On Some Interesting Classes of Analytic Functions Related to Univalent Functions

Abstract

Let $\overline{\mathcal{A}}$ be the new general class of functions $f\left(z\right)$ of the form $f\left(z\right)=z+{\displaystyle \sum _{k=1}^{\infty }}{a}_{1+\frac{k}{2}}{z}^{1+\frac{k}{2}}$ that are analytic in the open unit disc $\mathbb{U}.$ In the present paper, for $f\left(z\right)\in \overline{\mathcal{A}},$ we consider classes $\overline{{\mathcal{S}}^{\ast }}\left(\alpha \right),$ $\overline{\mathcal{C}}\left(\alpha \right)$ and $\overline{\mathcal{R}}\left(\alpha \right)$ and obtain some interesting properties of $f\left(z\right)\in \overline{\mathcal{A}}$ concerning $\overline{{\mathcal{S}}^{\ast }}\left(\alpha \right),$ $\overline{\mathcal{C}}\left(\alpha \right)$ and $\overline{\mathcal{R}}\left(\alpha \right)$, applying subordinations of $f\left(z\right).$

1. Introduction

Let $\overline{\mathcal{A}}$ be the class of functions $f\left(z\right)$ of the form

$$f\left(z\right)=z+\sum _{k=1}^{\infty }{a}_{1+\frac{k}{2}}{z}^{1+\frac{k}{2}}$$

(1)

that are analytic in the open unit disc $\mathbb{U}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}.$ Here, we take the principal value for $\sqrt{z}.$ Let $\overline{{\mathcal{S}}^{\ast }}\left(\alpha \right)$ denote the subclass of $\overline{\mathcal{A}}$ consisting of $f\left(z\right)$, which satisfies

$$Re\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>\alpha ,(z\in \mathbb{U})$$

(2)

for some real $\alpha$ $(0\le \alpha <1).$ Also, $\overline{\mathcal{C}}\left(\alpha \right)$ is the subclass of $\overline{\mathcal{A}}$ consisting of $f\left(z\right)$, which satisfies

$$Re\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}>\alpha ,(z\in \mathbb{U})$$

(3)

for some real $\alpha$ $(0\le \alpha <1).$ With the definitions for $\overline{{\mathcal{S}}^{\ast }}\left(\alpha \right)$ and $\overline{\mathcal{C}}\left(\alpha \right),$ we know that $f\left(z\right)\in \overline{\mathcal{C}}\left(\alpha \right)$ if and only if $z{f}^{\prime }\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right),$ and $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right)$ if and only if ${\int }_0^z\frac{f\left(t\right)}{t}dt\in \overline{\mathcal{C}}\left(\alpha \right).$ If we take $a_{1+\frac{k}{2}}=0$ for all odd $k=1,3,5,\cdots ,$ then $f\left(z\right)$ can be written by

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k,$$

(4)

and we write that $f\left(z\right)\in \mathcal{A}.$ If $f\left(z\right)\in \mathcal{A}$ satisfies the condition (2), then we say that $f\left(z\right)\in {\mathcal{S}}^{\ast }\left(\alpha \right),$ and if $f\left(z\right)\in \mathcal{A}$ satisfies the condition (3), then we say that $f\left(z\right)\in \mathcal{C}\left(\alpha \right).$ It is well known that

$$f\left(z\right)=\frac{z}{{\left(1-z\right)}^{2(1-\alpha )}}=z+\sum _{k=2}^{\infty }\frac{{\prod }_{j=2}^k(j-2\alpha )}{(k-1)!}{z}^k$$

is the extremal function for the class ${\mathcal{S}}^{\ast }\left(\alpha \right)$ and that

$$f\left(z\right)=\left\{\begin{array}{cc}\frac{1-{(1-z)}^{2\alpha -1}}{2\alpha -1}=z+\sum _{k=2}^{\infty }\frac{{\prod }_{j=2}^k(j-2\alpha )}{k!}{z}^k,\hfill & (\alpha \ne \frac{1}{2})\hfill \\ \\ -\log (1-z)=z+\sum _{k=2}^{\infty }\frac{1}{k}{z}^k,\hfill & (\alpha =\frac{1}{2})\hfill \end{array}\right.$$

is the extremal function for the class $\mathcal{C}\left(\alpha \right).$ For such extremal functions $f\left(z\right)\in \mathcal{A},$ we consider a function $f\left(z\right)\in \overline{\mathcal{A}}$ given by

$$f\left(z\right)=\frac{z}{{\left(1-\sqrt{z}\right)}^{2(1-\alpha )}}=z+\sum _{k=1}^{\infty }\frac{{\prod }_{j=1}^k(j+1-2\alpha )}{k!}{z}^{1+\frac{k}{2}}.$$

(5)

The function $f\left(z\right)$ given by (5) satisfies

$$Re\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}=Re\left\{\frac{1-\alpha \sqrt{z}}{1-\sqrt{z}}\right\}>\frac{1+\alpha }{2},(z\in \mathbb{U})$$

and $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\frac{1+\alpha }{2}\right).$ If we consider $f\left(z\right)\in \overline{\mathcal{A}}$ given by

$$f^{\prime }\left(z\right)=\frac{1}{{\left(1-\sqrt{z}\right)}^{2(1-\alpha )}},$$

then $f\left(z\right)$ satisfies

$$Re\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}=Re\left\{\alpha +\frac{1-\alpha }{1-\sqrt{z}}\right\}>\frac{1+\alpha }{2},(z\in \mathbb{U})$$

and $f\left(z\right)\in \overline{\mathcal{C}}\left(\frac{1+\alpha }{2}\right).$

Next, we consider a function $f\left(z\right)$ given by

$$f\left(z\right)=\frac{z}{{\left(1-\sqrt{z}\right)}^{4(1-\alpha )}}=z+\sum _{k=1}^{\infty }\frac{{\prod }_{j=4}^{k+3}(j-4\alpha )}{k!}{z}^{1+\frac{k}{2}}$$

(6)

for $0\le \alpha <1.$ The function $f\left(z\right)$ given by (6) satisfies

$$Re\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}=Re\left\{\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}}\right\}>\alpha ,(z\in \mathbb{U})$$

and $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right).$ Also, if we take

$$f\left(z\right)=\frac{1}{(1-2\alpha )(3-4\alpha )}\left(1-\frac{1+(4-3\alpha )\sqrt{z}}{{\left(1-\sqrt{z}\right)}^{3-4\alpha }}\right)=z+\sum _{k=1}^{\infty }\frac{2{\prod }_{j=k}^{k+3}(j-4\alpha )}{k!(k+2)}{z}^{1+\frac{k}{2}}$$

(7)

for $0\le \alpha <1,$ we see that

$$f^{\prime }\left(z\right)=\frac{1}{{\left(1-\sqrt{z}\right)}^{4(1-\alpha )}},$$

and

$$Re\left\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right\}=Re\left\{\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}}\right\}>\alpha ,(z\in \mathbb{U}).$$

Therefore, the function $f\left(z\right)$ given by (7) belongs to the class $\overline{\mathcal{C}}\left(\alpha \right).$ Further, we consider a function

$$\begin{array}{cc}\hfill f\left(z\right)& =-(1-2\alpha )z-4(1-\alpha )\sqrt{z}-4(1-\alpha )\log (1-\sqrt{z})\hfill \\ \hfill & =z+4(1-\alpha )\sum _{k=1}^{\infty }\left(\frac{1}{k+2}\right)z^{1+\frac{k}{2}}\hfill \end{array}$$

(8)

for $0\le \alpha <1.$ The function $f\left(z\right)$ given by (8) satisfies

$$Re{f}^{\prime }\left(z\right)=Re\left(\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}}\right)>\alpha ,(z\in \mathbb{U}).$$

(9)

Thus, we say that $f\left(z\right)\in \overline{\mathcal{R}}\left(\alpha \right)$ if $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies the inequality (9).

For analytic functions $f\left(z\right)$ and $F\left(z\right)$ in $\mathbb{U},$ we say that $f\left(z\right)$ is subordinate to $F\left(z\right),$ written as $f\left(z\right)\prec F\left(z\right)$ $(z\in \mathbb{U}),$ if there exists a function $w\left(z\right)$ analytic in $\mathbb{U}$ with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1$ $(z\in \mathbb{U}),$ and such that $f\left(z\right))=F(w\left(z\right)).$

With the definition for subordinations, if $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U},0\le \alpha <1)$$

then $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right),$ if $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U},0\le \alpha <1)$$

then $f\left(z\right)\in \overline{\mathcal{C}}\left(\alpha \right),$ and if $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$f^{\prime }\left(z\right)\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U},0\le \alpha <1)$$

then $f\left(z\right)\in \overline{\mathcal{R}}\left(\alpha \right).$

The function $f\left(z\right)$ given by (1) was considered by Owa [1] and Owa et al. [2]. In [1], he considered the coefficient inequalities for the function $f\left(z\right)$ and in [2], the authors considered the coefficient inequalities for functions in general subclasses. In the present paper, we consider some interesting properties of $f\left(z\right)\in \overline{\mathcal{A}}$ concerning $\overline{{\mathcal{S}}^{\ast }}\left(\alpha \right),$ $\overline{\mathcal{C}}\left(\alpha \right)$ and $\overline{\mathcal{R}}\left(\alpha \right)$ applying subordinations of $f\left(z\right).$

2. Subordination Properties

We consider some subordination properties for $f\left(z\right)\in \overline{\mathcal{A}}.$ We need the following result by Miller and Mocanu [3]:

Lemma 1.

Let β and γ be complex numbers. Let $h\left(z\right)$ be convex (univalent) in $\mathbb{U}$ such that

$$Re\left(\beta h\right(z)+\gamma )>0,(z\in \mathbb{U}).$$

(10)

If $p\left(z\right)$ is analytic in $\mathbb{U}$ with $p\left(0\right)=h\left(0\right)$ and

$$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{\beta p\left(z\right)+\gamma }\prec h\left(z\right),(z\in \mathbb{U}),$$

(11)

then

$$p\left(z\right)\prec h\left(z\right),(z\in \mathbb{U}).$$

(12)

Furthermore, if the Briot–Bouquet differential equation

$$g\left(z\right)+\frac{z{g}^{\prime }\left(z\right)}{\beta g\left(z\right)+\gamma }=h\left(z\right),(g\left(0\right)=h\left(0\right))$$

(13)

has a univalent solution $g\left(z\right),$ then

$$p\left(z\right)\prec g\left(z\right)\prec h\left(z\right),(z\in \mathbb{U})$$

(14)

and $g\left(z\right)$ is said to be the best dominant of (11).

Theorem 1.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies $f\left(z\right)\in \overline{\mathcal{C}}\left(\alpha \right)$ for $0<\alpha <1,$ then $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right).$

Proof. 

Let us consider a function $p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ and a function

$$h\left(z\right)=\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(0<\alpha <1)$$

for $f\left(z\right)\in \overline{\mathcal{A}}.$ Letting $\beta =1$ and $\gamma =0$ in Lemma 1, we have

$$Reh\left(z\right)=\alpha >0,(z\in \mathbb{U}).$$

This shows that

$$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec h\left(z\right),(z\in \mathbb{U})$$

by $f\left(z\right)\in \overline{\mathcal{C}}\left(\alpha \right).$ Then, we have that

$$p\left(z\right)=\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}),$$

that is, $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right).$ □

Theorem 2.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$f^{\prime }\left(z\right)+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)-\gamma }\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U})$$

(15)

for $0\le \alpha <1$ and $\gamma \ge -\alpha ,$ then $f\left(z\right)\in \overline{\mathcal{R}}\left(\alpha \right).$

Proof. 

We consider $p\left(z\right)=f^{\prime }\left(z\right),$ $\beta =1$ and

$$h\left(z\right)=\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U})$$

in Lemma 1. Then,

$$Re\left(h\right(z)-\gamma )>0,(z\in \mathbb{U})$$

and

$$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}=f^{\prime }\left(z\right)+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)-\gamma }\prec h\left(z\right)$$

by (15). Thus, we have $p\left(z\right)=f^{\prime }\left(z\right)\in \overline{\mathcal{R}}\left(\alpha \right).$ □

Further, letting $\beta =1,$ $\gamma \ge -\alpha ,$ $p\left(z\right)=\frac{f\left(z\right)}{z}$ and

$$h\left(z\right)=\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U})$$

in Lemma 1, we obtain the following theorem.

Theorem 3.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$\frac{f\left(z\right)}{z}+\frac{z{f}^{\prime }\left(z\right)+\gamma z}{f\left(z\right)+\gamma z}\prec \frac{2(1-\alpha \sqrt{z})}{1-\sqrt{z}},(z\in \mathbb{U})$$

then

$$\frac{f\left(z\right)}{z}\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

(16)

To consider our next results, we have to use here the following lemma due to Miller and Mocanu [4].

Lemma 2.

Let $F\left(z\right)$ be analytic in $\mathbb{U}$ and $G\left(z\right)$ be analytic in $\mathbb{U}$ and the boundary of $\mathbb{U}$ with $F\left(0\right)=G\left(0\right).$ If $F\left(z\right)$ is not subordinate to $G\left(z\right),$ then there exist points $z_0\in \mathbb{U}$ and ${\xi }_0\in \partial \mathbb{U},$ and a real $m\ge 1$ for which $F\left(\right|z|<|z_0\left|\right)\subset G\left(\mathbb{U}\right),$

$$\begin{array}{cc}\hfill & \left(i\right)F\left(z_0\right)=G\left({\xi }_0\right)\hfill \\ \hfill & \left(ii\right)z_0{F}^{\prime }\left(z_0\right)=m{\xi }_0{G}^{\prime }\left({\xi }_0\right).\hfill \end{array}$$

Now, we derive the following theorem with Lemma 2.

Theorem 4.

Let ${\gamma }_0$ be the solution of

$$\gamma \pi =\frac{3}{2}\pi -Ta{n}^{-1}\left(\frac{\gamma }{2}\right)$$

(17)

and let

$$\beta =\gamma +\frac{2}{\pi }Ta{n}^{-1}\left(\frac{\gamma }{2}\right),(0<\gamma \le {\gamma }_0).$$

(18)

If $p\left(z\right)$ is analytic in $\mathbb{U}$ with $p\left(0\right)={(1-\alpha )}^{\gamma }$ and

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\left((1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta },(z\in \mathbb{U})$$

(19)

then

$$p\left(z\right)\prec {\left((1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\gamma },(z\in \mathbb{U})$$

(20)

where $0\le \alpha <1.$

Proof. 

We follow the same methodology as the proof by Miller and Mocanu (Theorem 5 in [5]). Since (17) gives us that

$$\frac{2}{\pi }Ta{n}^{-1}\left(\frac{\gamma }{2}\right)=3-2\gamma ,$$

we see

$$\gamma \le 3-\beta ,(0<\gamma \le {\gamma }_0).$$

(21)

Consider

$$h\left(z\right)={\left((1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta }$$

and

$$g\left(z\right)={\left((1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\gamma }.$$

Then $\left|\arg h\left(z\right)\right|<\frac{\pi }{2}\beta$ and $\left|\arg g\left(z\right)\right|<\frac{\pi }{2}\gamma .$ For such functions, we have to prove that $p\left(z\right)\prec g\left(z\right)(z\in \mathbb{U}).$ Since $p\left(0\right)=g\left(0\right)={(1-\alpha )}^{\gamma },$ using Lemma 11, if $p\left(z\right)$ is not subordinate to $g\left(z\right),$ then there exist points $z_0\in \mathbb{U}$ and ${\xi }_0\in \partial \mathbb{U},$ and a real $m\ge 1$ for which $p\left(\right|z|<|z_0\left|\right)\subset g\left(\mathbb{U}\right),$

$$p\left(z_0\right)=g\left({\xi }_0\right)$$

(22)

and

$$z_0{p}^{\prime }\left(z_0\right)=m{\xi }_0{g}^{\prime }\left({\xi }_0\right).$$

(23)

Let $p\left(z_0\right)\ne 0,$ then $g\left({\xi }_0\right)\ne 0$ and $\sqrt{{\xi }_0}\ne \pm 1.$ For ${\xi }_0\in \partial \mathbb{U},$ we consider

$$ir=\frac{1+\sqrt{{\xi }_0}}{1-\sqrt{{\xi }_0}},({\xi }_0\in \partial \mathbb{U}).$$

Then, using (22) and (23), we see

$$\begin{array}{cc}\hfill p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right)& =g\left({\xi }_0\right)+m{\xi }_0{g}^{\prime }\left({\xi }_0\right)\hfill \\ \hfill & =g\left({\xi }_0\right)\left(1+m\frac{{\xi }_0{g}^{\prime }\left({\xi }_0\right)}{g\left({\xi }_0\right)}\right)\hfill \\ \hfill & ={\left(i(1-\alpha )r\right)}^{\gamma }\left(1+m\gamma \frac{\sqrt{{\xi }_0}}{1-\sqrt{{\xi }_0}}\right)\hfill \\ \hfill & ={\left(i(1-\alpha )r\right)}^{\gamma }\left(1+im\gamma \frac{1+r^2}{4r}\right).\hfill \end{array}$$

This implies that

$$\arg (p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right))=\frac{\pi }{2}\gamma +Ta{n}^{-1}\left(\frac{m\gamma \left(r+\frac{1}{r}\right)}{4}\right)$$

and

$$\frac{\pi }{2}\gamma +Ta{n}^{-1}\left(\frac{\gamma }{2}\right)\le \left|\arg (p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right))\right|\le \frac{\pi }{2}\gamma +\frac{\pi }{2}.$$

Further, applying (17) and (18), we see

$$\frac{\pi }{2}\beta \le \left|\arg (p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right))\right|\le 2\pi -\frac{\pi }{2}\beta .$$

Since $\left|\arg h\left(z\right)\right|<\frac{\pi }{2}\beta ,$ this contradicts the condition (19). Therefore, the subordination (20) is true when $p\left(z_0\right)\ne 0$ $(z_0\in \mathbb{U}).$ If $p\left(z_0\right)=0$ $(z_0\in \mathbb{U})$ following the same reasoning as in the proof by Miller and Mocanu (Theorem 5 in [5]), we prove

$$\frac{\pi }{2}\beta \le \left|z_0{p}^{\prime }\left(z_0\right)\right|\le 2\pi -\frac{\pi }{2}\beta$$

with (21). □

Remark 1.

We know that

$$Re\left(\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}}\right)>\alpha ,(z\in \mathbb{U}).$$

If we take

$$\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}}-\alpha =(1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}},$$

we consider functions in (19) and (20).

Letting $\gamma =2$ in Theorem 4, we have the following corollary:

Corollary 1.

If $p\left(z\right)$ is analytic in $\mathbb{U}$ with $p\left(0\right)={(1-\alpha )}^2$ and

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec {\left((1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\frac{5}{2}},(z\in \mathbb{U})$$

then

$$p\left(z\right)\prec {\left((1-\alpha )\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^2,(z\in \mathbb{U}).$$

Considering our next result, we need the following lemma due to Nunokawa, Owa, and Sokol [6]:

Lemma 3.

Let $p\left(z\right)$ be analytic in $\mathbb{U}$ with $p\left(0\right)=1$ and $p\left(z\right)\ne 0$ in $\mathbb{U}.$ If there exists a point $z_0\in \mathbb{U}$ such that

$$\left|\arg p\left(z\right)\right|<\frac{\pi }{2}\beta ,\left(\right|z|<|z_0|<1)$$

(24)

and

$$|\arg p\left(z_0\right)|=\frac{\pi }{2}\beta$$

(25)

for some real $\beta >0,$ then

$$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=ik\beta ,$$

(26)

where

$$k\ge \frac{a^2+1}{2a}\ge 1,(argp\left(z_0\right)=\frac{\pi }{2}\beta )$$

(27)

and

$$k\le -\frac{a^2+1}{2a}\le -1,(argp\left(z_0\right)=-\frac{\pi }{2}\beta ),$$

(28)

where

$${\left(p\left(z_0\right)\right)}^{\frac{1}{\beta }}=\pm ia,(a>0).$$

(29)

Theorem 5.

Let $p\left(z\right)$ be analytic in $\mathbb{U}$ with $p\left(0\right)=1$ and $p\left(z\right)\ne 0$ in $\mathbb{U}.$ If

$$p\left(z\right)+\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta }+\frac{2\beta \sqrt{z}}{1-z},(z\in \mathbb{U})$$

(30)

then

$$p\left(z\right)\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta },(z\in \mathbb{U})$$

(31)

where $0<\beta \le 1.$

Proof. 

Let us suppose that $p\left(z\right)$ does not satisfy the subordination (31), then there exists a point $z_0\in \mathbb{U}$ such that

$$\left|\arg p\left(z\right)\right|<\frac{\pi }{2}\beta ,\left(\right|z|<|z_0|<1)$$

and

$$|\arg p\left(z_0\right)|<\frac{\pi }{2}\beta .$$

Applying Lemma 3, we have

$$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=ik\beta ,$$

where k satisfies (27) and (28). If $\arg p\left(z_0\right)=\frac{\pi }{2}\beta ,$ we see that

$$p\left(z_0\right)+\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}={\left(ia\right)}^{\beta }+ik\beta .$$

At this time, for boundary point $z=e^{i\theta },$ we know that

$${\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta }+\frac{2\beta \sqrt{z}}{1-z}={\left(i\frac{sin\frac{\theta }{2}}{1-cos\frac{\theta }{2}}\right)}^{\beta }+i\frac{\beta }{sin\frac{\theta }{2}}.$$

Therefore, if the subordination (30) is satisfied,

$$a=\frac{sin\frac{\theta }{2}}{1-cos\frac{\theta }{2}}$$

and

$$k\ge \frac{a^2+1}{2a}=\frac{1}{sin\frac{\theta }{2}}.$$

This implies that $p\left(z\right)$ does not satisfy the subordination (30). If $\arg p\left(z_0\right)=-\frac{\pi }{2}\beta ,$ then considering the same way for $\arg p\left(z_0\right)=\frac{\pi }{2}\beta ,$ we also say that $p\left(z\right)$ does not satisfy the subordination (30). This completes the proof of the theorem. □

Corollary 2.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$f^{\prime }\left(z\right)+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta }+\frac{2\beta \sqrt{z}}{1-z},(z\in \mathbb{U})$$

then

$$f^{\prime }\left(z\right)\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta },(z\in \mathbb{U})$$

where $0<\beta \le 1.$

Corollary 3.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta }+\frac{2\beta \sqrt{z}}{1-z},(z\in \mathbb{U})$$

then

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta },(z\in \mathbb{U})$$

where $0<\beta \le 1.$

Further, we have the following corollary.

Corollary 4.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$1+\frac{2f^{″}\left(z\right)+z{f}^{‴}\left(z\right)}{f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta }+\frac{2\beta \sqrt{z}}{1-z},(z\in \mathbb{U})$$

then

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{\beta },(z\in \mathbb{U})$$

where $0<\beta \le 1.$

Next, we derive the following theorem.

Theorem 6.

Let $p\left(z\right)$ be analytic in $\mathbb{U}$ with $p\left(0\right)=-1$ and $p\left(z\right)\ne 0$ in $\mathbb{U}.$ If

$$\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}-p\left(z\right)\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U})$$

(32)

for some $n=0,1,2,3,\cdots ,$ then

$$-p\left(z\right)\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U}).$$

(33)

Proof. 

We consider that $p\left(z\right)$ does not satisfy the subordination (33). Using Lemma 2, we know that there exist points $z_0\in \mathbb{U}$ and ${\xi }_0\in \partial \mathbb{U},$ and a real $m\ge 1$ for which $p\left(\right|z|<|z_0\left|\right)\subset g\left(\mathbb{U}\right)$

$$\begin{array}{cc}\hfill & \left(i\right)-p\left(z_0\right)=g\left({\xi }_0\right)\hfill \\ \hfill & \left(ii\right)z_0{p}^{\prime }\left(z_0\right)=m{\xi }_0{g}^{\prime }\left({\xi }_0\right).\hfill \end{array}$$

where

$$g\left(z\right)={\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U}).$$

Thus, we have

$$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}-p\left(z_0\right)=g\left({\xi }_0\right)+m\frac{{\xi }_0{g}^{\prime }\left({\xi }_0\right)}{g\left({\xi }_0\right)}.$$

Let us suppose that ${\xi }_0=e^{i\theta }.$ Then, we have

$$\begin{array}{cc}\hfill \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}-p\left(z_0\right)& ={\left(\frac{1+e^{i\frac{\theta }{2}}}{1-e^{i\frac{\theta }{2}}}\right)}^{2n+1}+m(2n+1)\left(\frac{e^{i\frac{\theta }{2}}}{1-e^{i\theta }}\right)\hfill \\ \hfill & ={\left(i\right)}^{2n+1}{\left(\frac{sin\frac{\theta }{2}}{1-cos\frac{\theta }{2}}\right)}^{2n+1}+im(2n+1)\frac{1}{2sin\frac{\theta }{2}}\hfill \\ \hfill & =i\left\{{\left(\frac{sin\frac{\theta }{2}}{1-cos\frac{\theta }{2}}\right)}^{2n+1}+m(2n+1)\frac{1}{2sin\frac{\theta }{2}}\right\}.\hfill \end{array}$$

Since

$$Re\left(\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}-p\left(z_0\right)\right)=0,$$

$p\left(z\right)$ does not satisfy the subordination (32). Thus, $p\left(z\right)$ should satisfy the subordination given in (33). □

We state the following corollaries with the above theorem.

Corollary 5.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$f^{\prime }\left(z\right)+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U})$$

for some $n=0,1,2,\cdots ,$ then

$$f^{\prime }\left(z\right)\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U}).$$

Corollary 6.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U})$$

for some $n=0,1,2,\cdots ,$ then

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U}).$$

Corollary 7.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$\frac{f\left(z\right)}{z}+\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U})$$

for some $n=0,1,2,\cdots ,$ then

$$\frac{f\left(z\right)}{z}\prec {\left(\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^{2n+1},(z\in \mathbb{U}).$$

Finally, we derive the following theorem.

Theorem 7.

Let $p\left(z\right)$ be analytic in $\mathbb{U}$ with $p\left(0\right)=1.$ If $p\left(z\right)$ satisfies

$$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U})$$

(34)

for some real $\alpha (0\le \alpha <1),$ then

$$p\left(z\right)\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

(35)

Proof. 

We suppose that $p\left(z\right)$ is not subordinate to a function

$$g\left(z\right)=\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

Applying Lemma 2, we say that there exist points $z_0\in \mathbb{U}$ and ${\xi }_0\in \partial \left(\mathbb{U}\right),$ and a real $m\ge 1$ for which $p\left(\right|z|<|z_0\left|\right)\subset g\left(\mathbb{U}\right),$

$$p\left(z_0\right)=g\left({\xi }_0\right)$$

and

$$z_0{p}^{\prime }\left(z_0\right)=m{\xi }_0{g}^{\prime }\left({\xi }_0\right).$$

This gives us the following:

$$\begin{array}{cc}\hfill p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right)& =g\left({\xi }_0\right)+m{\xi }_0{g}^{\prime }\left({\xi }_0\right)\hfill \\ \hfill & =\frac{1+(1-2\alpha )\sqrt{{\xi }_0}}{1-\sqrt{{\xi }_0}}\left\{1+\frac{m}{2}\left(\frac{(1-2\alpha )\sqrt{{\xi }_0}}{1+(1-2\alpha )\sqrt{{\xi }_0}}+\frac{\sqrt{{\xi }_0}}{1-\sqrt{{\xi }_0}}\right)\right\}.\hfill \end{array}$$

Letting ${\xi }_0=e^{i\theta }$ $(0\le \theta <2\pi ),$ we see

$$\begin{array}{cc}\hfill p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right)& =\frac{1+(1-2\alpha )e^{i\frac{\theta }{2}}}{1-e^{i\frac{\theta }{2}}}\hfill \\ \hfill & +\frac{m}{2}\left(\frac{(1-2\alpha )e^{i\frac{\theta }{2}}}{1-e^{i\frac{\theta }{2}}}+\frac{e^{i\frac{\theta }{2}}(1+(1-2\alpha )e^{i\frac{\theta }{2}})}{{(1-e^{i\frac{\theta }{2}})}^2}\right).\hfill \end{array}$$

Thus, we see

$$\begin{array}{c}\hfill Re(p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right))=\alpha -\frac{m}{4}\left(1-2\alpha +\frac{1+(1-2\alpha )cos\frac{\theta }{2}}{1-cos\frac{\theta }{2}}\right).\end{array}$$

Consider

$$g\left(t\right)=\frac{1+(1-2\alpha )t}{1-t},(t=cos\frac{\theta }{2}).$$

Then

$$g^{\prime }\left(t\right)=\frac{2(1-\alpha )}{{(1-t)}^2}>0$$

and $g\left(t\right)>\alpha .$ Thus, we have

$$Re(p\left(z_0\right)+z_0{p}^{\prime }\left(z_0\right))<\alpha -\frac{m(1-\alpha )}{4}<\alpha .$$

(36)

Since $Reg\left(z\right)>\alpha$ $(z\in \mathbb{U}),$ (36) contradicts our condition (34). Therefore, $p\left(z\right)$ satisfies the subordination (35). □

Considering $p\left(z\right)=\frac{f\left(z\right)}{z},$ we have the following corollary.

Corollary 8.

If $f\left(z\right)\in \overline{\mathcal{A}}$ belongs to the class $\overline{\mathcal{R}}\left(\alpha \right),$ then

$$\frac{f\left(z\right)}{z}\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

Letting $p\left(z\right)=f^{\prime }\left(z\right),$ we have the following corollary:

Corollary 9.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

for some real $\alpha (0\le \alpha <1),$ then $f\left(z\right)\in \overline{\mathcal{R}}\left(\alpha \right).$

Further, we have the following:

Corollary 10.

If $f\left(z\right)\in \overline{\mathcal{A}}$ satisfies

$$Re\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\left(2+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)\right\}\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

for some real $\alpha (0\le \alpha <1),$ then $f\left(z\right)\in \overline{{\mathcal{S}}^{\ast }}\left(\alpha \right).$

Problem 1.

We consider a function $p\left(z\right)$ by

$$p\left(z\right)+z{p}^{\prime }\left(z\right)=\frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

(37)

in Theorem 7. Then, we know that

$$\begin{array}{cc}\hfill zp\left(z\right)& ={\int }_0^z\frac{1+(1-2\alpha )\sqrt{t}}{1-\sqrt{t}}dt\hfill \\ \hfill & =2{\int }_{1-\sqrt{z}}^1\left\{\frac{2(1-\alpha )}{u}+(4\alpha -3)+(1-2\alpha )u\right\}du\hfill \end{array}$$

(38)

with $\sqrt{t}=1-u.$ It follows that

$$zp\left(z\right)=z\left\{(2\alpha -1)-\frac{4(1-\alpha )}{\sqrt{z}}-\frac{4(1-\alpha )}{\sqrt{z}}\log (1-\sqrt{z})\right\}$$

(39)

and that

$$p\left(z\right)=(2\alpha -1)-\frac{4(1-\alpha )}{\sqrt{z}}-\frac{4(1-\alpha )}{\sqrt{z}}\log (1-\sqrt{z}).$$

(40)

Applying Theorem 7, we say that $p\left(z\right)$ given by (40) satisfies the following:

$$p\left(z\right)\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

(41)

To check the subordination (41), we have to check the following:

$$(2\alpha -1)-\frac{4(1-\alpha )}{\sqrt{z}}-\frac{4(1-\alpha )}{\sqrt{z}}\log (1-\sqrt{z})\prec \frac{1+(1-2\alpha )\sqrt{z}}{1-\sqrt{z}},(z\in \mathbb{U}).$$

(42)

How can we see the above subordination (42)?

3. Conclusions

In this paper, we consider $f\left(z\right)$ given by

$$f\left(z\right)=z+\sum _{k=1}^{\infty }{a}_{1+\frac{k}{2}}{z}^{1+\frac{k}{2}}$$

that is analytic in the open unit disc $\mathbb{U}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ and some interesting properties of $f\left(z\right)\in \overline{\mathcal{A}}$ concerning interesting classes $\overline{{\mathcal{S}}^{\ast }}\left(\alpha \right),$ $\overline{\mathcal{C}}\left(\alpha \right)$ and $\overline{\mathcal{R}}\left(\alpha \right)$ applying subordinations of $f\left(z\right).$ Our $f\left(z\right)$ functions, as given above, represent a novel consideration. So, we believe that our results in this paper will provide a new direction and contribute to developing a new perspective in the studies of geometric function theory.