{align*} $$
because of (3.2) but this contradicts hypothesis (3.1). This shows that
$$ \begin{align*} \left|\arg s(z)\right| <\frac{\gamma\pi}{2}, \quad z\in\mathbb D \end{align*} $$
because
$\arg s(0)=0$
and the hypothesis that there exists a point
$z_1\mathbb D$
with
$|\arg s(z_1)|=\pi \gamma /2$
gives the above contradiction.
Probably, the
$\gamma $
in (3.2) is not the smallest bound for
$|\arg q(z)/z|$
. The smallest bound gives the following theorem but not in explicit form.
Theorem 3.2 (See [Reference Hallenbeck and Ruscheweyh7])
Let the function h be analytic and convex univalent in
$\mathbb D$
with
$h(0)=a.$
Let also
$p(z)=a+b_{n}z^n+b_{n+1}z^{n+1}+ \cdots $
be analytic in
$\mathbb D.$
If
$$ \begin{align*} p(z)+\frac{zp'(z)}{c}\prec h(z),\quad z\in \mathbb D \end{align*} $$
for
$\mathrm {Re}c\geq 0,\ c\neq 0,$
then
$$ \begin{align*} p(z)\prec q_n(z) \prec h(z),\quad z\in \mathbb D, \end{align*} $$
where
$q_n(z)=\frac {c}{nz^{c/n}}\int \limits _{0}^{z}t^{{c/n}-1}h(t)\mathrm { d}t.$
Moreover, the function
$q_n(z)$
is convex univalent and is the best dominant in the sense that if
$p(z)\prec q(z)$
, then
$q_n(z)\prec q(z).$
Therefore, for
$p(z)+\frac {zp'(z)}{c}=q'(z)$
,
$c=1$
, we may apply the following version of the above Theorem 3.2.
Theorem 3.3 Let
$q'(z)=1+b_{n}z^n+b_{n+1}z^{n+1}+ \cdots $
be analytic in
$\mathbb D.$
If
$$ \begin{align*} q'(z)\prec h(z)=\left(\frac{1+z}{1-z}\right)^{\alpha}, \quad z\in \mathbb D,\quad\left[or~ if \quad\left|\arg q'(z)\right|<\frac{\alpha\pi}{2}, \quad z\in \mathbb D\right], \end{align*} $$
then
$$ \begin{align} q(z)/z\prec q_n(z) \prec h(z),\quad z\in \mathbb D, \end{align} $$
where
$q_n(z)=\frac {1}{nz^{1/n}}\int \limits _{0}^{z}t^{{1/n}-1}h(t)\mathrm { d}t.$
Moreover, the function
$q_n(z)$
is convex univalent and is the best dominant in the sense that if
$q(z)\prec p(z)$
, then
$q_n(z)\prec p(z).$
If we assume
$n=1$
, then (3.3) gives
$$ \begin{align*} q(z)/z\prec q_1(z) =\frac{1}{z} \int\limits_{0}^{z}\left(\frac{1+t}{1-t}\right)^{\alpha}\mathrm{d}t, \quad z\in \mathbb D. \end{align*} $$
From this we may find, for a given
$\alpha $
, the smallest
$\gamma $
such that
$$ \begin{align} \left|\arg\left\{\frac{1}{z}\int\limits_{0}^{z}\left(\frac{1+t}{1-t}\right)^{\alpha}\mathrm{ d}t\right\}\right| <\frac{\pi\gamma}{2},\quad z\in \mathbb D. \end{align} $$
This
$\gamma $
is probably smaller (and better) than the
$\gamma $
in (3.2) but to calculate
$\gamma $
in (3.4) is a hard problem.
Corollary 3.4 If
$q(z)\in \mathcal H$
,
$q(z)= 0$
,
$q(z)/z\neq 0,$
and
$q'(z)\neq 0$
in
$\mathbb D$
, and
$$ \begin{align*} \left|\arg q'(z)\right|<\frac{\alpha\pi}{2},\quad z\in\mathbb D, \end{align*} $$
for some
$\alpha>0$
, such that
$\alpha +\gamma <2$
, where
$\gamma>0$
is the smallest positive root of the equation
$$ \begin{align*} \frac{\pi\gamma}{2}+\tan^{-1}\gamma=\frac{\alpha\pi}{2}, \end{align*} $$
then
$$ \begin{align*} \left|\arg\frac{zq'(z)}{q(z)}\right|<\frac{\pi(\alpha+\gamma)}{2},\quad z\in\mathbb D. \end{align*} $$
Proof From Theorem 3.1, we have
$$ \begin{align*} \left|\arg\frac{q(z)}{z}\right|<\frac{\pi\gamma}{2}<\frac{\alpha\pi}{2},\quad z\in\mathbb D. \end{align*} $$
Therefore, we obtain
$$ \begin{align*} \left|\arg\frac{zq'(z)}{q(z)}\right| &\leq\left|\arg\left\{q'(z)\right\}\right|+\left|\arg\frac{q(z)}{z}\right|\\ &<\frac{\alpha\pi}{2}+\frac{\pi\gamma}{2}\\ &=\frac{\pi(\alpha+\gamma)}{2},\quad z\in \mathbb D.\\[-34pt] \end{align*} $$
If we take
$q(z)=f(z)$
,
$f\in \mathcal A$
, then Corollary 3.4 provides a sufficient condition for the function f to be strongly starlike of some order (see (1.3)).
Corollary 3.5 If
$f\in \mathcal A$
,
$f(z)/z\neq 0$
in
$\mathbb D$
, and
$$ \begin{align*} \left|\arg f'(z)\right|<\frac{\alpha\pi}{2},\quad z\in\mathbb D, \end{align*} $$
for some
$\alpha>0$
, such that
$\alpha +\gamma <2$
, where
$\gamma>0$
is the smallest positive root of the equation
$$ \begin{align} \frac{\pi\gamma}{2}+\tan^{-1}\gamma=\frac{\alpha\pi}{2}, \end{align} $$
then
$$ \begin{align*} \left|\arg \frac{zf'(z)}{f(z)}\right|<\frac{\pi(\alpha+\gamma)}{2},\quad z\in\mathbb D \end{align*} $$
or f is in the class
$\mathcal {SS}(\beta )$
of strongly starlike functions of order
$\beta =\alpha +\gamma $
(see (1.3)).
Therefore, the dimensions of strongly starlikeness of order
$\beta =\alpha +\gamma $
in the class
$ \mathcal A$
satisfies the condition
$$ \begin{align} D{\mathcal{SS}(\beta)}\{|\arg f'(z)|<\varepsilon\pi/2\}\geq\alpha, \end{align} $$
where
$\gamma $
is described in (3.5). For example, if we consider
$$ \begin{align*} \alpha=\frac{\sqrt{3}+1}{3}, \end{align*} $$
then from (3.5), we obtain
$$ \begin{align*} \gamma=\frac{\sqrt{3}}{3},\quad \beta=\alpha+\gamma=\frac{2\sqrt{3}+1}{3}<2 \end{align*} $$
and so (3.6) becomes
$$ \begin{align*} D{\mathcal{SS}((2\sqrt{3}+1)/3)}\{|\arg f'(z)| <\varepsilon\pi/2\}\geq\frac{\sqrt{3}+1}{3}. \end{align*} $$
Corollary 3.6 If
$f\in \mathcal A$
,
$f"(z)\neq 0,$
and
$f'(z)/z\neq 0$
in
$\mathbb D$
, and
$$ \begin{align*} \left|\arg f"(z)\right|<\frac{\alpha\pi}{2},\quad z\in\mathbb D, \end{align*} $$
for some
$\alpha>0$
, such that
$\alpha +\gamma <2$
, where
$\gamma $
is the smallest positive root of the equation
$$ \begin{align*} \frac{\pi\gamma}{2}+\tan^{-1}\gamma=\frac{\alpha\pi}{2}, \end{align*} $$
then
$$ \begin{align} \left|\arg \frac{zf"(z)}{f'(z)-1}\right| <\frac{\pi(\alpha+\gamma)}{2},\quad z\in\mathbb D. \end{align} $$
This means that
$f'(z)$
is strongly starlike of order
$\alpha +\gamma $
with respect to
$1=f'(0)$
.
Proof If we take
$q(z)=f'(z)-1$
,
$f\in \mathcal A$
, then Corollary 3.4 gives
$$ \begin{align*} \left|\arg\frac{zq'(z)}{q(z)}\right| =\left|\arg\frac{zf"(z)}{f'(z)-1}\right| <\frac{\pi(\alpha+\gamma)}{2},\quad z\in\mathbb D.\\[-34pt] \end{align*} $$
If we consider
$$ \begin{align*} \alpha=\sqrt{2}-\frac{3}{4}, \end{align*} $$
then
$$ \begin{align*} \gamma=\sqrt{2}-1,\quad \beta=\alpha+\gamma=2\sqrt{2}-\frac{7}{4}<2 \end{align*} $$
and so Corollary 3.6 becomes
$$ \begin{align*} \left|\arg f"(z)\right|<\frac{\alpha\pi}{2}\quad \Rightarrow \quad \left|\arg\frac{zf"(z)}{f'(z)-1}\right| <\frac{\pi(2\sqrt{2}-7/4)}{2}. \end{align*} $$
If we take
$\alpha =1,$
then Theorem 3.1 becomes the following corollary.
Corollary 3.7 If
$q(z)\in \mathcal H$
,
$q(0)=0,$
and
$$ \begin{align*} \left|\arg q'(z)\right|<\frac{\pi}{2},\quad z\in\mathbb D \end{align*} $$
and
$$ \begin{align*} \frac{q(z)}{z}\neq 0,\quad z\in\mathbb D, \end{align*} $$
then
$$ \begin{align*} \left|\arg \frac{q(z)}{z}\right|<\frac{\pi\gamma}{2}<\frac{\pi}{2},\quad z\in\mathbb D, \end{align*} $$
where
$\gamma =0.63832\ldots $
is the smallest positive root of the equation
$$ \begin{align*} \frac{\pi\gamma}{2}+\tan^{-1}\gamma=\frac{\pi}{2}. \end{align*} $$
Theorem 3.8 If
$f\in \mathcal A$
and
$$ \begin{align*} |\arg f'(z)|<\frac{\alpha\pi}{2},\quad z\in \mathbb D, \end{align*} $$
where
$\alpha = 0.6165513972\ldots $
is the smallest positive root of the equation
$$ \begin{align*} \pi(1-\alpha)+\tan^{-1}(1-\alpha)=\frac{\pi}{2}, \end{align*} $$
then
$$ \begin{align*} \left|\arg \frac{zf'(z)}{f(z)}\right|<\pi/2,\quad z\in \mathbb D \end{align*} $$
or
$f\in \mathcal S^*$
.
Proof If we take
$q(z)=f(z)$
,
$f\in \mathcal A$
, then from Theorem 3.1, the condition
$$ \begin{align} |\arg f'(z)|<\frac{\alpha\pi}{2},\quad z\in \mathbb D \end{align} $$
implies
$$ \begin{align} \left|\arg \frac{f(z)}{z}\right|<\frac{\pi\gamma}{2}<\frac{\alpha\pi}{2},\quad z\in\mathbb D, \end{align} $$
where
$\gamma $
is the smallest positive root of the equation
$$ \begin{align} \frac{\pi\gamma}{2}+\tan^{-1}\gamma=\frac{\alpha\pi}{2}. \end{align} $$
This gives
$$ \begin{align*} \left|\arg \frac{zf'(z)}{f(z)}\right| &\leq\left|\arg f'(z)\right|+\left|\arg \frac{f(z)}{z}\right|\\ &<\frac{\alpha\pi}{2}+\frac{\pi\gamma}{2}\\ &=\pi\gamma+\tan^{-1}\gamma,\quad z\in \mathbb D. \end{align*} $$
Then, we can see that if
$\gamma $
is such that
$$ \begin{align*} \pi\gamma+\tan^{-1}\gamma=\frac{\pi}{2} \end{align*} $$
(so
$\gamma =0.38344860\ldots $
), and if
$\frac {\alpha \pi }{2}+\frac {\pi \gamma }{2}=\frac {\pi }{2}$
, which is equivalent to
$$ \begin{align*} \alpha=1-\gamma= 0.6165513972\ldots, \end{align*} $$
then the inequality
$$ \begin{align*} |\arg f'(z)|<\frac{\alpha\pi}{2},\quad z\in \mathbb D \end{align*} $$
is a sufficient condition for the starlikeness and so
$D{\mathcal S^*}\{|\arg \{f'(z)\}| <\varepsilon \pi /2\}\geq 0.6165513972\ldots $
.
Theorem 3.8 says that the dimension of starlikeness in the class
$ \mathcal A$
satisfies
$$ \begin{align*} D{\mathcal S^*}\{|\arg f'(z)|<\varepsilon\pi/2\}\geq 0.6165513972\ldots. \end{align*} $$
The number
$0.6165513972\ldots $
cannot be extended to
$1$
because there exists a function
$g\in \mathcal A$
with
$|\arg \{g'(z)\}|<\pi /2$
which is not starlike (see [Reference Goodman5, p. 123]). Therefore, the dimension of starlikeness in the class
$ \mathcal A$
satisfies the condition
$$ \begin{align*} 0.6165513972\ldots\leq D{\mathcal S^*}\{|\arg\{f'(z)\}|<\varepsilon\pi/2\}<1. \end{align*} $$
Conjecture. The exact value of
$D{\mathcal S^*}\{|\arg f'(z)| <\varepsilon \pi /2\}$
is
$\sqrt {3}/2$
.
Corollary 3.9 If
$g\in \mathcal A$
,
$g'(z)\neq 0$
in
$\mathbb D$
and
$$ \begin{align} \left|\arg\left(zg'(z)\right)'\right|<\alpha\pi/2,\quad z\in \mathbb D, \end{align} $$
where
$\alpha = 0.6165513972\ldots $
is the smallest positive root of the equation
$$ \begin{align*} \pi(1-\alpha)+\tan^{-1}(1-\alpha)=\frac{\pi}{2}, \end{align*} $$
then
$$ \begin{align*} \left|\arg\left\{1+\frac{zg"(z)}{g'(z)}\right\}\right|<\frac{\pi}{2},\quad z\in\mathbb D, \end{align*} $$
or
$g\in \mathcal K$
.
Proof Let
$f(z)=zg'(z)$
. Applying (3.11) and Theorem 3.8, we obtain
$$ \begin{align*} \left|\arg\frac{zf'(z)}{f(z)}\right|= \left|\arg\left\{1+\frac{zg"(z)}{g'(z)}\right\}\right|<\pi/2,\quad z\in \mathbb D.\\[-34pt] \end{align*} $$
Therefore, Corollary 3.9 says that the dimension of convexity in the class
$ \mathcal A$
satisfies
$$ \begin{align*} D{\mathcal K}\{|\arg(zg'(z))'|<\varepsilon\pi/2\}\geq 0.6165513972\ldots. \end{align*} $$
The number
$0.6165513972\ldots $
cannot be extended to
$1$
because there exists a function
$f_0(z)\in \mathcal A$
with
$|\arg f_0'(z)|<\pi /2$
which is not starlike. If we consider
$g_0(z)\in \mathcal A$
such that
$zg_0'(z)=f_0(z)$
, then
$g_0(z)$
is not convex but satisfies
$|\arg \left (zg_0'(z)\right )'|<\pi /2$
. Therefore, the dimension of convexity under condition
$|\arg \left (zg'(z)\right )'|<\alpha \pi /2$
satisfies the inequalities
$$ \begin{align*} 1>D{\mathcal K}\{|\arg (zg'(z))'|<\varepsilon\pi/2\}\geq 0.6165513972\ldots. \end{align*} $$
Recall that
$f\in \mathcal A$
is in the class
$\mathcal B(\beta )$
of the Bazilevic̆ functions of type
$\beta $
if there exists a function
$g\in \mathcal S^*$
such that
$$ \begin{align} \left|\arg\left\{\frac{zf'(z)}{f^{1-\beta}(z)g^{\beta}(z)}\right\}\right| <\frac{\pi}{2},\quad z\in \mathbb D. \end{align} $$
Theorem 3.10 Let
$f\in \mathcal A$
be such that
$$ \begin{align} \left|\arg f'(z)\right|<\alpha_1\pi/2,\quad z\in \mathbb D, \end{align} $$
for some
$\alpha _1\in (0,2]$
. If there exists a function
$g\in \mathcal S^*$
such that
$$ \begin{align} \left|\arg g'(z)\right|<\alpha_2\pi/2,\quad z\in \mathbb D \end{align} $$
for some
$\alpha _2\in (0,2]$
, then f is a Bazilevic̆ function of type
$\beta $
, whenever
$\beta $
satisfies
$$ \begin{align} \alpha+\gamma_1|\beta-1|+\beta\gamma_2<1, \end{align} $$
where
$\gamma _t$
,
$t=1,2$
, is the smallest positive root of the equation
$$ \begin{align*} \frac{\pi\gamma_t}{2}+\tan^{-1}\gamma_t=\frac{\alpha_t\pi}{2}. \end{align*} $$
Proof By (3.8)–(3.10), we have from (3.13)
$$ \begin{align*} \left|\arg \frac{f(z)}{z}\right|<\gamma_1\pi/2,\quad z\in \mathbb D. \end{align*} $$
In the same way, by (3.14), we have
$$ \begin{align*} \left|\arg \frac{g(z)}{z}\right|<\gamma_2\pi/2,\quad z\in \mathbb D. \end{align*} $$
Therefore, we have
$$ \begin{align*} \left|\arg\frac{zf'(z)}{f^{1-\beta}(z)g^{\beta}(z)}\right| &=\left|\arg\left\{f'(z)\left(\frac{f(z)}{z}\right)^{\beta-1}\left(\frac{z}{g(z)}\right)^{\beta}\right\}\right|\\ &\leq\left|\arg f'(z)\right| +\left|(\beta-1)\arg\frac{f(z)}{z}\right| +\left|\beta\arg\frac{z}{g(z)}\right|\\ &<\frac{\alpha\pi}{2}+\frac{\gamma_1\pi}{2}|\beta-1|+\frac{\beta\gamma_2\pi}{2}\\ &=\frac{\pi}{2}(\alpha+\gamma_1|\beta-1|+\beta\gamma_2), \quad z\in \mathbb D.\\[-34pt] \end{align*} $$
For
$\beta =1,$
the class
$\mathcal B(\beta )$
of the Bazilevic̆ functions becomes a subclass of well-known class
$\mathcal C$
of close-to-convex functions.
Corollary 3.11 Let
$f\in \mathcal A$
be such that
$$ \begin{align} \left|\arg f'(z)\right|<\alpha_1\pi/2,\quad z\in \mathbb D, \end{align} $$
for some
$\alpha _1\in (0,2]$
. If there exists a function
$g\in \mathcal S^*$
such that
$$ \begin{align} \left|\arg g'(z)\right|<\alpha_2\pi/2,\quad z\in \mathbb D, \end{align} $$
where
$$ \begin{align*} \alpha_2\frac{\pi}{2}=(1-\alpha_1)\frac{\pi}{2}+\tan^{-1}(1-\alpha_1), \end{align*} $$
then f is a close-to-convex function.
Proof By (3.8)–(3.10), we have from (3.17)
$$ \begin{align*} \left|\arg \frac{g(z)}{z}\right|<(1-\alpha_1)\pi/2,\quad z\in \mathbb D. \end{align*} $$
Therefore, we have
$$ \begin{align*} \left|\arg \frac{zf'(z)}{g(z)}\right| &=\left|\arg\left\{f'(z)\left(\frac{z}{g(z)}\right)\right\}\right|\\ &\leq\left|\arg f'(z) \right| +\left|\arg\frac{z}{g(z)}\right|\\ &<\frac{\alpha_1\pi}{2}+\frac{(1-\alpha_1)\pi}{2}\\ &=\frac{\pi}{2}, \quad z\in \mathbb D.\\[-34pt] \end{align*} $$
Acknowledgements
The authors are thankful to the referee for pointing out some useful suggestions.
Author contributions
The authors declare that they have equal contribution in the article.
Competing interests
The authors declare that they have no financial interests. The authors have no competing interest to declare that are relevant to the content of this article.
Data availability statement
Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.
Funding
Not applicable.
Compliance with ethical standards
The authors did not receive support from any organization for the submitted work.
Ethical Approval
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