## 1 Introduction

Let A be the class of functions of the form

$f\left(z\right)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n},$
(1.1)

which are analytic in the open unit disc $E=\left\{z:|z|<1\right\}$. Let S, K, ${S}^{\ast }$, and C be the subclasses of A which consist of univalent, close-to-convex, starlike (with respect to origin), and convex functions, respectively. For recent developments, extensions, and applications, see [125] and the references therein.

A function f in A is said to be uniformly convex in E if f is a univalent convex function along with the property that, for every circular arc γ contained in E, with center ξ also in E, the image curve $f\left(\gamma \right)$ is a convex arc. The class of uniformly convex functions is denoted by $UCV$. The corresponding class $UST$ is defined by the relation that $f\in UCV$ if, and only if, $z{f}^{\prime }\in UST$. It is well known [13] that $f\in UCV$ if, and only if

$|\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}|<\mathrm{\Re }\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}\phantom{\rule{1em}{0ex}}\left(z\in E\right).$

Uniformly starlike and convex functions were first introduced by Goodman [3] and then studied by various other authors. If $f,g\in A$, we say f is subordinate to g in E, written as $f\prec g$ or $f\left(z\right)\prec g\left(z\right)$, if there exists a Schwarz function $w\left(z\right)$ such that $f\left(z\right)=g\left(w\left(z\right)\right)$ for $z\in E$.

For $0\le \beta <1$, the class $P\left(\beta \right)$ consists of functions $p\left(z\right)$ analytic in E with $p\left(0\right)=1$ such that $\mathrm{\Re }p\left(z\right)>\beta$ for $z\in E$, and, with $\beta =0$, we obtain the well-known class P of Carathéodory functions with positive real part.

For $k\in \left[0,\mathrm{\infty }\right)$, the conic regions ${\mathrm{\Omega }}_{k}$ are defined as follows, see [5]:

${\mathrm{\Omega }}_{k}=\left\{u+iv:u>k\sqrt{{\left(u-1\right)}^{2}+{v}^{2}}\right\}.$

For fixed k, ${\mathrm{\Omega }}_{k}$ represents the conic regions bounded, successively, by the imaginary axis ($k=0$), the right branch of a hyperbolic ($0) and a parabola ${v}^{2}=2u-1$ ($k=1$). When $k>1$, the domain becomes a bounded domain being the interior of the ellipse.

We shall consider the case when $k\in \left[0,1\right]$. Related to the domain ${\mathrm{\Omega }}_{k}$, the following functions ${p}_{k}\left(z\right)$, $k\in \left[0,1\right]$, play the role of extremal functions mapping in E onto ${\mathrm{\Omega }}_{k}$:

${p}_{k}\left(z\right)=\left\{\begin{array}{cc}\frac{1+z}{1-z}\hfill & \left(k=0\right),\hfill \\ 1+\frac{2}{{\pi }^{2}}{\left(log\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2}\hfill & \left(k=1\right),\hfill \\ 1+\frac{2}{1-{k}^{2}}{sinh}^{2}\left[\left(\frac{2}{\pi }arccosk\right)arctanh\sqrt{z}\right]\hfill & \left(0
(1.2)

These functions are univalent in E and belong to the class P. Using the subordination concept, we define the class $P\left({p}_{k}\right)$ as follows.

Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1$. Then $p\in P\left({p}_{k}\right)$ if, and only if, $p\prec {p}_{k}$ in E and ${p}_{k}\left(z\right)$ are given by (1.2).

The conic domains ${\mathrm{\Omega }}_{k}$ can be generalized as given by

${\mathrm{\Omega }}_{k,\beta }=\left(1-\beta \right){\mathrm{\Omega }}_{k}+\beta ,$

with the corresponding extremal function

${p}_{k,\beta }\left(z\right)=\left(1-\beta \right){p}_{k}+\beta \phantom{\rule{1em}{0ex}}\left(0\le \beta <1,k\in \left[0,1\right]\right).$

It can easily be seen that the analytic function $p\left(z\right)$, with $p\left(0\right)=1$, belongs to the class $P\left({p}_{k,\beta }\right)$ if $p\left(z\right)\prec {p}_{k,\beta }\left(z\right)$ in E.

It is easy to verify that $P\left({p}_{k,\beta }\right)$ is a convex set. It is known [6] that

$P\left({p}_{k}\right)\subset P\left(\frac{k}{k+1}\right)\subset P,$

and, for $p\in P\left({p}_{k}\right)$, we have

$|argp\left(z\right)|\le \sigma \frac{\pi }{2},$

where

$\sigma =\frac{2}{\pi }arctan\frac{1}{k}.$
(1.3)

So we can write $p\left(z\right)={h}^{\sigma }\left(z\right)$, $h\in P$.

Also

$P\left({p}_{k,\beta }\right)\subset P\left(\frac{k+\beta }{k+1}\right)\subset P.$

Sakaguchi [24] introduced and studied the class ${S}_{s}^{\ast }$ of starlike functions with respect to symmetrical points. The class ${S}_{s}^{\ast }$ includes the classes of convex and odd starlike functions with respect to the origin. It was shown [24] that a necessary and sufficient condition for $f\in {S}_{s}^{\ast }$ to be univalent and starlike with respect to symmetrical points in E is that

$\left(\frac{2z{f}^{\prime }\left(z\right)}{f\left(z\right)-f\left(-z\right)}\right)\in P,\phantom{\rule{1em}{0ex}}z\in E.$

Das and Singh [2] defined the classes ${C}_{s}$ of convex functions with respect to symmetrical points and showed that a necessary and sufficient condition for $f\in {C}_{s}$ is that

$\frac{2{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{{\left(f\left(z\right)-f\left(-z\right)\right)}^{\prime }}\in P,\phantom{\rule{1em}{0ex}}z\in E.$

It is also well known [2] that $f\in {C}_{s}$ if, and only if, $z{f}^{\prime }\in {S}_{s}^{\ast }$.

We now define the following.

Definition 1.1 Let $f\in A$. The f is said to be in the class $k-S{T}_{s}\left(\beta \right)$ if, and only if,

$\frac{2z{f}^{\prime }\left(z\right)}{\left(f\left(z\right)-f\left(-z\right)\right)}\in P\left({p}_{k,\beta }\right),\phantom{\rule{1em}{0ex}}z\in E.$

It can easily be seen that

$k-S{T}_{s}\left(\beta \right)\subset {S}_{s}^{\ast }\subset {S}_{s}^{\ast },\phantom{\rule{2em}{0ex}}{\beta }_{1}=\frac{k+\beta }{k+1}.$

Also, for $\beta =0=k$, the class $k-S{T}_{s}\left(\beta \right)$ reduces to ${S}_{s}^{\ast }$.

The class $k-UC{V}_{s}\left(\beta \right)$ is defined as follows.

Definition 1.2 Let $f\in A$. Then $f\in k-UC{V}_{s}\left(\beta \right)$ if, and only if $z{f}^{\prime }\in k-S{T}_{s}\left(\beta \right)$ for $z\in E$.

We note that

$k-UC{V}_{s}\left(\beta \right)\subset {C}_{s}\left({\beta }_{1}\right)\subset {C}_{s},\phantom{\rule{2em}{0ex}}{\beta }_{1}=\frac{k+\beta }{k+1}.$

Definition 1.3 Let $f\in A$. Then $f\in k-U{K}_{s}\left(\beta \right)$ if, and only if, there exists $g\in k-S{T}_{s}\left(\beta \right)$ such that

$\left(\frac{2z{f}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}\right)\in P\left({p}_{k,\beta }\right),\phantom{\rule{1em}{0ex}}z\in E.$

Since $P\left({p}_{k,\beta }\right)\subset P\left({\beta }_{1}\right)\subset P$, ${\beta }_{1}=\frac{k+\beta }{k+1}$, and $k-S{T}_{s}\left(\beta \right)\subset {S}_{s}^{\ast }$, we note that

$k-U{K}_{s}\left(\beta \right)\subset {K}_{s}\subset K,$

where ${k}_{S}$ consists of close-to-convex functions with respect to symmetrical starlike functions.

From the definition, it is clear that $k-U{K}_{s}\left(\beta \right)$ consists of univalent functions.

For $k=0$, $\beta =0$ and $f\left(z\right)=g\left(z\right)$, $k-U{K}_{s}\left(\beta \right)$ reduces to the class ${S}_{s}^{\ast }$.

## 2 Preliminary results

We shall need the following lemmas to prove our main results.

Lemma 2.1 [15]

Let $q\left(z\right)$ be a convex function in E with $q\left(0\right)=1$ and let another function $h:E\to \mathbb{C}$ be with $\mathrm{\Re }h\left(z\right)>0$. Let $p\left(z\right)$ be analytic in E with $p\left(0\right)=1$ such that

$\left(p\left(z\right)+h\left(z\right)z{p}^{\prime }\left(z\right)\right)\prec q\left(z\right),\phantom{\rule{1em}{0ex}}z\in E.$

Then $p\left(z\right)\prec q\left(z\right)$, $z\in E$.

Lemma 2.2 Let $N\left(z\right)$, $D\left(z\right)$ be analytic in E with

$N\left(0\right)=0=D\left(z\right)$

and let $D\in {S}^{\ast }$ for $z\in E$. Then $\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}\in P\left({p}_{k,\beta }\right)$ implies that $\frac{N\left(z\right)}{D\left(z\right)}\in P\left({p}_{k,\beta }\right)$ for $z\in E$.

Proof Let

$\frac{N\left(z\right)}{D\left(z\right)}=p\left(z\right).$

Then

$\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}=p\left(z\right)+h\left(z\right)\left(z{p}^{\prime }\left(z\right)\right),\phantom{\rule{1em}{0ex}}h\left(z\right)=\frac{1}{{h}_{0}\left(z\right)},$

where

${h}_{0}\left(z\right)=\frac{z{D}^{\prime }\left(z\right)}{D\left(z\right)}\in P.$

Since $\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}\in P\left({p}_{k,\beta }\right)$, we have

$\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}=\left(p\left(z\right)+h\left(z\right)\left(z{p}^{\prime }\left(z\right)\right)\right)\prec {p}_{k,\beta }\left(z\right),\phantom{\rule{1em}{0ex}}z\in E.$

We now use Lemma 2.1 and this implies that

This proves that $\frac{N\left(z\right)}{D\left(z\right)}\in P\left({p}_{k,\beta }\right)$ for $z\in E$. □

The following lemma is an easy extension of a result proved in [5].

Lemma 2.3 Let $k\in \left[0,\mathrm{\infty }\right)$ and ${\gamma }_{1}$, ${\delta }_{1}$ be any complex numbers with ${\gamma }_{1}\ne 0$ and let $\mathrm{\Re }\left\{\frac{{\gamma }_{1}k}{k+1}+{\delta }_{1}\right\}>\beta$. If $h\left(z\right)$ is analytic in E, $h\left(0\right)=1$ and it satisfies

$\left(h\left(z\right)+\frac{z{h}^{\prime }\left(z\right)}{{\gamma }_{1}h\left(z\right)+{\delta }_{1}}\right)\prec {p}_{k,\beta }\left(z\right),$
(2.1)

and ${q}_{k,\beta }\left(z\right)$ is an analytic solution of

$\left({q}_{k,\beta }\left(z\right)+\frac{z{q}_{k,\beta }^{\prime }\left(z\right)}{{\gamma }_{1}{q}_{k,\beta }\left(z\right)+{\delta }_{1}}\right)={p}_{k,\beta }\left(z\right),$

then ${q}_{k,\beta }$ is univalent and

$h\left(z\right)\prec {q}_{k,\beta }\left(z\right)\prec {p}_{k,\beta }\left(z\right),$

and ${q}_{k,\beta }\left(z\right)$ is the best dominant of (2.1).

## 3 The class $k-S{T}_{s}\left(\beta \right)$

In this section, we shall study some basic properties of the class $k-S{T}_{s}\left(\beta \right)$.

Theorem 3.1 Let $f\in k-S{T}_{s}\left(\beta \right)$. Then the odd function

$\mathrm{\Psi }\left(z\right)=\frac{1}{2}\left[f\left(z\right)-f\left(-z\right)\right],$
(3.1)

belongs to $k-ST\left(\beta \right)$ in E.

In particular $\mathrm{\Psi }\left(z\right)$ is an odd starlike function of order ${\beta }_{1}=\frac{k+\beta }{k+1}$ in E.

Proof Logarithmic differentiation of (3.1) and simple computation yield

Since $P\left({p}_{k,\beta }\right)$ is a convex set, it follows that $\frac{z{\mathrm{\Psi }}^{\prime }\left(z\right)}{\mathrm{\Psi }\left(z\right)}\in P\left({p}_{k,\beta }\right)$ and thus $\mathrm{\Psi }\in k-ST\left(\beta \right)$ in E. □

As a special case, we note that, for $k=0=\beta$, $\frac{1}{2}\left[f\left(z\right)-f\left(-z\right)\right]=\mathrm{\Psi }\left(z\right)\in {S}^{\ast }$ in E, and hence $\frac{z{f}^{\prime }}{\mathrm{\Psi }}\in P$. We now discuss a geometric property for $f\in k-S{T}_{s}\left(\beta \right)$. Here we investigate the behavior of the inclusion of the tangent at a point $w\left(\theta \right)=f\left(r{e}^{i\theta }\right)$ to the image ${\mathrm{\Gamma }}_{r}$ of the circle ${C}_{r}=\left\{z:|z|=r\right\}$, $0\le r<1$, $\theta \in \left[0,2\pi \right]$, under the mapping by means of a function f from the class $f\in k-S{T}_{s}\left(\beta \right)$.

Let

$\mathrm{\Phi }\left(\theta \right)=\frac{\pi }{2}+\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)=arg\frac{\partial }{\partial \theta }f\left(r{e}^{i\theta }\right),$

and, for ${\theta }_{2}>{\theta }_{1}$, ${\theta }_{1},{\theta }_{2}\in \left[0,2\pi \right]$,

$\mathrm{\Phi }\left({\theta }_{2}\right)-\mathrm{\Phi }\left({\theta }_{1}\right)={\theta }_{2}+arg{f}^{\prime }\left(r{e}^{i{\theta }_{2}}\right)-{\theta }_{1}-arg{f}^{\prime }\left(r{e}^{i{\theta }_{1}}\right).$

Now, since

$\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)=\theta +\mathrm{\Re }\left\{-iln{f}^{\prime }\left(r{e}^{i\theta }\right)\right\},$

then

$\frac{\partial }{\partial \theta }\left(\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)\right)=\mathrm{\Re }\left\{1+\frac{r{e}^{i\theta }{f}^{″}\left(r{e}^{i\theta }\right)}{{f}^{\prime }\left(r{e}^{i\theta }\right)}\right\}.$

Hence

${\int }_{{\theta }_{1}}^{{\theta }_{2}}\frac{\partial }{\partial \theta }\left(\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)\right)\phantom{\rule{0.2em}{0ex}}d\theta ={\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{1+\frac{r{e}^{i\theta }{f}^{″}\left(r{e}^{i\theta }\right)}{{f}^{\prime }\left(r{e}^{i\theta }\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta .$

Also, on the other hand,

$\begin{array}{rl}{\int }_{{\theta }_{1}}^{{\theta }_{2}}\frac{\partial }{\partial \theta }\left(\theta +arg{f}^{\prime }\left(r{e}^{i\theta }\right)\right)\phantom{\rule{0.2em}{0ex}}d\theta & ={\theta }_{2}+arg{f}^{\prime }\left(r{e}^{i{\theta }_{2}}\right)-{\theta }_{1}-arg{f}^{\prime }\left(r{e}^{i{\theta }_{1}}\right)\\ =\mathrm{\Phi }\left({\theta }_{2}\right)-\mathrm{\Phi }\left({\theta }_{1}\right).\end{array}$

So, the integral on the left side of the last inequality characterizes the increment of the angle of the inclination of the tangent to the curve ${\mathrm{\Gamma }}_{r}$ between the points $w\left({\theta }_{2}\right)$ and $w\left({\theta }_{1}\right)$ for ${\theta }_{2}>{\theta }_{1}$.

We have the following necessary condition for $f\in k-S{T}_{s}\left(\beta \right)$.

Theorem 3.2 Let $f\in k-S{T}_{s}\left(\beta \right)$. Then, with $z=r{e}^{i\theta }$ and $0\le {\theta }_{1}<{\theta }_{2}\le 2\pi$, $0\le \beta <1$ and $0\le k\le 1$, we have

${\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{{f}^{\prime }\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta >-\sigma \pi +2{cos}^{-1}\left\{\frac{2\left(1-\beta \right)}{1-\left(1-2\beta \right){r}^{2}}\right\}+{\beta }_{1}\left({\theta }_{2}-{\theta }_{1}\right),$

where σ is given by (1.3) and ${\beta }_{1}=\frac{k+\beta }{k+1}$.

Proof Since $\frac{{f}^{\prime }\left(z\right)}{{\mathrm{\Psi }}^{\prime }\left(z\right)}\in P\left({p}_{k,\beta }\right)$, $\mathrm{\Psi }\left(z\right)=\frac{1}{2}\left[f\left(z\right)-f\left(-z\right)\right]$ and $\mathrm{\Psi }\in k-UCV\left(\beta \right)\subset C\left(\beta \right)$.

We can write

${f}^{\prime }\left(z\right)={\left({\mathrm{\Psi }}_{1}^{\prime }\left(z\right)\right)}^{1-{\beta }_{1}}{h}^{\sigma }\left(z\right),\phantom{\rule{1em}{0ex}}{\mathrm{\Psi }}_{1}\in C,h\in P\left(\beta \right),$

and this gives us, with $z=r{e}^{i\theta }$, $0\le r<1$, $0\le {\theta }_{1}<{\theta }_{2}\le 2\pi$,

$\begin{array}{rl}{\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{{f}^{\prime }\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta =& \left(1-{\beta }_{1}\right){\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{\frac{{\left(z{\mathrm{\Psi }}_{1}^{\prime }\left(z\right)\right)}^{\prime }}{{\mathrm{\Psi }}_{1}^{\prime }\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta \\ +\sigma {\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\frac{2{h}^{\prime }\left(z\right)}{h\left(z\right)}\phantom{\rule{0.2em}{0ex}}d\theta +{\beta }_{1}\left({\theta }_{2}-{\theta }_{1}\right).\end{array}$
(3.2)

For $h\in P\left(\beta \right)$, we observe that

$\begin{array}{rl}\frac{\partial }{\partial \theta }argh\left(r{e}^{i\theta }\right)& =\frac{\partial }{\partial \theta }\mathrm{\Re }\left\{-ilnh\left(r{e}^{i\theta }\right)\right\}\\ =\mathrm{\Re }\left\{r{e}^{i\theta }\frac{{h}^{\prime }\left(r{e}^{i\theta }\right)}{h\left(r{e}^{i\theta }\right)}\right\}.\end{array}$

Therefore

${\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{\frac{r{e}^{i\theta }{h}^{\prime }\left(r{e}^{i\theta }\right)}{h\left(r{e}^{i\theta }\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta =argh\left(r{e}^{i{\theta }_{2}}\right)-argh\left(r{e}^{i{\theta }_{1}}\right),$

and

$\underset{h\in P\left(\beta \right)}{max}|{\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{\frac{r{e}^{i\theta }{h}^{\prime }\left(r{e}^{i\theta }\right)}{h\left(r{e}^{i\theta }\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta |=\underset{h\in P\left(\beta \right)}{max}|argh\left(r{e}^{i{\theta }_{2}}\right)-argh\left(r{e}^{i{\theta }_{1}}\right)|.$

We can write

$\frac{1}{1-\beta }\left[h\left(z\right)-\beta \right]=p\left(z\right),\phantom{\rule{1em}{0ex}}p\in P,$

and for $|z|=r<1$, it is well known that

$|p\left(z\right)-\frac{1+{r}^{2}}{1-{r}^{2}}|\le \frac{2r}{1-{r}^{2}}.$

From this, we have

$|h\left(z\right)-\frac{1+\left(1-2\beta \right){r}^{2}}{1-{r}^{2}}|\le \frac{2\left(1-\beta \right)r}{1-{r}^{2}}.$

Thus the values of h are contained in the circle of Apollonius whose diameter is the line segment from $\frac{1-\left(1-2\beta \right)r}{1+r}$ to $\frac{1+\left(1-2\beta \right)r}{1-r}$ and has the radius $\frac{2\left(1-\beta \right)r}{1-{r}^{2}}$. So $|argh\left(z\right)|$ attains its maximum at points where a ray from origin is tangent to the circle, that is, when

$argh\left(z\right)=±{sin}^{-1}\left(\frac{2\left(1-\beta \right)r}{1-\left(1-2\beta \right){r}^{2}}\right).$
(3.3)

From (3.3), we observe that

$\begin{array}{rl}\underset{h\in P\left(\beta \right)}{max}|{\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{r{e}^{i\theta }\frac{{h}^{\prime }\left(r{e}^{i\theta }\right)}{h\left(r{e}^{i\theta }\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta |& \le 2{sin}^{-1}\left(\frac{2\left(1-\beta \right)r}{1-\left(1-2\beta \right){r}^{2}}\right)\\ =\pi -2{cos}^{-1}\left(\frac{2\left(1-\beta \right)r}{1-\left(1-2\beta \right){r}^{2}}\right).\end{array}$
(3.4)

Also, for ${\mathrm{\Psi }}_{1}\in C$,

${\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{1+r{e}^{i\theta }\frac{{\mathrm{\Psi }}_{1}^{″}\left(r{e}^{i\theta }\right)}{{\mathrm{\Psi }}_{1}^{\prime }\left(r{e}^{i\theta }\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta \ge 0.$
(3.5)

Using (3.4) and (3.5) in (3.2), we obtain the required result. □

We note the following special cases:

1. 1.

For $k=0$, $0\le {\theta }_{1}<{\theta }_{2}\le 2\pi$, $z=r{e}^{i\theta }$, it follows from Theorem 3.2 that

${\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta >-\pi \phantom{\rule{1em}{0ex}}\left(z\in E\right).$

This is a necessary and sufficient condition for f to be close-to-convex (hence univalent) in E; see [7]. This also shows that $S{T}_{s}\left(\beta \right)\subset K$.

1. 2.

For $k=1$ ${\int }_{{\theta }_{1}}^{{\theta }_{2}}\mathrm{\Re }\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}\phantom{\rule{0.2em}{0ex}}d\theta >-\frac{\pi }{2}$.

2. 3.

When $k\in \left[0,1\right]$, it is obvious that $\sigma \in \left(0,1\right]$. In this case, the class $k-S{T}_{s}\left(\beta \right)$ consists of strongly close-to-convex functions of order σ in the sense of Pommerenke [20, 21].

Theorem 3.3 (Integral representation)

Let $f\in k-S{T}_{s}\left(\beta \right)$. Then

${f}^{\prime }\left(z\right)=\frac{1}{2}p\left(z\right)exp{\int }_{0}^{z}\frac{1}{t}\left[p\left(t\right)+p\left(-t\right)-2\right]\phantom{\rule{0.2em}{0ex}}dt,$

where $p\in P\left({p}_{k,\beta }\right)$, $z\in E$.

Proof Since $f\in k-S{T}_{s}\left(\beta \right)$, we can write

$\frac{2z{f}^{\prime }\left(z\right)}{f\left(z\right)-f\left(-z\right)}=p\left(z\right),\phantom{\rule{1em}{0ex}}p\in P\left({p}_{k,\beta }\right).$

This gives us

$\frac{2{\left[f\left(z\right)-f\left(-z\right)\right]}^{\prime }}{f\left(z\right)-f\left(-z\right)}-\frac{1}{z}=\frac{1}{2}\left[p\left(z\right)-p\left(-z\right)-2\right]$

and the result follows when we integrate. □

When $k=0$, $\beta =0$, we obtain the result for the class ${S}_{s}^{\ast }$ given in [5].

We now study the class $k-S{T}_{s}\left(\beta \right)$ under a certain integral operator.

Theorem 3.4 Let $g\in k-S{T}_{s}\left(\beta \right)$ and let for $m=1,2,3,\dots ,G$ be defined by

$G\left(z\right)=\frac{m+1}{2{z}^{m}}{\int }_{0}^{z}{t}^{m-1}\left\{g\left(t\right)-g\left(-t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt.$
(3.6)

Then $G\left(z\right)$ belongs to $k-S{T}_{s}\left(\beta \right)$ in E.

Proof Let

$J\left(z\right)={\int }_{0}^{z}{t}^{m-1}\frac{g\left(t\right)-g\left(-t\right)}{2}\phantom{\rule{0.2em}{0ex}}dt.$

Since $g\in k-S{T}_{s}\left(\beta \right)$, $\frac{1}{2}\left\{g\left(z\right)-g\left(-z\right)\right\}\in k-ST\left(\beta \right)\subset {S}^{\ast }\left({\beta }_{1}\right)\subset {S}^{\ast }$, and ${\beta }_{1}=\frac{k+\beta }{k+1}$. Therefore it can easily be verified that $J\left(z\right)$ is $\left(m+1\right)$-valently starlike in E.

We can write (3.6) as

${z}^{m}G\left(z\right)=\left(m+1\right)J\left(z\right),$

and, differentiating logarithmically, we have

$\frac{z{G}^{\prime }\left(z\right)}{G\left(z\right)}=\frac{z{J}^{\prime }\left(z\right)-mJ\left(z\right)}{J\left(z\right)}=\frac{N\left(z\right)}{D\left(z\right)},$

say, where $N\left(0\right)=D\left(0\right)=0$ and D is $\left(m+1\right)$-valently starlike.

Let

$\frac{N\left(z\right)}{D\left(z\right)}=h\left(z\right).$

Then

$\begin{array}{rl}\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}& =h\left(z\right)+\frac{z{h}^{\prime }\left(z\right)}{{h}_{0}\left(z\right)},\phantom{\rule{1em}{0ex}}{h}_{0}\left(z\right)=\frac{z{D}^{\prime }\left(z\right)}{D\left(z\right)}\in P\\ =h\left(z\right)+{H}_{0}\left(z\right)\left(z{h}^{\prime }\left(z\right)\right),\phantom{\rule{1em}{0ex}}{H}_{0}=\frac{1}{{h}_{0}}\in P.\end{array}$
(3.7)

Since

$\begin{array}{rl}\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}& =\frac{{\left(z{h}^{\prime }\left(z\right)\right)}^{\prime }-m{J}^{\prime }\left(z\right)}{{J}^{\prime }\left(z\right)}\\ =\left\{\frac{{\left(z{J}^{\prime }\left(z\right)\right)}^{\prime }}{{J}^{\prime }\left(z\right)}-m\right\}\in P\left({p}_{k,\beta }\right).\end{array}$

We now apply Lemma 2.2 to obtain

$\frac{N\left(z\right)}{D\left(z\right)}=\frac{z{G}^{\prime }\left(z\right)}{G\left(z\right)}\in P\left({p}_{k,\beta }\right),\phantom{\rule{1em}{0ex}}z\in E.$

This proves that $G\in k-ST\left(\beta \right)$ in E. □

Theorem 3.5 Let $f,g\in k-S{T}_{s}\left(\beta \right)$ and let F be defined by the following integral operator:

$F\left(z\right)=\left(\gamma +\frac{1}{\delta }\right){z}^{1-\frac{1}{\delta }}{\int }_{0}^{z}{t}^{\frac{1}{\delta }-2}{\left[\frac{f\left(t\right)-f\left(-t\right)}{2}\right]}^{\frac{1}{1+\gamma }}\left[\frac{g\left(t\right)-g\left(-t\right)}{2}\right]\phantom{\rule{0.2em}{0ex}}dt,$
(3.8)

where $z\in E$, $\delta >0$, $\gamma \ge 0$ and $\left[\frac{k\left(1+\gamma \right)}{k+1}+\left(\frac{1}{\delta }-1\right)\right]>\beta$. Then $F\left(z\right)$ belongs to $k-ST\left(\beta \right)$ for $z\in E$.

When $g\left(z\right)=z$, $\gamma =0$, we obtain a generalized form of the Bernardi operator; see [1]. Also for $g\left(z\right)=z$, $\gamma =0$, and $\delta =\frac{1}{2}$, we have the well-known integral operator studied by Libera [11] who showed that it preserves the geometric properties of convexity, starlikeness, and close-to-convexity.

Proof Let $\frac{f\left(z\right)-f\left(-z\right)}{2}={\mathrm{\Psi }}_{1}\left(z\right)$, $\frac{g\left(z\right)-g\left(-z\right)}{2}={\mathrm{\Psi }}_{2}\left(z\right)$. Then ${\mathrm{\Psi }}_{1},{\mathrm{\Psi }}_{2}\in k-ST\left(\beta \right)$ in E. We can write (3.8) as

$F\left(z\right)=\left(\gamma +\frac{1}{\delta }\right){z}^{1-\frac{1}{\delta }}{\int }_{0}^{z}{t}^{\frac{1}{\delta }-2}{\left({\mathrm{\Psi }}_{1}\left(t\right)\right)}^{\frac{1}{1+\gamma }}\left({\mathrm{\Psi }}_{2}\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}dt.$
(3.9)

Differentiating (3.9) logarithmically, and with $p\left(z\right)=\frac{z{F}^{\prime }\left(z\right)}{F\left(z\right)}$, we have

$\frac{\gamma }{1+\gamma }\frac{z{\mathrm{\Psi }}_{1}^{\prime }}{{\mathrm{\Psi }}_{1}\left(z\right)}+\frac{1}{1+\gamma }\frac{z{\mathrm{\Psi }}_{2}^{\prime }}{{\mathrm{\Psi }}_{2}\left(z\right)}=p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\left(1+\gamma \right)p\left(z\right)+\left(\frac{1}{\delta }-1\right)}.$
(3.10)

Since, for $i=1,2$, ${\mathrm{\Psi }}_{i}\in k-ST\left(\beta \right)$, $\frac{z{\mathrm{\Psi }}_{1}^{\prime }\left(z\right)}{{\mathrm{\Psi }}_{1}}={h}_{1}\left(z\right)$, $\frac{z{\mathrm{\Psi }}_{2}^{\prime }\left(z\right)}{{\mathrm{\Psi }}_{2}}={h}_{2}\left(z\right)$ both belong to $P\left({p}_{k,\beta }\right)$ in E, and $P\left({p}_{k,\beta }\right)$ is a convex set. Therefore

$\left(\frac{\gamma }{1+\gamma }{h}_{1}\left(z\right)+\frac{1}{1+\gamma }{h}_{2}\left(z\right)\right)\in P\left({p}_{k,\beta }\right),\phantom{\rule{1em}{0ex}}z\in E.$
(3.11)

From (3.10) and (3.11), it follows that

$\left(p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\left(1+\gamma \right)p\left(z\right)+\left(\frac{1}{\delta }-1\right)}\right)\prec {p}_{k,\beta }\left(z\right).$

We now apply Lemma 2.3 which gives us

$p\left(z\right)\prec {q}_{k,\beta }\left(z\right)\prec {p}_{k,\beta }\left(z\right).$

Thus $F\in k-ST\left(\beta \right)$ and the proof is complete. □

## 4 The class $k-U{K}_{s}\left(\beta \right)$

Here we shall study some properties of the class $k-U{K}_{s}\left(\beta \right)$ which consists of k-uniformly close-to-convex functions.

Let $L\left(r,f\right)$ denote the length of the image of the circle $|z|=r$ under f. We prove the following.

Theorem 4.1 Let $f\in k-U{K}_{s}\left(\beta \right)$. Then, for $0, $k\in \left[0,1\right]$,

$L\left(r,f\right)=O\left(1\right){\left(\frac{1}{1-r}\right)}^{\sigma -{\beta }_{1}},\phantom{\rule{1em}{0ex}}{\beta }_{1}<\frac{\sigma }{2},$

where ${\beta }_{1}=\frac{k+\beta }{k+1}$ and σ is given by (1.3), and $O\left(1\right)$ is a constant depending only on k, β.

Proof For $f\in k-U{K}_{s}\left(\beta \right)$, we can write

$z{f}^{\prime }\left(z\right)=\mathrm{\Psi }\left(z\right){h}^{\sigma }\left(z\right),\phantom{\rule{1em}{0ex}}h\in P,\mathrm{\Psi }\in {S}^{\ast }\left({\beta }_{1}\right),$
(4.1)

and $\mathrm{\Psi }\left(z\right)=\left\{g\left(z\right)-g\left(-z\right)\right\}$, $g\in k-S{T}_{s}\left(\beta \right)$.

Since $\mathrm{\Psi }\in {S}^{\ast }\left({\beta }_{1}\right)$ and is odd, there exists an odd starlike function ${\mathrm{\Psi }}_{1}\left(z\right)$ such that

$\mathrm{\Psi }\left(z\right)=z{\left(\frac{{\mathrm{\Psi }}_{1}\left(z\right)}{z}\right)}^{1-{\beta }_{1}}=z{\left(\frac{{\mathrm{\Psi }}_{1}\left(z\right)}{z}\right)}^{\frac{1-{\beta }_{1}}{k+1}}.$

Thus, with $z=r{e}^{i\theta }$,

$L\left(r,f\right)={\int }_{0}^{2\pi }|z{f}^{\prime }\left(z\right)|\phantom{\rule{0.2em}{0ex}}d\theta ={\int }_{0}^{2\pi }|{z}^{{\beta }_{1}}{\left({\mathrm{\Psi }}_{1}\left(z\right)\right)}^{1-{\beta }_{1}}{h}^{\sigma }\left(z\right)|\phantom{\rule{0.2em}{0ex}}d\theta ,$

and using Hölder’s inequality, we have

$L\left(r,f\right)\le 2\pi {r}^{{\beta }_{1}}{\left(\frac{1}{2\pi }{\int }_{0}^{2\pi }|{\mathrm{\Psi }}_{1}\left(z\right){|}^{\left(1-\beta \right)\left(\frac{z}{z-\sigma }\right)}\phantom{\rule{0.2em}{0ex}}d\theta \right)}^{\frac{2-\sigma }{z}}{\left(\frac{1}{2\pi }{\int }_{0}^{2\pi }|h\left(z\right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\theta \right)}^{\frac{\sigma }{2}}.$
(4.2)

For $h\in P$, it is well known [20] that

$\frac{1}{2\pi }{\int }_{0}^{2\pi }|h\left(z\right){|}^{2}\phantom{\rule{0.2em}{0ex}}d\theta \le \frac{1+3{r}^{2}}{1-{r}^{2}}.$
(4.3)

Using (4.3) and subordination for odd starlike functions in (4.2), it follows that

$\begin{array}{rl}L\left(r,f\right)& \le C\left({\beta }_{1},\sigma \right){\left(\frac{1}{1-{r}^{2}}\right)}^{\left[\left(1-{\beta }_{1}\right)\left(\frac{2}{2-\sigma }\right)-1\right]{\left[\frac{1+3{r}^{2}}{1-r}\right]}^{\frac{\sigma }{2}}}\\ =O\left(1\right){\left(\frac{1}{1-r}\right)}^{\sigma -{\beta }_{1}},\end{array}$

where C and $O\left(1\right)$ are constants depending only on ${\beta }_{1}$ and σ. This completes the proof. □

We now discuss the growth rate of coefficients of $f\in k-U{K}_{s}\left(\beta \right)$.

Theorem 4.2 Let $f\in k-U{K}_{s}\left(\beta \right)$ and be given by (1.1). Then

${a}_{n}=O\left(1\right){n}^{\sigma -{\beta }_{1}-1},\phantom{\rule{1em}{0ex}}n\ge 1,{\beta }_{1}<\frac{\sigma }{2},$

where $O\left(1\right)$ is a constant depending only on σ and ${\beta }_{1}$ and σ, ${\beta }_{1}$ are as given in Theorem  4.1.

Proof For $z=r{e}^{i\theta }$, $n\ge 1$, Cauchy’s Theorem gives us

$\begin{array}{rl}n|{a}_{n}|& =\frac{1}{2\pi {r}^{n+1}}|{\int }_{0}^{2\pi }z{f}^{\prime }\left(z\right){e}^{-in\theta }\phantom{\rule{0.2em}{0ex}}d\theta |\\ \le \frac{1}{2\pi {r}^{n+1}}{\int }_{0}^{2\pi }|z{f}^{\prime }\left(z\right)|\phantom{\rule{0.2em}{0ex}}d\theta \\ =\frac{1}{2\pi {r}^{n}}L\left(r,f\right).\end{array}$

With $r=\left(1-\frac{1}{n}\right)$, we use Theorem 4.1 and obtain the required result. □

Theorem 4.3 Let $f\in k-U{K}_{s}\left(\beta \right)$ and let F be defined by

$F\left(z\right)=\frac{m+1}{2{z}^{m}}{\int }_{0}^{z}{t}^{m-1}\left\{f\left(t\right)-f\left(-t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt.$
(4.4)

Then $F\in k-U{K}_{s}\left(\beta \right)$ in E. That is, the class $k-U{K}_{s}\left(\beta \right)$ is preserved under the integral operator (4.4).

Proof Since $f\in k-U{K}_{s}\left(\beta \right)$, we can write

$\left\{\frac{2z{f}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}\right\}\in P\left({p}_{k,\beta }\right),\phantom{\rule{1em}{0ex}}g\in k-S{T}_{s}\left(\beta \right)\subset {S}_{S}^{\ast }\left({\beta }_{1}\right).$

Let $G\left(z\right)=\frac{1}{2}\left\{{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right\}$ and be defined by (3.5). By Theorem 3.4, ${g}_{1}\in k-ST\left(\beta \right)$ and $G\in k-{S}_{s}T\left(\beta \right)\subset {S}_{s}^{\ast }\left({\beta }_{1}\right)$. Let $G=z{G}_{1}^{\prime }$. Then we can write

${G}_{1}^{\prime }\left(z\right)=\frac{1}{2}{\left[z{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right]}^{\prime },\phantom{\rule{1em}{0ex}}{G}_{1}\in k-UC{V}_{s}\left(\beta \right).$

Thus, from (4.4) and $g=z{g}_{1}^{\prime }$, ${g}_{1}\in {C}_{s}\left({\beta }_{1}\right)$, we have

$\begin{array}{rl}\frac{2{F}^{\prime }\left(z\right)}{{\left[{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right]}^{\prime }}=& \frac{{z}^{m}\left\{f\left(z\right)-f\left(-z\right)\right\}-m{\int }_{0}^{z}{t}^{m-1}\left\{f\left(t\right)-f\left(-t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt}{{z}^{m}\left\{{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right\}-m{\int }_{0}^{z}{t}^{m-1}\left\{{g}_{1}\left(t\right)-{g}_{1}\left(-t\right)\right\}\phantom{\rule{0.2em}{0ex}}dt}\\ =& \frac{N\left(z\right)}{D\left(z\right)},\end{array}$

say. We note that $N\left(0\right)=D\left(0\right)=0$, and for ${g}_{1}\in {C}_{S}\left({\beta }_{1}\right)$,

$\begin{array}{rl}\frac{{\left(z{D}^{\prime }\left(z\right)\right)}^{\prime }}{{D}^{\prime }\left(z\right)}& =m+\frac{{\left\{z{\left[{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right]}^{\prime }\right\}}^{\prime }}{{\left\{{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right\}}^{\prime }}\\ =m+{h}_{1}\left(z\right),\phantom{\rule{1em}{0ex}}{h}_{1}\in P\left({\beta }_{1}\right).\end{array}$

Since $P\left({\beta }_{1}\right)$ is a convex set, $D\in {C}_{s}\left({\beta }_{1}\right)\subset {S}^{\ast }$ in E. We thus have

$\frac{{N}^{\prime }\left(z\right)}{{D}^{\prime }\left(z\right)}=\frac{1}{2}\left[\frac{2z{f}^{\prime }\left(z\right)}{{\left[{g}_{1}\left(z\right)-{g}_{1}\left(-z\right)\right]}^{\prime }}+\frac{2\left(-z\right){f}^{\prime }\left(-z\right)}{{\left[{g}_{1}\left(-z\right)-{g}_{1}\left(z\right)\right]}^{\prime }}\right]\in P\left({p}_{k,\beta }\right).$

Now, using Lemma 2.2, it follows that

This proves that $F\in k-U{K}_{S}\left(\beta \right)$ in E. □

We study a partial converse of the above result as follows.

Theorem 4.4 Let $\left(\frac{2z{f}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}\right)\prec {p}_{k}\left(z\right)$ in E and let

${F}_{1}\left(z\right)=\frac{1}{1+m}{z}^{1-m}{\left({z}^{m}f\left(z\right)\right)}^{\prime },\phantom{\rule{1em}{0ex}}m=1,2,3,\dots .$
(4.5)

Then ${F}_{1}\in {K}_{s}$ for $|z|<{r}_{1}$, where

${r}_{1}=\left\{\frac{m+1}{\left(2-{\beta }_{1}\right)+\sqrt{{\left(z-{\beta }_{1}\right)}^{2}+\left(m+1\right)\left(m-1+2{\beta }_{1}\right)}}\right\},\phantom{\rule{1em}{0ex}}{\beta }_{1}=\frac{k+\beta }{k+1}.$
(4.6)

Proof We shall need the following well-known results for $p\in P\left(\alpha \right)$, $0\le \alpha <1$; see [4]:

$\frac{1-\left(1-2\alpha \right)r}{1+r}\le |p\left(z\right)|\le \frac{1+\left(1-2\alpha \right)r}{1-r},$
(4.7)
$|{p}^{\prime }\left(z\right)|\le \frac{2\left[\mathrm{\Re }p\left(z\right)-\alpha \right]}{1-{r}^{2}}.$
(4.8)

Since $f\in k-U{K}_{s}\left(\beta \right)$, there exists $g\in {S}_{s}^{\ast }\left({\beta }_{1}\right)$ such that, for $z\in E$.

$\left(\frac{2z{f}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}\right)=p\left(z\right),\phantom{\rule{1em}{0ex}}p\in P\left({p}_{k}\right)\subset P\left(\alpha \right),\alpha =\frac{k}{k+1}.$

From (4.5), we have

${F}_{1}\left(z\right)=\frac{1}{1+m}\left[mf\left(z\right)+z{f}^{\prime }\left(z\right)\right],$

and this gives us

$\begin{array}{rl}\frac{2z{F}_{1}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}& =\frac{1}{m+1}\left[\frac{2m{f}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}+\frac{2z{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{g\left(z\right)-g\left(-z\right)}\right]\\ =\frac{1}{m+1}\left[mp\left(z\right)+z{p}^{\prime }\left(z\right)+p\left(z\right)h\left(z\right)\right],\end{array}$

where

$h\left(z\right)=\frac{z{\mathrm{\Psi }}^{\prime }\left(z\right)}{\mathrm{\Psi }\left(z\right)}\in P\left({\beta }_{1}\right),\phantom{\rule{1em}{0ex}}\mathrm{\Psi }\left(z\right)=g\left(z\right)-g\left(-z\right).$

Now, using (4.7) and (4.8), we have

$\begin{array}{rl}\mathrm{\Re }\left\{\frac{2z{F}_{1}^{\prime }\left(z\right)}{g\left(z\right)-g\left(-z\right)}\right\}& \ge \frac{\left(\mathrm{\Re }p\left(z\right)-\alpha \right)}{1+m}\left\{m+\frac{1-\left(1-2{\beta }_{1}\right)r}{1+r}-\frac{2r}{1-{r}^{2}}\right\}\\ =\frac{\mathrm{\Re }p\left(z\right)-\alpha }{1+m}\left[\frac{T\left(r\right)}{1-{r}^{2}}\right],\end{array}$
(4.9)

where

$T\left(r\right)=\left(m+1\right)-2\left(2-{\beta }_{1}\right)r+\left(-m-2{\beta }_{1}+1\right){r}^{2}.$

We note that $T\left(0\right)=1+m>0$ and $T\left(1\right)=-3<0$. So there exists ${r}_{1}\in \left(0,1\right)$. The right hand side of (4.9) is positive for $|z|<{r}_{1}$, where ${r}_{1}$ is given by (4.6). This implies that $F\in {K}_{s}$ for $|z|<{r}_{1}$ and the proof is complete. □

We have the following special cases.

1. 1.

For $k=0=\beta$, $f\in {K}_{s}$. Then ${F}_{1}$, defined by (4.5) belongs to ${K}_{s}$ for $|z|<{r}_{0}=\frac{1+m}{2+\sqrt{3+{m}^{2}}}$.

2. 2.

When $m=1$ and ${\beta }_{1}=0$ (that is, $k=0=\beta$), then ${F}_{1}\left(z\right)=\frac{{\left(zf\left(z\right)\right)}^{\prime }}{2}$ belongs to the same class for $|z|<\frac{1}{2}$. This result has been proved by Livingston [12] for convex and starlike functions.