A New Operator for Meromorphic Functions

Abstract

Let $\Sigma$ be the class of functions $f\left(z\right)$ of the form $f\left(z\right)=\frac{1}{z}+{\sum }_{k=0}^{\infty }{a}_k{z}^k$, which are analytic in the punctured disk. Using the differentiations and integrations, new operator $D^{n}f\left(z\right)$ is introduced for $f\left(z\right)\in \Sigma$. The object of the present paper is to discuss some interesting properties for $D^{n}f\left(z\right)$ and some properties concerned with different boundary points of the open unit disk. Moreover, some simple examples for our results are shown.

1. Introduction

Let $\Sigma$ be the class of functions $f\left(z\right)$ of the following form:

$$f\left(z\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{a}_k{z}^k$$

(1)

which are analytic in the punctured disk ${\mathbb{U}}_0=\{z\in \mathbb{C}:0<|z|<1\}.$ For $f\left(z\right)\in \Sigma ,$ Uralegaddi and Somanatha [1] consider the following operator:

$$\begin{array}{l}{D}^{0}f\left(z\right)=f\left(z\right),\\ D^{1}f\left(z\right)=Df\left(z\right)=\frac{1}{z}+\sum _{k=0}^{\infty }(k+2)a_k{z}^k,\\ D^{2}f\left(z\right)=D\left(Df\left(z\right)\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{(k+2)}^2{a}_k{z}^k,\end{array}$$

and

$$\begin{array}{c}\hfill D^{n}f\left(z\right)=D\left(D^{n-1}f\left(z\right)\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{(k+2)}^n{a}_k{z}^k,\end{array}$$

(2)

for $n\in \mathbb{N}=\{1,2,3,\dots \}$ by using the expansions of functions. Here, we introduce operators $D^{n}f\left(z\right)$ and $D^{-n}f\left(z\right)$ by using the differentation and integration as follows:

$$\begin{array}{l}{D}^{0}f\left(z\right)=f\left(z\right),\\ D^{1}f\left(z\right)=Df\left(z\right)=\frac{1}{z}{\left(z^{2}f\left(z\right)\right)}^{\prime }=\frac{1}{z}+\sum _{k=0}^{\infty }(k+2)a_k{z}^k,\\ D^{2}f\left(z\right)=D\left(Df\left(z\right)\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{(k+2)}^2{a}_k{z}^k,\end{array}$$

and

$$\begin{array}{c}\hfill D^{n}f\left(z\right)=D\left(D^{n-1}f\left(z\right)\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{(k+2)}^n{a}_k{z}^k,\end{array}$$

(3)

for $n\in \mathbb{N}.$

Our operator $D^{n}f\left(z\right)$ in (3) is the same as $D^{n}f\left(z\right)$ in (2) due to Uralegaddi and Somanatha [1]. Moreover, we define the following:

$$\begin{array}{l}{D}^{-1}f\left(z\right)=\frac{1}{z^2}{\int }_0^{z}tf\left(t\right)dt=\frac{1}{z}+\sum _{k=0}^{\infty }\left(\frac{1}{k+2}\right)a_k{z}^k,\\ \hfill D^{-2}f\left(z\right)=D^{-1}\left(D^{-1}f\left(z\right)\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{\left(\frac{1}{k+2}\right)}^2{a}_k{z}^k\end{array}$$

and

$$\begin{array}{c}\hfill D^{-n}f\left(z\right)=D^{-1}\left(D^{-n+1}f\left(z\right)\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{\left(\frac{1}{k+2}\right)}^n{a}_k{z}^k\end{array}$$

(4)

for $n\in \mathbb{N}.$

With the above operators $D^{n}f\left(z\right)$ and $D^{-n}f\left(z\right),$ we define $D^{n}f\left(z\right)$ as follows:

$$\begin{array}{c}\hfill D^{n}f\left(z\right)=\frac{1}{z}+\sum _{k=0}^{\infty }{(k+2)}^n{a}_k{z}^k\end{array}$$

(5)

for $n\in \mathbb{Z}=\{\dots ,-2,-1,0,1,2,\dots \}.$

Example 1.

Let us consider function $f\left(z\right)\in \Sigma$ such that the following is the case:

$$\begin{array}{cc}\hfill D^{n}f\left(z\right)& =\frac{{(1-z)}^{2(1-\alpha )}}{z}\hfill \\ \hfill & =\frac{1}{z}+\sum _{k=0}^{\infty }\frac{{\prod }_{j=0}^k(2-2\alpha -j)}{(k+1)!}{(-1)}^k{z}^k\hfill \end{array}$$

(6)

for $0\le \alpha <1.$ Then, we have the following.

$$\begin{array}{c}\hfill D^{n-1}f\left(z\right)=\frac{1-{(1-z)}^{3-2\alpha }}{(3-2\alpha )z^2}\end{array}$$

(7)

$$\begin{array}{c}\hfill D^{n+1}f\left(z\right)=\frac{(1-(3-2\alpha )z){(1-z)}^{1-2\alpha }}{z}.\end{array}$$

(8)

We also introduce the subordinations of functions by Pommerenke [2]. Let $f\left(z\right)$ and $g\left(z\right)$ be analytic in the open unit disk $\mathbb{U}.$ Then, we say that function $f\left(z\right)$ is subordinate to $g\left(z\right),$ written $f\left(z\right)\prec g\left(z\right),$ if there exists an analytic function $w\left(z\right)$ in $\mathbb{U}$ such that $w\left(0\right)=0,$$\left|w\right(z\left)\right|<1(z\in \mathbb{U}),$ and $f\left(z\right)=g\left(w\right(z\left)\right)$ for all $z\in \mathbb{U}.$ In particular, if $g\left(z\right)$ is univalent in $\mathbb{U},$ then $f\left(z\right)\prec g\left(z\right)$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathbb{U}\right)\subseteq g\left(\mathbb{U}\right).$

2. Properties of the Operator ${\mathit{D}}^{\mathit{n}}\mathit{f}\left(\mathit{z}\right)$

Discussing our problems for $D^{n}f\left(z\right),$ we have to recall here the following lemma due to Miller and Mocanu [3,4] (refining the old one in Jack [5]).

Lemma 1.

Let function $w\left(z\right)$ given by the following:

$$w\left(z\right)=a_k{z}^k+a_{k+1}{z}^{k+1}+\dots ,(k\in \mathbb{N})$$

(9)

be analytic in $\mathbb{U}$ with $w\left(0\right)=0.$ If $\left|w\right(z\left)\right|$ attains its maximum value on the circle $\left|z\right|=r<1$ at a point $z_0(0<|z_0|<1),$ then there exists a real number $m\ge k$ such that the following is the case.

$$\frac{z_0{w}^{\prime }\left(z_0\right)}{w\left(z_0\right)}=m$$

(10)

$$Re\left(1+\frac{z_0{w}^{\prime \prime }\left(z_0\right)}{w^{\prime }\left(z_0\right)}\right)\ge m.$$

(11)

Applying the above lemma, we derive the following theorem.

Theorem 1.

A function $f\left(z\right)\in \Sigma$ satisfies the following:

$$z{D}^{n}f\left(z\right)\prec \frac{1-(3-2\alpha )z}{1-z},(z\in \mathbb{U})$$

(12)

if and only if

$$Re\left(z{D}^{n}f\left(z\right)\right)<2-\alpha ,(z\in \mathbb{U}),$$

(13)

where $0\le \alpha <1$ and $n\in \mathbb{Z}.$

Proof. 

We consider a function $f\left(z\right)\in \Sigma$, which satisfies subordination (12). Then, there exists an analytic function $w\left(z\right)$ in $\mathbb{U}$ such that $w\left(0\right)=0,$$\left|w\right(z\left)\right|<1(z\in \mathbb{U})$ and the following is the case.

$$z{D}^{n}f\left(z\right)=\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)},(z\in \mathbb{U}).$$

(14)

It follows from (14) that the following is the case.

$$\left|w\left(z\right)\right|=\left|\frac{1-z{D}^{n}f\left(z\right)}{(3-2\alpha )-z{D}^{n}f\left(z\right)}\right|<1,(z\in \mathbb{U}).$$

(15)

This implies that the following is the case.

$$\mathrm{Re}\left(z{D}^{n}f\left(z\right)\right)<2-\alpha ,(z\in \mathbb{U}).$$

Conversely, if $f\left(z\right)$ satisfies (13), then we take an analytic function $w\left(z\right)$ such that $w\left(0\right)=0,$$\left|w\right(z\left)\right|<1(z\in \mathbb{U})$ and the following is the case.

$$w\left(z\right)=\frac{1-z{D}^{n}f\left(z\right)}{(3-2\alpha )-z{D}^{n}f\left(z\right)},(z\in \mathbb{U}).$$

(16)

It follows from (16) that the following is the case:

$$z{D}^{n}f\left(z\right)=\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)},(z\in \mathbb{U})$$

that is, we obtain the following.

$$z{D}^{n}f\left(z\right)\prec \frac{1-(3-2\alpha )z}{1-z},(z\in \mathbb{U}).$$

(17)

□

Taking $n=1$ in Theorem 1, we have the following corollary.

Corollary 1.

A function $f\left(z\right)\in \Sigma$ satisfies the following:

$$2f\left(z\right)+z{f}^{\prime }\left(z\right)\prec \frac{1-(3-2\alpha )z}{1-z},(z\in \mathbb{U})$$

(18)

if and only if the following is the case

$$Re(2f\left(z\right)+z{f}^{\prime }\left(z\right))<2-\alpha ,(z\in \mathbb{U}),$$

(19)

where $0\le \alpha <1.$

Theorem 2.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$Re\left(\frac{D^{n+2}f\left(z\right)}{D^{n+1}f\left(z\right)}\right)<\frac{9-9\alpha +2{\alpha }^2}{2(2-\alpha )},(z\in \mathbb{U})$$

(20)

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

$$\mathrm{Re}\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U})$$

(21)

where $n\in \mathbb{Z}.$

Proof. 

We consider a function $w\left(z\right)$ by the following:

$$\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}=\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)},(w\left(z\right)\ne 1)$$

(22)

for $f\left(z\right)\in \Sigma$ and $0\le \alpha <1.$ Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ and $w\left(0\right)=0.$ It follows from (22) that the following is the case:

$$\frac{D^{n+2}f\left(z\right)}{D^{n+1}f\left(z\right)}=\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)}-\frac{(3-2\alpha )z{w}^{\prime }\left(z\right)}{1-(3-2\alpha )w\left(z\right)}+\frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)},$$

(23)

because

$$z{\left(D^{n}f\left(z\right)\right)}^{\prime }=D^{n+1}f\left(z\right)-2D^{n}f\left(z\right)$$

(24)

and

$$z{\left(D^{n+1}f\left(z\right)\right)}^{\prime }=D^{n+2}f\left(z\right)-2D^{n+1}f\left(z\right).$$

(25)

Therefore, our condition (20) implies the following.

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{D^{n+2}f\left(z\right)}{D^{n+1}f\left(z\right)}\right)& =\mathrm{Re}\left\{\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)}-\frac{(3-2\alpha )z{w}^{\prime }\left(z\right)}{1-(3-2\alpha )w\left(z\right)}+\frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}\right\}\hfill \\ \hfill & <\frac{9-9\alpha +2{\alpha }^2}{2(2-\alpha )},(z\in \mathbb{U}).\hfill \end{array}$$

(26)

We suppose that there exists a point $z_0\in \mathbb{U}$ such that the following is the case.

$${\max }_{\left|z\right|\le |z_0|}\left|w\left(z\right)\right|=|w\left(z_0\right)|=1,(w\left(z_0\right)\ne 1).$$

(27)

Then, Lemma 1 says that $w\left(z_0\right)=e^{i\theta }(0\le \theta <2\pi )$ and the following is the case.

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right),(m\ge 1).$$

(28)

It follows from the above that the following is the case.

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{D^{n+2}f\left(z_0\right)}{D^{n+1}f\left(z_0\right)}\right)& =\mathrm{Re}\left\{\frac{1-(3-2\alpha )e^{i\theta }}{1-e^{i\theta }}-\frac{(3-2\alpha )m{e}^{i\theta }}{1-(3-2\alpha )e^{i\theta }}+\frac{m{e}^{i\theta }}{1-e^{i\theta }}\right\}\hfill \\ \hfill & =2-\alpha -\frac{(3-2\alpha )m(\cos \theta -(3-2\alpha \left)\right)}{1+{(3-2\alpha )}^2-2(3-2\alpha )\cos \theta }-\frac{m}{2}\hfill \\ \hfill & >2-\alpha +\frac{(1-\alpha )m}{2(2-\alpha )}\hfill \\ \hfill & \ge \frac{9-9\alpha +2{\alpha }^2}{2(2-\alpha )}.\hfill \end{array}$$

(29)

This contradicts condition (26). Thus, we say that there is no $z_0(0<|z_0|<1)$ such that $\left|w\right(z_0|=1.$ This shows that the following is the case:

$$\left|w\left(z\right)\right|=\left|\frac{1-\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}}{\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-(3-2\alpha )}\right|<1,(z\in \mathbb{U}),$$

(30)

that is, we obtain the following.

$$\mathrm{Re}\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U}).$$

This completes the proof of the theorem. □

Making $n=0$ in Theorem 2, we have the following corollary.

Corollary 2.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$Re\left(\frac{4f\left(z\right)+5z{f}^{\prime }\left(z\right)+z^2{f}^{\prime \prime }\left(z\right)}{2f\left(z\right)+z{f}^{\prime }\left(z\right)}\right)<\frac{9-9\alpha +2{\alpha }^2}{2(2-\alpha )},(z\in \mathbb{U})$$

(31)

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

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

(32)

Example 2.

If we consider function $f\left(z\right)\in \Sigma$ given by the following:

$$\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}=\frac{1-(3-2\alpha )z}{1-z},(z\in \mathbb{U})$$

for $0\le \alpha <1,$ then we know that

$$Re\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U})$$

and

$$Re\left(\frac{D^{n+2}f\left(z\right)}{D^{n+1}f\left(z\right)}\right)<\frac{9-9\alpha +2{\alpha }^2}{2(2-\alpha )},(z\in \mathbb{U}).$$

Remark 1.

Uralegaddi and Somanatha [1] proved that if $f\left(z\right)\in \Sigma$ satisfies the following:

$$\begin{array}{c}\hfill Re\left(\frac{D^{n+2}f\left(z\right)}{D^{n+1}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U})\end{array}$$

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

$$\begin{array}{c}\hfill Re\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U}).\end{array}$$

Since the following is the case

$$\begin{array}{c}\hfill \frac{9-9\alpha +2{\alpha }^2}{2(2-\alpha )}>2-\alpha ,(0\le \alpha <1),\end{array}$$

Theorem 2 is better than their result.

Next, we derive the following theorem.

Theorem 3.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$Re\left(\frac{D^{n+j+1}f\left(z\right)}{D^{n+j}f\left(z\right)}-\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)<\frac{1-\alpha }{2(2-\alpha )},(z\in \mathbb{U})$$

(33)

for some real $\alpha (0\le \alpha <1),$$n\in \mathbb{Z}$ and $j\in \mathbb{N},$ then the following is the case.

$$Re\left(\frac{D^{n+j}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U}).$$

(34)

Proof. 

We define a function $w\left(z\right)$ by the following:

$$\frac{D^{n+j}f\left(z\right)}{D^{n}f\left(z\right)}=\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)},(w\left(z\right)\ne 1)$$

(35)

for $f\left(z\right)\in \Sigma .$ Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ and $w\left(0\right)=0.$ It follows from (35) that the following is the case.

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{D^{n+j+1}f\left(z\right)}{D^{n+j}f\left(z\right)}-\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)& =\mathrm{Re}\left\{-\frac{(3-2\alpha )z{w}^{\prime }\left(z\right)}{1-(3-2\alpha )w\left(z\right)}+\frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}\right\}\hfill \\ \hfill & <\frac{1-\alpha }{2(2-\alpha )},(z\in \mathbb{U}).\hfill \end{array}$$

(36)

We suppose that there exists a point $z_0\in \mathbb{U}$ such that the following is the case.

$${\max }_{\left|z\right|\le |z_0|}\left|w\left(z\right)\right|=|w\left(z_0\right)|=1,(w\left(z_0\right)\ne 1).$$

(37)

Then, applying Lemma 1, we write that $w\left(z_0\right)=e^{i\theta }(0\le \theta <2\pi )$ and the following is the case.

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right),(m\ge 1).$$

(38)

Thus, we observe the following.

$$\begin{array}{ll}\mathrm{Re}\left(\frac{D^{n+j+1}f\left(z_0\right)}{D^{n+j}f\left(z_0\right)}-\frac{D^{n+1}f\left(z_0\right)}{D^{n}f\left(z_0\right)}\right)& =\mathrm{Re}\left\{-\frac{(3-2\alpha )mw\left(z_0\right)}{1-(3-2\alpha )w\left(z_0\right)}+\frac{mw\left(z_0\right)}{1-w\left(z_0\right)}\right\}\\ & =\mathrm{Re}\left\{-\frac{(3-2\alpha )m}{e^{-i\theta }-(3-2\alpha )}+\frac{m}{e^{-i\theta }-1}\right\}\\ & =\frac{(3-2\alpha )m\left(\right(3-2\alpha )-\cos \theta )}{1+{(3-2\alpha )}^2-2(3-2\alpha )\cos \theta }-\frac{m}{2}\\ & >\frac{(1-\alpha )m}{2(2-\alpha )}\\ & \ge \frac{(1-\alpha )}{2(2-\alpha )}.\end{array}$$

(39)

This contradicts condition (36). Therefore, $\left|w\right(z\left)\right|<1$ for all $z\in \mathbb{U}.$ This implies the following.

$$\mathrm{Re}\left(\frac{D^{n+j}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U}).$$

This completes the proof of the theorem. □

Setting $n=0$ and $j=1$ in Theorem 3, we have the following corollary.

Corollary 3.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$Re\left(\frac{4f\left(z\right)+5z{f}^{\prime }\left(z\right)+z^2{f}^{\prime \prime }\left(z\right)}{2f\left(z\right)+z{f}^{\prime }\left(z\right)}-\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\frac{9-5\alpha }{2(2-\alpha )},(z\in \mathbb{U})$$

(40)

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

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

(41)

3. Properties Concerning with Different Boundary Points

For s different boundary points $z_{\ell }(\ell =1,2,3,\dots ,s)$ with $|z_{\ell }|=1,$ we write the following:

$${\beta }_s=\frac{1}{s}\sum _{\ell =1}^{s}F\left(z_{\ell }\right),$$

(42)

where $F\left(z\right)=z{D}^{n}f\left(z\right),{\beta }_s\in e^{i\gamma }F\left(\mathbb{U}\right),{\beta }_s\ne 1,-\frac{\pi }{2}\le \gamma \le \frac{\pi }{2}.$

Now, we show the following theorem.

Theorem 4.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$\left|\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-1\right|<\frac{|e^{i\gamma }-{\beta }_s|\rho }{1+|e^{i\gamma }-{\beta }_s|\rho },(z\in \mathbb{U})$$

(43)

for some real ${\beta }_s$ with ${\beta }_s\ne 1$ such that $z_{\ell }(\ell =1,2,3,\dots ,s),$ and for some real $\rho >1,$ then the following is the case:

$$|z{D}^{n}f\left(z\right)-1|<|e^{i\gamma }-{\beta }_s|\rho ,(z\in \mathbb{U}),$$

(44)

where $n\in \mathbb{Z}.$

Proof. 

Define a function $w\left(z\right)$ by the following.

$$w\left(z\right)=\frac{e^{i\gamma }z{D}^{n}f\left(z\right)-{\beta }_s}{e^{i\gamma }-{\beta }_s}-1=\frac{z{D}^{n}f\left(z\right)-1}{1-e^{-i\gamma }{\beta }_s},(z\in \mathbb{U}),$$

(45)

Since the following is the case:

$$\begin{array}{cc}\hfill \frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}& =\frac{{\left(z{D}^{n}f\left(z\right)\right)}^{\prime }}{D^{n}f\left(z\right)}+1\hfill \\ \hfill & =\frac{(1-e^{-i\gamma }{\beta }_s)z{w}^{\prime }\left(z\right)}{1+(1-e^{-i\gamma }{\beta }_s)w\left(z\right)}+1,\hfill \end{array}$$

(46)

we have the following.

$$\begin{array}{cc}\hfill \left|\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-1\right|& =\left|\frac{(1-e^{-i\gamma }{\beta }_s)z{w}^{\prime }\left(z\right)}{1+(1-e^{-i\gamma }{\beta }_s)w\left(z\right)}\right|\hfill \\ \hfill & <\frac{|e^{i\gamma }-{\beta }_s|\rho }{1+|e^{i\gamma }-{\beta }_s|\rho },(z\in \mathbb{U}).\hfill \end{array}$$

(47)

We suppose that there exists a point $z_0(0<|z_0|<1)$ such that the following is the case.

$${\max }_{\left|z\right|\le |z_0|}\left|w\left(z\right)\right|=|w\left(z_0\right)|=\rho >1.$$

(48)

Then, we can write that $w\left(z_0\right)=\rho e^{i\theta }(0\le \theta <2\pi )$ and the following:

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right),(m\ge 1)$$

(49)

by Lemma 1. This provides us with the following.

$$\begin{array}{cc}\hfill \left|\frac{D^{n+1}f\left(z_0\right)}{D^{n}f\left(z_0\right)}-1\right|& =\left|\frac{(1-e^{-i\gamma }{\beta }_s)m\rho }{1+(1-e^{-i\gamma }{\beta }_s)\rho }\right|\hfill \\ \hfill & \ge \frac{|e^{i\gamma }-{\beta }_s|m\rho }{1+|e^{i\gamma }-{\beta }_s|\rho }\hfill \\ \hfill & \ge \frac{|e^{i\gamma }-{\beta }_s|\rho }{1+|e^{i\gamma }-{\beta }_s|\rho }.\hfill \end{array}$$

(50)

This contradicts condition (47). Thus, there is no $z_0(0<|z_0|<1)$ such that $\left|w\right(z_0|=\rho >1.$ This implies the following.

$$\left|w\left(z\right)\right|=\left|\frac{z{D}^{n}f\left(z\right)-1}{1-e^{-i\gamma }{\beta }_s}\right|<\rho ,(z\in \mathbb{U}).$$

(51)

This completes the proof of the theorem. □

Example 3.

We consider a function $f\left(z\right)\in \Sigma$ such that the following is the case:

$$\begin{array}{c}\hfill f\left(z\right)=\frac{1}{z}+a_j{z}^j(j=0,1,2,\dots )\end{array}$$

(52)

for $z\in \mathbb{U}$ with the following.

$$\begin{array}{c}\hfill 0<|a_j|<\frac{j}{{(j+2)}^{n+1}}.\end{array}$$

(53)

Then, we know the following:

$$\begin{array}{c}\hfill \left|\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}-1\right|=\left|\frac{(j+1){(j+2)}^n{a}_j{z}^{j+1}}{1+{(j+2)}^n{a}_j{z}^{j+1}}\right|<\frac{(j+1){(j+2)}^n\left|a_j\right|}{1-{(j+2)}^n\left|a_j\right|}\end{array}$$

(54)

for $z\in \mathbb{U}.$ Consider the following five boundary points $z_{\ell }(\ell =1,2,3,4,5)$ such that the following:

$$z_1=e^{-i\frac{arg\left(a_j\right)}{j+1}},$$

(55)

$$z_2=e^{i\frac{\pi -6arg\left(a_j\right)}{6(j+1)}},$$

(56)

$$z_3=e^{i\frac{\pi -4arg\left(a_j\right)}{4(j+1)}},$$

(57)

$$z_4=e^{i\frac{\pi -3arg\left(a_j\right)}{3(j+1)}}$$

(58)

and the following is obtained.

$$z_5=e^{i\frac{\pi -2arg\left(a_j\right)}{2(j+1)}}.$$

(59)

For the above boundary points, we observe the following:

$$z_1{D}^{n}f\left(z_1\right)=1+{(j+2)}^n{a}_j{e}^{-iarg\left(a_j\right)}=1+{(j+2)}^n\left|a_j\right|,$$

(60)

$$\begin{array}{cc}\hfill z_2{D}^{n}f\left(z_2\right)& =1+{(j+2)}^n{a}_j{e}^{i\frac{\pi -6arg\left(a_j\right)}{6}}\hfill \\ \hfill & =1+{(j+2)}^n\frac{\sqrt{3}+i}{2}\left|a_j\right|,\hfill \end{array}$$

(61)

$$\begin{array}{cc}\hfill z_3{D}^{n}f\left(z_3\right)& =1+{(j+2)}^n{a}_j{e}^{i\frac{\pi -4arg\left(a_j\right)}{4}}\hfill \\ \hfill & =1+{(j+2)}^n\frac{\sqrt{2}(1+i)}{2}\left|a_j\right|,\hfill \end{array}$$

(62)

$$\begin{array}{cc}\hfill z_4{D}^{n}f\left(z_4\right)& =1+{(j+2)}^n{a}_j{e}^{i\frac{\pi -3arg\left(a_j\right)}{3}}\hfill \\ \hfill & =1+{(j+2)}^n\frac{1+\sqrt{3}i}{2}\left|a_j\right|,\hfill \end{array}$$

(63)

and

$$\begin{array}{cc}\hfill z_5{D}^{n}f\left(z_5\right)& =1+{(j+2)}^n{a}_j{e}^{i\frac{\pi -2arg\left(a_j\right)}{2}}\hfill \\ \hfill & =1+{(j+2)}^{n}i\left|a_j\right|.\hfill \end{array}$$

(64)

Thus, ${\beta }_5$ is given by the following.

$$\begin{array}{cc}\hfill {\beta }_5& =\frac{1}{5}\sum _{\ell =1}^{5}F\left(z_{\ell }\right)\hfill \\ \hfill & =\frac{1}{5}\sum _{\ell =1}^5{z}_{\ell }{D}^{n}f\left(z_{\ell }\right)\hfill \\ \hfill & =1+\frac{{(j+2)}^n(3+\sqrt{2}+\sqrt{3})(1+i)}{10}\left|a_j\right|.\hfill \end{array}$$

(65)

This shows the following:

$$\begin{array}{c}\hfill |e^{i\gamma }-{\beta }_5|=\frac{{(j+2)}^n\sqrt{2}(3+\sqrt{2}+\sqrt{3})}{10}\left|a_j\right|.\end{array}$$

(66)

with $\gamma =0.$ For such ${\beta }_5$ and $\gamma ,$ we take $\rho >1$ satisfying the following equation.

$$\begin{array}{c}\hfill \frac{(j+1){(j+2)}^n\left|a_j\right|}{1-{(j+2)}^n\left|a_j\right|}\le \frac{|e^{i\gamma }-{\beta }_5|\rho }{1+|e^{i\gamma }-{\beta }_5|\rho }.\end{array}$$

(67)

Such ρ satisfies the following:

$$\begin{array}{cc}\hfill \rho & \ge \frac{10(j+1)}{\sqrt{2}(3+\sqrt{2}+\sqrt{3})(1+{(j+2)}^{n+1}|a_j\left|\right)}\hfill \\ \hfill & \ge \frac{10}{\sqrt{2}(3+\sqrt{2}+\sqrt{3})}\hfill \\ \hfill & >1\hfill \end{array}$$

(68)

with the following being the case.

$$\begin{array}{c}\hfill 0<|a_j|<\frac{j}{{(j+2)}^{n+1}}.\end{array}$$

(69)

For such ${\beta }_5$ and $\rho >1,$ we have the following.

$$\begin{array}{cc}\hfill |z{D}^{n}f\left(z\right)-1|& {=|(j+2)}^n{a}_j{z}^{j+1}|\hfill \\ \hfill & \le \frac{|e^{i\gamma }-{\beta }_5|\rho }{(j+1)+(j+2)|e^{i\gamma }-{\beta }_5|\rho }\hfill \\ \hfill & <|e^{i\gamma }-{\beta }_5|\rho .\hfill \end{array}$$

(70)

Next, our result follows.

Theorem 5.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$\left|z{D}^{n+1}f\left(z\right)-1\right|<2|e^{i\gamma }-{\beta }_s|\rho ,(z\in \mathbb{U})$$

(71)

for some ${\beta }_s$ with ${\beta }_s\ne 1$ such that $z_{\ell }(\ell =1,2,3,\dots ,s),$ and for some real $\rho >1,$ then the following is the case:

$$|z{D}^{n}f\left(z\right)-1|<|e^{i\gamma }-{\beta }_s|\rho ,(z\in \mathbb{U}),$$

(72)

where $n\in \mathbb{Z}.$

Proof. 

We define function $w\left(z\right)$ by the following.

$$\begin{array}{cc}\hfill w\left(z\right)& =\frac{e^{i\gamma }z{D}^{n}f\left(z\right)-{\beta }_s}{e^{i\gamma }-{\beta }_s}-1\hfill \\ \hfill & =\frac{e^{i\gamma }(z{D}^{n}f\left(z\right)-1)}{e^{i\gamma }-{\beta }_s}\hfill \\ \hfill & =\frac{e^{i\gamma }}{e^{i\gamma }-{\beta }_s}\left(\sum _{k=0}^{\infty }{(k+2)}^n{a}_k{z}^{k+1}\right).\hfill \end{array}$$

(73)

Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ with $w\left(0\right)=0,$ since the following is the case.

$$\begin{array}{c}\hfill z{D}^{n}f\left(z\right)=1+(1+e^{-i\gamma }{\beta }_s)w\left(z\right),\end{array}$$

(74)

$$\begin{array}{cc}\hfill z{D}^{n+1}f\left(z\right)& ={\left(z^2{D}^{n}f\left(z\right)\right)}^{\prime }\hfill \\ \hfill & =1+(1+e^{-i\gamma }{\beta }_s)(w\left(z\right)+z{w}^{\prime }\left(z\right)).\hfill \end{array}$$

(75)

It follows from (71) that the following is the case.

$$\begin{array}{cc}\hfill |z{D}^{n+1}f\left(z\right)-1|& =\left|(1+e^{-i\gamma }{\beta }_s)w\left(z\right)\left(1+\frac{z{w}^{\prime }\left(z\right)}{w\left(z\right)}\right)\right|\hfill \\ \hfill & <2|e^{i\gamma }-{\beta }_s|\rho .\hfill \end{array}$$

(76)

Suppose that there exists a point $z_0(0<|z_0|<1)$ such that the following is the case.

$${\max }_{\left|z\right|\le |z_0|}\left|w\left(z\right)\right|=|w\left(z_0\right)|=\rho >1.$$

(77)

Then, Lemma 1 implies that $w\left(z_0\right)=\rho e^{i\theta }(0\le \theta <2\pi )$ and the following is the case.

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right),(m\ge 1).$$

(78)

Thus, we see that the following is the case:

$$\begin{array}{cc}\hfill |z_0{D}^{n+1}f\left(z_0\right)-1|& =\left|(1+e^{-i\gamma }{\beta }_s)w\left(z_0\right)\left(1+\frac{z_0{w}^{\prime }\left(z_0\right)}{w\left(z_0\right)}\right)\right|\hfill \\ \hfill & =(1+m)|e^{i\gamma }-{\beta }_s|\rho \hfill \\ \hfill & \ge 2|e^{i\gamma }-{\beta }_s|\rho \hfill \end{array}$$

(79)

which contradicts (76). Therefore, $\left|w\right(z\left)\right|<\rho$ all $z\in \mathbb{U}.$ This shows us that the following is the case.

$$\left|w\left(z\right)\right|=\left|\frac{e^{i\gamma }(z{D}^{n}f\left(z\right)-1)}{e^{i\gamma }-{\beta }_s}\right|<\rho ,(z\in \mathbb{U}).$$

(80)

This completes the proof of the theorem. □

Taking $n=0$ in Theorem 5, we have the following corollary.

Corollary 4.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$|2zf\left(z\right)+z^2{f}^{\prime }\left(z\right)-1|<2|e^{i\gamma }-{\beta }_s|\rho ,(z\in \mathbb{U})$$

(81)

for some ${\beta }_s$ with ${\beta }_s\ne 1$ such that $z_{\ell }(\ell =1,2,3,\dots ,s),$ and for some real $\rho >1,$ then the following is the case.

$$|zf\left(z\right)-1|<|e^{i\gamma }-{\beta }_s|\rho ,(z\in \mathbb{U}).$$

(82)

Finally, we show the following theorem.

Theorem 6.

If $f\left(z\right)\in \Sigma$ satisfies the following:

$$Re\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)<2-\alpha +\frac{1-\alpha }{2(1-\alpha +c)},(z\in \mathbb{U})$$

(83)

for some real $\alpha (0\le \alpha <1)$ and $c>0,$ then we have the following:

$$Re\left(\frac{D^{n+1}F\left(z\right)}{D^{n}F\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U}),$$

(84)

where $n\in \mathbb{Z}$ and

$$F\left(z\right)=\frac{c}{z^{c+1}}{\int }_0^z{t}^{c}f\left(t\right)dt.$$

(85)

Proof. 

It follows from (85) that we have the following:

$$cf\left(z\right)=(c+1)F\left(z\right)+z{F}^{\prime }\left(z\right)$$

(86)

and

$$c{D}^{n}f\left(z\right)=(c-1)D^{n}F\left(z\right)+D^{n+1}F\left(z\right).$$

(87)

Thus, we have the following.

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)& =\mathrm{Re}\left\{\frac{\frac{D^{n+2}F\left(z\right)}{D^{n+1}F\left(z\right)}+(c-1)}{(c-1)\frac{D^{n}F\left(z\right)}{D^{n+1}F\left(z\right)}+1}\right\}\hfill \\ \hfill & <2-\alpha +\frac{1-\alpha }{2(1-\alpha +c)},(z\in \mathbb{U}).\hfill \end{array}$$

(88)

Now, we define a function $w\left(z\right)$ by the following.

$$\begin{array}{c}\hfill \frac{D^{n+1}F\left(z\right)}{D^{n}F\left(z\right)}=\frac{1-(3-2\alpha )w\left(z\right)}{1-w\left(z\right)},(w\left(z\right)\ne 1).\end{array}$$

(89)

Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ and $w\left(0\right)=0.$ Since the following is the case:

$$\begin{array}{c}\hfill z{\left(D^{n}F\left(z\right)\right)}^{\prime }=D^{n+1}F\left(z\right)-2D^{n}F\left(z\right)\end{array}$$

(90)

and

$$\begin{array}{c}\hfill z{\left(D^{n+1}F\left(z\right)\right)}^{\prime }=D^{n+2}F\left(z\right)-2D^{n+1}F\left(z\right),\end{array}$$

(91)

we obtain the following.

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{D^{n+1}f\left(z\right)}{D^{n}f\left(z\right)}\right)& =\mathrm{Re}\left\{\frac{\frac{D^{n+2}F\left(z\right)}{D^{n+1}F\left(z\right)}+(c-1)}{(c-1)\frac{D^{n}F\left(z\right)}{D^{n+1}F\left(z\right)}+1}\right\}\hfill \\ \hfill & =\mathrm{Re}\left\{3-2\alpha -\frac{2(1-\alpha )}{1-w\left(z\right)}+\frac{2(\alpha -1)z{w}^{\prime }\left(z\right)}{(1-w(z\left)\right)(c-(c+2-2\alpha \left)w\right(z\left)\right)}\right\}\hfill \\ \hfill & <2-\alpha +\frac{1-\alpha }{2(1-\alpha +c)},(z\in \mathbb{U}).\hfill \end{array}$$

(92)

Supposing that there exists a point $z_0(0<|z_0|<1)$ such that the following is the case.

$${\max }_{\left|z\right|\le |z_0|}\left|w\left(z\right)\right|=|w\left(z_0\right)|=1.$$

(93)

We can write that $w\left(z_0\right)=e^{i\theta }(0\le \theta <2\pi )$ and the following:

$$z_0{w}^{\prime }\left(z_0\right)=mw\left(z_0\right),(m\ge 1)$$

(94)

by Lemma 1. This provides us with the following:

$$\begin{array}{cc}\hfill \mathrm{Re}\left(\frac{D^{n+1}f\left(z_0\right)}{D^{n}f\left(z_0\right)}\right)& =\mathrm{Re}\left\{3-2\alpha -\frac{2(1-\alpha )}{1-e^{i\theta }}+\frac{2(\alpha -1)m{e}^{i\theta }}{(1-e^{i\theta })(c-(c+2-2\alpha )e^{i\theta })}\right\}\hfill \\ \hfill & =\mathrm{Re}\left\{2-\alpha +\frac{2(1-\alpha )m}{(1-e^{i\theta })((c+2-2\alpha )-c{e}^{-i\theta })}\right\}\hfill \\ \hfill & \ge 2-\alpha +\frac{1-\alpha }{2(1-\alpha +c)}\hfill \end{array}$$

(95)

for $0\le \theta <2\pi .$ This contradicts condition (92) of the theorem. Therefore, there is no $z_0(0<|z_0|<1)$ such that $\left|w\right(z_0\left)\right|=1$ in $\mathbb{U}.$ It follows from (89) that the following is the case:

$$\begin{array}{c}\hfill \left|w\left(z\right)\right|=\left|\frac{\frac{D^{n+1}F\left(z\right)}{D^{n}F\left(z\right)}-1}{\frac{D^{n+1}F\left(z\right)}{D^{n}F\left(z\right)}-(3-2\alpha )}\right|<1,(z\in \mathbb{U}),\end{array}$$

(96)

that is, we have the following.

$$\begin{array}{c}\hfill \mathrm{Re}\left(\frac{D^{n+1}F\left(z\right)}{D^{n}F\left(z\right)}\right)<2-\alpha ,(z\in \mathbb{U}).\end{array}$$

□