数理解析研究所講究録
第 1824 巻 2013 年 14-20

14New Family of Integral Operators of Meromorphic Functions

Aabed Mohammed1and Maslina Darus2
1,2School of Mathematical Sceinces,

Faculty of Science and Technology,
Universiti Kebangsaan Malaysia
43600 Bangi, Selangor D. Ehsan,
Malaysia
1 aabedukm@yahoo.com

2maslina@ukm.myAbstract. We define here an integral operator $I_{n}(f_{i},g_{i})(z)$ for meromorphic functionsin the punctured open unit disk. Some properties for this operator are derived.

Keywords: analytic function, meromorphic function,starlike function, convex function,
integral operator.

AMS Mathematics Subject Classification: $30C45$

2-corresponding author

1LetIntroduction$\Sigma$denote the class of functions of the fonn
$f(z)= \frac{1}{z}+\sum_{na_{-}}^{\infty}a_{m}z^{n}$

(1.1)

,

which are analytic in the punctured open unit disk
$U^{*}=\{z\in \mathbb{C}:0<|z|<1\}=U\backslash \{0\}$where$\mathbb{U}$

(1.2)

,

is the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}.$

We say that a function $f\in\Sigma$ is meromorphic starlike of order $\delta(0\leq\delta<1)$ , and, if it satisfies the inequality
belongs to the class$\Sigma^{\star}(\delta)$$- \mathfrak{R}(\frac{zf’(z)}{f(z)})>\delta$

.

(1.3)

A function
is a meromorphic convex function of order $\delta(0\leq\delta<1)$ , if
the following inequality$f\in\Sigma$$- \Re(1+\frac{zf"(z)}{f(z)})>\delta$

and we denote this class by$\Sigma_{k}(\delta)$

.

,

$f$

satisfies(1.4)

15, Wang et al. [13] (see also [14]) introduced and studied the subclassFor
consisting of functions $f(z)$ satisfying$f\in\Sigma$$\Sigma_{N}(\lambda)$of$\Sigma$$- \Re(\frac{zf"(z)}{f^{l}(z)}+1)<\lambda (\lambda>1, z\in \mathbb{U})$.

In the literature, several integral operators of meromorphic functions in the punctured
open unit disk have been investigated and studied by many authors (cf., e.g., [1-11]).

For $i=1,2,$
operator$\cdots,$ $n,$$c>0$ , and:as follows$\alpha_{i},$$I_{n}(f_{i}, g_{i})(z)$$\gamma_{i}\geq 0$, we now, introduce a generalized integral$\Sigma^{n}arrow\Sigma$,$I_{n}(f_{i},g_{i})(z)= \frac{c}{z^{c+1}}\int_{0}^{z}u^{c-1}\prod_{i=1}^{n}(uf_{i}(u))^{\alpha}i(-u^{2}g_{i’}(u))^{\gamma_{i}}du$where$f_{i},$$g_{i}\in\Sigma$. Indeed, by varying the parameters$c,$$\alpha_{i}$and$\gamma_{i}$, the operator(1.5)$I_{n}(f_{i}, g_{i})$reduces to the following well-known integral operators.
(i) for$\gamma_{i}=0$, we obtain the integral operator$H(z)=I_{n}(f_{i})(z)= \frac{c}{z^{c+1}}\int_{0}^{z}u^{c-1}\prod_{i=1}^{n}(uf_{i}(u))^{\alpha t}du$

,

(1.6)

introduced by Frasin [8].
(ii)For $c=1$ and$\gamma_{i}=0$, we obtain the integral operator$\mathcal{H}_{n}(z)=I_{n}(f_{i})(z)=\frac{1}{z^{2}}\int_{0}^{z}\prod_{i=1}^{n}(uf_{i}(u))^{\circ:}du$

,

(1.7)

introduced by Mohammed and Dams [9].
(iii)For $c=1$ and$\alpha_{i}=0$, we obtain the integral operator,$\mathcal{H}_{\gamma_{1},\ldots,\gamma_{n}}(z)=I_{n}(g_{i})(z)=\frac{1}{z^{2}}\int_{0}^{z}\prod_{i=1}^{n}(-u^{2}g_{i’}(u))^{\gamma:}du$introduced by Mohammed and Darus [10](iv)If $n=1,$$\alpha_{1}=1,$$f_{1}=f$and$\gamma_{1}=0$

we have the integral operator$I_{c}(f)(z)= \frac{c}{z^{c+1}}\int_{0}^{z}u^{c-1}f(u)du,$

(1.8)

16
which was studied by many authors (cf., e.g., [1, 2, 6]).
For the starlikeness of the integral operator
lowing Lemma.
Lemma 1.1([12]).Suppose that the function
dition:$I_{n}(f_{i},g_{i})$$\Psi$:, we have to recall here the fol-satisfies the following con-$\mathbb{C}^{2}arrow \mathbb{C}$$\Re\{\Psi(is, t)\}\leq 0, (s, t\in \mathcal{R};t\leq\frac{-(1+s^{2})}{2})$If the function $p(z)=1+p_{1}z+\ldots$ , is analytic in$\mathbb{U}$

and$\Re\{\Psi(p(x), zp’(x))\}>0, (z\in \mathbb{U})$then$\Re\{p(z)\}>0 (z\in \mathbb{U})$2.

,.

Main Results

In the next theorem, we place conditions for the meromorphically starlikeness of the inwhich is defined in (1.5).
tegral operator$I_{n}(f_{i},g_{i})(z)$Theorem 2.1. For $i=1,2,$$g_{i}\in\Sigma_{k}$, and$\sum_{i=1}^{n}(\alpha_{i}+\gamma_{1})=1$

and let $c>0$ . Ifbelongs to, then the geneml integral operator$\ldots,$$n$, let$f_{i},g_{i}\in\Sigma,$$\alpha_{i},$$f_{i}\in\Sigma^{*},$$\gamma_{i}\geq 0$$I_{n}(f_{i},g_{i})(z)$the meromophic starlike function class.

Proof. From (1.5) it follows that$z^{2}I_{n}’(f_{i},g_{i})(z)+(c+1)zI_{n}(f_{:},g_{t})(z)=c \prod_{arrow-1}^{n}(zf_{t}(z))^{\alpha}(-z^{2}g_{1}’(z))^{\gamma_{i}}$(2.1)

Differentiating both sides of (2.1) logarithmically and multiplying by , we obtain$z$$\frac{z^{2}I_{n}"(f_{i},g_{i})(z)+(c+3)zI_{n}’(f_{i},g_{i})(z)+(c+1)I_{n}(f_{i},g_{i})(z)}{zI_{n}’(f_{i},g_{i})(z)+(c+1)I_{n}(f_{t},g_{i})(z)}$$= \sum_{:=1}^{n}\alpha_{i}\frac{zf_{t’}(z)}{f.\cdot(z)}+\sum_{1=1}^{n}\gamma_{i}(\frac{zg_{1}"(z)}{g_{i’}(z)}+1)+\sum_{1=1}^{n}\alpha_{1}+\sum_{i=1}^{n}\gamma_{1}$Which is equivalent to$- \frac{z^{2}I_{n}"(f_{i},g_{1}\cdot)(z)+(c+2)zI_{n}’(f_{1}\cdot,g_{i})(z)}{zI_{n}’(f_{t},g_{i})(z)+(c+1)I_{n}(f_{i},g_{i})(z)}$

.

(2.2)

17
$= \sum_{i=1}^{tl}\alpha_{i}(-\frac{zf_{i’}(z)}{f_{i}(z)})+\sum_{i=1}^{n}\gamma_{i}(-\frac{zg_{i"}(z)}{g_{i}(z)}-1)+1-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$

.

(2.3)

.

(2.4)

We can write (2.3) as the following$= \sum_{i=1}^{n}\alpha_{i}(-\frac{zf_{i^{l}}(z)}{f_{i}(z)})+\sum_{i=1}^{n}\gamma_{i}(-\frac{zg_{i"}(z)}{g_{i}(z)}-1)+1-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$We define the regular function

$p$

in

$\mathbb{U}$by$p(z)=- \frac{zI_{n}’(f_{i},g_{i})(z)}{I_{n}(f_{i},g_{i})(z)}$

,

(2.5)

and $p(O)=1$ . Differentiating $p(z)$ logarithmically, we obtain$-p(z)+ \frac{zp’(z)}{p(z)}=1+\frac{zI_{n}"(f_{i},g_{i})(z)}{I_{n}’(f_{i},g_{i})(z)}$

.

(2.6)

From (2.4),(2.5) and (2.6) we obtain$p(z)+ \frac{zp’(z)}{-p(z)+c+1}=\sum_{i=1}^{n}\alpha_{i}(-\frac{zf_{i}(z)’}{f_{i}(z)})+\sum_{i=1}^{n}\gamma_{i}(-\frac{zg_{i"}(z)}{g_{1}\cdot(z)}-1)+1-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$

.

(2.7)

Let us put$\Psi(u, v)=u+\frac{v}{-u+c+1}$.

(2.8)

From the hyposithes of Theorem 2.1, (2.7) and (2.8) we obtain$\Re\{\Psi(p(z), zp’(z))\}=\sum_{i=1}^{n}\alpha_{i}(-\Re\frac{zf_{i}’(z)}{f_{i}(z)})+\sum_{i=1}^{n}\gamma_{i}\{\Re(-\frac{zg_{i"}(z)}{g_{i}(z)}-1)\}+1-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$$>1- \sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})=0$

.

(2.9)

Now we proceed to show that$\Re\{\Psi(is, t)\}\leq 0, (s, t\in \mathcal{R};t\leq\frac{-(1+s^{2})}{2})$.

Indeed, from (2.8), we have. (2.10)$\Re\{\Psi(is, t)\}=\Re\{is+\frac{t}{-is+c+1}\}=\frac{t(c+1)}{s^{2}+(c+1)^{2}}\leq-\frac{(1+s^{2})(c+1)}{2[s^{2}+(c+1)^{2}]}<0$

Thus,from (2.9),(2.10) and by using Lemma 1.1, we conclude that $\Re\{p(z)\}>0$ , and so$- \Re\{\frac{zI_{n}’(f_{i},g_{i})(z)}{I_{n}(f_{\dot{\iota}},g_{i})(z)}\}>0.$

18that is$I_{n}(f_{1},g_{i})(z)$is starlike.

Next, we place conditions for the integral operator

Theorem 2.2. For $i=1,2,$, and$\ldots,$

$n$

, let

$f_{i},$$g_{i}\in\Sigma,$$\alpha_{i},$$I_{n}(f_{i},g_{i})$$\gamma\geq 0$

to be in the classand let $c>0$ . If$\Sigma_{N}(\lambda)$$f_{1}\in\Sigma^{*}(\delta)$.
,$g_{i}\in\Sigma_{k}(\delta)$$\sum_{1=1}^{n}(\alpha_{*}\cdot+\gamma_{i})>\frac{c+1}{1-\delta}$

then$I_{n}(f_{i}, g_{i})(z)\in\Sigma_{N}(\lambda),$,(2.11)$\lambda>1.$

Proof. Equivalently, (2.3) can be written as

Therefore.

$+ \sum_{1=1}^{n}\alpha_{i}(-\frac{zf_{l}’(x)}{f.\cdot(z)})+\sum_{1=1}^{n}\gamma_{i}(-\ovalbox{\tt\small REJECT}_{g_{i}z)}^{:^{\prime/_{z}}}-1)+c+2-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$(2.13)Taking real part of both sides of (2.13), we obtain$- \sum_{1=1}^{n}(\alpha_{i}+\gamma_{i})]\}+\sum_{i=i}^{n}\alpha_{i}(-\mathfrak{R}\#_{z)}^{z_{i}’z})+\sum_{1=1}^{n}\gamma_{1}\Re(-9’oe"\mu_{z)}^{z}-1)+c+2$$- \sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$$\leq|\frac{(c+1)I_{n}f_{i}.’g_{i})(z\rangle}{zI_{\mathfrak{n}}(fg.)(z)}!,[\sum_{arrow-1}^{n}\alpha_{i}(-\frac{zf_{i}’(z)}{f_{i}(z)})+\sum_{=1}^{n}\gamma_{i}(-\ovalbox{\tt\small REJECT}’g_{i}\lrcorner’(z\lrcorner-1)+1$$- \sum_{i=1}^{n}(\alpha_{1}+\gamma_{1})]|+\sum_{i=1}^{n}\alpha_{i}(-\Re_{f(z)}^{\lrcorner_{\llcorner}’uz}z)+\sum_{1=1}^{n}\gamma_{i}\Re(-\ovalbox{\tt\small REJECT}_{9^{\underline{l"}\coprod z}(z)}-1)+c+2$$- \sum_{i=1}^{n}(\alpha_{1}+\gamma_{i})$.
(2.14)

19
Let$\lambda=|\frac{(c+1)I_{n}(f_{4},g_{i})(z)}{zI_{n}(f_{i},g.\cdot)(z)}[\sum_{i=1}^{n}\alpha_{i}(-\frac{zf_{l}’(z)}{f.(z)})+\sum_{i=1}^{n}\gamma_{i}(-\frac{zg"(z)}{g’(z)}-1)+1-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})]|$.

$+ \sum_{i=1}^{n}\alpha_{i}(-\Re\frac{zf_{i’}(z)}{f_{i}(z)})+\sum_{i=1}^{n}\gamma_{i}\Re(-\frac{zg_{1}"(z)}{g_{i}(z)}-1)+c+2-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$Since$| \frac{(c+1)I_{n}(f_{i,}g.)(z)}{zI_{n}(f.,g.\cdot)(z\rangle}[\sum_{i=1}^{n}\alpha_{i}(-\frac{zf_{i}’(z)}{f_{*}(z)})+\sum_{\iota=1}^{n}\gamma_{i}(-\frac{zg_{i"}(z)}{g_{*}^{l}(z)}-1)+1-\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})]|>0,$ $f_{i}\in$$\Sigma^{\star}(\delta),$$g_{i}\in\Sigma_{k}(\delta)$, then we have$\lambda>c+2-(1-\delta)\sum_{i=1}^{n}(\alpha_{i}+\gamma_{i})$Then, by the hypothesis (2.11), we have

If we set$\gamma_{i}=0$$\lambda>1$. Therefore,.

$I_{n}(f_{i},g_{i})(z)\in\Sigma_{N}(\lambda),$$\lambda>1.$

in Theorem 2.2, then we have [8, Theorem 2.6].

Further, Putting $c=1,$$\gamma_{i}=0$

Corollary 2.3. For $i=1,2,$

in Theorem 2.2, we get$\ldots,$

$n$

, let

$f_{i}\in\Sigma,$$\alpha_{i}\geq 0$. If$f_{i}\in\Sigma^{\star}(\delta)$, and$\sum_{i=1}^{n}\alpha_{i}>\frac{2}{1-\delta},$then$\mathcal{H}_{n}(z)\in\Sigma_{N}(\lambda),$$\lambda>1.$

In addition, taking $c=1,$Corollary 2.4. For $i=1,2,$$\alpha_{i}=0$$\ldots,$$n$in Theorem 2.2, we receive, let$g_{i}\in\Sigma,$$\gamma_{i}\geq 0$. If$g_{i}\in\Sigma_{k}(\delta)$, and$\sum_{i=1}^{n}\gamma_{i}>\frac{2}{1-\delta},$then$\mathcal{H}_{\gamma_{1},\ldots,\gamma_{n}}(z)\in\Sigma_{N}(\lambda),$$\lambda>1.$

Acknowledgement:
The work here is supported by UKM-DLP-2011-050 and LRGS $/TD/2011/UKM/ICT/03/02.$

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