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  • Open Access

Criteria for starlike and convex functions of order α

Journal of Inequalities and Applications20152015:28

https://doi.org/10.1186/s13660-015-0548-0

  • Received: 7 December 2014
  • Accepted: 2 January 2015
  • Published:

Abstract

Let \(\mathcal{A}_{n}\) (\(n\in\mathbb{N}\)) be the class of certain analytic functions \(f(z)\) in the open unit disk \(\mathbb{U}\) and \(\mathcal{P}_{n}(\lambda)\) be the subclass of \(\mathcal{A}_{n}\) consisting of \(f(z)\) which satisfy \(|f''(z)| \leqq \lambda\) (\(\lambda> 0\)) in \(\mathbb{U}\). Some properties for the class \(\mathcal{P}_{n}(\lambda)\), which are the improvements of the previous results due to Ponnusamy and Singh (Complex Var. Theory Appl. 34:276-291, 1997), are discussed.

Keywords

  • starlike function
  • convex function
  • strongly starlike function
  • subordination

MSC

  • 30C45

1 Introduction

Let \(\mathcal{A}_{n}\) denote the class of functions of the form
$$ f(z)=z+\sum^{\infty}_{k=n+1}a_{k}z^{k} \quad \bigl(n\in\mathbb{N}=\{1,2,3,\ldots\}\bigr), $$
(1.1)
which are analytic in the open unit disk \(\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}\), and let \(\mathcal{A}_{1}=\mathcal{A}\).
A function \(f(z) \in\mathcal{A}\) is said to be in the class \(\mathcal{S}^{*}(\alpha)\) in \(\mathbb{U}\) if it satisfies
$$ \operatorname{Re}\frac{zf'(z)}{f(z)}>\alpha \quad (z\in\mathbb{U}) $$
(1.2)
for some real α (\(\alpha<1\)). If \(f(z) \in \mathcal{S}^{*}(\alpha)\) with \(0 \leqq\alpha< 1\), then \(f(z)\) is said to be univalent and starlike of order α in \(\mathbb{U}\). We denote \(\mathcal{S}^{*}(0) = \mathcal{S}^{*}\). A function \(f(z) \in\mathcal{A}\) is said to be in the class \(\mathcal{C}(\alpha)\) if it satisfies
$$ \operatorname{Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)} \biggr\} >\alpha\quad (z\in \mathbb{U}) $$
(1.3)
for some real α (\(\alpha<1\)). If \(f(z) \in \mathcal{C}(\alpha)\) with \(0 \leqq\alpha< 1\), then \(f(z)\) is said to be univalent and convex of order α in \(\mathbb{U}\). We write \(\mathcal{C}(0)=\mathcal{C}\).

Let \(f(z)\) and \(g(z)\) be analytic in \(\mathbb{U}\). Then we say that \(f(z)\) is subordinate to \(g(z)\) in \(\mathbb{U}\), written \(f(z)\prec g(z)\), if there exists a function \(w(z)\) analytic in \(\mathbb{U}\) which satisfies \(w(0)=0\), \(|w(z)|<1\) (\(z \in \mathbb{U}\)) and \(f(z)=g(w(z))\) for \(z\in\mathbb{U}\). If \(g(z)\) is univalent in \(\mathbb{U}\), then the subordination \(f(z)\prec g(z)\) is equivalent to \(f(0)=g(0)\) and \(f(\mathbb{U})\subset g(\mathbb{U})\) (cf. Duren [1]).

A function \(f(z) \in\mathcal{A}\) is said to be strongly starlike of order β in \(\mathbb{U}\) if it satisfies
$$ \frac{zf'(z)}{f(z)}\prec \biggl( \frac{1+z}{1-z} \biggr)^{\beta} $$
(1.4)
for some real β (\(0<\beta\leqq1\)). We denote this class by \(\widetilde{\mathcal{S}}^{*}(\beta)\). Note that \(\widetilde{\mathcal {S}}^{*}(1)=\mathcal{S}^{*}\).
Define
$$ \mathcal{P}_{n}(\lambda)=\bigl\{ f(z) \in \mathcal{A}_{n}: \bigl\vert f''(z)\bigr\vert \leqq \lambda\ (\lambda> 0; z\in\mathbb{U})\bigr\} . $$
(1.5)
Mocanu [2] considered the problem of finding λ such that
$$f(z) \in\mathcal{P}_{n}(\lambda)\quad \text{implies}\quad f(z) \in \mathcal{S}^{*}. $$
Mocanu [2] has shown that:

Theorem A

([2])

If
$$\lambda=\frac{n(n+1)}{2n+1}\quad (n\in\mathbb{N}), $$
then \(\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}\).

Ponnusamy and Singh [3] proved the following results.

Theorem B

Let
$$\lambda_{n}= \frac{n(n+1)}{\sqrt{(n+1)^{2}+1}}\quad (n\in\mathbb{N}). $$
If \(0<\lambda\leqq\lambda_{n}\), then \(\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}(\beta)\), where
$$\beta=\beta_{n}(\lambda)=\left \{ \begin{array}{l@{\quad}l} \frac{(n+1)(n-\lambda)}{n(n+1)+\lambda},& \textit{if }0<\lambda\leqq \frac {n(n+1)}{n+2}, \\ \frac{n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})},& \textit{if }\frac{n(n+1)}{n+2}\leq \lambda\leq\lambda_{n}. \end{array} \right . $$

Theorem C

Let \(0<\beta\leqq 1\) and
$$\lambda'_{n}= \frac{n(n+1)\sin \frac{\pi\beta}{2}}{ \sqrt {1+(n+1)^{2}+2(n+1)\cos \frac{\pi\beta}{2}}} \quad (n\in\mathbb{N}). $$
If \(0<\lambda\leqq\lambda'_{n}\), then \(\mathcal{P}_{n}(\lambda)\subset \widetilde{\mathcal{S}}^{*}(\beta)\).

It is easy to verify that Theorem B and Theorem C are better than Theorem A in two different ways.

In this paper we generalize and refine the above theorems. Furthermore we find λ such that \(f(z)\in \mathcal{P}_{n}(\lambda)\) implies \(f(z)\in \mathcal{C}(\alpha)\) (\(\alpha<1\)). These results are sharp.

2 Main results

To derive our first result, we need the following lemma due to Hallenbeck and Ruscheweyh [4].

Lemma

Let \(g(z)\) be analytic and convex univalent in \(\mathbb{U}\) and \(f(z)=g(0)+\sum^{\infty}_{k=n}a_{k}z^{k}\) (\(n\in\mathbb{N}\)) be analytic in \(\mathbb{U}\). If \(f(z)\prec g(z)\), then
$$z^{-c}\int_{0}^{z}t^{c-1}f(t)\, dt\prec\frac{1}{n}z^{- \frac{c}{n}}\int_{0}^{z}t^{ \frac{c}{n}-1}g(t) \, dt, $$
where \(\operatorname{Re} (c)\geqq0\) and \(c\neq0\).

Now, we derive the following.

Theorem 1

Let \(0<\lambda<n(n+1)\) (\(n\in \mathbb{N}\)). If \(f(z)\in \mathcal{P}_{n}(\lambda)\), then
$$ \biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert < \frac{n\lambda}{n(n+1)-\lambda} \quad (z\in \mathbb{U}). $$
(2.1)
The bound \(\frac{n\lambda}{n(n+1)-\lambda}\) in (2.1) is sharp.

Proof

Let
$$f(z)=z+\sum^{\infty}_{k=n+1}a_{k}z^{k} \in \mathcal{P}_{n}(\lambda)\quad \text{and} \quad 0<\lambda<n(n+1) \quad (n\in \mathbb{N}). $$
Then we have
$$ zf''(z)=n(n+1)a_{n+1}z^{n}+\cdots \prec\lambda z. $$
(2.2)
Applying the lemma with \(c=1\), it follows from (2.2) that
$$\frac{1}{z}\int_{0}^{z}tf''(t) \, dt\prec \frac{\lambda}{n}z^{-\frac{1}{n}}\int_{0}^{z}t^{\frac{1}{n}} \, dt, $$
which yields
$$ f'(z)-\frac{f(z)}{z}\prec\frac{\lambda z}{n+1}, $$
(2.3)
and hence
$$ \biggl\vert f'(z)-\frac{f(z)}{z}\biggr\vert < \frac{\lambda}{n+1}\quad (z\in\mathbb{U}). $$
(2.4)
By (2.3) we can write
$$ f'(z)-\frac{f(z)}{z}=\frac{\lambda w(z)}{n+1}, $$
(2.5)
where \(w(z)\) is analytic in \(\mathbb{U}\) with \(w(0)=0\) and \(|w(z)|<1\) (\(z\in\mathbb{U}\)). Since
$$f'(z)-\frac{f(z)}{z}=na_{n+1}z^{n}+\cdots, $$
the function \(w(z)\) in (2.5) satisfies \(|w(z)|\leq|z|^{n}\) (\(z\in\mathbb{U}\)) by the Schwarz lemma. Also (2.5) leads to
$$ \int_{0}^{z} \biggl( \frac{f'(t)}{t}- \frac{f(t)}{t^{2}} \biggr)\, dt= \frac{\lambda}{n+1}\int_{0}^{z} \frac{w(t)}{t}\, dt. $$
(2.6)
In view of (2.6), we deduce that
$$\begin{aligned} \biggl\vert \frac{f(z)}{z}-1\biggr\vert &=\frac{\lambda}{n+1}\biggl\vert \int_{0}^{1} \frac{w(uz)}{u}\, du\biggr\vert \leqq\frac{\lambda}{n+1}\int_{0}^{1} \frac{|w(uz)|}{u}\, du \\ &\leqq\frac{\lambda |z|^{n}}{n+1}\int_{0}^{1}u^{n-1} \, du<\frac{\lambda}{n(n+1)} \end{aligned}$$
and so
$$ \biggl\vert \frac{f(z)}{z}\biggr\vert >1- \frac{\lambda}{n(n+1)}>0\quad (z \in\mathbb{U}). $$
(2.7)
Now, by using (2.4) and (2.7), we find that
$$\begin{aligned} \biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert &=\biggl\vert \frac{z}{f(z)}\biggr\vert \biggl\vert f'(z)- \frac{f(z)}{z}\biggr\vert \\ &< \frac{ \frac{\lambda}{n+1}}{1- \frac{\lambda}{n(n+1)}}= \frac{n\lambda}{n(n+1)-\lambda} \end{aligned}$$
for \(z\in\mathbb{U}\), which shows (2.1).
For sharpness, we consider the function
$$ f(z)=z+ \frac{\lambda}{n(n+1)}z^{n+1}\quad (z\in\mathbb{U}) $$
(2.8)
for \(0<\lambda<n(n+1)\). Obviously \(f(z)\in \mathcal{P}_{n}(\lambda)\). Furthermore we have
$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert =\biggl\vert \frac{ \frac{\lambda}{n+1}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\vert \rightarrow\frac{n\lambda }{n(n+1)-\lambda} $$
as \(z\rightarrow e^{ \frac{\pi i}{n}}\). This completes the proof of Theorem 1. □

Next, we prove the following.

Theorem 2

Let \(0<\lambda<n(n+1)\) (\(n\in \mathbb{N}\)). Then
$$\mathcal{P}_{n}(\lambda)\subset\mathcal{S}^{*}(\alpha), $$
where
$$ \alpha=\alpha_{n}(\lambda)= \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}. $$
(2.9)
The result is sharp, that is, the order α is best possible.

Proof

If \(f(z)\in\mathcal{P}_{n}(\lambda)\) and \(0<\lambda<n(n+1)\) (\(n\in\mathbb{N}\)), then an application of Theorem 1 yields
$$1-\operatorname{Re} \frac{zf'(z)}{f(z)}<\frac{n\lambda}{n(n+1)-\lambda}\quad (z\in\mathbb{U}). $$
Hence \(f(z)\in\mathcal{S}^{*}(\alpha)\) where \(\alpha=\alpha_{n}(\lambda)\) is given by (2.9).
For the function \(f(z)\in\mathcal{P}_{n}(\lambda)\) defined by (2.8), we have
$$\operatorname{Re} \frac{zf'(z)}{f(z)}=\operatorname{Re} \biggl\{ \frac{1+ \frac{\lambda }{n}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\} \rightarrow \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}=\alpha $$
as \(z\rightarrow e^{ \frac{\pi i}{n}}\). Therefore the order α cannot be increased. □

Remark 1

Let us compare Theorem 2 with Theorem B. Clearly
$$n(n+1)>\lambda_{n}\quad \text{and} \quad \alpha_{n}( \lambda)>\beta_{n}(\lambda)\quad \biggl(0<\lambda\leqq \frac{n(n+1)}{n+2} \biggr). $$
Also, for \(\frac{n(n+1)}{n+2}\leqq\lambda\leqq\lambda_{n}\), we have
$$\begin{aligned} \alpha_{n}(\lambda)-\beta_{n}(\lambda)&=\frac{(n+1)(n-\lambda )}{n(n+1)-\lambda} -\frac{n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})} \\ &=\frac{2(n+1)(n-\lambda)(n(n+1)+\lambda) -(n^{2}(n+1)^{2}-((n+1)^{2}+1)\lambda^{2})}{ 2(n^{2}(n+1)^{2}-\lambda^{2})} \\ &=\frac{n^{2}(n+1-\lambda)^{2}}{2(n^{2}(n+1)^{2}-\lambda^{2})}>0. \end{aligned}$$
Thus we conclude that Theorem 2 extends and improves Theorem B by Ponnusamy and Singh [3].
Taking
$$\lambda= \frac{n(n+1)}{2n+1}\quad \text{and}\quad \lambda=n, $$
Theorem 2 reduces to the following.

Corollary 1

For \(n\in\mathbb{N}\) we have
$$ \mathcal{P}_{n} \biggl( \frac{n(n+1)}{2n+1} \biggr)\subset\mathcal {S}^{*} \biggl( \frac{1}{2} \biggr) \quad \textit{and}\quad \mathcal{P}_{n}(n)\subset \mathcal{S}^{*}. $$
(2.10)
The results are sharp.

Further, applying Theorem 1, we derive the following.

Theorem 3

Let \(0<\beta\leqq1\) and
$$ \widetilde{\lambda}_{n}= \frac{n(n+1)\sin \frac{\pi\beta}{2}}{n+\sin \frac{\pi\beta}{2}}\quad (n\in \mathbb{N}). $$
(2.11)
If \(0<\lambda\leqq\widetilde{\lambda}_{n}\), then \(\mathcal{P}_{n}(\lambda)\subset \widetilde{\mathcal{S}}^{*}(\beta)\) and the bound \(\widetilde{\lambda}_{n}\) cannot be increased.

Proof

Let
$$0<\beta\leqq1,\qquad f(z)\in \mathcal{P}_{n}(\lambda) \quad \text{and} \quad 0<\lambda\leqq \widetilde{\lambda}_{n}, $$
where \(\widetilde{\lambda}_{n}\) is given by (2.11). Then \(\widetilde{\lambda}_{n}\leqq n\) and it follows from Theorem 1 that
$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert < \frac{n \widetilde{\lambda}_{n}}{n(n+1)- \widetilde{\lambda}_{n}} =\sin \frac{\pi\beta}{2}\quad (z\in\mathbb{U}). $$
This implies that
$$\biggl\vert \arg\frac{zf'(z)}{f(z)}\biggr\vert < \frac{\pi\beta}{2}\quad (z\in \mathbb{U}). $$
Hence \(f(z)\in \widetilde{\mathcal{S}}^{*}(\beta)\).
Furthermore, for the function \(f\in\mathcal{P}_{n}(\lambda)\) defined by (2.8) and \(\widetilde{\lambda}_{n}<\lambda<n(n+1)\), we have
$$\biggl\vert \frac{zf'(z)}{f(z)}-1\biggr\vert \rightarrow\frac{n\lambda }{n(n+1)-\lambda} > \frac{n \widetilde{\lambda}_{n}}{n(n+1)- \widetilde{\lambda}_{n}} =\sin\frac{\pi\beta}{2} $$
as \(z \rightarrow e^{ \frac{\pi i}{n}}\). This shows that \(f \notin \widetilde{S}^{*}(\beta)\) and so the proof of Theorem 3 is completed. □

Remark 2

Since \(\widetilde{\lambda}_{n}>\lambda_{n}'\) (cf. Theorem C) we see that Theorem 3 is better than Theorem C by Ponnusamy and Singh [3].

Finally we discuss the following.

Theorem 4

Let \(0<\lambda<n\) (\(n\in \mathbb{N}\)) and \(0<\sigma\leqq1\). If \(f(z)\in \mathcal{P}_{n}(\lambda)\), then
$$ \operatorname{Re} \biggl\{ \sigma \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)} \biggr\} >\alpha \quad (z\in\mathbb{U}), $$
(2.12)
where
$$ \alpha=\alpha_{n}(\sigma,\lambda)=\sigma \frac{n-(n+1)\lambda}{n-\lambda}+(1- \sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}. $$
(2.13)
The result is sharp, that is, the bound \(\alpha_{n}(\sigma,\lambda)\) cannot be increased.

Proof

Let \(f(z)\in\mathcal{P}_{n}(\lambda)\) and \(0<\lambda<n\). Then, by (2.2) (used in the proof of Theorem 1) and the Schwarz lemma, we can write
$$ zf''(z)=\lambda w(z)\quad (z\in\mathbb{U}), $$
(2.14)
where \(w(z)\) is analytic in \(\mathbb{U}\) and \(|w(z)|\leq |z|^{n}\) (\(z\in\mathbb{U}\)). Further, we deduce from (2.14) that
$$f'(z)-1=\int_{0}^{z}f''(t) \, dt=\lambda\int_{0}^{z} \frac{w(t)}{t}\, dt= \lambda\int_{0}^{1} \frac{w(uz)}{u}\, du, $$
which leads to
$$\begin{aligned} \bigl\vert f'(z)\bigr\vert &\geqq1-\lambda\int _{0}^{1} \frac{|w(uz)|}{u}\, du \\ &>1-\lambda|z|^{n}\int_{0}^{1}u^{n-1} \, du \\ &>1-\frac{\lambda}{n}>0 \quad (z\in\mathbb{U}). \end{aligned}$$
(2.15)
Also, by Theorem 2, we have
$$ \operatorname{Re} \frac{zf'(z)}{f(z)}>\frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}\quad (z\in \mathbb{U}). $$
(2.16)
Let us define the function \(g(z)\) by
$$ g(z)=\sigma \biggl(1+ \frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)}- \alpha, $$
(2.17)
where \(0<\sigma\leqq1\) and α is given by (2.13). Then \(g(z)\) is analytic in \(\mathbb{U}\) and
$$\begin{aligned} g(0)&=1-\alpha=1-\sigma\frac{n-(n+1)\lambda}{n-\lambda}-(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda} \\ &=\sigma\frac{n\lambda}{n-\lambda}+(1-\sigma) \frac{n\lambda}{n(n+1)-\lambda}>0. \end{aligned}$$
We claim that \(\operatorname{Re} g(z)>0\) for \(z\in\mathbb{U}\). Otherwise there exists a point \(z_{0}\in\mathbb{U}\) such that
$$ \operatorname{Re} g(z)>0 \quad \bigl(\vert z\vert <|z_{0}|\bigr) \quad \text{and}\quad \operatorname{Re} g(z_{0})=0. $$
(2.18)
Thus, in view of (2.15)-(2.18) and (2.13), we find that
$$\begin{aligned} \sigma\bigl\vert z_{0}f''(z_{0}) \bigr\vert &=\bigl\vert f'(z_{0})\bigr\vert \biggl\vert g(z_{0})+ \alpha-\sigma -(1-\sigma) \frac{z_{0}f'(z_{0})}{f(z_{0})}\biggr\vert \\ &\geqq\bigl\vert f'(z_{0})\bigr\vert \biggl\vert \operatorname{Re} g(z_{0})+ \alpha-\sigma-(1-\sigma)\operatorname{Re} \frac{z_{0}f'(z_{0})}{f(z_{0})}\biggr\vert \\ & > \biggl(1-\frac{\lambda}{n} \biggr) \biggl(\sigma-\alpha+(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda} \biggr) \\ &=\sigma\lambda>0. \end{aligned}$$
This contradicts the expression (2.14). Hence, we say that \(\operatorname{Re} g(z)>0\) (\(z\in\mathbb{U}\)) and (2.12) is proved.
For the function \(f(z)\in \mathcal{P}_{n}(\lambda)\) (\(0<\lambda<n\)) defined by (2.8), we get
$$\begin{aligned}& \operatorname{Re} \biggl\{ \sigma \biggl(1+\frac{zf''(z)}{f'(z)} \biggr)+(1-\sigma) \frac{zf'(z)}{f(z)} \biggr\} \\& \quad =\sigma \biggl(1+\operatorname{Re} \biggl\{ \frac{\lambda z^{n}}{1+ \frac{\lambda}{n}z^{n}} \biggr\} \biggr)+(1-\sigma) \operatorname{Re} \biggl\{ \frac{1+ \frac{\lambda}{n}z^{n}}{1+ \frac{\lambda}{n(n+1)}z^{n}} \biggr\} \\& \quad \rightarrow\sigma\frac{n-(n+1)\lambda}{n-\lambda}+(1-\sigma) \frac{(n+1)(n-\lambda)}{n(n+1)-\lambda}=\alpha \end{aligned}$$
as \(z\rightarrow e^{ \frac{\pi i}{n}}\). Therefore the bound α is best possible. □

Making \(\sigma=1\) in Theorem 4, we have the following.

Corollary 2

Let \(0<\lambda<n\) (\(n\in\mathbb{N}\)). Then
$$ \mathcal{P}_{n}(\lambda)\subset\mathcal{C} \biggl( \frac{n-(n+1)\lambda}{n-\lambda} \biggr). $$
(2.19)
The result is sharp. In particular, for \(n\in\mathbb{N}\), we have
$$ \mathcal{P}_{n} \biggl( \frac{n}{2n+1} \biggr)\subset\mathcal{C} \biggl( \frac{1}{2} \biggr),\qquad \mathcal{P}_{n} \biggl( \frac{n}{n+1} \biggr)\subset \mathcal{C}, $$
(2.20)
and the results are sharp.

Taking \(\sigma=\frac{1}{2}\) in Theorem 4, we obtain the following.

Corollary 3

Let \(0<\lambda<n\) (\(n\in\mathbb {N}\)). If \(f(z)\in \mathcal{P}_{n}(\lambda)\), then
$$ \operatorname{ Re} \biggl\{ 1+\frac{zf''(z)}{f'(z)}+\frac{zf'(z)}{f(z)} \biggr\} > \frac{n-(n+1)\lambda}{n-\lambda}+\frac{(n+1)(n-\lambda)}{n(n+1)-\lambda } \quad (z\in \mathbb{U}). $$
(2.21)
The result is sharp.

Declarations

Acknowledgements

This work was partially supported by the National Natural Science Foundation of China (Grant Nos. 11171045; 11471163). The authors would like to express deep appreciation to Professor Shigeyoshi Owa for enlightening discussions and help.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Changshu Institute of Technology, Changshu, Jiangsu, 215500, China
(2)
Department of Mathematics, Suzhou University, Suzhou, Jiangsu, 215006, China

References

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  3. Ponnusamy, S, Singh, V: Criteria for strongly starlike functions. Complex Var. Theory Appl. 34, 267-291 (1997) View ArticleMathSciNetGoogle Scholar
  4. Hallenbeck, DJ, Ruscheweyh, S: Subordination by convex functions. Proc. Am. Math. Soc. 52, 191-195 (1975) View ArticleMATHMathSciNetGoogle Scholar

Copyright

© Xu and Yang; licensee Springer 2015

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