On extensions of the Jack and open-door lemmas

Abstract

In the present paper, we apply the theory of differential subordination to extend the famous Jack lemma for analytic functions. Also, using this extension we will produce new versions of the open-door lemma for analytic functions, and utilizing them some simple conditions for starlikeness of analytic and meromorphic functions are obtained. Our results improve and extend the earlier results investigated in the literature.

1 Introduction and preliminaries

Let $\mathbb{U}=\{z \in \mathbb{C}: |z |<1\}$. The class of analytic functions in $\mathbb{U}$ is denoted by $\mathcal{H}$. We define

$$ \mathcal{H}[a,n]=\bigl\{ f\in \mathcal{H}: f(z)=a+ a_{n}z^{n}+ a_{n+1}z^{n+1}+ \cdots \bigr\} , $$

where n is a positive integer number and $a\in \mathbb{C}$. Suppose $n\in \mathbb{N}$, we introduce the subclass $\mathcal{A}_{n}$ of $\mathcal{H}$ as follows:

$$ \mathcal{A}_{n}=\bigl\{ f\in \mathcal{H}: f(z)= z+ a_{n+1}z^{n+1}+ a_{n+2}z^{n+2}+ \cdots \bigr\} . $$

In particular, we put $\mathcal{A}_{1}=\mathcal{A}$. Let $\mathcal{S}$ denote the subclass of $\mathcal{A}$ consisting of all univalent functions in $\mathbb{U}$. A function $f\in \mathcal{A}$ is said to be starlike of order $0\leq \gamma <1$, written $f\in S^{*}(\gamma )$, if it satisfies

$$ \mathfrak{Re} {\frac{zf'(z)}{f(z)}} >\gamma \quad (z \in \mathbb{U}). $$

In particular, we set $\mathcal{S}^{*}(0)\equiv \mathcal{S}^{*}$.

Let $\mathcal{P}$ be the class of analytic functions $p:\mathbb{U}\rightarrow \mathbb{C}$ of the form $p(z)=1+\sum_{n=1}^{\infty}p_{n}z^{n}$ with $\mathfrak{Re} p(z)>0$ for $z\in \mathbb{U}$. This class $\mathcal{P}$ is known as the Carathéodory class or the class of functions with positive real part [4, 5], pioneered by Carathéodory. The theory of Carathéodory functions plays a very crucial role in geometric function theory.

Suppose f and g are in $\mathcal{H}$. We say that the function f is subordinate to g, denoted by $f \prec g$, if there exists an analytic function ω in $\mathbb{U}$, with $\omega (0)=0$ and $|\omega (z)|\leq |z|<1$, such that $f(z)=g(\omega (z))$. Moreover, if g is a univalent function in $\mathbb{U}$, then $f\prec g$ if and only if $f(0)=0$ and $f(\mathbb{U})\subset g(\mathbb{U})$.

One of the most important results in the geometric function theory (GFT) is the Jack lemma [9]. By extending this (Jack) lemma in [12] Miller and Mocanu found a new way for the study of GFT. Moreover, they proved a famous lemma known as the open-door lemma that is very useful for the investigation of various mapping and geometric properties in GFT. General forms of the Jack and open-door lemmas, respectively, are as follows:

Theorem 1

(the Jack lemma [9])

Let f be analytic and a nonconstant function in the $\mathbb{U}$ with $f(0)=0$. If $|f(z)|$ attains its maximum value at the point $z_{0}$ with $|z_{0}|=r$, then

$$ \frac{z_{0}f'(z_{0})}{f(z_{0})}=k, $$

where $k \geq 1$ is a real number.

Let c be a complex number such that $\mathfrak{Re} {c} >0$ and $n \in \mathbb{N}$. Suppose that

$$ C_{n}=C_{n}(c)=\frac {n}{\mathfrak{Re} c} \biggl[ \vert c \vert \sqrt{1+ \frac {2\mathfrak{Re} c}{n}} +\mathfrak{Im} c \biggr]. $$

Also, let

$$ V(A,B)=\mathbb{C}\setminus \lbrace iy: y\leq A \text{or } y\geq B \rbrace \quad \text{with } A, B \in \mathbb{R} \text{ and } A< B. $$

Theorem 2

(the open-door lemma [12, 13])

Let c be a complex number with $\mathfrak{Re} c>0$ and n be an integer with $n\geq 1$. Suppose that a function $q \in \mathcal{H}[c,n]$ satisfies the condition

$$ q(z)+\frac {zq^{\prime }(z)}{q(z)} \in V\bigl(-C_{n}(c),C_{n}(c)\bigr) \quad (z \in \mathbb{U}), $$

(1)

then $\mathfrak{Re} q(z) >0$ for $z \in \mathbb{U}$.

Kuroki and Owa [10] corrected the assertion (1) in the following form:

$$ q(z)+\frac {zq^{\prime }(z)}{q(z)} \in V\bigl(-C_{n}(c),C_{n}(\bar{c}) \bigr) \quad \Longrightarrow\quad \operatorname{Re} q(z) >0. $$

(2)

Also, Li and Sugawa [11] have extended Theorem 2, so that $\vert \arg q \vert <\frac{\pi \alpha}{2}$ for a given $0<{\alpha \leq 1}$.

To obtain information about applications of the Jack and open-door lemmas with developed versions of them the reader may refer to the works, for example, [3, 6–8, 10, 11, 13–15, 18]. Recently Amani et al. [1, 2] have obtained some results for the analytic functions with fixed initial coefficient associated with the Jack lemma. Motivated by the works mentioned above, in this paper, we first extend the Jack and open-door lemmas for analytic functions and then present some applications.

The contents of this article are as follows. In Sect. 2, we will present the extension of the Jack and open-door lemmas for analytic functions and then we obtain some corollaries from them. In Sect. 3, we apply the results of Sect. 2, to obtain some sufficient conditions for starlikeness and Carathédory functions.

To prove the main results, we shall require the following definition and lemmas:

Definition 1

(see [13])

Let Q be the set of functions q that are analytic and injective on $\overline{\mathbb{U}}\backslash E(q)$, where

$$ E(q):= \Bigl\{ \zeta \in \partial \mathbb{U} : \lim_{z \rightarrow \zeta}q(z)= \infty \Bigr\} , $$

and are such that $q'(\zeta )\neq 0$ for $\zeta \in \partial \mathbb{U}\backslash E(q)$.

Lemma 1

(see [12])

Let $p(z)= a+a_{n}z^{n}+\cdots$ be analytic in $\mathbb{U}$ with $p(z)\not \equiv a$ and $n\geq 1$, and let $q\in Q$ with $q(0)=a$. If there exist points $z_{0}\in \mathbb{U}$ and $\zeta _{0}\in \partial \mathbb{U} \backslash E(q)$ such that $p(z_{0})=q(\zeta _{0})$ and $p(\mathbb{U}_{r_{0}})= p(\{z: |z|<|z_{0}|\})\subset q(\mathbb{U})$, where $r_{0}=|z_{0}|$, then there exists an $m\geq n\geq 1$ such that

$(i)$ $z_{0}p'(z_{0})= m\zeta _{0}q'(\zeta _{0})$; and

$(\mathit{ii})$ $\mathfrak{Re} \{1+\frac{z_{0}p''(z_{0})}{p'(z_{0})} \} \geq m \mathfrak{Re} \{1+ \frac{\zeta _{0}q''(\zeta _{0})}{q'(\zeta _{0})} \}$.

Lemma 2

(see [13])

Let $q\in Q$ with $q(0)=a$, and let $p(z)= a+a_{n}z^{n}+\cdots$ be analytic in $\mathbb{U}$ with $p(z)\not \equiv a$ and $n\geq 1$. If p is not subordinate to q, then there exist points $z_{0}= r_{0}e^{i\theta _{0}}\in \mathbb{U}$ and $\zeta _{0}\in \partial \mathbb{U} \backslash E(q)$, and an $m\geq n\geq 1$ for which $p(\mathbb{U}_{r_{0}}) \subset q(\mathbb{U})$,

$(i)$ $p(z_{0})= q(\zeta _{0})$;

$(\mathit{ii})$ $z_{0}p'(z_{0})= m\zeta _{0}q'(\zeta _{0})$; and

$(\mathit{iii})$ $\mathfrak{Re} \{1+\frac{z_{0}p''(z_{0})}{p'(z_{0})} \} \geq m \mathfrak{Re} \{1+ \frac{\zeta _{0}q''(\zeta _{0})}{q'(\zeta _{0})} \}$.

2 Main results

Initially in this section, we establish the extension of the Jack lemma [9] as follows:

Theorem 3

(extension of the Jack lemma)

Let $c=re^{it}$ with $-\frac{\pi \alpha}{\alpha +\lambda}< t< \frac{\pi \lambda}{\alpha +\lambda}$, where $0<\alpha \leq 1$ and $0<\lambda \leq 1$. Also, let $p\in \mathcal{H}[c^{\frac{\alpha +\lambda}{2}},n]$ with $p(z)\neq 0$ in $\mathbb{U}$. If there exist elements $z_{1}\in \mathbb{U}$ and $z_{2}\in \mathbb{U}$ such that $|z_{1}|=|z_{2}|= r$ and for all $z\in \mathbb{U}_{r}=\{z\in \mathbb{C}: |z|< r\}$

$$ -\frac{\pi \alpha}{2}= \arg p(z_{1})< \arg p(z)< \arg p(z_{2})= \frac{\pi \lambda}{2}, $$

then we have

$$ z_{1}p'(z_{1})= -i \frac{\lambda +\alpha}{2} k_{1} p(z_{1}) $$

(3)

and

$$ z_{2}p'(z_{2})= i \frac{\lambda +\alpha}{2} k_{2} p(z_{2}), $$

(4)

where

$$ k_{1}\geq n \biggl( \frac{1+\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}{\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr) $$

(5)

and

$$ k_{2}\geq n \biggl( \frac{1-\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}{\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr). $$

(6)

Proof

Let us define

$$ q(z)=\exp \biggl\{ \frac{\pi i(\lambda -\alpha )}{4} \biggr\} \biggl( \frac{c_{1}+\bar{c}_{1} z}{1-z} \biggr)^{\frac{\lambda +\alpha}{2}}, $$

with $c_{1}= c \exp \{ \frac{-\pi i (\lambda -\alpha )}{2(\lambda +\alpha )} \}$. It is easy to consider that q is analytic in $\mathbb{U}$, $q(0)=c^{\frac{\lambda +\alpha}{2}}$ and

$$ -\frac{\pi \alpha}{2}< \arg q(\mathbb{U})< \frac{\pi \lambda}{2}. $$

Moreover, $q\in Q$ and $E(q)={1}$. By the hypothesis and the properties of the function q, we have $p(z_{1})\in q(\partial \mathbb{U})$ and $p(z_{2})\in q(\partial \mathbb{U})$, also $p(\{z: |z|< r\})\subset q(\mathbb{U})$. Define

$$ p_{1}(z)=\exp \biggl\{ \frac{-\pi i (\lambda -\alpha )}{2(\lambda +\alpha )} \biggr\} \bigl\{ p(z) \bigr\} ^{\frac{2}{\lambda +\alpha}} \quad (z\in \mathbb{U}) $$

and

$$ q_{1}(z)= \frac{c_{1}+\bar{c_{1}} z}{1-z} \quad (z\in \mathbb{U}), $$

with $c_{1}= c \exp \{ \frac{-\pi i (\lambda -\alpha )}{2(\lambda +\alpha )} \}$. Then, it can be readily observed that $q_{1}\in Q$, $q_{1}(0)= p_{1}(0)$, $q_{1}(\mathbb{U})=\{w\in \mathbb{C}: \mathfrak{Re}w>0\}$ (note that $\mathfrak{Re}c_{1}>0$) and $p_{1}(\{z: |z|< r\})\subset q_{1}(\mathbb{U})$. Also, $p_{1}(z_{1})=-ix_{1}$ and $p_{1}(z_{2})=ix_{2}$, with $x_{1}, x_{2}>0$, thus there exist complex numbers $\zeta _{1}$ and $\zeta _{2}$ in $\partial \mathbb{U}$ such that $p_{1}(z_{1})= q_{1}(\zeta _{1})$ and $p_{1}(z_{2})= q_{1}(\zeta _{2})$. Calculating the inverse of $q_{1}$ and taking the derivative of $q_{1}$, we obtain

$$ q_{1}^{-1}(z)=\frac{z-c_{1}}{z+\bar{c_{1}}} \quad \text{and} \quad q'_{1}(z)=\frac{2\mathfrak{Re}c_{1}}{(1-z)^{2}}. $$

On the other hand, since $p\in \mathcal{H}[c^{\frac{\alpha +\lambda}{2}},n]$, we have $p_{1}\in \mathcal{H}[a,n]$ with

$$ a=c\exp \biggl\{ \frac{\pi i (\alpha -\lambda )}{2(\lambda +\alpha )} \biggr\} =c_{1}. $$

Hence, applying Lemma 1, we deduce that there exist numbers $m_{1}\geq n\geq 1$ and $m_{2}\geq n \geq 1$ such that

$$ z_{1}p'_{1}(z_{1})=m_{1} \zeta _{1}q'_{1}(\zeta _{1}) \quad \text{and} \quad z_{2}p'_{1}(z_{2})=m_{2} \zeta _{2}q'_{1}(\zeta _{2}). $$

Since $p_{1}(z_{1})= -i x_{1}$ with $x_{1}>0$ and $\zeta _{1}= q^{-1}_{1}(p_{1}(z_{1}))= \frac{ix_{1}+c_{1}}{ix_{1}-\bar{c_{1}}}$, we have

$$\begin{aligned} \frac{z_{1}p'(z_{1})}{p(z_{1})} &=\frac{\lambda +\alpha}{2} \frac{z_{1}p'_{1}(z_{1})}{p_{1}(z_{1})} \\ &=\frac{\lambda +\alpha}{2} \frac{m_{1}\zeta _{1}q'_{1}\zeta _{1})}{p_{1}(z_{1})} \\ & =m_{1}\frac{\lambda +\alpha}{2} \frac{ix_{1}+c_{1}}{ix_{1}-\bar{c_{1}}}\times \frac{1}{-ix_{1}} \times \frac{2\mathfrak{Re}c_{1}}{(1-\frac{ix_{1}+c_{1}}{ix_{1}-\bar{c_{1}}})^{2}} \\ &=m_{1}\frac{\lambda +\alpha}{2}\frac{1}{ix_{1}}\times \frac{x_{1}^{2}+2x_{1}\mathfrak{Im}c_{1}+ \vert c_{1} \vert ^{2}}{2\mathfrak{Re}c_{1}} \\ &=-im_{1} \biggl(\frac{\lambda +\alpha}{2} \biggr) \frac{x_{1}^{2}+2 \vert c \vert x_{1} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{1}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}. \end{aligned}$$

Set

$$ f(x)= \frac{x^{2}+2 \vert c \vert x \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x \cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \quad (x>0). $$

By a simple calculation, we can easily find that

$$ \min_{x>0} f(x) = f\bigl( \vert c \vert \bigr)= \frac{1+\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}{ \cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}. $$

Hence, we conclude that

$$ k_{1}= m_{1}f(x_{1})\geq n \biggl( \frac{1+\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}{\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr). $$

Then, assertions (3) and (5) hold. Now, similar to the trend of the last case, due to $p_{1}(z_{2})=ix_{2}$, with $x_{2}>0$ and $\zeta _{2}=q_{1}^{-1}(ix_{2})= \frac{ix_{2}-c_{1}}{ix_{2}+ \bar{c_{1}}}$ we can obtain

$$\begin{aligned} \frac{z_{2}p'(z_{2})}{p(z_{2})} &=\frac{\lambda +\alpha}{2} \frac{z_{2}p'_{1}(z_{2})}{p_{1}(z_{2})} \\ &=\frac{\lambda +\alpha}{2} \frac{m_{2}\zeta _{2}q'_{1}\zeta _{2})}{p_{1}(z_{2})} \\ &=m_{2}\frac{\lambda +\alpha}{2} \frac{ix_{2}-c_{1}}{ix_{2}+\bar{c_{1}}}\times \frac{1}{ix_{2}} \times \frac{2\mathfrak{Re}c_{1}}{(1-\frac{ix_{2}-c_{1}}{ix_{2}+\bar{c_{1}}})^{2}} \\ &=m_{2}\frac{\lambda +\alpha}{2}\frac{1}{ix_{2}}\times \frac{-x_{2}^{2}+2x_{2}\mathfrak{Im}c_{1}- \vert c_{1} \vert ^{2}}{2\mathfrak{Re}c_{1}} \\ &=im_{2} \biggl(\frac{\lambda +\alpha}{2} \biggr) \frac{x_{2}^{2}-2 \vert c \vert x_{2} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{2}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}. \end{aligned}$$

Set

$$ g(x)= \frac{x^{2}-2 \vert c \vert x \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x \cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \quad (x>0). $$

By computing, we have

$$ \min_{x>0} g(x) = g\bigl( \vert c \vert \bigr)= \frac{1-\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}{ \cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}. $$

Therefore, akin to the former reasoning, we can deduce assertions (4) and (6). □

Remark 1

By putting $\lambda =\alpha$ in Theorem 3 we obtain the extended version of the Nunokawa lemma [14], which is an extension of the Jack lemma.

By putting $\alpha =\lambda$ in Theorem 3 and by utilizing the same method as in the theorem mentioned above we obtain:

Corollary 1

Let $c=re^{it}$ be a complex number with $\mathfrak{Re}c>0$ and $0<\lambda \leq 1$. Also, let $p\in \mathcal{H}[c^{\lambda},n]$ with $p(z)\neq 0$ in $\mathbb{U}$. If there exists a point $z_{0}\in \mathbb{U}$ such that

$$ \bigl\vert \arg p(z) \bigr\vert < \frac{\lambda \pi}{2} \quad \textit{for } \vert z \vert < \vert z_{0} \vert $$

and $p(z_{0})^{\frac{1}{\lambda}}=\pm ia$, where $a>0$ and $0<\lambda \leq 1$, then we have

$$ z_{0} p'(z_{0})= im\lambda p(z_{0}), $$

where

$$ m\geq n \biggl( \frac{a^{2}-2a\mathfrak{Im}{c}+ \vert c \vert ^{2}}{2a\mathfrak{Re}{c}} \biggr) \quad \textit{when } \arg{p(z_{0})}=\frac{\lambda \pi}{2} $$

and

$$ m\leq -n \biggl( \frac{a^{2}+ 2a\mathfrak{Im}{c}+ \vert c \vert ^{2}}{2a\mathfrak{Re}{c}} \biggr) \quad \textit{when } \arg{p(z_{0})}=-\frac{\lambda \pi}{2}. $$

Remark 2

By setting $\lambda = 1$ and $n=1$ in Corollary 1, we find that this corollary extends Theorem 3 obtained in [15]. Also, by setting $c=1$ in Theorem 3 we obtain the result shown in [19] (see Theorem 1.3).

Theorem 4

(extension of the open-door lemma)

Let $c=re^{it}$ with $-\frac{\pi \alpha}{\alpha +\lambda}< t< \frac{\pi \lambda}{\alpha +\lambda}$, where $0<\alpha \leq 1$ and $0<\lambda \leq 1$. Also, let $p\in \mathcal{H}[c^{\frac{\alpha +\lambda}{2}},n]$ with $p(z)\neq 0$ in $\mathbb{U}$. If

$$ \gamma p(z)^{\frac{2}{\alpha +\lambda}}+\frac{2}{\alpha +\lambda} \frac{zp'(z)}{p(z)}\neq iy \quad (z \in \mathbb{U}) $$

for all $y\in \mathbb{R}$, where

$$ y\geq \frac{\sqrt{n}}{\cos{B}}\bigl(\sqrt{n+2 \vert c \vert \cos{B}}-\sqrt{n} \sin{B}\bigr) $$

or

$$ y\leq -\frac{\sqrt{n}}{\cos{B}}\bigl(\sqrt{n+2 \vert c \vert \cos{B}}+\sqrt{n} \sin{B}\bigr), $$

then we have

$$ -\frac{\alpha \pi}{2}< \arg{p(z)}< \frac{\lambda \pi}{2} \quad (z\in \mathbb{U}), $$

(7)

where $\gamma = \exp \{-i\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}\}$ and $B=t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}$.

Proof

Let us set

$$ p_{1}(z)=\exp \biggl\{ \frac{-\pi i (\lambda -\alpha )}{2(\lambda +\alpha )} \biggr\} \bigl\{ p(z) \bigr\} ^{\frac{2}{\lambda +\alpha}} \quad (z\in \mathbb{U}) $$

and

$$ q_{1}(z)= \frac{c_{1}+\bar{c_{1}} z}{1-z} \quad (z\in \mathbb{U}), $$

where $c_{1}= c \exp \{ \frac{-\pi i (\lambda -\alpha )}{2(\lambda +\alpha )} \}$. It is clear that $p_{1}\in \mathcal{H}[a,n]$, with $a=c\exp \{\frac{\pi i (\alpha -\lambda )}{2(\lambda +\alpha )} \}=c_{1}$ and $p_{1}(0)=q_{1}(0)$. Suppose $p(\mathbb{U})$ is not contained in the sector $\{w: -\frac{\pi \alpha}{2}< \arg{w}< \frac{\pi \lambda}{2}\}$. Then $p_{1}\mathbb{U}$) is not contained in the right half-plane $\mathfrak{Re}w>0$. On the other hand, we have $q_{1}(\mathbb{U})=\{w: \mathfrak{Re}w>0\}$, this implies that $p_{1}\nprec q_{1}$, then from Lemma 2 we conclude that there exist points $z_{1}\in \mathbb{U}$ and $\zeta _{1}\in \partial \mathbb{U}$ such that $p_{1}(z_{1})= q_{1}(\zeta _{1})$ and $z_{1}p'_{1}(z_{1})= m_{1} \zeta _{1}q'_{1}(\zeta _{1})$, where $m_{1}\geq n$. Consequently, $p_{1}(z_{1})=-ix_{1}$ or $p_{1}(z_{1})=ix_{2}$ with $x_{1},x_{2}>0$. Initially, let $p_{1}(z_{1})= -ix_{1}$, with $x_{1}>0$. As the argument of Theorem 3 we have

$$ \frac{z_{1}p'({z_{1}})}{p(z_{1})} =-im_{1} \biggl( \frac{\lambda +\alpha}{2} \biggr) \frac{x_{1}^{2}+2 \vert c \vert x_{1} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha})+ \vert c \vert ^{2}}{2 \vert c \vert x_{1}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}, $$

where $m_{1}\geq n \geq 1$. Then, this yields

$$\begin{aligned} &\mathfrak{Im} \biggl\{ \gamma p(z_{1})^{\frac{2}{\alpha +\lambda}}+ \frac{2}{\alpha +\lambda}\frac{zp'(z_{1})}{p(z_{1})} \biggr\} \\ &\quad=\mathfrak{Im} \biggl\{ -ix_{1}-im_{1} \frac{x_{1}^{2}+2 \vert c \vert x_{1} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{1}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr\} \\ &\quad=- \biggl(x_{1}+m_{1} \frac{x_{1}^{2}+2 \vert c \vert x_{1} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{1}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr) \\ &\quad\leq - \biggl(x_{1}+n \frac{x_{1}^{2}+2 \vert c \vert x_{1} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{1}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr). \end{aligned}$$

Assume

$$ f(x)= x+n \frac{x^{2}+2 \vert c \vert x\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \quad (x>0). $$

A simple calculation gives us that

$$ \min_{x>0} f(x) = f \biggl( \frac{ \vert c \vert \sqrt{n}}{\sqrt{n+2 \vert c \vert \cos{B}}} \biggr)= \frac{\sqrt{n}}{\cos{B}} \bigl(\sqrt{n+2 \vert c \vert \cos{B}}+\sqrt{n}\sin{B} \bigr) $$

and so it follows that

$$ \mathfrak{Im} \biggl\{ \gamma p(z_{1})^{\frac{2}{\alpha +\lambda}}+ \frac{2}{\alpha +\lambda}\frac{z_{1}p'(z_{1})}{p(z_{1})} \biggr\} \leq -\frac{\sqrt{n}}{\cos{B}}\bigl( \sqrt{n+2 \vert c \vert \cos{B}}+\sqrt{n}\sin{B}\bigr), $$

where $\gamma = \exp \{-i\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}\}$ and $B=t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}$. On the other hand, we have

$$ \mathfrak{Re} \biggl\{ \gamma p(z_{1})^{\frac{2}{\alpha +\lambda}}+ \frac{2}{\alpha +\lambda}\frac{z_{1}p'(z_{1})}{p(z_{1})} \biggr\} =0, $$

which contradicts the hypothesis. For the case $p_{1}(z_{1})=ix_{2}$, similar to the approach of Theorem 3 we have

$$ \frac{z_{1}p'({z_{1}})}{p(z_{1})} =im_{2} \biggl( \frac{\lambda +\alpha}{2} \biggr) \frac{x_{2}^{2}-2 \vert c \vert x_{2} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha})+ \vert c \vert ^{2}}{2 \vert c \vert x_{2}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}, $$

where $m_{2}\geq n\geq 1$. Then, this yields

$$\begin{aligned} &\mathfrak{Im} \biggl\{ \gamma p(z_{1})^{\frac{2}{\alpha +\lambda}}+ \frac{2}{\alpha +\lambda}\frac{zp'(z_{1})}{p(z_{1})} \biggr\} \\ &\quad=\mathfrak{Im} \biggl\{ ix_{2}+im_{2} \frac{x_{2}^{2}-2 \vert c \vert x_{2} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{2}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \biggr\} \\ &\quad=x_{2}+m_{2} \frac{x_{2}^{2}-2 \vert c \vert x_{2} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{2}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \\ &\quad\geq x_{2}+n \frac{x_{2}^{2}-2 \vert c \vert x_{2} \sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x_{2}\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})}. \end{aligned}$$

Suppose

$$ g(x)= x+n \frac{x^{2}-2 \vert c \vert x\sin (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})+ \vert c \vert ^{2}}{2 \vert c \vert x\cos (t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )})} \quad (x>0). $$

By computing we can deduce that

$$ \min_{x>0} g(x) = g \biggl( \frac{ \vert c \vert \sqrt{n}}{\sqrt{n+2 \vert c \vert \cos{B}}} \biggr)= \frac{\sqrt{n}}{\cos{B}} \bigl(\sqrt{n+2 \vert c \vert \cos{B}}-\sqrt{n}\sin{B} \bigr). $$

Thus, we have

$$ \mathfrak{Im} \biggl\{ \gamma p(z_{1})^{\frac{2}{\alpha +\lambda}}+ \frac{2}{\alpha +\lambda}\frac{zp'(z_{1})}{p(z_{1})} \biggr\} \geq \frac{\sqrt{n}}{\cos{B}}\bigl( \sqrt{n+2 \vert c \vert \cos{B}}-\sqrt{n}\sin{B}\bigr), $$

where $\gamma = \exp \{-i\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}\}$ and $B=t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}$. On the other hand, we have

$$ \mathfrak{Re} \biggl\{ \gamma p(z_{1})^{\frac{2}{\alpha +\lambda}}+ \frac{2}{\alpha +\lambda}\frac{z_{1}p'(z_{1})}{p(z_{1})} \biggr\} =0, $$

which contradicts the hypothesis. Hence, the assertion (7) is established. □

Remark 3

By putting $\alpha =\lambda =1$ in Theorem 4 we obtain the result [10] that is a corrected version of the open-door lemma. Also, it can readily be observed that Theorem 4 extends and improves Lemma 2.5b in [13].

Also, we can write the other version of the extension of the open-door lemma as follows:

Corollary 2

Let $c=re^{it}$ be a complex number with $\mathfrak{Re}c>0$. Also, let $0<\lambda \leq 1$ and $p\in \mathcal{H}[c^{\lambda},n]$ with $p(z)\neq 0$ in $\mathbb{U}$. If

$$ p(z)^{\frac{1}{\lambda}}+\frac{1}{\lambda}\frac{zp'(z)}{p(z)}\neq iy \quad (z\in \mathbb{U}) $$

for all $y\in \mathbb{R}$, where

$$ y\geq \frac{\sqrt{n}}{\cos{t}} \bigl(\sqrt{n+2 \vert c \vert \cos{t}}-\sqrt{n} \sin{t} \bigr) $$

or

$$ y\leq -\frac{\sqrt{n}}{\cos{t}} \bigl(\sqrt{n+2 \vert c \vert \cos{t}}+\sqrt{n} \sin{t} \bigr), $$

then

$$ -\frac{\lambda \pi}{2}< \arg{p(z)}< \frac{\lambda \pi}{2} \quad (z\in \mathbb{U}). $$

Proof

By putting $\alpha =\lambda$ in Theorem 4, the desired result is obtained. □

Corollary 3

Let $f\in \mathcal{A}_{n}$ with $f(z)f'(z)\neq 0$ in $\mathbb{U}\backslash \{0\}$ and let $\alpha +\lambda =\frac{2}{t_{1}}$, where $0<\alpha \leq 1$, $0<\lambda \leq 1$ and $t_{1}\geq 1$. If

$$ (\gamma -1)\frac{zf'(z)}{f(z)}+\biggl(1+\frac{zf''(z)}{f'(z)}\biggr)\neq iy \quad (z \in \mathbb{U}) $$

for all $y\in \mathbb{R}$, where

$$ y\geq \frac{\sqrt{n}}{\cos \{-\frac{\pi t_{1} (\lambda -\alpha )}{4} \}} \biggl(\sqrt{n+\cos \biggl\{ -\frac{\pi t_{1} (\lambda -\alpha )}{4} \biggr\} }-\sqrt{n}\sin \biggl\{ - \frac{\pi t_{1} (\lambda -\alpha )}{4} \biggr\} \biggr) $$

or

$$ y\leq - \frac{\sqrt{n}}{\cos \{-\frac{\pi t_{1} (\lambda -\alpha )}{4} \}} \biggl(\sqrt{n+\cos \biggl\{ -\frac{\pi t_{1} (\lambda -\alpha )}{4} \biggr\} }+\sqrt{n}\sin \biggl\{ - \frac{\pi t_{1} (\lambda -\alpha )}{4} \biggr\} \biggr), $$

then

$$ -\frac{\pi}{2}\alpha t_{1}< \arg{\frac{zf'(z)}{f(z)}}< \frac{\pi}{2} \lambda t_{1} \quad (z\in \mathbb{U}), $$

where $\gamma =\exp (-i\pi \frac{t_{1}(\lambda -\alpha )}{4})$.

Proof

Let $p(z)=(\frac{zf'(z)}{f(z)})^{\frac{1}{t_{1}}}$, then we have $p\in \mathcal{H}[1,n]$ with $p(z)\neq 0$ in $\mathbb{U}$. Now, making use of Theorem 4 and letting $p(z)=(\frac{zf'(z)}{f(z)})^{\frac{1}{t_{1}}}$, $c=1$, $t=0$, and $\alpha +\lambda =\frac{2}{t_{1}}$ in this theorem, the result is gained. □

Theorem 5

Let $c=re^{it}$ with $-\frac{\pi \alpha}{\alpha +\lambda}< t< \frac{\pi \lambda}{\alpha +\lambda}$, where $0<\alpha \leq 1$ and $0<\lambda \leq 1$. Also, let $p\in \mathcal{H}[c^{\frac{\alpha +\lambda}{2}},n]$ with $p(z)\neq 0$ in $\mathbb{U}$ and $n>\frac{2|c|}{\cos{B}}$. If

$$ \gamma p(z)^{\frac{2}{\alpha +\lambda}}-\frac{2}{\alpha +\lambda} \frac{zp'(z)}{p(z)}\neq iy \quad (z \in \mathbb{U}) $$

for all $y\in \mathbb{R}$, where

$$ y\geq \frac{\sqrt{n}}{\cos{B}}\bigl(\sqrt{n-2 \vert c \vert \cos{B}}+\sqrt{n} \sin{B}\bigr) $$

or

$$ y\leq -\frac{\sqrt{n}}{\cos{B}}\bigl(\sqrt{n-2 \vert c \vert \cos{B}}-\sqrt{n} \sin{B}\bigr), $$

then we have

$$ -\frac{\alpha \pi}{2}< \arg{p(z)}< \frac{\lambda \pi}{2} \quad (z\in \mathbb{U}), $$

where $\gamma = \exp \{-i\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}\}$ and $B=t-\pi \frac{\lambda -\alpha}{2(\lambda +\alpha )}$.

Proof

The proof of this theorem is similar to that of Theorem 4, so we omit the detailed proof. □

Corollary 4

Let $f(z)= \frac{1}{z}+a_{n} z^{n}+\cdots$ be a meromorphic function with $f'f\neq 0$ in $\mathbb{U}\backslash \{0\}$ and $n>1$. If

$$ -1-\frac{zf''(z)}{f'(z)}\neq iy \quad (z\in \mathbb{U}) $$

for all $y\in \mathbb{R}$, where

$$ y\geq \sqrt{n+1}(\sqrt{n-1}) $$

or

$$ y\leq -\sqrt{n+1}(\sqrt{n-1}), $$

then we have

$$ -\frac{\pi}{2}< \arg \biggl\{ -\frac{zf'(z)}{f(z)} \biggr\} < \frac{\pi}{2} \quad (z\in \mathbb{U}). $$

Proof

Let $p(z)=-\frac{zf'(z)}{f(z)}$, then $p\in \mathcal{H}[1, n+1]$. Differentiating logarithmically from p, we obtain

$$ p(z)-\frac{zp'(z)}{p(z)}=-1-\frac{zf''(z)}{f'(z)} \quad (z\in \mathbb{U}). $$

Then, setting $c=1$, $t=0$, $\alpha =\lambda =1$, and substituting n by $n+1$ in Theorem 5, the result is obtained. □

3 Applications related to extension of the Jack lemma

Corollary 5

Let $c\in \mathbb{C}$ and β be a real number. Let $(c^{\lambda}-\beta )^{\frac{1}{\lambda}}= re^{it}$ and $\mathfrak{Re}(c^{\lambda}-\beta )^{\frac{1}{\lambda}}>0$. Suppose $0<\lambda \leq 1$ and $p\in \mathcal{H}[c^{\lambda},n]$ with $p(z)\neq \beta$ in $\mathbb{U}$. If there exists a point $z_{0}\in \mathbb{U}$ such that

$$ \bigl\vert \arg \bigl(p(z)-\beta \bigr) \bigr\vert < \frac{\lambda \pi}{2} \quad \textit{for } \vert z \vert < \vert z_{0} \vert $$

and $(p(z_{0})-\beta )^{\frac{1}{\lambda}}=\pm ia$, where $a>0$, then we have

$$ \frac{z_{0} p'(z_{0})}{p(z_{0})-\beta}= im\lambda , $$

where

$$ m\geq n \biggl( \frac{a^{2}-2a\mathfrak{Im}{(c^{\lambda}-\beta )^{\frac{1}{\lambda}}}+ \vert c^{\lambda}-\beta \vert ^{\frac{2}{\lambda}}}{2a\mathfrak{Re}{(c^{\lambda}-\beta )^{\frac{1}{\lambda}}}} \biggr), \quad \textit{if } \arg \bigl\{ p(z_{0})-\beta \bigr\} = \frac{\lambda \pi}{2} $$

and

$$ m\leq -n \biggl( \frac{a^{2}+2a\mathfrak{Im}{(c^{\lambda}-\beta )^{\frac{1}{\lambda}}}+ \vert c^{\lambda}-\beta \vert ^{\frac{2}{\lambda}}}{2a\mathfrak{Re}{(c^{\lambda}-\beta )^{\frac{1}{\lambda}}}} \biggr), \quad \textit{if } \arg \bigl\{ p(z_{0})-\beta \bigr\} =- \frac{\lambda \pi}{2}. $$

Proof

Let us set $q(z)= p(z)-\beta$, then $q(z)\in \mathcal{H}[c_{1}^{\lambda},n]$ with $c_{1}=(c^{\lambda}-\beta )^{\frac{1}{\lambda}}$ and $q(z)\neq 0$ in $\mathbb{U}$. According to the hypothesis, $\mathfrak{Re}{c_{1}}>0$ and there exists a point $z_{0}\in \mathbb{U}$ such that $|\arg{q(z)}|<\frac{\lambda \pi}{2}$ for $|z|<|z_{0}|$ and $q(z_{0})^{\frac{1}{\lambda}}=\pm ia$ with $a>0$. Now, utilizing Corollary 1 for q, the proof is completed. □

By using Corollary 5, we obtain:

Corollary 6

Let $f\in \mathcal{A}_{n}$ with $\frac{f(z)}{z}\neq \beta$ in $\mathbb{U}$. Suppose $0\leq \beta <1$. If

$$ \frac{zf'(z)-f(z)}{f(z)-\beta z} \neq is \quad (z\in \mathbb{U}) $$

for all $s\in \mathbb{R}$, where $|s|\geq n$, then we have

$$ \mathfrak{Re} {\frac{f(z)}{z}}>\beta . $$

Proof

Let us define $p(z)=\frac{f(z)}{z}$, then $p\in \mathcal{H}[1,n]$. If there exists a point $z_{0}\in \mathbb{U}$ such that $\mathfrak{Re}{p(z)}>\beta$ for $|z|<|z_{0}|$ and $\mathfrak{Re}{p(z_{0})}=\beta$, then we have $|\arg{(p(z)-\beta )}|<\frac{\pi}{2}$ for $|z|<|z_{0}|$ and $p(z_{0})-\beta = \pm ia$, where $a>0$. Therefore, through Corollary 5, we have

$$ \frac{z_{0}f'(z_{0})-f(z_{0})}{f(z_{0})-\beta z_{0}}= \frac{z_{0} p'(z_{0})}{p(z_{0})-\beta}= im \quad (z\in \mathbb{U}), $$

where we have

$$ m\geq n \biggl(\frac{a^{2}-(1-\beta )^{2}}{2a(1-\beta )} \biggr) \geq n \quad \text{when } p(z_{0})-\beta =ia $$

and

$$ m\leq -n \biggl(\frac{a^{2}-(1-\beta )^{2}}{2a(1-\beta )} \biggr) \leq -n \quad \text{when } p(z_{0})-\beta = -ia. $$

This contradicts the hypothesis and so the proof is completed. □

Also, using the same method as Corollary 5, we can obtain:

Corollary 7

Let $c\in \mathbb{C}$ and β be a real number. Let $(\beta -c)^{\frac{1}{\lambda}}= re^{it}$ and $\mathfrak{Re}(\beta -c)^{\frac{1}{\lambda}}>0$. Suppose $0<\lambda \leq 1$ and $p\in \mathcal{H}[c,n]$ with $p(z)\neq \beta$ in $\mathbb{U}$. If there exists a point $z_{0}\in \mathbb{U}$ such that

$$ \bigl\vert \arg \bigl(\beta -p(z)\bigr) \bigr\vert < \frac{\lambda \pi}{2} \quad \textit{for } \vert z \vert < \vert z_{0} \vert $$

and $(\beta -p(z_{0}))^{\frac{1}{\lambda}}=\pm ia$, where $a>0$, then we have

$$ \frac{z_{0} p'(z_{0})}{p(z_{0})-\beta}= im\lambda , $$

where

$$ m\geq n \biggl( \frac{a^{2}-2a\mathfrak{Im}{(\beta -c)^{\frac{1}{\lambda}}}+ \vert \beta -c \vert ^{\frac{2}{\lambda}}}{2a\mathfrak{Re}{(\beta -c)^{\frac{1}{\lambda}}}} \biggr) \quad \textit{when } \arg \bigl\{ \beta -p(z_{0})\bigr\} =\frac{\lambda \pi}{2} $$

and

$$ m\leq -n \biggl( \frac{a^{2}-2a\mathfrak{Im}{(\beta -c)^{\frac{1}{\lambda}}}+ \vert \beta -c \vert ^{\frac{2}{\lambda}}}{2a\mathfrak{Re}{(\beta -c)^{\frac{1}{\lambda}}}} \biggr) \quad \textit{when } \arg \bigl\{ \beta -p(z_{0})\bigr\} =-\frac{\lambda \pi}{2}. $$

Proof

Let us set $q(z)= \beta -p(z)$. Then, the continuation of the proof is similar to that of Corollary 5, and so we omit it. □

Similar to Corollary 6 and applying Corollary 7, we can gain the following corollary.

Corollary 8

Let $\beta >1$ and $f\in \mathcal{A}_{n}$ with $\frac{f(z)}{z}\neq \beta$ in $\mathbb{U}$. If

$$ \frac{zf'(z)-f(z)}{f(z)-\beta z} \neq is \quad (z\in \mathbb{U}) $$

for all $s\in \mathbb{R}$, where $|s|\geq n$, then we have

$$ \mathfrak{Re} {\frac{f(z)}{z}}< \beta . $$

Theorem 6

Let c, $\gamma >1$, $0\leq \beta <1$ and $0<\alpha \leq 1$ be real numbers. Also, let $c^{\alpha}-\beta >0$. If $p\in \mathcal{H}[c^{\alpha},n]$ with $p(z)\neq \beta$ in $\mathbb{U}$ and

$$ \bigl\vert \arg \bigl(p(z)- \beta +\gamma zp'(z)\bigr) \bigr\vert < \frac{\pi}{2}\biggl(\alpha + \frac{2}{\pi}\tan ^{-1}( \alpha \gamma n)\biggr) \quad (z\in \mathbb{U}), $$

then

$$ \bigl\vert \arg \bigl(p(z)-\beta \bigr) \bigr\vert < \frac{\pi}{2}\alpha \quad (z\in \mathbb{U}). $$

Proof

If there exists a point $z_{0}\in \mathbb{U}$ such that $|\arg (p(z)-\beta )|<\frac{\pi}{2}\alpha$ for $|z|<|z_{0}|$ and $|\arg (p(z_{0})-\beta _{1})|=\frac{\pi}{2}\alpha$, then from Corollary 5 we have

$$ \frac{z_{0}p'(z_{0})}{p(z_{0})-\beta}=i\alpha m, $$

where $|m|\geq n$. Thus, for the case $\arg (p(z_{0})-\beta )=\frac{\pi}{2}\alpha$ we have

$$\begin{aligned} \arg \bigl\{ p(z_{0})- \beta +\gamma z_{0}p'(z_{0}) \bigr\} &= \arg \biggl\{ \bigl(p(z_{0})- \beta \bigr) \biggl(1+\gamma \frac{z_{0}p'(z_{0})}{p(z_{0})-\beta}\biggr) \biggr\} \\ =\frac{\pi}{2}\alpha +\arg \{1+i\gamma \alpha m\}\geq \frac{\pi}{2} \alpha +\tan ^{-1}(\gamma \alpha n), \end{aligned}$$

which contradicts the assumption. Also, for the case $\arg (p(z_{0})-\beta _{1})=-\frac{\pi}{2}\alpha$ we have

$$\begin{aligned} \arg \bigl\{ p(z_{0})- \beta _{1}+\gamma z_{0}p'(z_{0})\bigr\} &= \arg \biggl\{ \bigl(p(z_{0})- \beta _{1}\bigr) \biggl(1+\gamma \frac{z_{0}p'(z_{0})}{p(z_{0})-\beta}\biggr) \biggr\} \\ & =-\frac{\pi}{2}\alpha +\arg \{1+i\gamma \alpha m\}\leq -\biggl( \frac{\pi}{2}\alpha +\tan ^{-1}(\gamma \alpha n)\biggr), \end{aligned}$$

which contradicts the assumption. Hence, the proof is completed. □

Remark 4

By letting $c= \gamma = \alpha = n=1$ in Theorem 6, we can obtain Theorem 3 in [17].

Theorem 7

Let $-\lambda < b <\lambda$, $\lambda >0$, and $k>0$. Also, let $p\in \mathcal{H}[1,n]$ with $p(z)\neq \frac{2\lambda}{b+\lambda}$ in $\mathbb{U}$. If for all $z\in \mathbb{U}$

$$ \mathfrak{Re} \biggl\{ p(z)+k\frac{zp'(z)}{p(z)} \biggr\} < \textstyle\begin{cases} nk\frac{\lambda +b}{2(\lambda -b)} & \textit{if } -\lambda < b\leq 0, n>\frac{2(\lambda -b)}{k(\lambda +b)}, \\ \frac{nk}{2}\frac{\lambda -b}{\lambda +b}+\frac{2\lambda}{\lambda +b} & \textit{if } \frac{\lambda}{1+kn}\leq b < \lambda, \\ \frac{nk}{2}\frac{\lambda -b}{\lambda +b} & \textit{if } 0< b< \frac{\lambda}{1+kn}, n>\frac{2(\lambda +b)}{k(\lambda -b)}, \end{cases} $$

then we have

$$ \biggl\vert p(z)-\frac{\lambda}{b+\lambda} \biggr\vert < \frac{\lambda}{b+\lambda}\quad (z \in \mathbb{U}). $$

Proof

Let us define

$$ q(z)=\frac{\lambda (1-z)}{\lambda -bz}. $$

One can readily check that $q\in Q$ with $q(0)=p(0)=1$ and q maps the open unit disc $\mathbb{U}$ onto the disk with the center $\frac{\lambda}{\lambda +b}$ and the radius $\frac{\lambda}{\lambda +b}$. Furthermore, we have

$$ q^{-1}(z)= \frac{\lambda (z-1)}{bz-\lambda} \quad \text{and} \quad q'(z)=\frac{\lambda (b-\lambda )}{(\lambda -bz)^{2}}. $$

We claim that $p\prec q$, otherwise if $p\nprec q$, then from Lemma 2, there exist points $z_{0}\in \mathbb{U}$ and $\zeta _{0}\in \partial \mathbb{U}$ such that $p(z_{0})=q(\zeta _{0})$ and $z_{0}p'(z_{0})= m\zeta _{0}q'(\zeta _{0})$ where $m\geq n$. As

$$ \zeta _{0}= q^{-1}\bigl(p(z_{0})\bigr)= \frac{\lambda (p(z_{0})-1)}{bp(z_{0})-\lambda}, $$

we have

$$ z_{0}p'(z_{0})=-m \frac {(1-p(z_{0}))(\lambda -bp(z_{0}))}{(\lambda -b)}. $$

Put

$$ p(z_{0})= \frac{\lambda}{\lambda +b}+ \frac{\lambda}{\lambda +b} e^{it}, $$

where t is a fixed real number. Through the relations obtained and with computing, we can conclude that

$$ \mathfrak{Re} \biggl\{ p(z_{0})+k\frac{z_{0}p'(z_{0})}{p(z_{0})} \biggr\} = \biggl( \frac{\lambda (\lambda -b(1+km))}{(\lambda +b)(\lambda -b)} \biggr) (1+ \cos{t})+mk\frac{\lambda +b}{2(\lambda -b)}. $$

(8)

In resumption of the argument, we consider the three cases. If $-\lambda < b\leq 0$, then equation (8) implies that

$$ \mathfrak{Re} \biggl\{ p(z_{0})+k\frac{z_{0}p'(z_{0})}{p(z_{0})} \biggr\} \geq nk \frac{\lambda +b}{2(\lambda -b)}, $$

which contradicts the hypothesis. Also, for the case $0< b<\lambda \leq b(1+kn)$, we put

$$ f(x)= mk\frac{\lambda +b}{2(\lambda -b)}+(1+x) \frac{\lambda}{\lambda +b}\frac{\lambda -b(1+km)}{\lambda -b} \quad (-1 \leq x \leq 1), $$

where $x=\cos{t}$, and thus

$$ f'(x)= \frac{\lambda}{\lambda +b}\frac{\lambda -b(1+km)}{\lambda -b} \leq 0 \quad (-1\leq x \leq 1). $$

Therefore, we have

$$ f(x)\geq f(1)=\frac{mk(\lambda -b)}{2(\lambda +b)}+ \frac{2\lambda}{(\lambda +b)}\geq \frac{nk(\lambda -b)}{2(\lambda +b)}+ \frac{2\lambda}{(\lambda +b)} \quad (-1\leq x \leq 1), $$

which contradicts the hypothesis. Finally, for the case $0< b<\frac{\lambda}{1+kn}$ we set

$$ g(x)= \frac{\lambda +b}{2}-\frac{\lambda b}{\lambda +b}- \frac{\lambda b}{\lambda +b} x \quad (-1 \leq x \leq 1), $$

where $x=\cos{t}$, and thus $g'(x)= -\frac{\lambda}{\lambda +b}<0$. Therefore, for all $-1\leq x\leq 1$ we have $g(x)\geq g(1)= \frac{(\lambda -b)^{2}}{2(\lambda +b)}>0$. Consequently,

$$\begin{aligned} \mathfrak{Re} \biggl\{ p(z_{0})+k\frac{z_{0}p'(z_{0})}{p(z_{0})} \biggr\} &= \frac{\lambda}{\lambda +b}(1+x)+ \biggl(\frac{mk}{\lambda -b} \biggr)g(x)\\ & \geq \frac{nk}{(\lambda -b)}\frac{(\lambda -b)^{2}}{2(\lambda +b)}= nk \frac{\lambda -b}{2(\lambda +b)}, \end{aligned}$$

which contradicts the assumption. This finishes the proof. □

Corollary 9

Let $-\lambda < b <\lambda$, $\lambda >0$ and $k>0$. Also, let $f\in \mathcal{A}_{n}$ with $\frac{zf'(z)}{f(z)}\neq \frac{2\lambda}{b+\lambda}$ in $\mathbb{U}$. If for all $z\in \mathbb{U}$

$$\begin{aligned} &\mathfrak{Re} \biggl\{ (1-k)\frac{zf'(z)}{f(z)}+k \biggl(1+ \frac{zf''(z)}{f'(z)} \biggr) \biggr\} \\ & \quad < \textstyle\begin{cases} nk\frac{\lambda +b}{2(\lambda -b)} & \textit{if } -\lambda < b\leq 0, n>\frac{2(\lambda -b)}{k(\lambda +b)}, \\ \frac {nk}{2}\frac{\lambda -b}{\lambda +b}+ \frac{2\lambda}{\lambda +b} & \textit{if } \frac{\lambda}{1+kn}\leq b < \lambda, \\ \frac{nk}{2}\frac{\lambda -b}{\lambda +b} & \textit{if } 0< b< \frac{\lambda}{1+kn}, n>\frac{2(\lambda +b)}{k(\lambda -b)}, \end{cases}\displaystyle \end{aligned}$$

then we have

$$ \biggl\vert \frac{zf'(z)}{f(z)}-\frac{\lambda}{b+\lambda} \biggr\vert < \frac{\lambda}{b+\lambda} \quad (z\in \mathbb{U}). $$

Proof

By letting $p(z)=\frac{zf'(z)}{f(z)}$ in Theorem 7, the result is obtained. □

Remark 5

The result gained in Corollary 9 improves and extends the result obtained in [16] (see Theorem 3.1 in [16]).

By putting $k=1$, $b=1$, $\lambda =3$, and $n=2$ in Corollary 9, we obtain:

Corollary 10

Let $f\in \mathcal{A}_{2}$ with $\frac{zf'(z)}{f(z)}\neq \frac{3}{2}$ in $\mathbb{U}$. If

$$ \mathfrak{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} < 2\quad (z\in \mathbb{U}), $$

then we have

$$ \biggl\vert \frac{zf'(z)}{f(z)}-\frac{3}{4} \biggr\vert < \frac{3}{4} \quad (z \in \mathbb{U}). $$

Remark 6

The result obtained in Corollary 10 improves and extends the result obtained in [16] (see Corollary 3.3 in [16]).

By putting $k=1$, $b=1$, $\lambda =3$, and $n=3$ in Corollary 9, we have:

Corollary 11

Let $f\in \mathcal{A}_{3}$ with $\frac{zf'(z)}{f(z)}\neq \frac{3}{2}$ in $\mathbb{U}$. If

$$ \mathfrak{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} < \frac{9}{4} \quad (z \in \mathbb{U}), $$

then we have

$$ \biggl\vert \frac{zf'(z)}{f(z)}-\frac{3}{4} \biggr\vert < \frac{3}{4} \quad (z \in \mathbb{U}). $$

By setting $k=1$, $b=3$, $\lambda =5$, and $n=1$ in Corollary 9, we have:

Corollary 12

Let $f\in \mathcal{A}_{1}$ with $\frac{zf'(z)}{f(z)}\neq \frac{5}{4}$ in $\mathbb{U}$. If

$$ \mathfrak{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} < \frac{11}{8} \quad (z \in \mathbb{U}), $$

then we have

$$ \biggl\vert \frac{zf'(z)}{f(z)}-\frac{5}{8} \biggr\vert < \frac{5}{8} \quad (z \in \mathbb{U}). $$

By setting $k=1$ and $b=0$ in Corollary 9, we obtain:

Corollary 13

Let $n>2$ and $f\in \mathcal{A}_{n}$ with $\frac{zf'(z)}{f(z)}\neq 2$ in $\mathbb{U}$. If

$$ \mathfrak{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} < \frac{n}{2} \quad (z \in \mathbb{U}), $$

then we have

$$ \biggl\vert \frac{zf'(z)}{f(z)}-1 \biggr\vert < 1 \quad (z\in \mathbb{U}). $$

Remark 7

The results obtained in Corollary 12 and Corollary 13 improve and extend the result obtained in [16] (see Corollary 3.2 in [16]).