Revisiting spontaneous silver nanoparticles formation: a factor influencing the determination of minimum inhibitory concentration values?

Karolína M. Šišková; Renáta Večeřová; Hana Kubičková; Magdaléna Bryksová; Klára Čépe; Milan Kolář; Karolína M. Šišková; Renáta Večeřová; Hana Kubičková; Magdaléna Bryksová; Klára Čépe; Milan Kolář

doi:10.3934/environsci.2015.3.607

AIMS Environmental Science

2015, Volume 2, Issue 3: 607-622. doi: 10.3934/environsci.2015.3.607

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Research article

Revisiting spontaneous silver nanoparticles formation: a factor influencing the determination of minimum inhibitory concentration values?

1.
Faculty of Science, Palacký University in Olomouc, Šlechtitelů 11, 78371 Olomouc, Czech Republic;
2.
Department of Microbiology, Faculty of Medicine and Dentistry, Palacký University in Olomouc, Hněvotínská 3, 77146 Olomouc, Czech Republic

Received: 06 May 2015 Accepted: 17 June 2015 Published: 23 June 2015

The present study gives evidence that silver nanoparticles (AgNPs) are spontaneously formed from Ag⁺ ions in Mueller-Hinton broth, which is frequently used as a standard cultivation medium for many types of bacteria. Silver ions often serve as a reference in the determination of minimum inhibitory concentration (MIC) values of engineered AgNPs. It is thus a question if the MIC values determined for engineered AgNPs are not influenced by the presence of spontaneously formed AgNPs. Furthermore, as shown here, the addition of augmented concentrations of selected amino acids, namely glutamic acid and glutamine, can change the growth and characteristic features of spontaneously formed AgNPs. For the sake of a direct comparison, the influence of the two selected amino acids on characteristics and MIC values determination of engineered AgNPs has been also investigated. The determined MIC values of all investigated systems (i.e., with and without the presence of engineered AgNPs) and their mutual comparison demonstrated that MIC values are slightly influenced by the actual composition of a cultivation medium for bacterial growth. On the other hand, the actual composition of a cultivation medium is crucial for the final characteristics of AgNPs. The changes in characteristic features of spontaneously formed as well as engineered AgNPs are most probably induced by the covalent bonding of amino acids to AgNPs surface which is proven by vibrational spectroscopic techniques.

Keywords:

Citation: Karolína M. Šišková, Renáta Večeřová, Hana Kubičková, Magdaléna Bryksová, Klára Čépe, Milan Kolář. Revisiting spontaneous silver nanoparticles formation: a factor influencing the determination of minimum inhibitory concentration values?[J]. AIMS Environmental Science, 2015, 2(3): 607-622. doi: 10.3934/environsci.2015.3.607

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Abstract

1. Introduction and definitions

Let $ \mathfrak{A} $ denote the family of all functions $ f $ which are analytic in the open unit disc $ \mathcal{U} = \left \{ z:\left \vert z\right \vert < 1\right \} $ and satisfying the normalization

$\begin{equation} f(z) = z+\sum\limits_{n = 2}^{\infty } a_{n}z^{n}, \end{equation}$ $

(1.1)

while by $ \mathcal{S} $ we mean the class of all functions in $ \mathfrak{A} $ which are univalent in $ \mathcal{U}. $ Also let $ \mathcal{S}^{\ast } $ and $ \mathcal{C} $ denote the familiar classes of starlike and convex functions, respectively. If $ f $ and $ g $ are analytic functions in $ \mathcal{U }, $ then we say that $ f $ is subordinate to $ g, $ denoted by $ f\prec g, $ if there exists an analytic Schwarz function $ w $ in $ \mathcal{U} $ with $ w\left(0\right) = 0 $ and $ \left \vert w(z)\right \vert < 1 $ such that $ f(z) = g\left(w(z)\right). $ Moreover if the function $ g $ is univalent in $ \mathcal{U} $, then

$\begin{equation*} f\left( z\right) \prec g\left( z\right) \Leftrightarrow f(0) = g(0)\text{ and } f(\mathcal{U})\subset g(\mathcal{U}). \end{equation*}$ $

For arbitrary fixed numbers $ A, $ $ B $ and $ b $ such that $ A $, $ B $ are real with $ -1\leq B < A\leq 1 $ and $ b\in \mathbb{C} \backslash \{0\} $, let $ \mathcal{P}[b, A, B] $ denote the family of functions

$\begin{equation} p\left( z\right) = 1+\sum\limits_{n = 1}^{\infty} p_{n}\, z^{n}, \end{equation}$ $

(1.2)

analytic in $ \mathcal{U} $ such that

$\begin{equation*} 1+\frac{1}{b}\left \{ p(z)-1\right \} \prec \frac{1+Az}{1+Bz}. \end{equation*}$ $

Then, $ p\in \mathcal{P}[b, A, B] $ can be written in terms of the Schwarz function $ w $ by

$\begin{equation*} p(z) = \frac{b\left( 1+Aw(z)\right) +(1-b)\left( 1+Bw(z)\right) }{1+Bw(z)}. \end{equation*}$ $

By taking $ b = 1-\sigma $ with $ 0\leq \sigma < 1, $ the class $ \mathcal{P} [b, A, B] $ coincides with $ \mathcal{P}[\sigma, A, B], $ defined by Polatoğ lu ^[17,18] (see also ^[2]) and if we take $ b = 1, $ then $ \mathcal{ P}[b, A, B] $ reduces to the familiar class $ \mathcal{P}[A, B] $ defined by Janowski ^[10]. Also by taking $ A = 1, $ $ B = -1 $ and $ b = 1 $ in $ \mathcal{P} [b, A, B] $, we get the most valuable and familiar set $ \mathcal{P} $ of functions having positive real part. Let $ \mathcal{S}^{\ast }[A, B, b] $ denote the class of univalent functions $ g $ of the form

$\begin{equation} g(z) = z+\sum \limits_{n = 2}^{\infty }b_{n}z^{n}, \end{equation}$ $

(1.3)

in $ \mathcal{U} $ such that

$\begin{equation*} 1 + \frac{1}{b} \left \{ \frac{zg^{\prime }(z)}{g(z)} -1\right \} \prec \frac{1+Az}{1+Bz}, \text{ }-1\leq B \lt A\leq 1, \quad z\in \mathcal{U}. \end{equation*}$ $

Then $ S^* [A, B] : = S^* [A, B, 1] $ and the subclass $ \mathcal{S}^{\ast}[1,-1,1] $ coincides with the usual class of starlike functions.

The set of Bazilevič functions in $ \mathcal{U} $ was first introduced by Bazilevič ^[7] in 1955. He defined the Bazilevič function by the relation

$\begin{equation*} f(z) = \left \{ \left( \alpha +i\beta \right) \int \limits_{0}^{z}g^{\alpha }(t)p(t)t^{i\beta -1}dt\right \} ^{\frac{1}{\alpha +i\beta }}, \end{equation*}$ $

where $ p\in \mathcal{P}, $ $ g\in \mathcal{S}^{\ast }, $ $ \beta $ is real and $ \alpha > 0 $. In 1979, Campbell and Pearce ^[8] generalized the Bazilevi č functions by means of the differential equation

$\begin{equation*} 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}+\left( \alpha +i\beta -1\right) \frac{zf^{\prime }(z)}{f(z)} = \alpha \frac{zg^{\prime }(z)}{g(z)}+ \frac{zp^{\prime }(z)}{p(z)}+i\beta , \end{equation*}$ $

where $ \alpha +i\beta \in \mathbb{C} -\left \{ \text{negative integers}\right \}. $ They associate each generalized Bazilevič functions with the quadruple $ (\alpha, \beta, g, p). $

Now we define the following subclass.

Definition 1.1. Let $ g $ be in the class $ \mathcal{S}^{\ast }[A, B] $ and let $ p\in \mathcal{P} [b, A, B] $. Then a function $ f $ of the form $ \left(1.1 \right) $ is said to belong to the class of generalized Bazilevič function associated with the quadruple $ (\alpha, \beta, g, p) $ if $ f $ satisfies the differential equation

where $ \alpha + i\beta \in \mathbb{C} - \{{\text {negative integers}} \} $.

The above differential equation can equivalently be written as

$\begin{equation*} \frac{zf^{\prime }(z)}{f(z)} = \left( \frac{g(z)}{z}\right) ^{\alpha }\left( \frac{z}{f(z)}\right) ^{\alpha +i\beta }p(z), \end{equation*}$ $

$\begin{equation*} \frac{z^{1-i\beta }f^{\prime }(z)}{f^{1-(\alpha +i\beta )}g^{\alpha }(z)} = p(z), \text{ }z\in \mathcal{U}. \end{equation*}$ $

Since $ p\in \mathcal{P}[b, A, B], $ it follows that

$\begin{equation*} 1+\frac{1}{b}\left \{ \frac{z^{1-i\beta }f^{\prime }(z)}{f^{1-(\alpha +i\beta )}g^{\alpha }(z)}-1\right \} \prec \frac{1+Az}{1+Bz}, \end{equation*}$ $

where $ g\in \mathcal{S}^{\ast }[A, B]. $

Several research papers have appeared recently on classes related to the Janowski functions, Bazilevič functions and their generalizations, see ^{[3,4,5,13,16,21,22]}.

2. Lemmas

The following are some results that would be useful in proving the main results.

Lemma 2.1. Let $ p\in \mathcal{P}[b, A, B] $ with $ b\neq 0, $ $ -1\leq B < A\leq 1, $ and has the form $ \left(1.2\right). $ Then for $ z = re^{i\theta }, $

$\begin{equation*} \frac{1}{2\pi }\int_{0}^{2\pi }\left \vert p(re^{i\theta })\right \vert ^{2}d\theta \leq \frac{1+\left[ \left \vert b\right \vert ^{2}(A-B)^{2}-1 \right] r^{2}}{1-r^{2}}. \end{equation*}$ $

Proof. The proof of this lemma is straightforward but we include it for the sake of completeness. Since $ p\in \mathcal{P}[b, A, B] $, we have

$\begin{equation*} p(z) = b\tilde{p}(z)+(1-b), \quad \tilde{p}\in \mathcal{P}[A, B]. \end{equation*}$ $

Let $ \tilde{p}(z) = 1+\sum_{n = 1}^{\infty }c_{n}\, z^{n} $. Then

$\begin{equation*} 1+\sum \limits_{n = 1}^{\infty }p_{n}\, z^{n} = b\left( 1+\sum \limits_{n = 1}^{\infty }c_{n}\, z^{n}\right) +(1-b). \end{equation*}$ $

Comparing the coefficients of $ z^{n} $, we have

$\begin{equation*} p_{n} = bc_{n}. \end{equation*}$ $

Since $ \left \vert c_{n}\right \vert \leq A-B $ ^[20], it follows that $ \left \vert p_{n}\right \vert \leq \left \vert b\right \vert (A-B) $ and so

$\begin{align*} \frac{1}{2\pi }\int_{0}^{2\pi }\left \vert p(re^{i\theta })\right \vert ^{2}d\theta & = \frac{1}{2\pi }\int_{0}^{2\pi }\left \vert \sum \limits_{n = 0}^{\infty }p_{n}r^{n}e^{in\theta }\right \vert ^{2}d\theta \\ & = \frac{1}{2\pi }\int_{0}^{2\pi }\left( \sum\limits_{n = 0}^{\infty }\left \vert p_{n}\right \vert ^{2}r^{2n}\right) d\theta \\ & = \sum\limits_{n = 0}^{\infty }\left \vert p_{n}\right \vert ^{2}r^{2n} \\ &\leq 1+\left \vert b\right \vert ^{2}(A-B)^{2}\sum\limits_{n = 1}^{\infty }r^{2n} \\ & = 1+\left \vert b\right \vert ^{2}(A-B)^{2}\frac{r^{2}}{1-r^{2}} \\ & = \frac{1+\left( \left \vert b\right \vert ^{2}(A-B)^{2}-1\right) r^{2}}{ 1-r^{2}}. \end{align*}$ $

Thus the proof is complete.

Lemma 2.2. ^[1] Let $ \Omega $ be the family of analytic functions $ \omega $ on $ \mathcal{U} $, normalized by $ \omega (0) = 0 $, satisfying the condition $ \left \vert \omega (z)\right \vert < 1 $. If $ \omega \in \Omega $ and

$\begin{equation*} \omega (z) = \omega _{1}z + \omega _{2}z^{2} + \cdots, \quad \left( z\in \mathcal{U}\right) , \end{equation*}$ $

then for any complex number $ t $,

$\begin{equation*} \left \vert \omega _{2}-t\omega _{1}^{2}\right \vert \leq \max \left \{ 1, \left \vert t\right \vert \right \}. \end{equation*}$ $

The above inequality is sharp for $ \omega (z) = z $ or $ \omega (z) = z^{2} $.

Lemma 2.3. Let $ p(z) = 1+\sum \limits_{n = 1}^{\infty }p_{n}z^{n}\in \mathcal{P}[b, A, B], $ $ b\in \mathbb{C}\backslash \{0\}, $ $ -1\leq B < A\leq 1. $ Then for any complex number $ \mu $,

$\begin{align*} \left \vert p_{2}-\mu p_{1}^{2}\right \vert &\leq \left \vert b\right \vert (A-B)\max \left \{ 1, \left \vert \mu b(A-B)+B\right \vert \right \} \\ & = \begin{cases} \left \vert b\right \vert \left( A-B\right), & {\mathit{\text{if}}}\; \left \vert \mu b\left( A-B\right) +B\right \vert \leq 1, \\ \left \vert b\right \vert \left( A-B\right) \left \vert \mu b\left( A-B\right) +B\right \vert, & {\mathit{\text{if}}}\; \left \vert \mu b\left( A-B\right) +B\right \vert \geq 1. \end{cases} \end{align*}$ $

This result is sharp.

Proof. Let $ p\in \mathcal{P}[b, A, B] $. Then we have

$\begin{equation*} 1+\frac{1}{b}\left \{ p(z)-1\right \} \prec \frac{1+Az}{1+Bz}, \end{equation*}$ $

or, equivalently

$\begin{equation*} p(z)\prec \frac{1+\left[ bA+(1-b)B\right] z}{1+Bz} = 1+b(A-B)\sum \limits_{n = 1}^{\infty }(-B)^{n-1}z^{n}, \end{equation*}$ $

which would further give

$\begin{align*} 1+p_{1}z+p_{2}z^{2}+\cdots & = 1+b(A-B)\omega (z)+b(A-B)(-B)\omega ^{2}(z)+\cdots \\ & = 1+b(A-B)\left( \omega _{1}z+\omega _{2}z^{2}+\cdots \right) \\ &{\ } \qquad +b(A-B)(-B)\left( \omega _{1}z+\omega _{2}z^{2}+\cdots \right) ^{2}+\cdots \\ & = 1+b(A-B)\omega _{1}z+b(A-B)\left \{ \omega _{2}-B\omega _{1}^{2}\right \} z^{2}+\cdots. \end{align*}$ $

Comparing the coefficients of $ z $ and $ z^{2} $, we obtain

$\begin{align*} p_{1} & = b(A-B)\omega _{1} \\ p_{2} & = b(A-B)\omega _{2}-b(A-B)B\omega _{1}^{2}. \end{align*}$ $

By a simple computation,

$\begin{equation*} \left \vert p_{2}-\mu p_{1}^{2}\right \vert = \left \vert b\right \vert (A-B)\left \vert \omega _{2}-\left( \mu b(A-B)+B\right) \omega _{1}^{2}\right \vert . \end{equation*}$ $

Now by using Lemma 2.2 with $ t = \mu b(A-B)+B $, we get the required result. Equality holds for the functions

$\begin{align*} p_{\circ}(z) & = \frac{1+\left( bA+(1-b)B\right) z^{2}}{1+Bz^{2}} = 1+b(A-B)z^{2}+b(A-B)(-B)z^{4}+\cdots, \\ p_{1}(z) & = \frac{1+\left( bA+(1-b)B\right) z}{1+Bz} = 1+b(A-B)z+b(A-B)(-B)z^{2}+\cdots. \end{align*}$ $

Now we prove the following result by using a method similar to the one in Libera ^[12].

Lemma 2.4. Suppose that $ N $ and $ D $ are analytic in $ \mathcal{U} $ with $ N(0) = D(0) = 0 $ and $ D $ maps $ \mathcal{U} $ onto a many sheeted region which is starlike with respect to the origin. If $ \frac{N^{\prime }(z)}{D^{\prime }(z) }\in \mathcal{P}[b, A, B] $, then

$\begin{equation*} \frac{N(z)}{D(z)}\in \mathcal{P}[b, A, B]. \end{equation*}$ $

Proof. Let $ \frac{N^{\prime }(z)}{D^{\prime }(z)}\in \mathcal{P}[b, A, B]. $ Then by using a result due to Attiya ^[6], we have

$\begin{equation*} \left \vert \frac{N^{\prime }(z)}{D^{\prime }(z)}-c(r)\right \vert \leq d\left( r\right) , \qquad \left \vert z\right \vert \lt r, \quad 0 \lt r \lt 1, \end{equation*}$ $

where $ c(r) = \frac{1-B\left[B+b\left(A-B\right) \right] r^{2}}{1-B^{2}r^{2}} $ and $ d\left(r\right) = \frac{\left \vert b\right \vert \left(A-B\right) r^{2}}{1-B^{2}r^{2}}. $ We choose $ A\left(z\right) $ such that $ \left \vert A\left(z\right) \right \vert < d\left(r\right) $ and

$\begin{equation*} A\left( z\right) D^{\prime }(z) = N^{\prime }(z)-c\left( r\right) D^{\prime }(z). \end{equation*}$ $

Now for a fixed $ z_{0} $ in $ \mathcal{U} $, consider the line segment $ L $ joining $ 0 $ and $ D\left(z_{0}\right) $ which remains in one sheet of the starlike image of $ \mathcal{U} $ by $ D. $ Suppose that $ L^{-1} $ is the pre-image of $ L $ under $ D $. Then

$\begin{align*} \big \vert N(z_{0})-c\left( r\right) D(z_{0})\big \vert & = \left \vert \int_{0}^{z_0}\Big( N^{\prime}(t)-c(r) D^{\prime }(t)\Big) dt\right \vert \\ & = \left \vert \int_{L^{-1}} A(t) D^{\prime}(t)\, dt\right \vert \\ & \lt d(r) \int_{L^{-1}}\left \vert dD(t)\right \vert \\ & = d(r) D(z_{0}). \end{align*}$ $

This implies that

$\begin{equation*} \left \vert \frac{N(z_{0})}{D(z_{0})}-c\left( r\right) \right \vert \lt d\left( r\right) . \end{equation*}$ $

Therefore

$\begin{equation*} \frac{N(z)}{D(z)}\in \mathcal{P}[b, A, B]. \end{equation*}$ $

For $ A = -B = b = 1 $, we have the following result due to Libera ^[12].

Lemma 2.5. If $ N $ and $ D $ are analytic in $ \mathcal{U} $ with $ N(0) = D(0) = 0 $ and $ D $ maps $ \mathcal{U} $ onto a many sheeted region which is starlike with respect to the origin, then

$\begin{equation*} \frac{N^{\prime }(z)}{D^{\prime }(z)}\in \mathcal{P}\mathit{\text{ implies }}\frac{ N(z)}{D(z)}\in \mathcal{P}. \end{equation*}$ $

Lemma 2.6. ^[14] If $ -1\leq B < A\leq 1, \beta _{1} > 0 $ and the complex number $ \gamma $ satisfies $ {Re}\left \{ \gamma \right \} \geq -\beta_{1}\left(1-A\right) /\left(1-B\right), $ then the differential equation

$\begin{equation*} q\left( z\right) +\frac{zq^{\prime }\left( z\right) }{\beta _{1}q\left( z\right) +\gamma } = \frac{1+Az}{1+Bz}, \quad z\in \mathcal{U}, \end{equation*}$ $

has a univalent solution in $ \mathcal{U} $ given by

$\begin{equation*} q\left( z\right) = \begin{cases} \frac{z^{\beta_{1} + \gamma}\left( 1+Bz\right)^{\beta_{1}\left( A-B\right)/B} }{\beta_{1} \int_{0}^{z} t^{\beta_{1}+\gamma -1}\left( 1+Bt\right)^{\beta_{1}\left( A-B\right) /B}dt} - \frac{\gamma }{\beta _{1}}, & B\neq 0, \\ \frac{z^{\beta_{1}+\gamma}\, e^{\beta_{1}Az}}{\beta_{1} \int_{0}^{z} t^{\beta_{1}+\gamma -1}\, e^{\beta_{1}At} dt} - \frac{\gamma}{\beta_{1}}, & B = 0. \end{cases} \end{equation*}$ $

If $ p\left(z\right) = 1+p_{1}z+p_{2}z^{2}+\cdots $ is analytic in $ \mathcal{U} $ and satisfies

$\begin{equation*} p\left( z\right) +\frac{zp^{\prime }\left( z\right) }{\beta _{1}p\left( z\right) +\gamma }\prec \frac{1+Az}{1+Bz}, \end{equation*}$ $

then

$\begin{equation*} p\left( z\right) \prec q\left( z\right) \prec \frac{1+Az}{1+Bz}, \end{equation*}$ $

and $ q\left(z\right) $ is the best dominant.

3. Some auxiliary results

Before proving the results for the generalized Bazilevič functions, let us discuss a few results related to the function $ g \in S^*[A, B] $.

Theorem 3.1. Let $ g\in \mathcal{S}^{\ast }[A, B] $ and of the form $ \left(1.3\right) $. Then for any complex number $ \mu, $

$\begin{equation*} \left \vert b_{3}-\mu b_{2}^{2}\right \vert \leq \frac{(A-B)}{2}\max \left \{ 1, \; \left \vert 2(A-B)\mu -(A-2B))\right \vert \right \} . \end{equation*}$ $

Proof. The proof of the result is the same as of Lemma 2.3. The result is sharp and equality holds for the function defined by

$\begin{equation*} g_{1}(z) = \begin{cases} z(1+Bz^{2})^{\frac{A-B}{2B}} = z+\frac{1}{2}\left( A-B\right) z^{3}+\cdots, & B\neq 0, \\ ze^{\frac{A}{2}z^{2}} = z+\frac{A}{2}z^{3}+\cdots , & B = 0, \end{cases} \end{equation*}$ $

$\begin{align*} g_{2}(z) & = \begin{cases} z(1+Bz)^{\frac{A-B}{B}}, & B\neq 0, \\ ze^{Az}, & B = 0, \end{cases} \\ & = \begin{cases} z+\left( A-B\right) z^{2}+\frac{1}{2}\left( A-B\right) \left( A-2B\right) z^{3}+\cdots , & B\neq 0, \\ z+Az^{2}+\frac{1}{2}A^{2}z^{3}+\cdots , & B = 0. \end{cases} \end{align*}$ $

Theorem 3.2. Let $ g\in \mathcal{S}^{\ast }[A, B]. $ Then for $ c > 0, $ $ \alpha > 0 $ and $ \beta $ any real number,

$\begin{equation} G^{\alpha }(z) = \frac{c+\alpha +i\beta }{z^{c+i\beta }}\int_{0}^{z}t^{c+i \beta -1}g^{\alpha }(t)\, dt, \end{equation}$ $

(3.1)

is in $ \mathcal{S}^{\ast }[A, B] $. In addition

$\begin{equation*} {Re}\frac{zG^{\prime }(z)}{G(z)} \gt \delta = \underset{\left \vert z\right \vert = 1}{\min}\, {Re}\, q(z), \end{equation*}$ $

where

$\begin{equation*} q(z) = \begin{cases} \frac{1}{\alpha }\frac{\alpha +i\beta +c}{_{2}F_{1}\left( 1; \;\alpha \left( 1-\frac{A}{B}\right); \;\alpha +i\beta +c+1; \;\frac{Bz}{ 1+Bz}\right) } -\left( c+i\beta \right), & B\neq 0, \\ \frac{1}{\alpha }\frac{\alpha +i\beta +c}{_{1}F_{1}\left( 1; \;\alpha +i\beta +c+1; \; -\alpha Az\right) }-\left( c+i\beta \right) , & B = 0. \end{cases} \end{equation*}$ $

Proof. From $ \left(3.1\right), $ we have

$\begin{equation*} z^{c+i\beta }G^{\alpha }(z) = \left( c+\alpha +i\beta \right) \int_{0}^{z}t^{c+i\beta -1}g^{\alpha }(t)dt. \end{equation*}$ $

Differentiating and rearranging gives

$\begin{equation} \left( c+\alpha +i\beta \right) \frac{g^{\alpha }(z)}{G^{\alpha }(z)} = (c+i\beta )+\alpha p(z), \end{equation}$ $

(3.2)

where $ p(z) = \frac{zG^{\prime }(z)}{G(z)} $. Then differentiating $ \left(3.2\right) $ logarithmically, we have

$\begin{equation*} \frac{zg^{\prime }(z)}{g(z)} = p(z)+\frac{zp^{\prime }(z)}{\alpha p(z)+(c+i\beta )}. \end{equation*}$ $

Since $ g\in \mathcal{S}^{\ast }[A, B] $, it follows that

$\begin{equation*} p(z)+\frac{zp^{\prime }(z)}{\alpha p(z)+(c+i\beta )}\prec \frac{1+Az}{1+Bz}. \end{equation*}$ $

Now by using Lemma 2.6, for $ \beta _{1} = \alpha $ and $ \gamma = c+i\beta, $ we obtain

$\begin{equation*} p(z)\prec q(z)\prec \frac{1+Az}{1+Bz}, \end{equation*}$ $

where

$\begin{equation*} q(z) = \begin{cases} \frac{z^{c+\alpha +i\beta }\left( 1+Bz\right)^{\alpha \left( A-B\right) /B}}{\alpha {\int_{0}^{z} }t^{c+\alpha +i\beta -1}\left( 1+Bt\right)^{\alpha \left( A-B\right) /B}dt} -\frac{c+i\beta }{\alpha }, & B\neq 0, \\ \frac{z^{c+\alpha +i\beta }e^{\alpha Az}} {\alpha \int_{0}^{z} t^{c+\alpha +i\beta -1}e^{\alpha At}dt} -\frac{c+i\beta }{\alpha }, & B = 0. \end{cases} \end{equation*}$ $

Now by using the properties of the familiar hypergeometric functions proved in ^[15], we have

$\begin{equation*} q(z) = \begin{cases} \frac{1}{\alpha} \frac{\alpha +i\beta +c}{_{2}F_{1}\left( 1; \;\alpha \left( 1-\frac{A}{B }\right) ; \;\alpha +i\beta +c+1; \;\frac{Bz}{1+Bz}\right)} -\left( c+i\beta \right), & B \neq 0, \\ \frac{\alpha +i\beta +c}{_{1}F_{1}\left( 1; \;\alpha +i\beta +c+1; \;-\alpha Az\right)} -\left( c+i\beta \right), & B = 0. \end{cases} \end{equation*}$ $

This implies that

$\begin{equation*} p(z)\prec q(z) = \begin{cases} \frac{1}{\alpha} \frac{\alpha +i\beta +c}{_{2}F_{1}\left( 1; \;\alpha \left( 1-\frac{A}{B }\right); \;\alpha +i\beta +c+1; \;\frac{Bz}{1+Bz}\right)} - \left( c+i\beta \right) , & B\neq 0, \\ \frac{1}{\alpha} \frac{\alpha +i\beta +c}{_{1}F_{1}\left( 1; \;\alpha +i\beta +c+1; \; -\alpha Az\right) } -\left( c+i\beta \right) , & B = 0, \end{cases} \end{equation*}$ $

and

$\begin{equation*} {Re}\frac{zG^{\prime }(z)}{G(z)} = {Re}\, p(z) \gt \delta = \underset{ \left \vert z\right \vert = 1}{\min }\; {Re}\, q(z). \end{equation*}$ $

Theorem 3.3. Let $ g\in \mathcal{S}^{\ast }[A, B]. $ Then

$\begin{equation*} S(z) = \int_{0}^{z}t^{c+i\beta -1}g^{\alpha }(t)\, dt, \end{equation*}$ $

is $ (\alpha +c) $-valent starlike, where $ \alpha > 0, $ $ c > 0 $ and $ \beta $ is a real number.

Proof. Let $ D_{1}(z) = zS^{\prime }(z) = z^{c+i\beta }g^{\alpha }(z) $ and $ N_{1}(z) = S(z). $ Then

$\begin{align*} {Re}\frac{zD_{1}^{\prime }(z)}{D_{1}(z)} & = {Re}\left \{ (c+i\beta) +\alpha \frac{zg^{\prime }(z)}{g(z)}\right \} \\ & = c+\alpha \, {Re}\frac{zg^{\prime }(z)}{g(z)}. \end{align*}$ $

Since $ g\in \mathcal{S}^{\ast }[A, B]\subset \mathcal{S}^{\ast }\left(\frac{ 1-A}{1-B}\right) $, see ^[10], it follows that

$\begin{equation*} {Re}\frac{zD_{1}^{\prime }(z)}{D_{1}(z)} \gt c+\alpha \left( \frac{1-A}{1-B} \right) \gt 0. \end{equation*}$ $

Also

$\begin{equation*} {Re}\frac{D_{1}^{\prime }(z)}{N_{1}^{\prime }(z)} = {Re}\left \{ (c+i\beta )+\alpha \frac{zg^{\prime }(z)}{g(z)}\right \} \gt c+\alpha \left( \frac{1-A}{ 1-B}\right) \gt 0. \end{equation*}$ $

Now by using Lemma 2.5$, $ we have

$\begin{equation*} {Re}\frac{D_{1}(z)}{N_{1}(z)} \gt 0\text{ or }{Re}\frac{zS^{\prime }(z)}{S(z)} \gt 0. \end{equation*}$ $

By the mean value theorem for harmonic functions,

$\begin{equation*} \left. {Re}\frac{zS^{\prime }(z)}{S(z)}\right|_{z = 0} = \frac{1}{2\pi } \int_{0}^{2\pi} {Re}\frac{re^{i\theta }S^{\prime }(re^{i\theta })}{ S(re^{i\theta })}\, d\theta. \end{equation*}$ $

Therefore

$\begin{align*} \int_{0}^{2\pi } {Re}\frac{re^{i\theta }S^{\prime }(re^{i\theta })}{ S(re^{i\theta })}\, d\theta & = \left. 2\pi {Re}\left \{ c+i\beta +\alpha \frac{zg^{^{\prime }}(z)}{g(z)}\right \} \right| _{z = 0} \\ & = 2\pi (c+\alpha ). \end{align*}$ $

Now by using a result due to [9,p 212], we have that $ S $ is $ (c+\alpha) $-valent starlike function.

4. Main results

Now we are ready to discuss some results related to the defined generalized Bazilevič functions.

Theorem 4.1. Let $ f $ be a generalized Bazilevič function associated by the quadruple $ \left(\alpha, \beta, g, p\right), $ where $ g\in \mathcal{S} ^{\ast }[A, B] $ of the form $ \left(1.3\right) $ and $ p\in \mathcal{P} [b, A, B] $ of the form $ \left(1.2\right) $. Then for $ c > 0, $

$\begin{equation} F(z) = \left[ \frac{c+\alpha +i\beta }{z^{c}}\int_{0}^{z}t^{c-1}f^{\alpha +i\beta }(t)dt\right] ^{\frac{1}{\alpha +i\beta }} \end{equation}$ $

(4.1)

is a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, G, p), $ where $ G\in \mathcal{S}^{\ast }[A, B, \delta] $, as defined by $ \left(3.1\right). $

Proof. From $ \left(4.1\right) $, we have

$\begin{equation*} F^{\alpha +i\beta }(z) = \frac{c+\alpha +i\beta }{z^{c}}\int_{0}^{z} t^{c-1}(f(t))^{\alpha +i\beta }dt. \end{equation*}$ $

This implies that

$\begin{equation*} z^{c}F^{\alpha +i\beta }(z) = \left( c+\alpha +i\beta \right) \int_{0}^{z}t^{c-1}(f(t))^{\alpha +i\beta }dt. \end{equation*}$ $

Differentiate both sides and rearrange, we get

$\begin{equation*} cz^{c-1}F^{\alpha +i\beta }(z)+(\alpha +i\beta )z^{c}F^{\alpha +i\beta -1}(z)F^{\prime }(z) = \left( c+\alpha +i\beta \right) z^{c-1}(f(z))^{\alpha +i\beta }, \end{equation*}$ $

and

$\begin{equation*} \frac{z^{1-i\beta }F^{\prime }(z)}{F^{1-(\alpha +i\beta )}\left( z\right) } = \frac{1}{\alpha +i\beta }\left \{ (c+\alpha +i\beta )z^{-i\beta }f^{\alpha +i\beta }(z)-cz^{-i\beta }F^{\alpha +i\beta }(z)\right \} . \end{equation*}$ $

Now from $ (3.1) $, we have

$\begin{align*} & \quad \frac{z^{1-i\beta }F^{\prime }(z)}{F^{1-(\alpha +i\beta )}G^{\alpha }(z)} \\ & = \frac{\frac{1}{\alpha +i\beta }\left \{ (c+\alpha +i\beta )z^{-i\beta }f^{\alpha +i\beta }(z)-cz^{-i\beta -c}(c+\alpha +i\beta )\int_{0}^{z}t^{c-1}(f(t))^{\alpha +i\beta }dt\right \} }{\frac{(c+\alpha +i\beta )}{z^{c+i\beta }}\int_{0}^{z}t^{c+i\beta -1}g^{\alpha }(t)dt} \\ & = \frac{\frac{1}{\alpha +i\beta }\left \{ (z^{c}f^{\alpha +i\beta }(z)-c\int_{0}^{z}t^{c-1}(f(t))^{\alpha +i\beta }dt\right \} }{ \int_{0}^{z}t^{c+i\beta -1} g^{\alpha }(t)dt} \\ &: = \frac{N(z)}{D(z)}. \end{align*}$ $

With this, note that

$\begin{align*} \frac{N^{\prime }(z)}{D^{\prime }(z)} & = \frac{\frac{1}{\alpha +i\beta } \left \{ (cz^{c-1}f^{\alpha +i\beta }(z) + (\alpha +i\beta) z^c f^{\alpha +i\beta -1}(z) f^{\prime }(z) - cz^{c-1}(f(z))^{\alpha +i\beta }\right \} }{ z^{c+i\beta -1} g^{\alpha }(z)} \\ & = \frac{z^{1-i\beta }f^{\prime }(z)}{f^{1-(\alpha +i\beta )}(z)g^{\alpha }(z)}, \end{align*}$ $

which implies $ \frac{N^{\prime }(z)}{D^{\prime }(z)} \in \mathcal{P}[b, A, B] $. By Theorem 3.3, we know that $ D(z) = \int_{0}^{z}t^{c+i\beta -1}g^{\alpha }(t)dt $ is $ (\alpha +c) $-valent starlike. Therefore by using Lemma 2.4, we obtain

$\begin{equation*} \frac{z^{1-i\beta }F^{\prime }(z)}{F^{1-(\alpha +i\beta )}(z)G^{\alpha }(z)} \in \mathcal{P}[b, A, B]. \end{equation*}$ $

This is the equivalent form of Definition 1.1. Hence the result follows.

Corollary 4.2. Let $ A = 1, B = -1 $ and $ \beta = 0 $ in Theorem 4.1. Then

$\begin{equation*} G^{\alpha }(z) = \frac{(\alpha +c)}{z^{c}}\int_{0}^{z}t^{c-1}g^{\alpha }(t)dt \end{equation*}$ $

belong to $ S^{\ast }\left(\delta _{1}\right) $, where

$\begin{equation*} \delta _{1} = \frac{-(1+2c)+\sqrt{(1+2c)^{2}+8\alpha }}{4\alpha }, (\mathit{\text{see [16]}}). \end{equation*}$ $

Hence $ G $ is starlike when $ g\in \mathcal{S}^{\ast }, $ and

$\begin{equation*} F^{\alpha }(z) = \frac{(\alpha +c)}{z^{c}}\int_{0}^{z}t^{c-1}g^{\alpha }(t)dt \end{equation*}$ $

belongs to the class of Bazilevič functions associated by the quadruple $ (\alpha, 0, G, p). $

Theorem 4.3. Let $ f $ of the form $ \left(1.1\right) $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, g, p) $, with $ g\in \mathcal{S}^{\ast }[A, B] $ of the form $ \left(1.3\right) $ and $ p\in \mathcal{P}[b, A, B] $ of the form $ \left(1.2\right) $. Then

$\begin{equation*} \left \vert a_{3}-\frac{3+\alpha +i\beta }{2(2+\alpha +i\beta )} a_{2}^{2}\right \vert \leq \frac{A-B}{2\left \vert 2+\alpha +i\beta \right \vert }\left[ \alpha +\left \vert b\right \vert \max \left \{ 2, \, \left \vert b(A-B)+2B\right \vert \right \} \right]. \end{equation*}$ $

This inequality is sharp.

Proof. Since $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, g, p) $, we have

$\begin{equation} 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}+(\alpha +i\beta -1)\frac{ zf^{\prime }(z)}{f(z)} = \alpha \frac{zg^{\prime }(z)}{g(z)}+\frac{zp^{\prime }(z)}{p(z)}+i\beta . \end{equation}$ $

(4.2)

As $ f, \ g $ and $ p $ respectively have the form $ \left(1.1\right), \left(1.3\right) $ and $ \left(1.2\right), $ it is easy to get

$\begin{align*} 1+\frac{zf^{\prime \prime }(z)}{f^{\prime }(z)} & = 1+2a_{2}z+\left( 6a_{3}-4a_{2}^{2}\right) z^{2}+\cdots, \\ \frac{zf^{\prime }(z)}{f(z)} & = 1+a_{2}z+\left( 2a_{3}-a_{2}^{2}\right) z^{2}+\cdots, \\ \frac{zg^{\prime }(z)}{g(z)} & = 1+b_{2}z+\left( 2b_{3}-b_{2}^{2}\right) z^{2}+\cdots, \\ \frac{zp^{\prime }(z)}{p(z)} & = p_{1}z+\left( 2p_{2}-p_{1}^{2}\right) z^{2}+\cdots. \end{align*}$ $

Putting these values in $ \left(4.2\right) $ and comparing the coefficients of $ z, $ we obtain

$\begin{equation} (1+\alpha +i\beta )a_{2} = \alpha b_{2}+p_{1}. \end{equation}$ $

(4.3)

Similarly by comparing the coefficients of $ z^{2} $ and rearranging, we have

$\begin{equation} 2(2+\alpha +i\beta )a_{3} = \alpha (2b_{3}-b_{2}^{2})+2p_{2}-p_{1}^{2}+a_{2}^{2}(3+\alpha +i\beta ). \end{equation}$ $

(4.4)

From $ \left(4.4\right) $, we have

$\begin{align*} \left \vert a_{3}-\frac{3+\alpha +i\beta }{2(2+\alpha +i\beta )} a_{2}^{2}\right \vert & = \left \vert \frac{\alpha \left( b_{3}-\frac{1}{2} b_{2}^{2}\right) +(p_{2}-\frac{1}{2}p_{1}^{2})}{2+\alpha +i\beta }\right \vert \\ &\leq \frac{\alpha \left \vert b_{3}-\frac{1}{2}b_{2}^{2}\right \vert }{ \left \vert 2+\alpha +i\beta \right \vert }+\frac{\left \vert p_{2}-\frac{1}{ 2}p_{1}^{2}\right \vert }{\left \vert 2+\alpha +i\beta \right \vert }. \end{align*}$ $

Now by using Theorem 3.1 and Lemma 2.3, both with $ \mu = \frac{1}{2} $, we obtain

$\begin{equation*} \left \vert b_{3}-\frac{1}{2}b_{2}^{2}\right \vert \leq \frac{A-B}{2}\max \{1, |B|\} = \frac{A-B}{2}, \end{equation*}$ $

and

$\begin{equation*} \left \vert p_{2}-\frac{1}{2}p_{1}^{2}\right \vert \leq \left \vert b\right \vert (A-B)\max \left \{ 1, \frac{1}{2}\big \vert b(A-B)+2B\big \vert \right \} . \end{equation*}$ $

Therefore, we have

$\begin{equation*} \left \vert a_{3}-\frac{3+\alpha +i\beta }{2(2+\alpha +i\beta )} a_{2}^{2}\right \vert \leq \frac{A-B}{2\left \vert 2+\alpha +i\beta \right \vert }\left[ \alpha + 2\left \vert b\right \vert \max \left \{ 1, \, \frac{1 }{2}\big \vert b(A-B)+2B\big \vert \right \} \right]. \end{equation*}$ $

The equality

$\begin{equation*} \left \vert b_{3}-\frac{1}{2}b_{2}^{2}\right \vert = \frac{A-B}{2} \end{equation*}$ $

for $ B\neq 0 $ can be obtained for

$\begin{equation*} g(z) = \begin{cases} z(1+Bz)^{\frac{A-B}{B}} = z+\left( A-B\right) z^{2}+\frac{1}{2}\left( A-B\right) \left( A-2B\right) z^{3}+\cdots, \\ z(1+Bz^{2})^{\frac{A-B}{2B}} = z+\frac{1}{2}\left( A-B\right) z^{3}+\cdots. \end{cases} \end{equation*}$ $

Similarly, the equality

$\begin{equation*} \left \vert b_{3}-\frac{1}{2}b_{2}^{2}\right \vert = \frac{A}{2} \end{equation*}$ $

for $ B = 0 $ can be obtained for the function $ g_{\ast }(z) = $ $ ze^{\frac{A}{2} z^{2}} = z+\frac{A}{2}z^{3}+\cdots. $ Also equality for the functional $ \left \vert p_{2}-\frac{1}{2}p_{1}^{2}\right \vert $ can be obtained by the functions

$\begin{equation*} p_{\circ }(z) = \frac{1+\left( bA+(1-b)B\right) z}{1+Bz}\; \text{ or } \;p_{1}(z) = \frac{ 1+\left( bA+(1-b)B\right) z^{2}}{1+Bz^{2}}. \end{equation*}$ $

Corollary 4.4. For $ A = 1, $ $ B = -1 $ and $ b = 1 $, we have the result proved in ^[8]:

$\begin{equation*} \left \vert a_{3}-\frac{3+\alpha +i\beta }{2(2+\alpha +i\beta )} a_{2}^{2}\right \vert \leq \frac{\alpha +2}{\left \vert 2+\alpha +i\beta \right \vert }. \end{equation*}$ $

For $ \alpha = 1 $, $ \beta = 0 $, we have $ f\in \mathcal{K} $, the class of close-to-convex functions, and

$\begin{equation*} \left \vert a_{3}-\frac{2}{3}a_{2}^{2}\right \vert \leq 1. \end{equation*}$ $

The latter result has been proved in ^[11].

Theorem 4.5. Let $ f $ of the form $ \left(1.1\right) $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, \beta, g, p) $, with $ g\in \mathcal{S}^{\ast }[A, B] $ and of the form $ \left(1.3\right) $ and $ p\in \mathcal{P}[b, A, B] $ of the form $ \left(1.2\right) $. Then

(i)

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{(A-B)(\alpha +\left \vert b\right \vert )}{\left \vert 1+\alpha +i\beta \right \vert }. \end{equation*}$ $

(ii) If $ \alpha = 0, $ then

$\begin{equation*} \left \vert a_{3}\right \vert \leq \frac{\left \vert b\right \vert (A-B)}{ \left \vert 2+i\beta \right \vert }\max \left \{ 1, \left \vert \frac{b(A-B)}{ 2}\left( 1-\frac{(3+i\beta )}{\left( 1+i\beta \right) ^{2}}\right) +B\right \vert \right \} . \end{equation*}$ $

Both the above inequalities are sharp.

Proof. (ⅰ) From $ \left(4.3\right) $, we have

$\begin{equation*} (1+\alpha +i\beta )a_{2} = \alpha b_{2}+p_{1}. \end{equation*}$ $

This implies that

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{\alpha \left \vert b_{2}\right \vert +\left \vert p_{1}\right \vert }{\left \vert 1+\alpha +i\beta \right \vert }. \end{equation*}$ $

By using the coefficient bound for $ \mathcal{S}^{\ast }[A, B] $ along with the coefficient bound of $ \mathcal{P}[b, A, B] $, we have

$\begin{equation*} \left \vert b_{2}\right \vert \leq A-B\text{ and }\left \vert p_{1}\right \vert \leq \left \vert b\right \vert (A-B). \end{equation*}$ $

This implies that

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{\left( \alpha +\left \vert b\right \vert \right) (A-B)}{\left \vert 1+\alpha +i\beta \right \vert }. \end{equation*}$ $

Equality can be obtained by the functions

$\begin{equation*} g_{\circ }(z) = z(1+Bz)^{\frac{A-B}{B}}, \text{ }B\neq 0\text{ and }p_{\circ }(z) = \frac{1+[bA+(1-b)B]z}{1+Bz}. \end{equation*}$ $

(ⅱ) Let $ \alpha = 0. $ Then from $ \left(4.3\right) $ and $ \left(4.4 \right) $, we have

$\begin{align*} (2+i\beta )a_{3} & = p_{2}-\frac{1}{2}p_{1}^{2}+\frac{(3+i\beta )p_{1}^{2}}{ 2\left( 1+i\beta \right) ^{2}} \\ & = p_{2}-\frac{1}{2}\left( 1-\frac{(3+i\beta )}{\left( 1+i\beta \right) ^{2}} \right) p_{1}^{2}. \end{align*}$ $

This implies

$\begin{equation*} \left \vert a_{3}\right \vert = \frac{1}{\left \vert 2+i\beta \right \vert } \left \vert p_{2}-\mu p_{1}^{2}\right \vert , \end{equation*}$ $

where $ \mu = \frac{1}{2}-\frac{(3+i\beta)}{2\left(1+i\beta \right) ^{2}}. $ Now by using Lemma 2.3, we obtain

$\begin{equation*} \left \vert a_{3}\right \vert \leq \frac{\left \vert b\right \vert (A-B)}{ \left \vert 2+i\beta \right \vert }\max \left \{ 1, \left \vert \frac{b(A-B)}{ 2}\left( \frac{-2-\beta ^{2}+i\beta )}{\left( 1+i\beta \right) ^{2}}\right) +B\right \vert \right \} . \end{equation*}$ $

Sharpness can be attained by the functions

$\begin{align*} p_{0}(z) & = \frac{1+\left( bA+(1-b)B\right) z^{2}}{1+Bz^{2}} = 1+b(A-B)z^{2}+b(A-B)(-B)z^{4}+\cdots, \\ p_{1}(z) & = \frac{1+\left( bA+(1-b)B\right) z}{1+Bz} = 1+b(A-B)z+b(A-B)(-B)z^{2}+\cdots. \end{align*}$ $

Corollary 4.6. For $ A = 1, $ $ B = -1 $ and $ b = 1 $, we have

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{2\left( \alpha +1\right) }{\left \vert 1+\alpha +i\beta \right \vert }, \end{equation*}$ $

and

$\begin{equation*} \left \vert a_{3}\right \vert = \frac{2}{\left \vert 2+i\beta \right \vert } \max \left \{ 1, \left \vert \frac{(3+i\beta )}{\left( 1+i\beta \right) ^{2}} \right \vert \right \} . \end{equation*}$ $

In the final part of this paper, we look at some results for the generalized Bazilevič functions associated with $ \beta = 0 $.

Let $ C_{r} $ denote the closed curve which is the image of the circle $ \left \vert z\right \vert = r < 1 $ under the mapping $ w = $ $ f(z) $, and $ L_{r}\left(f(z)\right) $ denote the length of $ C_{r} $. Also let $ M(r) = {\max} _{\left \vert z\right \vert = r} \left \vert f(z)\right \vert \ $and $ m(r) = { \min}_{\left \vert z\right \vert = r} \left \vert f(z)\right \vert $. We now prove the following result.

Theorem 4.7. Let $ f $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $. Then for $ B\neq 0 $,

$\begin{equation*} L_{r}(f(z))\leq \begin{cases} C(\alpha, b, A, B)M^{1-\alpha }(r) \left[ 1 - (1-r) ^{\alpha \left( \frac{A-B}{B } \right)} \right], & 0 \lt \alpha \leq 1, \\ C(\alpha, b, A, B)m^{1-\alpha}(r) \left[ 1 - (1-r) ^{\alpha \left( \frac{A-B}{B} \right)} \right], & \alpha \gt 1, \end{cases} \end{equation*}$ $

where

$\begin{equation*} C(\alpha, b, A, B) = 2\pi |b| B \left[ (A-B) + \frac{1}{\alpha} \right]. \end{equation*}$ $

Proof. As $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $, we have

$\begin{equation*} zf^{\prime }(z) = f^{1-\alpha }(z)g^{\alpha }(z)p(z), \end{equation*}$ $

where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P}[b, A, B] $. Since for $ z = re^{i\theta } $, $ 0 < r < 1 $,

$\begin{equation*} L_{r}(f(z)) = \int_{0}^{2\pi }\left \vert zf^{\prime }(z)\right \vert d\theta , \end{equation*}$ $

we have for $ 0 < \alpha \leq 1 $,

$\begin{align*} & L_{r}(f(z)) \\ & = \int_{0}^{2\pi }\left \vert f^{1-\alpha }(z)g^{\alpha }(z)p(z)\right \vert d\theta, \\ &\leq M^{1-\alpha }(r)\int_{0}^{2\pi }\int_{0}^{r}\big \vert \alpha g^{\prime }\left( z\right) g^{\alpha - 1}\left( z\right) p\left( z\right) + g^{\alpha} \left( z\right) p^{\prime }\left( z\right)\big \vert ds\, d\theta, \\ &\leq M^{1-\alpha }(r)\left \{ \int_{0}^{2\pi }\int_{0}^{r}\frac{\alpha |g^{\alpha }(z)| }{s}\big \vert h\left( z\right) p(z)\big \vert ds\, d\theta + \int_{0}^{2\pi }\int_{0}^{r} \frac{|g^{\alpha }(z)|}{s}\big \vert zp^{\prime }(z)\big \vert ds\, d\theta \right \} , \end{align*}$ $

where $ \frac{zg^{\prime }(z)}{g(z)} = h\left(z\right) \in \mathcal{P}[A, B] $. Now by using the distortion theorem for Janowski starlike functions when $ B\neq 0 $ (see ^[10]) and the Cauchy-Schwarz inequality, we have

$\begin{multline*} L_{r}(f(z))\leq M^{1-\alpha }(r) \\ \times \int_{0}^{r}\frac{s^{\alpha -1}}{\left( 1-\left \vert B\right \vert s\right) ^{\alpha \frac{B-A}{B}}}\left \{ \alpha \sqrt{ \int_{0}^{2\pi }\left \vert h\left( z\right) \right \vert ^{2}d\theta} \sqrt{ \int_{0}^{2\pi }\left \vert p(z)\right \vert ^{2}d\theta} +\int_{0}^{2\pi }\left \vert zp^{\prime }(z)\right \vert d\theta \right \}ds. \end{multline*}$ $

Now by using Lemma 2.1 for both the classes $ \mathcal{P}[b, A, B] $ and $ \mathcal{P}[A, B] $, along with the result

$\begin{equation*} \int_0^{2\pi} \left \vert zp^{\prime }(z)\right \vert \leq \frac{\left \vert b\right \vert \left( A-B\right) r}{ 1 - B^{2}r^{2}}, \end{equation*}$ $

for $ p\in \mathcal{P}[b, A, B] $ (see ^[19]), we can write

$\begin{multline*} L_{r}(f(z)) \leq 2\pi M^{1-\alpha }(r) \\ \times \int_{0}^{r} \frac{s^{\alpha -1}}{\left( 1-\left \vert B\right \vert s\right) ^{\alpha \frac{B-A}{B}}} \left \{ \alpha \sqrt{ \frac{1+\left[ (A-B)^{2}-1\right] s^{2}}{1-s^{2}}} \sqrt{ \frac{1+\left[ \left \vert b\right \vert ^{2}(A-B)^{2}-1\right] s^{2}}{1-s^{2}}} \right. \\ + \frac{\left \vert b\right \vert \left( A-B\right) s}{ 1-B^{2}s^{2}} \Bigg \} ds. \end{multline*}$ $

Since $ 1-\left \vert B\right \vert r\geq 1-r $ and $ 1- B^2 r^2\geq 1-r^2 $,

$\begin{align*} L_{r}\left( f(z)\right) &\leq 2\pi M^{1-\alpha }(r) \left[ \left \vert b\right \vert \left( A-B\right) ^{2}+\left \vert b\right \vert \left( A-B\right) \right] \int_0^r \frac{1}{(1-s)^{\alpha \frac{B-A}{B}+1}} ds \\ & = C(\alpha, b, A, B)M^{1-\alpha }(r) \left[ 1 - (1-r) ^{\alpha \left( \frac{A-B }{B} \right)} \right], \end{align*}$ $

where $ C(\alpha, b, A, B) = 2\pi |b| [(A-B) + (1/ \alpha)] B $.

When $ \alpha > 1, $ we can prove similarly as above to get

$\begin{equation*} L_{r}\left( f(z)\right) \leq C(\alpha, b, A, B)m^{1- \alpha}(r) \left[ 1 - (1-r) ^{\alpha \left( \frac{A-B}{B} \right)} \right]. \end{equation*}$ $

Corollary 4.8. For $ g\in \mathcal{S}^{\ast } $ and $ p\in \mathcal{P}(b), $ we have

$\begin{equation*} L_{r}(f(z))\leq \begin{cases} 2\pi |b| \left( 2 + \frac{1}{\alpha}\right) M^{1-\alpha }(r) \left[ \frac{1}{ (1-r)^{2\alpha}} - 1\right], & 0 \lt \alpha \leq 1, \\ 2\pi |b| \left( 2 + \frac{1}{\alpha}\right) m^{1- \alpha}(r) \left[ \frac{1}{ (1-r)^{2\alpha}} - 1\right], & \alpha \gt 1. \end{cases} \end{equation*}$ $

Theorem 4.9. Let $ f $ be a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p) $, where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P} [b, A, B] $. Then for $ B\neq 0 $,

$\begin{equation*} \left \vert a_{n}\right \vert \leq \begin{cases} \frac{1}{n} |b| B \left( A-B + \frac{1}{\alpha}\right) \lim_{r \rightarrow 1^{-}} M^{1-\alpha }(r), & 0 \lt \alpha \leq 1, \\ \frac{1}{n} |b| B \left( A-B + \frac{1}{\alpha}\right) \lim_{r \rightarrow 1^{-}} m^{1-\alpha }(r), & \alpha \gt 1. \end{cases} \end{equation*}$ $

Proof. By Cauchy's theorem for $ z = re^{i\theta }, $ $ n\geq 2, $ we have

$\begin{equation*} na_{n} = \frac{1}{2\pi r^{n}}\int_{0}^{2\pi } zf^{\prime }(z)e^{-in\theta }d\theta . \end{equation*}$ $

Therefore

$\begin{align*} n\left \vert a_{n}\right \vert &\leq \frac{1}{2\pi r^{n}}\int_{0}^{2\pi }\left \vert zf^{\prime }(z)\right \vert d\theta , \\ & = \frac{1}{2\pi r^{n}} L_{r}(f(z)). \end{align*}$ $

By using Theorem 4.7 for the case $ 0 < \alpha \leq 1, $ we have

$\begin{equation*} n\left \vert a_{n}\right \vert \leq \frac{1}{2\pi r^{n}} \left( 2\pi |b| B \left( A-B + \frac{1}{\alpha} \right) M^{1-\alpha}(r) \left[ 1 - (1-r)^{\alpha \frac{A-B}{B}}\right] \right). \end{equation*}$ $

Hence, by taking $ r $ approaches $ 1^{-} $,

$\begin{equation*} \left \vert a_{n}\right \vert \leq \frac{1}{n} |b| B \left( A-B + \frac{1}{ \alpha}\right) \lim\limits_{r \rightarrow 1^{-}} M^{1-\alpha }(r). \end{equation*}$ $

For $ \alpha > 1, $ we have

$\begin{equation*} \left \vert a_{n}\right \vert \leq \frac{1}{n} |b| B \left( A-B + \frac{1}{ \alpha}\right) \lim\limits_{r \rightarrow 1^{-}} m^{1-\alpha }(r). \end{equation*}$ $

Theorem 4.10. Let $ f $ be a generalized Bazilevič function represented by the quadruple $ (\alpha, 0, g, p), $ where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P }[b, A, B]. $ Then for $ B\neq 0 $,

$\begin{align*} \left \vert f(z)\right \vert ^{\alpha }&\leq \alpha \frac{ (1-B^{2})+(A-B)\left( \left \vert b\right \vert -B\, Re(b)\right) }{1-B} r^{\alpha}\, _{2}F_{1}\left( \alpha \left( 1-\frac{A}{B}\right) +1;\alpha ;\alpha + 1;\left \vert B\right \vert r\right). \end{align*}$ $

Proof. Since $ f $ is a generalized Bazilevič function associated by the quadruple $ (\alpha, 0, g, p), $ by definition, we have

$\begin{equation*} \frac{zf^{\prime }(z)}{f^{1-\alpha }(z)g^{\alpha }(z)} = p(z), \end{equation*}$ $

where $ g\in \mathcal{S}^{\ast }[A, B] $ and $ p\in \mathcal{P}[b, A, B]. $ This implies that

$\begin{equation*} f^{\alpha }(z) = \alpha \int_{0}^{z}t^{-1}g^{\alpha }(t)p(t)dt, \end{equation*}$ $

and so

$\begin{align*} |f(z)|^{\alpha } &\leq \alpha \int_{0}^{|z|}|t^{-1}||g^{\alpha }(t)||p(t)|d|t|, \\ & = \alpha \int_{0}^{r} s^{-1}|g^{\alpha }(t)||p(t)|ds. \end{align*}$ $

Now by using the results

$\begin{equation*} |g(z)| \leq r(1-\left \vert B\right \vert r)^{\frac{A-B}{B}}, B\neq 0, \text{ (see [10]), } \end{equation*}$ $

and

$\begin{equation*} \left \vert p(z)\right \vert \leq \frac{1+\left \vert b\right \vert (A-B)r-B \left[ (A-B)Re\{b\}+B\right] r^{2}}{1-\left \vert B\right \vert ^{2}r^{2}}, \text{ (see [6]), } \end{equation*}$ $

we have

$\begin{align*} |f(z)|^{\alpha } &\leq \alpha \int_{0}^{r} s^{-1}\frac{s^{\alpha }}{(1-\left \vert B\right \vert s)^{\alpha \left( 1-\frac{A}{B}\right) }}\, \frac{ 1+|b|(A-B)s - B\left[ (A-B)Re\{b\}+B\right] s^{2}}{1-\left \vert B\right \vert ^{2}s^{2}}ds \\ &\leq \alpha \frac{(1-B^{2})+(A-B)\left( \left \vert b\right \vert -B Re\{b\} \right) }{{1+\left \vert B\right \vert} }\int_{0}^{r} s^{\alpha -1}(1-\left \vert B\right \vert s)^{-\alpha \left( 1-\frac{A}{B}\right) -1}ds. \end{align*}$ $

Putting $ s = ru, $ we have

$\begin{align*} |f(z)|^{\alpha } &\leq \alpha \frac{(1-B^{2})+(A-B) \left( \left \vert b\right \vert - B{Re\{b\}}\right) }{1-B}r^{\alpha }\int_{0}^{1}u^{\alpha -1}(1 - |B|ru)^{-\alpha \left( 1-\frac{A}{B}\right) -1}du \\ & = \frac{(1-B^{2})+(A-B)\left( \left \vert b\right \vert -BRe\{b\} \right) }{1-B}r^{\alpha }\text{ }_{2}F_{1}\left( \alpha \left( 1-\frac{A}{B}\right) +1;\alpha ;\alpha +1;\left \vert B\right \vert r\right), \end{align*}$ $

where $ {_{2}}F_{1} \left(a; b;c; z\right) $ is the hypergeometric function.

Corollary 4.11. For $ g\in \mathcal{S}^{\ast \mathit{\text{}}} $and $ p\in \mathcal{P}, $ we have

$\begin{equation*} |f(z)|^{\alpha }\leq 2\alpha r^{\alpha }\mathit{\text{}}_{2}F_{1}\left( 2\alpha +1;\alpha ;\alpha +1;r\right). \end{equation*}$ $

Acknowledgments

The research for the fourth author is supported by the USM Research University Individual Grant (RUI) 1001/PMATHS/8011038.

Conflict of interest

The authors declare that they have no conflict of interests.

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