New Class of Specific Functions with Fractional Derivatives

Abstract

Let ${\mathcal{A}}_n$ be the class of specific analytic functions $f\left(z\right)=z+{\displaystyle \sum _{k=1}^{\infty }}{a}_{1+{\displaystyle \frac{k}{n}}}{z}^{1+{\displaystyle \frac{k}{n}}}(n\in \mathbb{N}=\{1,2,3,\dots \})$ in the open unit disk $\mathbb{U}.$ For $f\left(z\right)\in {\mathcal{A}}_n,$ fractional derivatives $D_z^{\lambda }f\left(z\right)$ and $D_z^{j+\lambda }f\left(z\right)$$(0\le \lambda <1,j\in \mathbb{N})$ are defined by using Gamma functions. Applying such fractional derivatives, we introduce a new subclass ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta )$ of ${\mathcal{A}}_n.$ In this paper, we establish sufficient conditions for $f\left(z\right)$ for ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ),$ coefficient inequalities for $|a_{1+{\displaystyle \frac{1}{n}}}|$ and $|a_{1+{\displaystyle \frac{k}{n}}}|$$(k=2,3,4,\dots )$ of $f\left(z\right)\in {\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ),$ and some interesting argument properties of fractional derivatives for $f\left(z\right)\in {\mathcal{A}}_n$ through an example.

1. Introduction

Let $\mathcal{A}$ be the class of functions $f\left(z\right)$ that are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$ and normalized by the conditions $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=1.$ For this class, we introduce a subclass ${\mathcal{A}}_n$ of $\mathcal{A}$ consisting of specific functions $f\left(z\right)$ given by

$$f\left(z\right)=z+\sum _{k=1}^{\infty }{a}_{1+{\displaystyle \frac{k}{n}}}{z}^{1+{\displaystyle \frac{k}{n}}}(n\in \mathbb{N}=\{1,2,3,\dots \}).$$

This class was introduced by Güney, Breaz and Owa [1]. They considered $f\left(z\right)\in {\mathcal{A}}_n$ given by

$$f\left(z\right)={\displaystyle \frac{z}{{\left(1-\sqrt[n]{z}\right)}^{2n(1-\alpha )}}}=z+\sum _{k=1}^{\infty }{\displaystyle \frac{{\prod }_{\ell =1}^k(\ell +2n(1-\alpha )-1)}{k!}}{z}^{1+{\displaystyle \frac{k}{n}}}(n\in \mathbb{N}=\{1,2,3,\dots \})$$

for $0\le \alpha <1$ and showed that

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

For $f\left(z\right)\in {\mathcal{A}}_n,$ we define the fractional integral of order $\lambda$ as follows:

$$D_z^{-\lambda }f\left(z\right)={\displaystyle \frac{1}{\Gamma \left(\lambda \right)}}{\int }_0^z{\displaystyle \frac{f\left(t\right)}{{(z-t)}^{1-\lambda }}}dt(\lambda >0),$$

where $\Gamma \left(\lambda \right)$ is the Gamma function.

The fractional derivative of order $\lambda$$(0\le \lambda <1)$ for $f\left(z\right)\in {\mathcal{A}}_n$ is defined by

$$D_z^{\lambda }f\left(z\right)={\displaystyle \frac{1}{\Gamma (1-\lambda )}}{\displaystyle \frac{d}{dz}}{\int }_0^z{\displaystyle \frac{f\left(t\right)}{{(z-t)}^{\lambda }}}dt.$$

Further, $D_z^{j+\lambda }f\left(z\right)$$(j\in \mathbb{N})$ is defined by

$$D_z^{j+\lambda }f\left(z\right)={\displaystyle \frac{d^j}{d{z}^j}}\left(D_z^{\lambda }f\left(z\right)\right).$$

The definitions $D_z^{-\lambda }f\left(z\right),$$D_z^{\lambda }f\left(z\right)$ and $D_z^{j+\lambda }f\left(z\right)$ were given by Owa [2,3]. With the above definitions, we know that

$$D_z^{0}f\left(z\right)=f\left(z\right),$$

$$D_z^{j}f\left(z\right)=f^{\left(j\right)}\left(z\right)$$

and

$$D_z^{j+\lambda }f\left(z\right)=D_z^{\lambda }\left(f^{\left(j\right)}\left(z\right)\right).$$

Thus, we have

$$D_z^{\lambda }f\left(z\right)={\displaystyle \frac{1}{\Gamma (2-\lambda )}}{z}^{1-\lambda }+\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}{a}_{1+{\displaystyle \frac{k}{n}}}{z}^{1+{\displaystyle \frac{k}{n}}-\lambda }$$

and

$$\begin{array}{c}\hfill D_z^{j+\lambda }f\left(z\right)={\displaystyle \frac{{\prod }_{\ell =1}^j(2-\lambda -\ell )}{\Gamma (2-\lambda )}}{z}^{1-\lambda -j}+\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(\prod _{\ell =1}^j\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)a_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}+1-\lambda -j}\end{array}$$

for $j\in \mathbb{N}.$ With the above the fractional derivative, we introduce the subclass ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta )$ of ${\mathcal{A}}_n$ as follows:

$$Re\left(e^{i\alpha }{\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}}\right)>\beta cos\alpha (z\in \mathbb{U})$$

(1)

with some real $\alpha$$(|\alpha |<{\displaystyle \frac{\pi }{2}})$ and $\beta$$(0\le \beta <1),$ where $j\in \mathbb{N}$ and $0\le \lambda <1.$

If $j=1,$ $\lambda =0$ and $\alpha =0$ in (1), then we have

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

for the class ${\mathcal{A}}_n(1,0,0,\beta ).$ Functions $f\left(z\right)\in {\mathcal{A}}_n$ in the class ${\mathcal{A}}_n(1,0,0,\beta )$ were discussed by Güney, Breaz and Owa [1]. Let us consider a function $f\left(z\right)\in {\mathcal{A}}_n$ given by

$$D_z^{j-1+\lambda }f\left(z\right)={\displaystyle \frac{z^{2-\lambda -j}}{\Gamma (2-\lambda ){\left(1-\sqrt[n]{z}\right)}^t}}$$

(2)

with

$$t=2n(2-\lambda -j)(1-\beta )e^{-i\alpha }cos\alpha (j\in \mathbb{N}).$$

Then, $f\left(z\right)$ satisfies

$${\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{D_z^{j-1+\lambda }f\left(z\right)}}=(2-\lambda -j)+{\displaystyle \frac{t}{n}}\left({\displaystyle \frac{\sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)$$

and

$$\begin{array}{cc}\hfill Re\left(e^{i\alpha }{\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}}\right)& =cos\alpha +2(1-\beta )cos\alpha Re\left({\displaystyle \frac{\sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)\hfill \\ \hfill & >cos\alpha -(1-\beta )cos\alpha \hfill \\ \hfill & =\beta cos\alpha (z\in \mathbb{U}).\hfill \end{array}$$

Therefore, $f\left(z\right)\in {\mathcal{A}}_n$ given by (2) belongs to the class ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta )$. Accordingly, it is very important to provide example functions $f\left(z\right)$ for the new class ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$

2. Conditions of $\mathit{f}\left(\mathit{z}\right)$ for the Class ${\mathcal{A}}_{\mathit{n}}(\mathit{j},\mathit{\lambda },\mathit{\alpha },\mathit{\beta })$

We first consider the following condition on $f\left(z\right)$ for the class ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$

Theorem 1. 

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

$$\left|{\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}}-1\right|<(1-\beta )cos\alpha (z\in \mathbb{U})$$

(3)

for some real α$(|\alpha |<{\displaystyle \frac{\pi }{2}}),$$0\le \beta <1,$$0\le \lambda <1$ and $j\in \mathbb{N},$ then $f\left(z\right)\in {\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$

Proof. 

We define a function $w\left(z\right)$ as follows:

$${\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}}-1=(1-\beta )cos\alpha w\left(z\right)(z\in \mathbb{U})$$

for $f\left(z\right)\in {\mathcal{A}}_n$ which satisfies (3). Then, we see that $w\left(z\right)$ is analytic in $\mathbb{U},$ that $w\left(0\right)=0$ and that $|w(z)|<1$$(z\in \mathbb{U}).$ For such a function $w\left(z\right),$ we have

$$\begin{array}{cc}\hfill Re\left(e^{i\alpha }{\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}}\right)& =Re\left(e^{i\alpha }(1+(1-\beta )cos\alpha w\left(z\right))\right)\hfill \\ \hfill & =cos\alpha +(1-\beta )cos\alpha Re\left(e^{i\alpha }w\left(z\right)\right)\hfill \\ \hfill & \ge cos\alpha -(1-\beta )cos\alpha |e^{i\alpha }w\left(z\right)|\hfill \\ \hfill & >\beta cos\alpha (z\in \mathbb{U}).\hfill \end{array}$$

This shows us that $f\left(z\right)\in {\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$ □

If we consider $j=1$ and $\lambda =0$ in Theorem 1, then we have the following corollary.

Corollary 1. 

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

$$\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right|<(1-\beta )cos\alpha (z\in \mathbb{U})$$

for some real α$(|\alpha |<{\displaystyle \frac{\pi }{2}})$ and $0\le \beta <1,$ then

$$Re\left(e^{i\alpha }{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>\beta cos\alpha (z\in \mathbb{U}).$$

Next, we derive the following theorem.

Theorem 2. 

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

$$\Gamma (1-\lambda )\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(1-\lambda +{\displaystyle \frac{k}{n(1-\beta )}}sec\alpha \right)\left|a_{1+{\displaystyle \frac{k}{n}}}\right|\le 1$$

(4)

for some real α$(|\alpha |<{\displaystyle \frac{\pi }{2}}),$$0\le \beta <1$ and $0\le \lambda <1,$ then $f\left(z\right)\in {\mathcal{A}}_n(1,\lambda ,\alpha ,\beta ).$ If $f\left(z\right)\in {\mathcal{A}}_n$ satisfies

$$\begin{array}{cc}\hfill {\displaystyle \frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}& \sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(2-\lambda -j+{\displaystyle \frac{k}{n(1-\beta )}}sec\alpha \right)\hfill \\ \hfill & \times \left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)\left|a_{1+{\displaystyle \frac{k}{n}}}\right|\le 1\hfill \end{array}$$

(5)

for some real α$(|\alpha |<{\displaystyle \frac{\pi }{2}}),$$0\le \beta <1,$$0\le \lambda <1$ and $j=2,3,\dots ,$ then $f\left(z\right)\in {\mathcal{A}}_n(1,\lambda ,\alpha ,\beta ).$

Proof. 

For $f\left(z\right)\in {\mathcal{A}}_n$ and $j\ge 2,$ we have that

$$\begin{array}{cc}\hfill z{D}_z^{j+\lambda }f\left(z\right)-& (2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)={\displaystyle \frac{{\prod }_{\ell =1}^j(2-\lambda -\ell )}{\Gamma (2-\lambda )}}{z}^{2-\lambda -j}\hfill \\ \hfill & +\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(\prod _{\ell =1}^j\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)a_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}+2-\lambda -j}\hfill \\ \hfill & -(2-\lambda -j)\left\{{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}{z}^{2-\lambda -j}\right.\hfill \\ \hfill & \left.+\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)a_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}+2-\lambda -j}\right\}\hfill \\ \hfill & =\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left({\displaystyle \frac{k}{n}}\right)\left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)a_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}+2-\lambda -j}.\hfill \end{array}$$

It follows from the above that

$$\begin{array}{cc}\hfill & \left|{\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}}-1\right|\hfill \\ \hfill & =\left|{\displaystyle \frac{\frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}{\sum }_{k=1}^{\infty }\frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}\left(\frac{k}{n}\right)\left({\prod }_{\ell =1}^{j-1}\left(\frac{k}{n}+2-\lambda -\ell \right)\right)a_{1+\frac{k}{n}}{z}^{\frac{k}{n}}}{1+\frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}{\sum }_{k=1}^{\infty }\frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}\left({\prod }_{\ell =1}^{j-1}\left(\frac{k}{n}+2-\lambda -\ell \right)\right)a_{1+\frac{k}{n}}{z}^{\frac{k}{n}}}}\right|\hfill \\ \hfill & <{\displaystyle \frac{\frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}{\sum }_{k=1}^{\infty }\frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}\left(\frac{k}{n}\right)\left({\prod }_{\ell =1}^{j-1}\left(\frac{k}{n}+2-\lambda -\ell \right)\right)\left|a_{1+\frac{k}{n}}\right|}{1-\frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\sum }_{k=1}^{\infty }\frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}\left({\prod }_{\ell =1}^{j-1}\left(\frac{k}{n}+2-\lambda -\ell \right)\right)\left|a_{1+\frac{k}{n}}\right|}}\hfill \end{array}$$

for $z\in \mathbb{U}$. Thus, we know by Theorem 1 that if $f\left(z\right)$ satisfies

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left({\displaystyle \frac{k}{n}}\right)\left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)\left|a_{1+{\displaystyle \frac{k}{n}}}\right|\hfill \\ \hfill & \le (1-\beta )cos\alpha \left\{1-{\displaystyle \frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}}\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)\left|a_{1+{\displaystyle \frac{k}{n}}}\right|\right\},\hfill \end{array}$$

(6)

then $f\left(z\right)\in {\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$ Finally, we see that the inequality (6) implies (5).

If $j=1,$ we also have that

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{z{D}_z^{1+\lambda }f\left(z\right)}{(1-\lambda )D_z^{\lambda }f\left(z\right)}}-1\right|<{\displaystyle \frac{\Gamma (1-\lambda ){\sum }_{k=1}^{\infty }\frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}\left(\frac{k}{n}\right)\left|a_{1+\frac{k}{n}}\right|}{1-\Gamma (2-\lambda ){\sum }_{k=1}^{\infty }\frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}\left|a_{1+\frac{k}{n}}\right|}}.\end{array}$$

Noting that if $f\left(z\right)$ satisfies

$$\begin{array}{cc}\hfill & \Gamma (1-\lambda )\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left({\displaystyle \frac{k}{n}}\right)\left|a_{1+{\displaystyle \frac{k}{n}}}\right|\hfill \\ \hfill & \le (1-\beta )cos\alpha \left(1-\Gamma (2-\lambda )\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left|a_{1+{\displaystyle \frac{k}{n}}}\right|\right),\hfill \end{array}$$

we obtain the inequality in (4). Further, if we consider a function $f\left(z\right)\in {\mathcal{A}}_n$ given by

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

with

$$\begin{array}{c}\hfill c_{1+{\displaystyle \frac{k}{n}}}={\displaystyle \frac{\Gamma \left(\frac{k}{n}+2-\lambda \right)e^{i\theta }}{\Gamma (1-\lambda )\Gamma \left(\frac{k}{n}+2\right)\left(1-\lambda +\frac{k}{n(1-\beta )}sec\alpha \right)k(k+1)}}(0\le \theta <2\pi ),\end{array}$$

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

$$\begin{array}{cc}\hfill \Gamma (1-\lambda )& \sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(1-\lambda +{\displaystyle \frac{k}{n(1-\beta )}}sec\alpha \right)\left|c_{1+{\displaystyle \frac{k}{n}}}\right|\hfill \\ \hfill & =\sum _{k=1}^{\infty }{\displaystyle \frac{1}{k(k+1)}}=\sum _{k=1}^{\infty }\left({\displaystyle \frac{1}{k}}-{\displaystyle \frac{1}{k+1}}\right)=1.\hfill \end{array}$$

If we take a function $f\left(z\right)\in {\mathcal{A}}_n$ given by

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

with

$$\begin{array}{cc}\hfill c_{1+{\displaystyle \frac{k}{n}}}& ={\displaystyle \frac{\left({\prod }_{\ell =1}^j(2-\lambda -\ell )\right)\Gamma \left(\frac{k}{n}+2-\lambda \right)e^{i\theta }}{\Gamma (2-\lambda )\Gamma \left(\frac{k}{n}+2\right)\left(2-\lambda -j+\frac{k}{n(1-\beta )}sec\alpha \right)}}\hfill \\ \hfill & \times {\displaystyle \frac{1}{\left({\prod }_{\ell =1}^{j-1}\left(\frac{k}{n}+2-\lambda -\ell \right)\right)k(k+1)}}(0\le \theta <2\pi ),\hfill \end{array}$$

then $f\left(z\right)$ satisfies the equality in (5). □

3. Coefficient Inequalities of $\mathit{f}\left(\mathit{z}\right)$ for ${\mathcal{A}}_{\mathit{n}}(\mathit{j},\mathit{\lambda },\mathit{\alpha },\mathit{\beta })$

We would like to discuss coefficient inequalities of $f\left(z\right)$ for the class ${\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$ We introduce the lemma established by Carathéodory [4].

Lemma 1. 

If a function $p\left(z\right)$ given by

$$\begin{array}{c}\hfill p\left(z\right)=1+\sum _{k=1}^{\infty }{c}_k{z}^k\end{array}$$

is analytic in $\mathbb{U}$ and $Rep\left(z\right)>\alpha$$(z\in \mathbb{U})$ with $0\le \alpha <1,$ then

$$\begin{array}{c}\hfill |c_k|\le 2(1-\alpha )(z\in \mathbb{U}).\end{array}$$

(7)

The equality in (7) is satisfied by $p\left(z\right)$ such that

$$\begin{array}{c}\hfill p\left(z\right)={\displaystyle \frac{1+(1-2\alpha )z}{1-z}}=1+2(1-\alpha )\sum _{k=1}^{\infty }{z}^k.\end{array}$$

With the above lemma, we see the following lemma.

Lemma 2. 

If a function $p\left(z\right)$ given by

$$\begin{array}{c}\hfill p\left(z\right)=1+\sum _{k=1}^{\infty }{c}_{{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}}(n\in \mathbb{N})\end{array}$$

is analytic in $\mathbb{U}$ and $Rep\left(z\right)>\alpha$$(z\in \mathbb{U})$ with $0\le \alpha <1,$ then

$$\begin{array}{c}\hfill |c_{{\displaystyle \frac{k}{n}}}|\le 2(1-\alpha )(k\in \mathbb{N}).\end{array}$$

(8)

The equality in (8) is satisfied by $p\left(z\right)$ such that

$$\begin{array}{c}\hfill p\left(z\right)={\displaystyle \frac{1+(1-2\alpha )\sqrt[n]{z}}{1-\sqrt[n]{z}}}=1+2(1-\alpha )\sum _{k=1}^{\infty }{z}^{{\displaystyle \frac{k}{n}}}.\end{array}$$

Applying Lemma 2, we prove the following theorem.

Theorem 3. 

If $f\left(z\right)\in {\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ),$ then

$$\begin{array}{c}\hfill \left|a_{1+{\displaystyle \frac{1}{n}}}\right|\le {\displaystyle \frac{2n\left|e^{i\alpha }-\beta cos\alpha \right|\Gamma \left(\frac{1}{n}+2-\lambda \right){\prod }_{\ell =1}^j|2-\lambda -\ell |}{\Gamma (2-\lambda )\Gamma \left(\frac{1}{n}+2\right){\prod }_{\ell =1}^{j-1}\left|\frac{1}{n}+2-\lambda -\ell \right|}}\end{array}$$

(9)

and

$$\begin{array}{cc}\hfill \left|a_{1+{\displaystyle \frac{k}{n}}}\right|\le & {\displaystyle \frac{2n\left|e^{i\alpha }-\beta cos\alpha \right|\Gamma \left(\frac{k}{n}+2-\lambda \right){\prod }_{\ell =1}^j|2-\lambda -\ell |}{k\Gamma (2-\lambda )\Gamma \left(\frac{k}{n}+2\right){\prod }_{\ell =1}^{j-1}\left|\frac{k}{n}+2-\lambda -\ell \right|}}\hfill \\ \hfill & \times \prod _{m=1}^{k-1}\left(1+{\displaystyle \frac{2n}{m}}\left|e^{i\alpha }-\beta cos\alpha \right||2-\lambda -j|\right)\hfill \end{array}$$

(10)

for $k=2,3,4,\dots ,$ where $j=2,3,4,\dots .$ The equalities in (9) and (10) are satisfied by $f\left(z\right)\in {\mathcal{A}}_n$ such that

$$\begin{array}{c}\hfill D_z^{j-1+\lambda }f\left(z\right)={\left(z{\left(1-\sqrt[n]{z}\right)}^t\right)}^{2-\lambda -j}\end{array}$$

with

$$\begin{array}{c}\hfill t=2n\left(\beta cos\alpha e^{-i\alpha }-1\right).\end{array}$$

Proof. 

Let us consider a function $p\left(z\right)$ as follows:

$$\begin{array}{cc}\hfill p\left(z\right)& ={\displaystyle \frac{e^{i\alpha }\frac{z{D}_z^{j+\lambda }f\left(z\right)}{(2-\lambda -j)D_z^{j-1+\lambda }f\left(z\right)}-\beta cos\alpha }{e^{i\alpha }-\beta cos\alpha }}=1+\sum _{k=1}^{\infty }{c}_{{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}}\hfill \end{array}$$

(11)

for $f\left(z\right)\in {\mathcal{A}}_n(j,\lambda ,\alpha ,\beta ).$ Then, we know that $p\left(z\right)$ is analytic in $\mathbb{U},$$p\left(0\right)=1$ and that $Rep\left(z\right)>0$$(z\in \mathbb{U}).$ Thus, Lemma 2 implies that

$$\begin{array}{c}\hfill \left|c_{{\displaystyle \frac{k}{n}}}\right|\le 2(k\in \mathbb{N})\end{array}$$

(12)

and the equality in (12) is satisfied by

$$\begin{array}{c}\hfill p\left(z\right)={\displaystyle \frac{1+\sqrt[n]{z}}{1-\sqrt[n]{z}}}.\end{array}$$

By (11), we see

$$\begin{array}{cc}\hfill e^{i\alpha }z{D}_z^{j+\lambda }f\left(z\right)& -\beta (2-\lambda -j)cos\alpha D_z^{j-1+\lambda }f\left(z\right)\hfill \\ \hfill & =(2-\lambda -j)\left(e^{i\alpha }-\beta cos\alpha \right)p\left(z\right)D_z^{j-1+\lambda }f\left(z\right).\hfill \end{array}$$

It is clear that

$$\begin{array}{cc}\hfill e^{i\alpha }z{D}_z^{j+\lambda }f\left(z\right)& -\beta (2-\lambda -j)cos\alpha D_z^{j-1+\lambda }f\left(z\right)\hfill \\ \hfill & ={\displaystyle \frac{\left(e^{i\alpha }-\beta cos\alpha \right)}{\Gamma (2-\lambda )}}\prod _{\ell =1}^j(2-\lambda -\ell )z^{2-\lambda -j}\hfill \\ \hfill & +\sum _{k=1}^{\infty }\left\{\left(e^{i\alpha }\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)-\beta (2-\lambda -j)\right){\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\right.\hfill \\ \hfill & \left.\times \left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)\right)\right\}{a}_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}+2-\lambda -j}.\hfill \end{array}$$

(13)

Further, we calculate

$$\begin{array}{cc}\hfill & (2-\lambda -j)\left(e^{i\alpha }-\beta cos\alpha \right)p\left(z\right)D_z^{j-1+\lambda }f\left(z\right)\hfill \\ \hfill & =(2-\lambda -j)\left(e^{i\alpha }-\beta cos\alpha \right)\left(1+\sum _{k=1}^{\infty }{c}_{{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}}\right)\times \left({\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}{z}^{2-\lambda -j}\right.\hfill \\ \hfill & \left.+\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\left(\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}+1-\lambda -j}\right)\right)\hfill \\ \hfill & =(2-\lambda -j)(e^{i\alpha }-\beta cos\alpha )\left\{{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}{z}^{2-\lambda -j}\right.\hfill \\ \hfill & +\left(c_{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}+{\displaystyle \frac{\Gamma \left(\frac{1}{n}+2\right)}{\Gamma \left(\frac{1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\right)z^{{\displaystyle \frac{1}{n}}+2-\lambda -j}\hfill \\ \hfill & +\left(c_{{\displaystyle \frac{2}{n}}}{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}+c_{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{2}{n}+2\right)}{\Gamma \left(\frac{2}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{\Gamma \left(\frac{2}{n}+2\right)}{\Gamma \left(\frac{2}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{2}{n}}}\right)z^{{\displaystyle \frac{2}{n}}+2-\lambda -j}\hfill \\ \hfill & +\left(c_{{\displaystyle \frac{3}{n}}}{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}+c_{{\displaystyle \frac{2}{n}}}{\displaystyle \frac{\Gamma \left(\frac{1}{n}+2\right)}{\Gamma \left(\frac{1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\right.\hfill \\ \hfill & +c_{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{2}{n}+2\right)}{\Gamma \left(\frac{2}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{2}{n}}}\hfill \\ \hfill & \left.+{\displaystyle \frac{\Gamma \left(\frac{3}{n}+2\right)}{\Gamma \left(\frac{3}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{3}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{3}{n}}}\right)z^{{\displaystyle \frac{3}{n}}+2-\lambda -j}\hfill \\ \hfill & +\dots +\left(c_{{\displaystyle \frac{k}{n}}}{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}\right.\hfill \\ \hfill & +c_{{\displaystyle \frac{k-1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{1}{n}+2\right)}{\Gamma \left(\frac{1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\hfill \\ \hfill & +c_{{\displaystyle \frac{k-2}{n}}}{\displaystyle \frac{\Gamma \left(\frac{2}{n}+2\right)}{\Gamma \left(\frac{2}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{2}{n}}}\hfill \\ \hfill & +\dots +c_{{\displaystyle \frac{2}{n}}}{\displaystyle \frac{\Gamma \left(\frac{k-2}{n}+2\right)}{\Gamma \left(\frac{k-2}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k-2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k-2}{n}}}\hfill \\ \hfill & +c_{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{k-1}{n}+2\right)}{\Gamma \left(\frac{k-1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k-1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k-1}{n}}}\hfill \\ \hfill & \left.\left.+{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k}{n}}}\right)z^{{\displaystyle \frac{k}{n}}+2-\lambda -j}+\dots \right\}.\hfill \end{array}$$

(14)

Therefore, for $k=1,$ we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{e^{i\alpha }\Gamma \left(\frac{1}{n}+2\right)}{n\Gamma \left(\frac{1}{n}+2-\lambda \right)}}& \prod _{\ell =1}^{j-1}\left({\displaystyle \frac{1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\hfill \\ \hfill & =c_{{\displaystyle \frac{1}{n}}}(2-\lambda -j)\left(e^{i\alpha }-\beta cos\alpha \right){\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}.\hfill \end{array}$$

Using $|c_{{\displaystyle \frac{1}{n}}}|\le 2$ for $p\left(z\right),$ we have

$$\begin{array}{c}\hfill \left|a_{1+{\displaystyle \frac{1}{n}}}\right|\le {\displaystyle \frac{2n|e^{i\alpha }-\beta cos\alpha |\Gamma \left(\frac{1}{n}+2-\lambda \right){\prod }_{\ell =1}^{j-1}|2-\lambda -\ell |}{\Gamma (2-\lambda )\Gamma \left(\frac{1}{n}+2\right){\prod }_{\ell =1}^{j-1}\left|\frac{1}{n}+2-\lambda -\ell \right|}}.\end{array}$$

(15)

This proves the coefficient inequality (9) for $a_{1+{\displaystyle \frac{1}{n}}}.$ □

Considering the coefficient for $z^{{\displaystyle \frac{2}{n}}+2-\lambda -j}$ in (13) and (14), we prove

$$\begin{array}{cc}\hfill {\displaystyle \frac{2e^{i\alpha }\Gamma \left(\frac{2}{n}+2\right)}{n\Gamma \left(\frac{2}{n}+2-\lambda \right)}}& \prod _{\ell =1}^{j-1}\left({\displaystyle \frac{2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{2}{n}}}\hfill \\ \hfill & =(2-\lambda -j)\left(e^{i\alpha }-\beta cos\alpha \right)\left(c_{{\displaystyle \frac{2}{n}}}{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}\right.\hfill \\ \hfill & \left.+c_{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{1}{n}+2\right)}{\Gamma \left(\frac{1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\right).\hfill \end{array}$$

Therefore, using (8) and (15), it follows that

$$\begin{array}{cc}\hfill \left|a_{1+{\displaystyle \frac{2}{n}}}\right|\le & {\displaystyle \frac{2n|e^{i\alpha }-\beta cos\alpha |\Gamma \left(\frac{2}{n}+2-\lambda \right){\prod }_{\ell =1}^{j-1}|2-\lambda -\ell |}{2\Gamma (2-\lambda )\Gamma \left(\frac{2}{n}+2\right){\prod }_{\ell =1}^{j-1}\left|\frac{2}{n}+2-\lambda -\ell \right|}}\hfill \\ \hfill & \times \left(1+2n|2-\lambda -j|\left|e^{i\alpha }-\beta cos\alpha \right|\right).\hfill \end{array}$$

This shows the coefficient inequality for $a_{1+{\displaystyle \frac{2}{n}}}$ in (10).

Furthermore, we have from (13) and (14) that

$$\begin{array}{cc}\hfill {\displaystyle \frac{k{e}^{i\alpha }\Gamma \left(\frac{k}{n}+2\right)}{n\Gamma \left(\frac{k}{n}+2-\lambda \right)}}& \prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k}{n}}}\hfill \\ \hfill & =(2-\lambda -j)\left(e^{i\alpha }-\beta cos\alpha \right)\left(c_{{\displaystyle \frac{k}{n}}}{\displaystyle \frac{{\prod }_{\ell =1}^{j-1}(2-\lambda -\ell )}{\Gamma (2-\lambda )}}\right.\hfill \\ \hfill & +c_{{\displaystyle \frac{k-1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{1}{n}+2\right)}{\Gamma \left(\frac{1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{1}{n}}}\hfill \\ \hfill & +c_{{\displaystyle \frac{k-2}{n}}}{\displaystyle \frac{\Gamma \left(\frac{2}{n}+2\right)}{\Gamma \left(\frac{2}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{2}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{2}{n}}}\hfill \\ \hfill & \left.+\dots +c_{{\displaystyle \frac{1}{n}}}{\displaystyle \frac{\Gamma \left(\frac{k-1}{n}+2\right)}{\Gamma \left(\frac{k-1}{n}+2-\lambda \right)}}\prod _{\ell =1}^{j-1}\left({\displaystyle \frac{k-1}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k-1}{n}}}\right).\hfill \end{array}$$

Thus, by mathematical induction, we also prove that

$$\begin{array}{cc}\hfill \left|a_{1+{\displaystyle \frac{k}{n}}}\right|\le & {\displaystyle \frac{2n|e^{i\alpha }-\beta cos\alpha |\Gamma \left(\frac{k}{n}+2-\lambda \right){\prod }_{\ell =1}^j|2-\lambda -\ell |}{k\Gamma (2-\lambda )\Gamma \left(\frac{k}{n}+2\right){\prod }_{\ell =1}^{j-1}\left|\frac{k}{n}+2-\lambda -\ell \right|}}\hfill \\ \hfill & \times \prod _{m=1}^{k-1}\left(1+{\displaystyle \frac{2n}{m}}\left|e^{i\alpha }-\beta cos\alpha \right||2-\lambda -\ell |\right)\hfill \end{array}$$

for $k=2,3,4,\dots .$

Finally, considering a function $p\left(z\right)$ given by

$$\begin{array}{c}\hfill p\left(z\right)={\displaystyle \frac{1+\sqrt[n]{z}}{1-\sqrt[n]{z}}}=1+2\sum _{k=1}^{\infty }{z}^{{\displaystyle \frac{k}{n}}},\end{array}$$

Equation (11) implies that

$$\begin{array}{cc}\hfill {\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{D_z^{j-1+\lambda }f\left(z\right)}}& =(2-\lambda -j)\left(e^{-i\alpha }\beta cos\alpha +(1-e^{-i\alpha }\beta cos\alpha ){\displaystyle \frac{1+\sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)\hfill \\ \hfill & =(2-\lambda -j){\displaystyle \frac{1+(1-2e^{-i\alpha }\beta cos\alpha )\sqrt[n]{z}}{1-\sqrt[n]{z}}}\hfill \end{array}$$

and that

$$\begin{array}{c}\hfill {\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{D_z^{j-1+\lambda }f\left(z\right)}}=(2-\lambda -j)\left({\displaystyle \frac{1}{z}}+2(1-e^{-i\alpha }\beta cos\alpha ){\displaystyle \frac{\sqrt[n]{z}}{z(1-\sqrt[n]{z})}}\right).\end{array}$$

This gives us

$$\begin{array}{c}\hfill {\left(log{D}_z^{j-1+\lambda }f\left(z\right)\right)}^{\prime }=(2-\lambda -j)\left({(logz)}^{\prime }+2n(\beta cos\alpha e^{-i\alpha }-1){\left(log(1-\sqrt[n]{z})\right)}^{\prime }\right).\end{array}$$

Thus, we obtain the following:

$$\begin{array}{c}\hfill D_z^{j-1+\lambda }f\left(z\right)={\left(z{(1-\sqrt[n]{z})}^t\right)}^{2-\lambda -j}\end{array}$$

with

$$\begin{array}{c}\hfill t=2n(\beta cos\alpha e^{-i\alpha }-1).\end{array}$$

Taking $\lambda =0$ in Theorem 3, we see the following corollary.

Corollary 2. 

If $f\left(z\right)\in {\mathcal{A}}_n(j,0,\alpha ,\beta ),$ then

$$\begin{array}{c}\hfill \left|a_{1+{\displaystyle \frac{1}{n}}}\right|\le {\displaystyle \frac{2n\left|e^{i\alpha }-\beta cos\alpha \right|{\prod }_{\ell =1}^j|2-\ell |}{{\prod }_{\ell =1}^{j-1}\left|\frac{1}{n}+2-\ell \right|}}\end{array}$$

and

$$\begin{array}{cc}\hfill \left|a_{1+{\displaystyle \frac{k}{n}}}\right|\le & {\displaystyle \frac{2n\left|e^{i\alpha }-\beta cos\alpha \right|{\prod }_{\ell =1}^j|2-\ell |}{k{\prod }_{\ell =1}^{j-1}\left|\frac{k}{n}+2-\ell \right|}}\times \prod _{m=1}^{k-1}\left(1+{\displaystyle \frac{2n}{m}}\left|e^{i\alpha }-\beta cos\alpha \right||2-j|\right)\hfill \end{array}$$

for $k=2,3,4,\dots ,$ where $j=2,3,4,\dots .$

4. Argument Properties of Fractional Derivatives

For analytic functions $f\left(z\right)$ and $g\left(z\right)$ in $\mathbb{U}$, $f\left(z\right)$ is said to be subordinate to $g\left(z\right)$ in $\mathbb{U},$ written $f\left(z\right)\prec g\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}),$ such that $f\left(z\right)=g\left(w\right(z\left)\right)$$(z\in \mathbb{U}).$ If $g\left(z\right)$ is univalent in $\mathbb{U},$ then $f\left(z\right)\prec g\left(z\right)$$(z\in \mathbb{U})$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathbb{U}\right)\subset g\left(\mathbb{U}\right)$ [5]. For subordinations, Miller and Mocanu [6] gave the following lemma.

Lemma 3. 

Let ${\beta }_0=1.21872\dots$ be the solution of $\beta \pi ={\displaystyle \frac{3}{2}}\pi -Ta{n}^{-1}\beta$ and $\alpha =\alpha \left(\beta \right)=\beta +2Ta{n}^{-1}\left({\displaystyle \frac{\beta }{\pi }}\right)$ for $0<\beta <{\beta }_0.$ If $g\left(z\right)$ is analytic in $\mathbb{U}$ with $g\left(0\right)=1,$ then

$$\begin{array}{c}\hfill g\left(z\right)+z{g}^{\prime }\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\alpha }(z\in \mathbb{U})\end{array}$$

(16)

gives

$$\begin{array}{c}\hfill g\left(z\right)\prec {\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }(z\in \mathbb{U}).\end{array}$$

(17)

It follows that the subordination given in (16) implies that

$$\begin{array}{c}\hfill |arg(g\left(z\right)+z{g}^{\prime }\left(z\right))|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U})\end{array}$$

and the subordination (17) implies that

$$\begin{array}{c}\hfill |argg\left(z\right)|<{\displaystyle \frac{\pi }{2}}\beta (z\in \mathbb{U}).\end{array}$$

We also introduce the next lemma, which was established by Nunokawa [7].

Lemma 4. 

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

$$\begin{array}{c}\hfill |argg\left(z\right)|<{\displaystyle \frac{\pi }{2}}\alpha \left(\right|z|<|z_0\left|\right)\end{array}$$

and

$$\begin{array}{c}\hfill |argg\left(z_0\right)|={\displaystyle \frac{\pi }{2}}\alpha \end{array}$$

for some real $\alpha >0,$ then we have

$$\begin{array}{c}\hfill {\displaystyle \frac{z_0{g}^{\prime }\left(z_0\right)}{g\left(z_0\right)}}=ik\alpha ,\end{array}$$

where

$$\begin{array}{c}\hfill k\ge {\displaystyle \frac{1}{2}}\left(a+{\displaystyle \frac{1}{a}}\right)(forargg\left(z_0\right)={\displaystyle \frac{\pi }{2}}\alpha ),\end{array}$$

(18)

$$\begin{array}{c}\hfill k\le {\displaystyle \frac{1}{2}}\left(a+{\displaystyle \frac{1}{a}}\right)(forargg\left(z_0\right)=-{\displaystyle \frac{\pi }{2}}\alpha ),\end{array}$$

(19)

and

$$\begin{array}{c}\hfill g{\left(z_0\right)}^{{\displaystyle \frac{1}{\alpha }}}=\pm ia(a>0).\end{array}$$

Now we have the following theorem.

Theorem 4. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }}}+j+\lambda \right)\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U})\end{array}$$

(20)

for some $\alpha >0,$$0\le \lambda <1$ and $j\in \mathbb{N},$ then

$$\begin{array}{c}\hfill \left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha \beta (z\in \mathbb{U}),\end{array}$$

where

$$\begin{array}{c}\hfill \beta ={\displaystyle \frac{2}{\pi }}{\int }_0^{1}Si{n}^{-1}\left({\displaystyle \frac{2\rho }{1+{\rho }^2}}\right)d\rho =0.55872876\dots .\end{array}$$

(21)

Proof. 

For $f\left(z\right)\in A_n$ satisfying (20), we define the function $g\left(z\right)$ as follows:

$$\begin{array}{cc}\hfill g\left(z\right)& ={\displaystyle \frac{\Gamma (2-\lambda )D_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )z^{1-j-\lambda }}}\hfill \\ \hfill & =1+{\displaystyle \frac{\Gamma (2-\lambda )}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\sum _{k=1}^{\infty }{\displaystyle \frac{\Gamma \left(\frac{k}{n}+2\right)}{\Gamma \left(\frac{k}{n}+2-\lambda \right)}}\prod _{\ell =1}^j\left({\displaystyle \frac{k}{n}}+2-\lambda -\ell \right)a_{1+{\displaystyle \frac{k}{n}}}{z}^{{\displaystyle \frac{k}{n}}}.\hfill \end{array}$$

(22)

Then, $g\left(z\right)$ is analytic in $\mathbb{U}$ with $g\left(0\right)=1$ and $g\left(z\right)\ne 0$$(z\in \mathbb{U})$, as follows from (20). For the above-defined $g\left(z\right)$, we see that

$$\begin{array}{cc}\hfill argg\left(z\right)& =arg\left\{{\displaystyle \frac{1}{z}}\left(zg\left(z\right)\right)\right\}\hfill \\ \hfill & =arg\left\{{\displaystyle \frac{1}{z}}{\int }_0^z{\left(tg\left(t\right)\right)}^{\prime }dt\right\}\hfill \\ \hfill & =arg\left\{{\displaystyle \frac{1}{z}}{\int }_0^r{\displaystyle \frac{d}{d\rho }}\left(\rho e^{i\theta }g\left(\rho e^{i\theta }\right)\right)e^{i\theta }d\rho \right\}\hfill \\ \hfill & =arg\left\{{\int }_0^r\left(g\left(\rho e^{i\theta }\right)+\rho e^{i\theta }{\displaystyle \frac{d}{d\rho }}g\left(\rho e^{i\theta }\right)\right)d\rho \right\},\hfill \end{array}$$

where $z=r{e}^{i\theta }$ and $t=\rho e^{i\theta }.$ Let

$$\begin{array}{c}\hfill 0={\rho }_0<{\rho }_1<{\rho }_2<\dots <{\rho }_{n-1}<{\rho }_n=r,\sum _{\ell =1}^n{\rho }_{\ell }=\rho \end{array}$$

and

$$\begin{array}{c}\hfill {\rho }_{\ell }-{\rho }_{\ell -1}=\delta (\ell =1,2,3,\dots ,n).\end{array}$$

Then, we have

$$\begin{array}{cc}\hfill |argg(z\left)\right|& =\left|arg\left\{\underset{n\to \infty }{lim}\sum _{\ell =1}^n\delta (g\left({\rho }_{\ell }{e}^{i\theta }\right)+{\rho }_{\ell }{e}^{i\theta }{g}^{\prime }\left({\rho }_{\ell }{e}^{i\theta }\right))\right\}\right|\hfill \\ \hfill & \le \underset{n\to \infty }{lim}\sum _{\ell =1}^n\delta \left|arg(g\left({\rho }_{\ell }{e}^{i\theta }\right)+{\rho }_{\ell }{e}^{i\theta }{g}^{\prime }\left({\rho }_{\ell }{e}^{i\theta }\right))\right|.\hfill \end{array}$$

Since

$$\begin{array}{c}\hfill |argg\left(z\right)+z{g}^{\prime }\left(z\right)|=\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+j+\lambda \right)\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha \end{array}$$

by (20), using Lemma 4, we obtain the following:

$$\begin{array}{cc}\hfill |argg(z\left)\right|& =\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\right)\right|\hfill \\ \hfill & \le \underset{n\to \infty }{lim}\sum _{\ell =1}^n\delta \left|arg{\left({\displaystyle \frac{1+{\rho }_{\ell }{e}^{i\theta }}{1-{\rho }_{\ell }{e}^{i\theta }}}\right)}^{\alpha }\right|\hfill \\ \hfill & =\alpha {\int }_0^{r}Si{n}^{-1}\left({\displaystyle \frac{2\rho }{1+{\rho }^2}}\right)d\rho \hfill \\ \hfill & <\alpha {\int }_0^{1}Si{n}^{-1}\left({\displaystyle \frac{2\rho }{1+{\rho }^2}}\right)d\rho ={\displaystyle \frac{\pi }{2}}\alpha \beta .\hfill \end{array}$$

This completes the proof of the theorem. □

Taking $j=1$ and $\lambda =0$ in Theorem 4, we have the following corollary.

Corollary 3. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|arg\left(f^{\prime }\left(z\right)+z{f}^{″}\left(z\right)\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}),\end{array}$$

for some $\alpha >0,$ then

$$\begin{array}{c}\hfill \left|arg(f^{\prime }\left(z\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha \beta (z\in \mathbb{U}),\end{array}$$

where β is given by (21).

Remark 1. 

If we take $\alpha ={\displaystyle \frac{3}{2}},$ then we have

$$\begin{array}{c}\hfill \alpha \beta ={\displaystyle \frac{3}{\pi }}\left(1-{\displaystyle \frac{2}{\pi }}log2\right)=0.533546701\dots .\end{array}$$

Next, applying Lemma 4, as developed by Nunokawa [7], we prove the following theorem.

Theorem 5. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|arg\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+{\displaystyle \frac{\Gamma (2-\lambda )D_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )z^{1-j-\lambda }}}+j+\lambda -1\right)\right|<Ta{n}^{-1}\beta \left(\alpha \right)(z\in \mathbb{U})\end{array}$$

(23)

for some α$(0<\alpha <1),$$0\le \lambda <1$ and $j\in \mathbb{N},$ then

$$\begin{array}{c}\hfill \left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}),\end{array}$$

where

$$\begin{array}{c}\hfill \beta \left(\alpha \right)={\displaystyle \frac{\alpha }{2}}\left\{{\left({\displaystyle \frac{1+\alpha }{1-\alpha }}\right)}^{{\displaystyle \frac{1-\alpha }{2}}}+{\left({\displaystyle \frac{1-\alpha }{1+\alpha }}\right)}^{{\displaystyle \frac{1+\alpha }{2}}}\right\}sec\left({\displaystyle \frac{\pi }{2}}\alpha \right)+tan\left({\displaystyle \frac{\pi }{2}}\alpha \right).\end{array}$$

(24)

Proof. 

We consider a function $g\left(z\right)$ given by (22) for $f\left(z\right)\in A_n.$ Then, $g\left(z\right)$ is analytic in $\mathbb{U}$ with $g\left(0\right)=1.$ For the above-defined $g\left(z\right),$ we see that

$$\begin{array}{c}\hfill {\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}+g\left(z\right)={\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+{\displaystyle \frac{\Gamma (2-\lambda )D_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )z^{1-j-\lambda }}}+j+\lambda -1.\end{array}$$

We suppose that there exists a point $z_0\in \mathbb{U}$ such that

$$\begin{array}{c}\hfill |argg\left(z\right)|=\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha \left(\right|z|<|z_0\left|\right)\end{array}$$

and

$$\begin{array}{c}\hfill |argg\left(z_0\right)|=\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z_0\right)}{z_0^{1-j-\lambda }}}\right)\right|={\displaystyle \frac{\pi }{2}}\alpha .\end{array}$$

Then, Lemma 4 gives us the following:

$$\begin{array}{c}\hfill {\displaystyle \frac{z_0{g}^{\prime }\left(z_0\right)}{g\left(z_0\right)}}={\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f\left(z_0\right)}}+j+\lambda -1=ik\alpha ,\end{array}$$

where k is given by (18) and (19). If $argg\left(z_0\right)={\displaystyle \frac{\pi }{2}}\alpha ,$ then $g\left(z_0\right)=a^{\alpha }{e}^{i{\displaystyle \frac{\pi }{2}}\alpha }$ and

$$\begin{array}{cc}\hfill arg\left({\displaystyle \frac{z_0{g}^{\prime }\left(z_0\right)}{g\left(z_0\right)}}+g\left(z_0\right)\right)& =arg\left({\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f\left(z_0\right)}}+{\displaystyle \frac{\Gamma (2-\lambda )D_z^{j+\lambda }f\left(z_0\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )z_0^{1-j-\lambda }}}+j+\lambda -1\right)\hfill \\ \hfill & =arg\left(ik\alpha +a^{\alpha }{e}^{i{\displaystyle \frac{\pi }{2}}\alpha }\right)\hfill \\ \hfill & =arg\left\{a^{\alpha }cos\left({\displaystyle \frac{\pi }{2}}\alpha \right)+i\left(k\alpha +a^{\alpha }sin\left({\displaystyle \frac{\pi }{2}}\alpha \right)\right)\right\}\hfill \\ \hfill & \ge arg\left\{a^{\alpha }cos\left({\displaystyle \frac{\pi }{2}}\alpha \right)+i\left({\displaystyle \frac{\alpha }{2}}\left(a+{\displaystyle \frac{1}{a}}\right)+a^{\alpha }sin\left({\displaystyle \frac{\pi }{2}}\alpha \right)\right)\right\}\hfill \\ \hfill & =Ta{n}^{-1}\left\{{\displaystyle \frac{\alpha }{2}}\left(a^{1-\alpha }+a^{-1-\alpha }\right)sec\left({\displaystyle \frac{\pi }{2}}\alpha \right)+tan\left({\displaystyle \frac{\pi }{2}}\alpha \right)\right\}\hfill \\ \hfill & \ge Ta{n}^{-1}\left\{{\displaystyle \frac{\alpha }{2}}\left({\left({\displaystyle \frac{1+\alpha }{1-\alpha }}\right)}^{{\displaystyle \frac{1-\alpha }{2}}}+{\left({\displaystyle \frac{1-\alpha }{1+\alpha }}\right)}^{{\displaystyle \frac{1+\alpha }{2}}}\right)sec\left({\displaystyle \frac{\pi }{2}}\alpha \right)+tan\left({\displaystyle \frac{\pi }{2}}\alpha \right)\right\}\hfill \\ \hfill & =Ta{n}^{-1}\beta \left(\alpha \right).\hfill \end{array}$$

This contradicts the condition (23) of the theorem. If $argg\left(z_0\right)=-{\displaystyle \frac{\pi }{2}}\alpha ,$ then we also have that

$$\begin{array}{cc}\hfill arg\left({\displaystyle \frac{z_0{g}^{\prime }\left(z_0\right)}{g\left(z_0\right)}}+g\left(z_0\right)\right)& =arg\left({\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f\left(z_0\right)}}+{\displaystyle \frac{\Gamma (2-\lambda )D_z^{j+\lambda }f\left(z_0\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )z_0^{1-j-\lambda }}}+j+\lambda -1\right)\hfill \\ \hfill & \le -Ta{n}^{-1}\left\{{\displaystyle \frac{\alpha }{2}}\left({\left({\displaystyle \frac{1+\alpha }{1-\alpha }}\right)}^{{\displaystyle \frac{1-\alpha }{2}}}+{\left({\displaystyle \frac{1-\alpha }{1+\alpha }}\right)}^{{\displaystyle \frac{1+\alpha }{2}}}\right)sec\left({\displaystyle \frac{\pi }{2}}\alpha \right)+tan\left({\displaystyle \frac{\pi }{2}}\alpha \right)\right\}\hfill \\ \hfill & =-Ta{n}^{-1}\beta \left(\alpha \right).\hfill \end{array}$$

This also contradicts the condition (23) of the theorem for $g\left(z_0\right)=-{\displaystyle \frac{\pi }{2}}\alpha .$ Therefore, we say that there is no $z_0\in \mathbb{U}$ such that

$$\begin{array}{c}\hfill |argg\left(z\right)|=\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha \left(\right|z|<|z_0\left|\right)\end{array}$$

and

$$\begin{array}{c}\hfill |argg\left(z_0\right)|=\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z_0\right)}{z_0^{1-j-\lambda }}}\right)\right|={\displaystyle \frac{\pi }{2}}\alpha .\end{array}$$

This implies that

$$\begin{array}{c}\hfill |argg\left(z\right)|=\left|arg\left({\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z^{1-j-\lambda }}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\end{array}$$

□

Taking $j=1$ and $\lambda =0$ in Theorem 5, we have the following corollary.

Corollary 4. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|arg\left({\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}+f^{\prime }\left(z\right)\right)\right|<Ta{n}^{-1}\beta \left(\alpha \right)(z\in \mathbb{U}),\end{array}$$

for some real α$(0<\alpha <1),$ then

$$\begin{array}{c}\hfill \left|arg{f}^{\prime }\left(z\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}),\end{array}$$

where $\beta \left(\alpha \right)$ is given by (24).

Next, we consider $f\left(z\right)\in A_n$ given by

$$\begin{array}{c}\hfill {\displaystyle \frac{\Gamma (2-\lambda )z^{j+\lambda }{D}_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}={\int }_0^z{\left({\displaystyle \frac{1+\sqrt[n]{t}}{1-\sqrt[n]{t}}}\right)}^{\alpha }dt(z\in \mathbb{U})\end{array}$$

with $0\le \alpha <1$, $0\le \lambda <1$ and $j\in \mathbb{N}.$ Then $f\left(z\right)$ satisfies

$$\begin{array}{cc}\hfill & \left|arg{\left({\displaystyle \frac{\Gamma (2-\lambda )z^{j+\lambda }{D}_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\right)}^{\prime }\right|\hfill \\ \hfill & =\left|arg\left(z^{j+\lambda }\left((j+\lambda ){\displaystyle \frac{D_z^{j+\lambda }f\left(z\right)}{z}}+D_z^{j+1+\lambda }f\left(z\right)\right)\right)\right|\hfill \\ \hfill & =\alpha \left|arg\left({\displaystyle \frac{1+\sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\hfill \end{array}$$

To discuss such an argument problem, we need the result devised by Fejér and Riesz [8] (or by Tsuji [9]).

Lemma 5. 

Let a function $f\left(z\right)$ be analytic in $\left|z\right|\le 1.$ Then, $f\left(z\right)$ satisfies

$$\begin{array}{c}\hfill {\int }_{-1}^1{\left|f\left(z\right)\right|}^{\alpha }\left|dz\right|\le {\displaystyle \frac{1}{2}}{\int }_{\left|z\right|=1}{\left|f\left(z\right)\right|}^{\alpha }\left|dz\right|(\alpha >0),\end{array}$$

(25)

where the above integral on the left-hand side is considered along the real axis.

Remark 2. 

When we make the change of variables in Lemma 5, Equation (25) becomes

$$\begin{array}{c}\hfill {\int }_{-r}^r|f\left(\rho e^{i\theta }\right){|}^{\alpha }d\rho \le {\displaystyle \frac{r}{2}}{\int }_0^{2\pi }{\left|f\left(r{e}^{i\theta }\right)\right|}^{\alpha }d\theta .\end{array}$$

(26)

Further, Gwynme [10] gave the following lemma.

Lemma 6. 

Let $f\left(z\right)$ be a complex valued harmonic function defined on a neighborhood of a closed disc of radius one and center the origin in the complex plane. Then

$$\begin{array}{c}\hfill f\left(r{e}^{i\theta }\right)={\displaystyle \frac{1}{2\pi }}{\int }_0^{2\pi }f\left(e^{i\rho }\right){\displaystyle \frac{1-r^2}{1+r^2-2rcos(\theta -\rho )}}d\rho \end{array}$$

and

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2\pi }}{\int }_0^{2\pi }f\left(e^{i\rho }\right){\displaystyle \frac{1-r^2}{1+r^2-2rcos\rho }}d\rho =1.\end{array}$$

Now, we derive the following theorem.

Theorem 6. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right|\le {\displaystyle \frac{\alpha }{2}}(z\in \mathbb{U})\end{array}$$

(27)

for some α$(0\le \alpha <1),$ then

$$\begin{array}{c}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}),\end{array}$$

where $j\in \mathbb{N}$ and $0\le \lambda <1.$

Proof. 

We note that

$$\begin{array}{c}\hfill log\left({\displaystyle \frac{\Gamma (2-\lambda )z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\right)=log\left|{\displaystyle \frac{\Gamma (2-\lambda )z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\right|+iarg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\end{array}$$

(28)

and

$$\begin{array}{cc}\hfill log\left({\displaystyle \frac{\Gamma (2-\lambda )z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\right)& ={\int }_0^z\left(log{\left({\displaystyle \frac{\Gamma (2-\lambda )t^{j-1+\lambda }{D}_z^{j+\lambda }f\left(t\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}\right)}^{\prime }\right)dt\hfill \\ \hfill & ={\int }_0^z\left({\displaystyle \frac{j-1+\lambda }{t}}+{\displaystyle \frac{D_z^{j+1+\lambda }f\left(t\right)}{D_z^{j+\lambda }f\left(t\right)}}\right)dt.\hfill \end{array}$$

(29)

It follows from (28) and (29) that

$$\begin{array}{cc}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|& =\left|Im\left({\int }_0^z\left({\displaystyle \frac{j-1+\lambda }{t}}+{\displaystyle \frac{D_z^{j+1+\lambda }f\left(t\right)}{D_z^{j+\lambda }f\left(t\right)}}\right)dt\right)\right|\hfill \\ \hfill & \le \left|Im\left({\int }_0^r\left({\displaystyle \frac{j-1+\lambda }{\rho e^{i\theta }}}+{\displaystyle \frac{D_z^{j+1+\lambda }f\left(\rho e^{i\theta }\right)}{D_z^{j+\lambda }f\left(\rho e^{i\theta }\right)}}\right)e^{i\theta }d\rho \right)\right|\hfill \\ \hfill & ={\int }_0^r\left|Im\left({\displaystyle \frac{j-1+\lambda }{\rho }}+{\displaystyle \frac{e^{i\theta }{D}_z^{j+1+\lambda }f\left(\rho e^{i\theta }\right)}{D_z^{j+\lambda }f\left(\rho e^{i\theta }\right)}}\right)\right|d\rho \hfill \\ \hfill & \le {\int }_{-r}^r\left|{\displaystyle \frac{D_z^{j+1+\lambda }f\left(\rho e^{i\theta }\right)}{D_z^{j+\lambda }f\left(\rho e^{i\theta }\right)}}\right|d\rho ,\hfill \end{array}$$

where $z=r{e}^{i\theta }$$(0\le \theta <2\pi )$, $0\le r<1$ and $0\le \rho \le r.$ Using (26) for $\alpha =1,$ we see that

$$\begin{array}{cc}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|& \le {\displaystyle \frac{r}{2}}{\int }_0^{2\pi }\left|{\displaystyle \frac{D_z^{j+1+\lambda }f\left(r{e}^{i\theta }\right)}{D_z^{j+\lambda }f\left(r{e}^{i\theta }\right)}}\right|d\theta \hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\int }_0^{2\pi }\left|{\displaystyle \frac{r{e}^{i\theta }{D}_z^{j+1+\lambda }f\left(r{e}^{i\theta }\right)}{D_z^{j+\lambda }f\left(r{e}^{i\theta }\right)}}\right|d\theta .\hfill \end{array}$$

(30)

Therefore, if $f\left(z\right)$ satisfies (27), then $f\left(z\right)$ satisfies

$$\begin{array}{cc}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|& <{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\hfill \end{array}$$

□

Taking $j=1$ and $\lambda =0$ in Theorem 6, we have the following corollary.

Corollary 5. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right|\le {\displaystyle \frac{\alpha }{2}}(z\in \mathbb{U})\end{array}$$

for some α$(0\le \alpha <1),$ then

$$\begin{array}{c}\hfill \left|arg{f}^{\prime }\left(z\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\end{array}$$

Further, on applying Lemma 6, we arrive at the following theorem.

Theorem 7. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right|<{\displaystyle \frac{\alpha }{2}}Re\left({\displaystyle \frac{1+\beta \sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)(z\in \mathbb{U})\end{array}$$

(31)

for some α$(0\le \alpha <1)$ and β$(\beta \ne -1),$ then

$$\begin{array}{c}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}),\end{array}$$

where $j\in \mathbb{N}$ and $0\le \lambda <1.$

Proof. 

From (30), we see that

$$\begin{array}{c}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|<{\displaystyle \frac{1}{2}}{\int }_0^{2\pi }\left|{\displaystyle \frac{r{e}^{i\theta }{D}_z^{j+1+\lambda }f\left(r{e}^{i\theta }\right)}{D_z^{j+\lambda }f\left(r{e}^{i\theta }\right)}}\right|d\theta \end{array}$$

for $f\left(z\right)\in A_n.$ It follows that (31) gives

$$\begin{array}{cc}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|& <{\displaystyle \frac{\alpha }{4}}{\int }_0^{2\pi }Re\left({\displaystyle \frac{1+\beta \sqrt[n]{r}{e}^{i\frac{\theta }{n}}}{1-\sqrt[n]{r}{e}^{i\frac{\theta }{n}}}}\right)d\theta \hfill \\ \hfill & ={\displaystyle \frac{\alpha }{4}}{\int }_0^{2\pi }\left({\displaystyle \frac{1+(\beta -1)\sqrt[n]{r}cos\left(\frac{\theta }{n}\right)-\beta {\left(\sqrt[n]{r}\right)}^2}{1-2\sqrt[n]{r}cos\left(\frac{\theta }{n}\right)-{\left(\sqrt[n]{r}\right)}^2}}\right)d\theta \hfill \\ \hfill & ={\displaystyle \frac{\alpha }{4}}{\int }_0^{2\pi }\left\{{\displaystyle \frac{1-\beta }{2}}+\left({\displaystyle \frac{1+\beta }{2}}\right){\displaystyle \frac{1-{\left(\sqrt[n]{r}\right)}^2}{1+{\left(\sqrt[n]{r}\right)}^2-2\sqrt[n]{r}cos\left(\frac{\theta }{n}\right)}}\right\}d\theta .\hfill \end{array}$$

(32)

Applying Lemma 6, we obtain

$$\begin{array}{c}\hfill {\int }_0^{2\pi }{\displaystyle \frac{1-{\left(\sqrt[n]{r}\right)}^2}{1+{\left(\sqrt[n]{r}\right)}^2-2\sqrt[n]{r}cos\left(\frac{\theta }{n}\right)}}d\theta =n{\int }_0^{{\displaystyle \frac{2\pi }{n}}}{\displaystyle \frac{1-{\left(\sqrt[n]{r}\right)}^2}{1+{\left(\sqrt[n]{r}\right)}^2-2\sqrt[n]{r}cos\rho }}d\rho \le 2\pi .\end{array}$$

Therefore, (32) becomes

$$\begin{array}{c}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\end{array}$$

□

Setting $j=1$ and $\lambda =0$ in Theorem 7, we obtain the following corollary.

Corollary 6. 

If $f\left(z\right)\in A_n$ satisfies

$$\begin{array}{c}\hfill \left|{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right|<{\displaystyle \frac{\alpha }{2}}Re\left({\displaystyle \frac{1+\beta \sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)(z\in \mathbb{U})\end{array}$$

for some α$(0\le \alpha <1)$ and β$(\beta \ne -1),$ then

$$\begin{array}{c}\hfill \left|arg{f}^{\prime }\left(z\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\end{array}$$

Example 1. 

We consider a function $f\left(z\right)\in A_n$ given by

$$\begin{array}{c}\hfill {\displaystyle \frac{\Gamma (2-\lambda )z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)}{{\prod }_{\ell =1}^j(2-\lambda -\ell )}}={\left({\displaystyle \frac{2}{2-\sqrt[n]{z}}}\right)}^{3\alpha }(z\in \mathbb{U})\end{array}$$

with $0<\alpha <1.$ If we write that

$$\begin{array}{c}\hfill w\left(z\right)={\displaystyle \frac{2}{2-\sqrt[n]{z}}}(z\in \mathbb{U}),\end{array}$$

$w\left(z\right)$ satisfies

$$\begin{array}{c}\hfill \left|w\left(z\right)-{\displaystyle \frac{4}{3}}\right|<{\displaystyle \frac{2}{3}}(z\in \mathbb{U})\end{array}$$

and

$$\begin{array}{c}\hfill \left|argw\left(z\right)\right|=\left|arg\left({\displaystyle \frac{2}{2-\sqrt[n]{z}}}\right)\right|<{\displaystyle \frac{\pi }{6}}(z\in \mathbb{U}).\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill \left|arg\left(z^{j-1+\lambda }{D}_z^{j+\lambda }f\left(z\right)\right)\right|=3\alpha \left|arg\left({\displaystyle \frac{2}{2-\sqrt[n]{z}}}\right)\right|<{\displaystyle \frac{\pi }{2}}\alpha (z\in \mathbb{U}).\end{array}$$

For such a function $f\left(z\right),$ we obtain

$$\begin{array}{cc}\hfill \left|{\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right|& =\left|{\displaystyle \frac{3\alpha }{n}}\left({\displaystyle \frac{\sqrt[n]{z}}{2-\sqrt[n]{z}}}\right)-(j-1+\lambda )\right|\hfill \\ \hfill & <{\displaystyle \frac{3\alpha }{n}}\left|{\displaystyle \frac{\sqrt[n]{z}}{2-\sqrt[n]{z}}}\right|+(j-1+\lambda )\hfill \\ \hfill & <{\displaystyle \frac{3\alpha }{n}}+(j-1+\lambda )(z\in \mathbb{U}).\hfill \end{array}$$

If we consider some β such that

$$\begin{array}{c}\hfill \beta \le {\displaystyle \frac{n-12}{n}}-{\displaystyle \frac{4(j-1+\lambda )}{\alpha }},\end{array}$$

then

$$\begin{array}{cc}\hfill \left|{\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right|& <{\displaystyle \frac{3\alpha }{n}}+(j-1+\lambda )\hfill \\ \hfill & \le {\displaystyle \frac{\alpha (1-\beta )}{4}}\hfill \\ \hfill & <{\displaystyle \frac{\alpha }{2}}Re\left({\displaystyle \frac{1+\beta \sqrt[n]{z}}{1-\sqrt[n]{z}}}\right)(z\in \mathbb{U}).\hfill \end{array}$$