Starlikeness of meromorphic functions involving certain differential inequalities

1. Introduction

Let \(\Sigma\) be the class of functions of the form

\[f(z)=\frac{1}{z}+\sum_0^\infty a_nz^n,\] which are analytic in the punctured unit disc \( \mathbb E_0=\mathbb E\setminus\{0\},\) where \(\mathbb E = \{z:|z|< 1\}. \) A function \(f \in \Sigma \) is said to be meromorphic starlike of order \(\alpha \) if \(f(z) \neq 0 \) for \(z \in \mathbb E_0 \) and \[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\alpha,\hspace{1.7cm}(\alpha< 1;z \in \mathbb E).\] The class of such functions is denoted by \(\mathcal {MS}^*(\alpha)\) and the class of meromorphic starlike functions is denoted by \(\mathcal {MS}^*=\mathcal {MS}^*(0)\).

In the theory of meromorphic functions, many authors have obtained different sufficient conditions for meromorphically starlike functions. Some of them are stated below:

Kargar et al., [1] proved the following results:

Theorem 1. Assume that \(f(z)\neq 0\) for \(\mathbb E_0.\) If \(f\in\Sigma(p)\) satisfies \[\left|\frac{1}{\sqrt[p]{ f(z)}} \left(\frac{f'(z)}{f(z)}+\right)+p\right| < p\lambda(\beta)|b(z)|, ~z\in\mathbb E_0,\] then \(f\) is a \(p\)-valently meromorphic strongly-starlike of order \(\beta\).

Theorem 2. Assume that \(f(z)\neq 0\) for \(\mathbb E_0.\) If \(f\in\Sigma\) satisfies \[\left|\left(\frac{f(z)}{z^{-\alpha}}\right)^\frac{1}{\alpha-1} \left(\frac{f'(z)}{f(z)}+\frac{\alpha}{z}\right)+1-\alpha\right| < \frac{2}{\sqrt 5}, ~z\in\mathbb E_0,\] then \(f\) is meromorphic starlike function of order \(\alpha\).

Goswami et al., [2] proved the following results:

Theorem 3. If \(f(z) \in \Sigma_p,n\) with \(f(z)\neq 0\) for all \(z\in\mathbb E_0\), satisfies the following inequality \[\left|\displaystyle [z^p f(z)]^{\frac{1}{\alpha-p}}\left(\frac{z f'(z)}{f(z)}+\alpha\right)+p-\alpha\right|< \frac{(n+1)(p-\alpha)}{\sqrt{(n+1)^2+1}}, z\in\mathbb E, \] for some real values of \(\alpha~(0\leq\alpha< p)\), then \(f\in\mathcal{MS}_{p,n}^*(\alpha).\)

Theorem 4.If \(f(z) \in \Sigma_p,n\) with \(f(z)\neq 0\) for all \(z\in\mathbb E_0\) satisfies the following inequality \[\left|\frac{ \gamma [z^p f(z)]^\gamma}{z}\left(\frac{z f'(z)}{f(z)}+p\right)\right| \leq\frac{(n+1)}{2\sqrt{(n+1)^2+1}}, z\in\mathbb E, \] for \(\gamma\leq-\displaystyle\frac{1}{p}\), then \(f\in\mathcal{MS}_{p,n}^*\left(p+\displaystyle\frac{1}{\gamma}\right).\)

In [3], Sahoo et al., investigated a new class \(\mathcal{U}_n(\alpha,\lambda,\mu),\) of non-Bazilevic analytic functions by

\[\mathcal{U}_n(\alpha,\lambda,\mu)=\left\{f\in\mathcal {A}_n:\left|(1-\alpha)\left(\frac{z}{f(z)}\right)^\mu+\alpha f'(z)\left(\frac{z}{f(z)}\right)^{\mu+1}-1\right|< \lambda, ~z\in\mathbb E\right\}.\] For different choices of \(\mu\) with \(\alpha=1\), many authors has studied this class which are included in [4,5,6]. In this paper, we define above class of non-Bazilevic functions in punctured unit disk and study a differential inequality to obtain certain new criteria for starlikeness of meromorphic functions.

2. Main results

To prove our main result, we shall make use of following lemma of Hallenback and Ruscheweyh [7].

Lemma 1. Let G be a convex function in \(\mathbb E\), with \(G(0)=a\) and let \(\gamma\) be a complex number, with \(\Re(\gamma)>0\). If \(F(z)=a+a_nz^n+a_{n+1}z^{n+1}+\ldots\) , is analytic in \(\mathbb E\) nd \(F\prec G\), then \[\frac{1}{z^\gamma} \int^z_0 F(w)w^{\gamma-1}dw\prec \frac{1}{nz^{\frac{\gamma}{n}}}\int^z_0 G(w)w^{\frac{\gamma}{n}-1}dw .\]

Theorem 5. Let \(\alpha, \beta, \delta\) be real numbers such that \(\displaystyle \alpha< \frac{2}{\delta-1},\) \(\beta>0\), \(0\leq\delta< 1\) and let

If \(f\in\Sigma\) satisfies the differential inequality then \[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\delta, ~z\in\mathbb E.\]

Proof. Let us define \[\left(\frac{1}{zf(z)}\right)^\beta=u(z),~z\in\mathbb E.\] Differentiate logarithmically, we obtain

Therefore, in view of (3), we have The use of Lemma 1 \(\left(taking~ \displaystyle\gamma =-\frac{\beta}{\alpha}\right)\) in (4) gives \[u(z)\prec1+\frac{\gamma Mz}{\gamma+1},\] or \[ |u(z)-1|< \frac{\beta M}{\beta-\alpha}< 1,\] therefore, we obtain Write \(\displaystyle -\frac{zf'(z)}{f(z)}=(1-\delta)w(z)+\delta,\) \(0\leq\delta< 1\) and therefore (2) reduces to \[|(1+\alpha) u(z)-\alpha u(z) [(1-\delta)w(z)+\delta]-1|< M.\] We need to show that \(\Re (w(z))>0, ~z\in\mathbb E.\) If possible, suppose that \(\Re (w(z))\ngtr 0, ~z\in\mathbb E,\) then there must exist a point \(z_0\in\mathbb E\) such that \(w(z_0)=ix,x\in\mathbb R.\) To prove the required result, it is now sufficient to prove that By making use of (3), we have Now (7) is true in view of (1) and therefore, (6) holds. Hence \(\Re(w(z))>0\), i.e., \[\displaystyle -\Re\left(\frac{zf'(z)}{f(z)}\right)>\delta, ~0\leq\delta< 1, ~z\in\mathbb E.\]

Remark 1. Let \(\alpha, \beta, \delta\) be real numbers such that \(\displaystyle \alpha< \frac{2}{\delta-1},\) \(0\leq\delta< 1\), \(\beta>0\) and if \( f(z)\in\Sigma\) satisfies \[\left|\left(\frac{1}{zf(z)}\right)^\beta \left(\frac{1}{\alpha}+1+\frac{zf'(z)}{f(z)}\right)-\frac{1}{\alpha}\right|< \displaystyle \frac{(\beta-\alpha)[2+\alpha(1-\delta)]}{\alpha^2[1+\beta(1-\delta)]},\] then \[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\delta, ~z\in\mathbb E.\]

Letting \(\alpha\rightarrow\infty\) in above remark, we get the following result:

Theorem 6. Let \(\beta, \delta\) be real numbers such that \(\beta>0, 0\leq\delta< 1\) and let \(f(z)\in\Sigma\) satisfy \[\left|\left(\frac{1}{zf(z)}\right)^\beta \left(1+\frac{zf'(z)}{f(z)}\right)\right|< \displaystyle \frac{1-\delta}{1+\beta(1-\delta)}\] then \[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\delta, ~z\in\mathbb E.\]

3. Deductions

Setting \(\beta=1\) in Theorem 5, we obtain

Corollary 1. Let \(\alpha\) and \(\delta\) be real numbers such that \(\displaystyle \alpha< \frac{2}{\delta-1},\) \(0\leq\delta< 1\) and suppose that \(f\in\Sigma\) satisfies \[\displaystyle \left|\frac{1}{zf(z)} \left(1+\alpha+\alpha\frac{zf'(z)}{f(z)}\right)-1\right|< \frac{(1-\alpha)(2+\alpha(1-\delta))}{\alpha(2-\delta)},~z\in\mathbb E,\] then \[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\delta~, z\in\mathbb E,\] i.e., \(f\in\mathcal{MS}^*(\delta),~z\in\mathbb E.\)

Writing \(\delta=0\) in above corollary, we get the following result:

Corollary 2. Let \(f\in\Sigma\) satisfy \[\displaystyle\left|\frac{1}{zf(z)} \left(1+\alpha+\alpha\frac{zf'(z)}{f(z)}\right)-1\right|< \frac{(1-\alpha)(2+\alpha)}{2\alpha},~z\in\mathbb E,\] then \(f\in\mathcal{MS}^*,~z\in\mathbb E\).

Setting \(\beta=1\) in Theorem 6, we get the following result:

Corollary 3. Let \(\delta\) be a real number such that \( 0\leq\delta< 1\) and let \(f(z)\in\Sigma\) satisfy \[\left|\frac{1}{zf(z)} \left(1+\frac{zf'(z)}{f(z)}\right)\right|< \displaystyle \frac{1-\delta}{2-\delta}\] then \[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\delta, ~z\in\mathbb E.\]

Setting \(\delta=0\) in above corollary, we get the following result:

Corollary 4. Let \(f(z)\in\Sigma\) satisfy \[\left|\frac{1}{zf(z)} \left(1+\frac{zf'(z)}{f(z)}\right)\right|< \displaystyle \frac{1}{2}\] then \(f\in\mathcal{MS}^*,~z\in\mathbb E\).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.