New Result of Analytic Functions Related to Hurwitz Zeta Function

Abstract

By using a linear operator, we obtain some new results for a normalized analytic function f defined by means of the Hadamard product of Hurwitz zeta function. A class related to this function will be introduced and the properties will be discussed.

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1. Introduction

A meromorphic function is a single-valued function, that is, analytic in all but possibly a discrete subset of its domain, and at those singularities it must go to infinity like a polynomial (i.e., these exceptional points must be poles and not essential singularities). A simpler definition states that a meromorphic function $f(z)$ is a function of the form $$\begin{array}{}\text{(1)}& f\left(z\right)=\frac{g\left(z\right)}{h\left(z\right)},\end{array}$$ where $g(z)$ and $h(z)$ are entire functions with $h(z)\ne \mathrm{0}$ (see [1, page 64]). A meromorphic function therefore may only have finite-order, isolated poles and zeros and no essential singularities in its domain. A meromorphic function with an infinite number of poles is exemplified by $\text{csc}(\mathrm{1}/z)$ on the punctured disk $U^{*}=\{z:\mathrm{0}<|z|<\mathrm{1}\}$ .

An equivalent definition of a meromorphic function is a complex analytic map to the Riemann sphere. For example, the Gamma function is meromorphic in the whole complex plane; see [1, 2].

In the present paper, we will derive some properties of univalent functions defined by means of the Hadamard product of Hurwitz Zeta function; a class related to this function will be introduced and the properties of the liner operator $L_a^t(\alpha ,\beta )f(z)$ will be discussed.

2. Preliminaries

Let $\Sigma$ denote the class of meromorphic functions $f(z)$ normalized by $$\begin{array}{}\text{(2)}& f\left(z\right)=\frac{\mathrm{1}}{z}+\sum _{n=\mathrm{1}}^{\infty }‍a_n{z}^n,\end{array}$$ which are analytic in the punctured unit disk $U^{*}$ . For $\mathrm{0}\le \beta$ , we denote by $S^{*}(\beta )$ and $k(\beta )$ the subclasses of $\Sigma$ consisting of all meromorphic functions which are, respectively, starlike of order $\beta$ and convex of order $\beta$ in $U^{*}$ .

For functions $f_j(z)$    $(j=\mathrm{1};\mathrm{2})$ defined by $$\begin{array}{}\text{(3)}& f_j\left(z\right)=\frac{\mathrm{1}}{z}+\sum _{n=\mathrm{1}}^{\infty }‍a_{n,j}{z}^n,\end{array}$$ we denote the Hadamard product (or convolution) of $f_{\mathrm{1}}(z)$ and $f_{\mathrm{2}}(z)$ by $$\begin{array}{}\text{(4)}& \left(f_{\mathrm{1}}*f_{\mathrm{2}}\right)=\frac{\mathrm{1}}{z}+\sum _{n=\mathrm{1}}^{\infty }‍a_{n,\mathrm{1}}{a}_{n,\mathrm{2}}{z}^n.\end{array}$$

Let us define the function $\tilde{\varphi }(\alpha ,\beta ;z)$ by $$\begin{array}{}\text{(5)}& \tilde{\varphi }\left(\alpha ,\beta ;z\right)=\frac{\mathrm{1}}{z}+\sum _{n=\mathrm{0}}^{\infty }‍\frac{{\left(\alpha \right)}_{n+\mathrm{1}}}{{\left(\beta \right)}_{n+\mathrm{1}}}{z}^n,\end{array}$$ for $\beta \ne \mathrm{0},-\mathrm{1},-\mathrm{2},\dots$ , and $\alpha \in ℂ/\{\mathrm{0}\}$ , where $(\lambda )n=\lambda (\lambda +\mathrm{1}{)}_{n+\mathrm{1}}$ is the Pochhammer symbol. We note that $$\begin{array}{}\text{(6)}& \tilde{\varphi }\left(\alpha ,\beta ;z\right)=\frac{\mathrm{1}}{z^{\mathrm{2}}}{F}_{\mathrm{1}}\left(\mathrm{1},\alpha ,\beta ;z\right),\end{array}$$ where $$\begin{array}{}\text{(7)}& {\hspace{0.17em}}_2{F}_{\mathrm{1}}\left(b,\alpha ,\beta ;z\right)=\sum _{n=\mathrm{0}}^{\infty }\frac{{\left(b\right)}_n{\left(\alpha \right)}_n}{{\left(\beta \right)}_n}\frac{z^n}{n!}\end{array}$$ is the well-known Gaussian hypergeometric function.

We recall here a general Hurwitz-Lerch-Zeta function, which is defined in [3, 4] by the following series: $$\begin{array}{}\text{(8)}& \mathrm{\Phi }\left(z,t,a\right)=\frac{\mathrm{1}}{a^t}+\sum _{n=\mathrm{1}}^{\infty }‍\frac{z^n}{{\left(n+a\right)}^t}\end{array}$$ ( $a\in ℂ/{\mathbb{Z}}_{\mathrm{0}}^{-}$ , ${\mathbb{Z}}_{\mathrm{0}}^{-}=\{\mathrm{0},-\mathrm{1},-\mathrm{2},\dots \}$ ; $t\in ℂ$ when $z\in U=U^{*}\subset \{\mathrm{0}\}$ ; $\Re (t)>\mathrm{1}$ when $z\in \partial U$ ).

Important special cases of the function $\mathrm{\Phi }(z,t,a)$ include, for example, the Riemann zeta function $\zeta (t)=\mathrm{\Phi }(\mathrm{1},t,\mathrm{1})$ , the Hurwitz zeta function $\zeta (t,a)=\mathrm{\Phi }(\mathrm{1},t,a)$ , the Lerch zeta function $l_t(\zeta )=\mathrm{\Phi }({\exp }^{\mathrm{2}\pi i\xi },t,\mathrm{1})$ , $(\xi \in ℝ,\Re (t)>\mathrm{1})$ , and the polylogarithm $L_t^i(z)=z\mathrm{\Phi }(z,t,a)$ . Recent results on $\mathrm{\Phi }(z,t,a)$ can be found in the expositions [5, 6]. By making use of the following normalized function we define $$\begin{array}{}\text{(9)}& G_{t,a}\left(z\right)={\left(\mathrm{1}+a\right)}^t\left[\mathrm{\Phi }\left(z,t,a\right)-a^t+\frac{\mathrm{1}}{z{\left(\mathrm{1}+a\right)}^t}\right]=\frac{\mathrm{1}}{z}+\sum _{n=\mathrm{1}}^{\infty }‍{\left(\frac{\mathrm{1}+a}{n+a}\right)}^t{z}^n,\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ Corresponding to the functions $G_{t,a}(z)$ and using the Hadamard product for $f(z)\in \Sigma$ , we define a new linear operator $L_{t,a}(\alpha ,\beta )$ on $\Sigma$ by the following series: $$\begin{array}{}\text{(10)}& L_a^t\left(\alpha ,\beta \right)f\left(z\right)=\varphi \left(\alpha ,\beta ;z\right)*G_{t,a}\left(z\right)=\frac{\mathrm{1}}{z}+\sum _{n=\mathrm{1}}^{\infty }‍\frac{{\left(\alpha \right)}_{n+\mathrm{1}}}{{\left(\beta \right)}_{n+\mathrm{1}}}{\left(\frac{\mathrm{1}+a}{n+a}\right)}^t{a}_n{z}^n.\phantom{22222222222222222}\left(z\in U^{*}\right).\end{array}$$ The meromorphic functions with the generalized hypergeometric functions were considered recently by many others; see, for example, [7–12].

It follows from (10) that $$\begin{array}{}\text{(11)}& z{\left(L_a^t\left(\alpha ,\beta \right)f\left(z\right)\right)}^{\prime }\hspace{1em}=\alpha \left(L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)-\left(\alpha +\mathrm{1}\right)L_a^t\left(\alpha ,\beta \right)f\left(z\right).\end{array}$$

In order to prove our main results, we recall the following lemma according to Yang [13].

Lemma 1.

Let $q(z)=\mathrm{1}+q_n{z}^n+q_{n+\mathrm{1}}{z}^{n+\mathrm{1}}+\cdots$ be analytic functions in $U=U^{*}\cup \{\mathrm{0}\}$ with $q(z)\ne \mathrm{0}$ for $z\in U$ . If $$\begin{array}{}\text{(12)}& \Re \left\{\mathrm{1}+a\frac{z{q}^{\prime }\left(z\right)}{q^{\mathrm{2}}\left(z\right)}\right\}<M,\hspace{1em}\left(z\in U\right),\end{array}$$ where $a>\mathrm{0}$ , and $$\begin{array}{}\text{(13)}& \mathrm{1}<M\le \frac{n\hspace{0.17em}a}{\mathrm{2}\log \mathrm{2}},\end{array}$$ then $$\begin{array}{}\text{(14)}& \Re \left\{\frac{\mathrm{1}}{q\left(z\right)}\right\}>\mathrm{1}-\frac{\mathrm{2}\left(M-\mathrm{1}\right)}{na}\log \mathrm{2},\hspace{1em}\left(z\in U\right).\end{array}$$ The bound in (14) is the best possible.

3. Main Results

We begin with the following theorem.

Theorem 2.

Let $\alpha +\mathrm{1}>\mathrm{0}$ , $L_a^t(\alpha ,\beta )f(z)/L_a^t(\alpha +\mathrm{1},\beta )f(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(15)}& \Re \{\mathrm{1}+\frac{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{\left(\alpha +\mathrm{1}\right)L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\left(\mathrm{1}+\frac{\alpha L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\right)\hspace{1em}\hspace{0.5em}-\frac{L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\phantom{\frac{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{\left(\alpha +\mathrm{1}\right)L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\left(\mathrm{1}+\frac{\alpha L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\right)}\}<M,\end{array}$$ where $$\begin{array}{}\text{(16)}& \mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{1}<M\le \frac{n}{\mathrm{2}\left(\alpha +\mathrm{1}\right)\log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(17)}& \Re \left\{\frac{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\right\}\hspace{1em}>\mathrm{1}-\frac{\mathrm{2}\left(\alpha +\mathrm{1}\right)\left(M-\mathrm{1}\right)}{n}\log \mathrm{2},\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ The bound in (17) is the best possible.

Proof.

Define the function $q(z)$ by $$\begin{array}{}\text{(18)}& q\left(z\right)=\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}.\end{array}$$ Then, clearly $q(z)=\mathrm{1}+q_n{z}^n+q_{n+\mathrm{1}}{z}^{n+\mathrm{1}}+\cdots$ analytic function in $U^{*}$ with $q(z)\ne \mathrm{0}$ for $z\in U^{*}$ . It follows from (18) and (11) that $$\begin{array}{}\text{(19)}& \frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)}=\frac{z{\left(L_a^t\left(\alpha ,\beta \right)f\left(z\right)\right)}^{\prime }}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}-\frac{z{\left(L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)}^{\prime }}{L_p^{*}\left(a+\mathrm{1},c\right)f\left(z\right)}\end{array}$$ by making use of the familiar identity (11) in (19), we obtain $$\begin{array}{}\text{(20)}& \frac{L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}=\frac{\mathrm{1}}{\alpha +\mathrm{1}}+\frac{\mathrm{1}}{\left(\alpha +\mathrm{1}\right)q\left(z\right)}-\frac{z{q}^{\prime }\left(z\right)}{\left(\alpha +\mathrm{1}\right)q\left(z\right)}\end{array}$$ or, equivalent, $$\begin{array}{}\text{(21)}& \mathrm{1}+\frac{\mathrm{1}}{\left(\alpha +\mathrm{1}\right)}\frac{z{q}^{\prime }\left(z\right)}{q^{\mathrm{2}}\left(z\right)}\hspace{1em}=\mathrm{1}+\frac{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{\left(\alpha +\mathrm{1}\right)L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\left(\mathrm{1}+\frac{\alpha L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\right)\hspace{1em}\hspace{1em}\begin{array}{c}-\frac{L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}.\end{array}\end{array}$$ Applying Lemma 1, with $a=\mathrm{1}/(\mathrm{1}+\alpha )$ , we get the required result.

Letting $\alpha =\beta =\mathrm{1}$ in Theorem 2, we have the following.

Corollary 3.

Let $G_{t,a}(z)/z{(G_{t,a}(z))}^{\prime }\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(22)}& \Re \{\mathrm{1}+\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}{\mathrm{2}{G}_{t,a}\left(z\right)}\left(\mathrm{1}+\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}{G_{t,a}\left(z\right)}\right)\hspace{1em}\hspace{0.5em}-\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }+\left(\mathrm{1}/\mathrm{2}\right){\left(G_{t,a}\left(z\right)\right)}^{\mathrm{\prime \prime }}}{G_{t,a}\left(z\right)}\phantom{\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}{\mathrm{2}{G}_{t,a}\left(z\right)}\left(\mathrm{1}+\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}{G_{t,a}\left(z\right)}\right)}\}<M,\end{array}$$ where $$\begin{array}{}\text{(23)}& \mathrm{1}<M\le \frac{n}{\mathrm{4}\log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(24)}& \Re \left\{\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}{G_{t,a}\left(z\right)}\right\}>\mathrm{1}-\frac{\mathrm{4}\left(M-\mathrm{1}\right)}{n}\log \mathrm{2},\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ The bound in (24) is the best possible.

Letting $M=\mathrm{1}+n/\mathrm{4}\log \mathrm{2}$ in Corollary 3, we have the following.

Corollary 4.

Let $G_{t,a}(z)/z{(G_{t,a}(z))}^{\prime }\ne \mathrm{0}$ and $t=\mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(25)}& \Re \left\{\mathrm{1}+\frac{z{f}^{\prime }\left(z\right)}{\mathrm{2}f\left(z\right)}\left(\mathrm{1}+\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)-\frac{z{f}^{\prime }\left(z\right)+\left(\mathrm{1}/\mathrm{2}\right)z^{\mathrm{2}}{f}^{\mathrm{\prime \prime }}\left(z\right)}{f\left(z\right)}\right\}\hspace{1em}<\mathrm{1}+\frac{n}{\mathrm{4}\log \mathrm{2}}.\end{array}$$ Then $f(z)$ is starlike in $U^{*}$ .

Theorem 5.

Let $\delta (\alpha +\mathrm{1})>\mathrm{0}$ , $z{L}_a^t(\alpha +\mathrm{1},\beta )f(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(26)}& \Re \left\{{\left(z{L}_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)}^{\delta }\left(\frac{L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)\right\}<M,\end{array}$$ where $$\begin{array}{}\text{(27)}& \mathrm{1}<M\le \frac{n}{\mathrm{2}\delta \left(\alpha +\mathrm{1}\right)\log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(28)}& \Re {\left(z{L}_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)}^{\delta }\hspace{1em}>\mathrm{1}-\frac{\mathrm{2}\delta \left(\alpha +\mathrm{1}\right)\left(M-\mathrm{1}\right)}{n}\log \mathrm{2},\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ The bound in (28) is the best possible.

Proof.

Define the function $q(z)$ by $$\begin{array}{}\text{(29)}& q\left(z\right)={\left(z{L}_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)}^{\delta }.\end{array}$$ Then, clearly $q(z)=\mathrm{1}+q_n{z}^n+q_{n+\mathrm{1}}{z}^{n+\mathrm{1}}+\cdots$ analytic function in $U^{*}$ with $q(z)\ne \mathrm{0}$ for $z\in U^{*}$ . It follows from (29) that $$\begin{array}{}\text{(30)}& \frac{z{q}^{\prime }\left(z\right)}{\delta q\left(z\right)}=\frac{z{\left(L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)}^{\prime }}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}-\mathrm{1}.\end{array}$$ by making use of the familiar identity (11) in (30), we get $$\begin{array}{}\text{(31)}& \frac{L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}-\mathrm{1}=\frac{\mathrm{1}}{\delta \left(\alpha +\mathrm{1}\right)}\frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)},\end{array}$$ or, equivalent $$\begin{array}{}\text{(32)}& \mathrm{1}+\frac{\mathrm{1}}{\delta \left(\alpha +\mathrm{1}\right)}\frac{z{q}^{\prime }\left(z\right)}{q^{\mathrm{2}}\left(z\right)}\hspace{1em}={\left(z{L}_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)\right)}^{\delta }\left(\frac{L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right).\end{array}$$ Applying Lemma 1, with $a=\mathrm{1}/(\mathrm{1}+\alpha )$ , we get the required result.

Letting $\alpha =\beta =\mathrm{1}$ in Theorem 5, we have

Corollary 6.

Let $\delta >\mathrm{0},G_{t,a}(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(33)}& \Re \left\{{\left(z{G}_{t,a}\left(z\right)\right)}^{\delta }\left(\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }+\left(\mathrm{1}/\mathrm{2}\right){\left(G_{t,a}\left(z\right)\right)}^{\mathrm{\prime \prime }}}{G_{t,a}\left(z\right)}\right)\right\}<M,\end{array}$$ where $$\begin{array}{}\text{(34)}& \mathrm{1}<M\le \frac{n}{\mathrm{4}\delta \log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(35)}& \Re {\left(z{G}_{t,a}\left(z\right)\right)}^{\delta }>\mathrm{1}-\frac{\mathrm{4}\delta \left(M-\mathrm{1}\right)}{n}\log \mathrm{2},\hspace{1em}\left(z\in U^{*}\right).\end{array}$$

The bound in (35) is the best possible.

Letting $\delta =\mathrm{1}$ , $M=\mathrm{1}+n/\mathrm{8}\log \mathrm{2}$ , and $t=\mathrm{0}$ in Corollary 6, we have the following.

Corollary 7.

Let $f^{\prime }(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(36)}& \Re \left\{zf{\left(z\right)}^{\prime }\left(\mathrm{1}+\frac{z{f}^{\mathrm{\prime \prime }}\left(z\right)}{\mathrm{2}{f}^{\prime }\left(z\right)}\right)\right\}<\mathrm{1}+\frac{n}{\mathrm{8}\log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(37)}& \Re \left\{zf{\left(z\right)}^{\prime }\right\}>\mathrm{0},\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ The result is sharp.

Theorem 8.

Let $\xi >\mathrm{0}$ , $z{(L_a^t(\alpha +\mathrm{1},\beta )f(z))}^{\prime }/L_a^t(\alpha ,\beta )f(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(38)}& \Re \{\mathrm{1}+{\left(\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)}^{\xi }\hspace{1em}\hspace{0.5em}\times \left(\frac{\left(\alpha +\mathrm{1}\right)L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)\hspace{1em}\hspace{0.5em}-\alpha {\left(\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)}^{\xi }-\mathrm{1}\}<M,\end{array}$$ where $$\begin{array}{}\text{(39)}& \mathrm{1}<M\le \mathrm{1}+\frac{n}{\mathrm{2}\xi \log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(40)}& \Re {\left(\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)}^{\xi }\hspace{1em}>\mathrm{1}-\frac{\mathrm{2}\xi \left(M-\mathrm{1}\right)}{n}\log \mathrm{2}\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ The bound in (40) is the best possible.

Proof.

Define the function $q(z)$ by $$\begin{array}{}\text{(41)}& q\left(z\right)={\left(\frac{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}\right)}^{\xi }.\end{array}$$ Then, clearly $q(z)=\mathrm{1}+q_n{z}^n+q_{n+\mathrm{1}}{z}^{n+\mathrm{1}}+\cdots$ analytic function in $U^{*}$ with $q(z)\ne \mathrm{0}$ for $z\in U^{*}$ . Also by a simple computation and by making use of the familiar identity (11), we find from (41) that $$\begin{array}{}\text{(42)}& \mathrm{1}+\frac{\mathrm{1}}{\xi }\frac{z{q}^{\prime }\left(z\right)}{q^{\mathrm{2}}\left(z\right)}=\mathrm{1}+{\left(\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)}^{\xi }\begin{array}{c}\times (\phantom{{\left(\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)}^{\xi }}\frac{\left(\alpha +\mathrm{1}\right)L_a^t\left(\alpha +\mathrm{2},\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\end{array}\hspace{1em}\hspace{0.5em}-\alpha {\left(\frac{L_a^t\left(\alpha ,\beta \right)f\left(z\right)}{L_a^t\left(\alpha +\mathrm{1},\beta \right)f\left(z\right)}\right)}^{\xi }-\mathrm{1}).\end{array}$$ Applying Lemma 1, with $a=\mathrm{1}/\xi$ , we get the required result.

Letting $\alpha =\beta =\mathrm{1}$ in Theorem 8, we have the following.

Corollary 9.

Let $\xi >\mathrm{0}$ , $z{(G_{t,a}(z))}^{\prime }/G_{t,a}(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(43)}& \Re \{\phantom{\left(\mathrm{1}+\frac{z\left(G_{t,a}\left(z\right)\right)}{\left(G_{t,a}\left(z\right)\right)}-{\left(\frac{G_{t,a}\left(z\right)}{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}\right)}^{\xi }\right)}\mathrm{1}+{\left(\frac{G_{t,a}\left(z\right)}{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}\right)}^{\xi }\hspace{1em}\hspace{1em}\times \left(\mathrm{1}+\frac{z{\left(G_{t,a}\left(z\right)\right)}^{\prime \prime }}{\left(G_{t,a}\left(z\right)\right)}-{\left(\frac{G_{t,a}\left(z\right)}{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}\right)}^{\xi }\right)\}<M,\end{array}$$ where $$\begin{array}{}\text{(44)}& \mathrm{1}<M\le \mathrm{1}+\frac{n}{\mathrm{2}\xi \log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(45)}& \Re {\left(\frac{G_{t,a}\left(z\right)}{z{\left(G_{t,a}\left(z\right)\right)}^{\prime }}\right)}^{\xi }>\mathrm{1}-\frac{\mathrm{2}\xi \left(M-\mathrm{1}\right)}{n}\log \mathrm{2},\hspace{1em}\left(z\in U\right).\end{array}$$ The bound in (45) is the best possible.

Letting $\xi =\mathrm{1}$ , $M=\mathrm{1}+n/\mathrm{2}\log \mathrm{2}$ , and $t=\mathrm{0}$ in Corollary 9, we have the following.

Corollary 10.

Let $z{f}^{\prime }(z)/f(z)\ne \mathrm{0}$ for $z\in U^{*}$ and suppose that $$\begin{array}{}\text{(46)}& \Re \left\{\mathrm{1}+{\left(\frac{f\left(z\right)}{z{f}^{\prime }\left(z\right)}\right)}^{\xi }\left(\mathrm{1}+\frac{z{f}^{\mathrm{\prime \prime }}\left(z\right)}{f^{\prime }\left(z\right)}-{\left(\frac{f\left(z\right)}{z{f}^{\prime }\left(z\right)}\right)}^{\xi }\right)\right\}\hspace{1em}<\mathrm{1}+\frac{n}{\mathrm{2}\log \mathrm{2}}.\end{array}$$ Then $$\begin{array}{}\text{(47)}& \Re \left(\frac{f\left(z\right)}{z{f}^{\prime }\left(z\right)}\right)>\mathrm{0},\hspace{1em}\left(z\in U^{*}\right).\end{array}$$ The result is sharp.

Conflict of Interests

The authors declare that they have no competing interests.

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