Antibiotic residues in raw and pasteurized milk in Iran: A systematic review and meta-analysis

Behnam Bahramian; Mahmood Alizadeh Sani; Mohammad Parsa-Kondelaji; Hedayat Hosseini; Yousef Khaledian; Mitra Rezaie; Behnam Bahramian; Mahmood Alizadeh Sani; Mohammad Parsa-Kondelaji; Hedayat Hosseini; Yousef Khaledian; Mitra Rezaie

doi:10.3934/agrfood.2022031

AIMS Agriculture and Food

2022, Volume 7, Issue 3: 500-519. doi: 10.3934/agrfood.2022031

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Review

Antibiotic residues in raw and pasteurized milk in Iran: A systematic review and meta-analysis

1.
Department of Nutrition, Faculty of Medicine, Mashhad University of Medical Sciences, Mashhad, Iran
2.
Medical Toxicology Research Center, Mashhad University of Medical Sciences, Mashhad, Iran
3.
Division of Food Safety and Hygiene, Department of Environmental Health Engineering, School of Public Health, Tehran University of Medical Sciences, Tehran, Iran
4.
Department of Hematology and Blood Banking, Faculty of Medical Sciences, Mashhad University of Medical Sciences, Mashhad, Iran
5.
Department of Food Science and Technology, National Nutrition and Food Technology Research Institute, Faculty of Nutrition Sciences and Food Technology, Shahid Beheshti University of Medical Sciences, Tehran, Iran
6.
Food Safety Research Center, Shahid Beheshti University of Medical Sciences, Tehran, Iran
7.
Department of Food Hygiene and Quality Control, Faculty of Veterinary, Shahrekord University, Shahrekord, Iran

Received: 11 April 2022 Revised: 30 May 2022 Accepted: 20 June 2022 Published: 12 July 2022

Improper use of antibiotics to treat or prevent infections, and as a stimulant for livestock growth, can affect public health and the dairy industry due to the spread of antibiotic residues in milk. This systematic review and meta-analysis aimed to investigate antibiotic residues in raw and pasteurized milk in Iran. Data were collected through searching the databases, including Scopus, PubMed, Science Direct, Web of Science, Google Scholar, SID, and Magiran using the following keywords: "pasteurized milk", "raw milk", "milk", "antibiotic residues", " antibiotic", and "Iran". Finally, 40 eligible studies were selected for the systematic review and meta-analysis. According to the reviewed studies, the prevalence of antibiotic residues in raw and pasteurized milk was 26% (95% CI: 20–33%) and 21% (95% CI: 15-27%), respectively. It seems that the control measures are inefficient in dairy industry and milk collection centers for the presence of antibiotic residues and the time of antibiotic withdrawal during milk delivery. Permanent control of milk in the collection centers by the responsible organizations along with implementation of Hazard Analysis Critical Control Point system in milk factories can be very effective in reducing antibiotic residues.

Keywords:

Citation: Behnam Bahramian, Mahmood Alizadeh Sani, Mohammad Parsa-Kondelaji, Hedayat Hosseini, Yousef Khaledian, Mitra Rezaie. Antibiotic residues in raw and pasteurized milk in Iran: A systematic review and meta-analysis[J]. AIMS Agriculture and Food, 2022, 7(3): 500-519. doi: 10.3934/agrfood.2022031

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Abstract

1. Introduction

Let $\mathcal{A}$ denote the class of analytic functions $f$ with the normalized condition $f\left(0 \right) = f'\left(0 \right) - 1 = 0$ in an open unit disk $E = \left\{ {z \in \mathbb{C}:\left| z \right| < 1} \right\}$ . The function $f \in \mathcal{A}$ has the Taylor-Maclaurin series expansion given by

$\begin{equation} f\left( z \right) = z + \sum\limits_{n = 2}^\infty {{a_n}} {z^n},\;z \in E. \end{equation}$

(1.1)

Further, we denote $S$ the subclass of $\mathcal{A}$ consisting of univalent functions in ${E.}$

Let $H$ denote the class of Schwarz functions $\omega$ which are analytic in ${E}$ such that $\omega \left(0 \right) = 0$ and $\left| {\omega \left(z \right)} \right| < 1.$ The function $\omega\in H$ has the series expansion given by

$\begin{equation} \omega \left( z \right) = \sum\limits_{k = 1}^\infty {{b_k}} {z^k},\;z \in E. \end{equation}$

(1.2)

We say that the analytic function $f$ is subordinate to another analytic function $g$ in $E$ , expressed as $f \prec g,$ if and only if there exists a Schwarz function $\omega \in H$ such that $f\left(z \right) = g\left({\omega \left(z \right)} \right)$ for all $z \in E.$ In particular, if $g$ is univalent in $E,$ then $f \prec g \Leftrightarrow f\left(0 \right) = g\left(0 \right)$ and $f\left(E \right) = g\left(E \right).$

A function $p$ analytic in $E$ with $p\left(0 \right) = 1$ is said to be in the class of Janowski if it satisfies

$\begin{equation} p\left( z \right) \prec \frac{{1 + Az}}{{1 + Bz}},\; - 1 \leqslant B < A \leqslant 1, \;z \in E, \end{equation}$

(1.3)

where

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

(1.4)

This class was introduced by Janowski ^[1] in 1973 and is denoted by $P\left({A, B} \right)$ . We note that $P\left({1, - 1} \right) \equiv P$ , the well-known class of functions with positive real part consists of functions $p$ with $\operatorname{Re} p\left(z \right) > 0$ and $p\left(0 \right) = 1$ .

A function $f\in \mathcal{A}$ is called star-like with respect to symmetric conjugate points in $E$ if it satisfies

$\begin{equation} \operatorname{Re} \left\{ {\frac{{2zf'\left( z \right)}}{{f\left( z \right) - \overline {f\left( {-\overline z } \right)} }}} \right\} > 0,\;z \in E. \end{equation}$

(1.5)

This class was introduced by El-Ashwah and Thomas ^[2] in 1987 and is denoted by ${S_{SC}^*}$ . In 1991, Halim ^[3] defined the class $S_{SC}^*\left(\delta \right)$ consisting of functions $f \in \mathcal{A}$ that satisfy

$\begin{equation} \operatorname{Re} \left\{ {\frac{{2zf'\left( z \right)}}{{f\left( z \right) - \overline {f\left( {-\overline z } \right)} }}} \right\} > \delta, \; {\rm{0}} \leqslant \delta < {\rm{1}},\;z \in E. \end{equation}$

(1.6)

In terms of subordination, in 2011, Ping and Janteng ^[4] introduced the class $S_{SC}^*\left({A, B} \right)$ consisting of functions $f \in \mathcal{A}$ that satisfy

$\begin{equation} \frac{{2zf'\left( z \right)}}{{f\left( z \right) - \overline {f\left( {-\overline z } \right)} }} \prec \frac{{1 + Az}}{{1 + Bz}},\; - 1 \leqslant B < A \leqslant 1,\;z \in E. \end{equation}$

(1.7)

It follows from (1.7) that $f \in S_{SC}^*\left({A, B} \right)$ if and only if

$\begin{equation} \frac{{2zf'\left( z \right)}}{{f\left( z \right) - \overline {f\left( {-\overline z } \right)} }} = \frac{{1 + A\omega \left( z \right)}}{{1 + B\omega \left( z \right)}} = p\left( z \right), \;\omega \in H. \end{equation}$

(1.8)

Motivated by the work mentioned above, for functions $f \in \mathcal{A},$ we now introduce the subclass of the tilted star-like functions with respect to symmetric conjugate points as follows:

Definition 1.1. Let $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ be the class of functions defined by

$\begin{equation} \left( {{e^{i\alpha }}\frac{{zf'(z)}}{{h\left( z \right)}} - \delta - i\sin \alpha } \right)\frac{1}{{{\tau_{\alpha \delta }}}} \prec \frac{{1 + Az}}{{1 + Bz}}, {}\;z \in E, \end{equation}$

(1.9)

where $h\left(z \right) = \frac{{f\left(z \right) - \overline {f\left({\overline { - z} } \right)} }}{2}, \; {\tau _{\alpha \delta }} = \cos \alpha - \delta, \; 0 \leqslant \delta < 1, {}\left| \alpha \right| < \frac{\pi }{2}, {}$ and ${} - {{1}} \leqslant B < A \leqslant 1.$

By definition of subordination, it follows from (1.9) that there exists a Schwarz function $\omega$ which satisfies $\omega \left(0 \right) = 0$ and $\left| {\omega \left(z \right)} \right| < 1$ , and

$\begin{equation} \left( {{e^{i\alpha }}\frac{{zf'(z)}}{{h\left( z \right)}} - \delta - i\sin \alpha } \right)\frac{1}{{{\tau _{\alpha \delta }}}} = \frac{{1 + A\omega \left( z \right)}}{{1 + B\omega \left( z \right)}} = p\left( z \right), \;\omega \in H. \end{equation}$

(1.10)

We observe that for particular values of the parameters $\alpha, \; \delta, \; A$ , and $B$ , the class $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ reduces to the following existing classes:

(a) For $\alpha = \delta = 0, A = 1$ and $B = - 1,$ the class $S_{SC}^*\left({0, 0, 1, -1} \right) = S_{SC}^*$ introduced by El-Ashwah and Thomas ^[2].

(b) For $\alpha = 0, A = 1$ and $B = - 1,$ the class $S_{SC}^*\left({0, \delta, 1, -1} \right) = S_{SC}^*\left(\delta \right)$ introduced by Halim ^[3].

(c) For $\alpha = \delta = 0,$ the class $S_{SC}^*\left({0, 0, A, B} \right) = S_{SC}^*\left({A, B} \right)$ introduced by Ping and Janteng ^[4].

It is obvious that $S_{SC}^*\left({0} \right) \equiv S_{SC}^*$ , $S_{SC}^*\left({1, - 1} \right) \equiv S_{SC}^*$ , and $S_{SC}^*\left({0, 0, 1, - 1} \right) \equiv S_{SC}^*.$ Aside from that, in recent years, several authors obtained many interesting results for various subclasses of star-like functions with respect to other points, i.e., symmetric points and conjugate points. This includes, but is not limited to, these properties: coefficient estimates, Hankel and Toeplitz determinants, Fekete-Szegö inequality, growth and distortion bounds, and logarithmic coefficients. We may point interested readers to recent advances in these subclasses as well as their geometric properties, which point in a different direction than the current study, for example ^{[5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21]}.

In this paper, we obtained some interesting properties for the class $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ given in Definition 1.1. The paper is organized as follows: the authors presented some preliminary results in Section 2 and obtained the estimate for the general Taylor-Maclaurin coefficients $\left| {{a_n}} \right|, \; n \geqslant 2$ , the upper bounds of the second Hankel determinant $\left| {{H_2}\left(2 \right)} \right| = \left| {{a_2}{a_4} - {a_3}^2} \right|$ , the Fekete-Szegö inequality $\left| {{a_3} - \mu {a_2}^2} \right|$ with complex parameter $\mu$ , and the distortion and growth bounds for functions belonging to the class $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ as the main results in Section 3. In addition, some consequences of the main results from Section 3 were presented in Section 4. Finally, the authors offer a conclusion and some suggestions for future study in Section 5.

2. Preliminary results

We need the following lemmas to derive our main results.

Lemma 2.1. ^[22] If the function $p$ of the form $p\left(z \right) = 1 + \sum\limits_{n = 1}^\infty {{p_n}{z^n}}$ is analytic in $E$ and $p\left(z \right) \prec \frac{{1 + Az}}{{1 + Bz}}$ , then

$\begin{equation*} \left| {{p_n}} \right| \leqslant A - B,{\mathit{\rm{}}} - 1 \leqslant B < A \leqslant 1,{\mathit{\rm{n}}} \in \mathbb{N}{\mathit{\rm{.}}} \end{equation*}$

Lemma 2.2. ^[23] Let $p \in P$ of the form $p\left(z \right) = 1 + \sum\limits_{n = 1}^\infty {{p_n}{z^n}}$ and $\mu \in \mathbb{C}.$ Then

$\left| {{p_n} - \mu {p_k}{p_{n - k}}} \right| \leqslant 2{\mathit{\text{max}}}\left\{ {1,\left| {2\mu - 1} \right|} \right\},{\mathit{\text{}}}1 \leqslant k \leqslant n - 1.$

If $\left| {2\mu - 1} \right| \geqslant 1,$ then the inequality is sharp for the function $p\left(z \right) = \frac{{1 + z}}{{1 - z}}$ or its rotations. If $\left| {2\mu - 1} \right| < 1,$ then the inequality is sharp for the function $p\left(z \right) = \frac{{1 + {z^n}}}{{1 - {z^n}}}$ or its rotations.

Lemma 2.3. ^[24] For a function $p \in P$ of the form $p\left(z \right) = 1 + \sum\limits_{n = 1}^\infty {{p_n}{z^n}}$ , the sharp inequality $\left| {{p_n}} \right| \leqslant 2$ holds for each $n \geqslant 1.$ Equality holds for the function $p\left(z \right) = \frac{{1 + z}}{{1 - z}}.$

3. Main results

This section is devoted to our main results. We begin by finding the upper bound of the Taylor-Maclaurin coefficients $\left| {{a_n}} \right|,$ $n \geqslant 2$ for functions belonging to $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ . Further, we estimate the upper bounds of the second Hankel determinant $\left| {{H_2}\left(2 \right)} \right|$ and the Fekete-Szegö inequality $\left| {{a_3} - \mu {a_2}^2} \right|$ with complex parameter $\mu$ , and the distortion and growth bounds for functions in the class $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ .

3.1. Coefficient estimates

Theorem 3.1. Let $f \in S_{SC}^*\left({\alpha, \delta, A, B} \right).$ Then for $n \geqslant 1,$

$\begin{equation} \left| {{a_{2n}}} \right| \leqslant \frac{\psi }{{n!{2^n}}}\prod\limits_{j = 1}^{n - 1} {\left( {\psi + 2j} \right)} \end{equation}$

(3.1)

and

$\begin{equation} \left| {{a_{2n + 1}}} \right| \leqslant \frac{\psi }{{n!{2^n}}}\prod\limits_{j = 1}^{n - 1} {\left( {\psi + 2j} \right)}, \end{equation}$

(3.2)

where $\psi = \left({A - B} \right){\tau _{\alpha \delta }}$ and ${\mathit{\text{}}}{\tau _{\alpha \delta }} = \cos \alpha - \delta.$

Proof. In view of (1.10), we have

$\begin{equation} {e^{i\alpha }}zf'(z) = h(z)\left( {{\tau _{\alpha \delta }}p\left( z \right) + \delta + i\sin \alpha } \right). \end{equation}$

(3.3)

Using (1.1) and (1.4), (3.3) yields

$\begin{equation} {e^{i\alpha }}\left( {z + 2{a_2}{z^2} + 3{a_3}{z^3} + 4{a_4}{z^4} + 5{a_5}{z^5} + \ldots } \right) \\ = {e^{i\alpha }}\left( {z + {a_3}{z^3} + {a_5}{z^5} + \ldots } \right) + {\tau _{\alpha \delta }}\\\left[ {{p_1}{z^2} + {p_2}{z^3} + \left( {{p_3} + {a_3}{p_1}} \right){z^4} + \left( {{p_4} + {a_3}{p_2}} \right){z^5} + \ldots } \right]. \end{equation}$

(3.4)

Comparing the coefficients of the like powers of $z^n$ , $n \geqslant 1$ on both sides of the series expansions of (3.4), we obtain

$\begin{equation} 2{a_2} = {\tau _{\alpha \delta }}{e^{ - i\alpha }}{p_1}, \end{equation}$

(3.5)

$\begin{equation} 2{a_3} = {\tau _{\alpha \delta }}{e^{ - i\alpha }}{p_2}, \end{equation}$

(3.6)

$\begin{equation} 4{a_4} = {\tau _{\alpha \delta }}{e^{ - i\alpha }}\left( {{p_3} + {a_3}{p_1}} \right), \end{equation}$

(3.7)

$\begin{equation} 4{a_5} = {\tau _{\alpha \delta }}{e^{ - i\alpha }}\left( {{p_4} + {a_3}{p_2}} \right), \end{equation}$

(3.8)

$\begin{equation} 2n{a_{2n}} = {\tau _{\alpha \delta }}{e^{ - i\alpha }}\left( {{p_{2n - 1}} + {a_3}{p_{2n - 3}} + {a_5}{p_{2n - 5}} + \ldots + {a_{2n - 1}}{p_1}} \right), \end{equation}$

(3.9)

and

$\begin{equation} 2n{a_{2n + 1}} = {\tau _{\alpha \delta }}{e^{ - i\alpha }}\left( {{p_{2n}} + {a_3}{p_{2n - 2}} + {a_5}{p_{2n - 4}} + \ldots + {a_{2n - 1}}{p_2}} \right). \end{equation}$

(3.10)

We prove (3.1) and (3.2) using mathematical induction. Using Lemma 2.1, from (3.5) $-$ (3.8), respectively, we obtain

$\begin{equation} \left| {{a_2}} \right| \leqslant \frac{{\left( {A - B} \right){\tau _{\alpha \delta }}}}{2}, \end{equation}$

(3.11)

$\begin{equation} \left| {{a_3}} \right| \leqslant \frac{{\left( {A - B} \right){\tau _{\alpha \delta }}}}{2}, \end{equation}$

(3.12)

$\begin{equation} \left| {{a_4}} \right| \leqslant \frac{{\left( {A - B} \right){\tau _{\alpha \delta }}\left( {\left( {A - B} \right){\tau _{\alpha \delta }} + 2} \right)}}{8}, \end{equation}$

(3.13)

and

$\begin{equation} {\left| {{a_5}} \right| \leqslant \frac{{\left( {A - B} \right){\tau _{\alpha \delta }}\left( {\left( {A - B} \right){\tau _{\alpha \delta }} + 2} \right)}}{8}} . \end{equation}$

(3.14)

It follows that (3.1) and (3.2) hold for $n = 1, 2.$

For simplicity, we denote $\psi = \left({A - B} \right){\tau _{\alpha \delta }}$ and from (3.1) in conjunction with Lemma 2.1 yields

$\begin{equation} \left| {{a_{2n}}} \right| \leqslant \frac{\psi}{{2n}}\left[ {1 + \sum\limits_{k = 1}^{n - 1} {\left| {{a_{2k + 1}}} \right|} } \right]. \end{equation}$

(3.15)

Next, we assume that (3.1) holds for $k = 3, 4, 5, \ldots, \left({n - 1} \right).$

From (3.15), we get

$\begin{equation} \left| {{a_{2n}}} \right| \leqslant \frac{\psi }{{2n}}\left[ {1 + \sum\limits_{k = 1}^{n - 1} {\frac{\psi }{{k!{2^k}}}\prod\limits_{j = 1}^{k - 1} {\left( {\psi + 2j} \right)} } } \right]. \end{equation}$

(3.16)

To complete the proof, it is sufficient to show that

$\begin{equation} \frac{\psi }{{2m}}\left[ {1 + \sum\limits_{k = 1}^{m - 1} {\frac{\psi }{{k!{2^k}}}\prod\limits_{j = 1}^{k - 1} {\left( {\psi + 2j} \right)} } } \right] = \frac{\psi }{{m!{2^m}}}\prod\limits_{j = 1}^{m - 1} {\left( {\psi + 2j} \right)}, \end{equation}$

(3.17)

for $m = 3, 4, 5, \ldots, n.$ It is easy to see that (3.17) is valid for $m = 3.$

Now, suppose that (3.17) is true for $m = 4, \ldots, \left({n - 1} \right).$ Then, from (3.16) we get

$\begin{equation*} \frac{\psi }{{2n}}\left[ {1 + \sum\limits_{k = 1}^{n - 1} {\frac{\psi }{{k!{2^k}}}\prod\limits_{j = 1}^{k - 1} {\left( {\psi + 2j} \right)} } } \right] \\ = \frac{{n - 1}}{n}\left[ {\frac{\psi }{{2\left( {n - 1} \right)}}\left( {1 + \sum\limits_{k = 1}^{n - 2} {\frac{\psi }{{k!{2^k}}}\prod\limits_{j = 1}^{k - 1} {\left( {\psi + 2j} \right)} } } \right)} \right] + \frac{\psi }{{2n}}\frac{\psi }{{\left( {n - 1} \right)!{2^{n - 1}}}}\prod\limits_{j = 1}^{n - 2} {\left( {\psi + 2j} \right)} \\ = \frac{{\left( {n - 1} \right)}}{n}\frac{\psi }{{\left( {n - 1} \right)!{2^{n - 1}}}}\prod\limits_{j = 1}^{n - 2} {\left( {\psi + 2j} \right)} + \frac{\psi }{{2n}}\frac{\psi }{{\left( {n - 1} \right)!{2^{n - 1}}}}\prod\limits_{j = 1}^{n - 2} {\left( {\psi + 2j} \right)} \\ = \frac{\psi }{{\left( {n - 1} \right)!{2^{n - 1}}}}\left( {\frac{{\psi + 2\left( {n - 1} \right)}}{{2n}}} \right)\prod\limits_{j = 1}^{n - 2} {\left( {\psi + 2j} \right)} \\ = \frac{\psi }{{n!{2^n}}}\prod\limits_{j = 1}^{n - 1} {\left( {\psi + 2j} \right)} . \end{equation*}$

Thus, (3.17) holds for $m = n$ and hence (3.1) follows. Similarly, we can prove (3.2) and is omitted. This completes the proof of Theorem 3.1.

3.2. Hankel determinant

Theorem 3.2. If $f \in S_{SC}^*\left({\alpha, \delta, A, B} \right),$ then

$\begin{equation} \left| {{H_2}(2)} \right| \leqslant \frac{{{\psi ^2}}}{{16}}\left[ 2\left| { 2{\Upsilon ^ + } + 1} \right| + \left| {\psi {e^{ - i\alpha }} - 4} \right| + {\Upsilon ^ + }\left| { \psi {e^{ - i\alpha }} + 2{\Upsilon ^ + }} \right| \right], \end{equation}$

(3.18)

where $\psi = \left({A - B} \right){\tau _{\alpha \delta }}$ , ${\mathit{\text{}}}{\tau _{\alpha \delta }} = \cos \alpha - \delta,$ and ${\Upsilon ^ + } = 1 + B$ .

Proof. By definition of subordination, there exists a Schwarz function $\omega$ which satisfies $\omega \left(0 \right) = 0$ and $\left| {\omega \left(z \right)} \right| < 1,$ and from (1.10), we have

(3.19)

Let the function

$p\left( z \right) = \frac{{1 + \omega \left( z \right)}}{{1 - \omega \left( z \right)}} = 1 + \sum\limits_{n = 1}^\infty {{p_n}{z^n}} .$

Then, we have

$\begin{equation} \omega \left( z \right) = \frac{{p\left( z \right) - 1}}{{p\left( z \right) + 1}}. \end{equation}$

(3.20)

Substituting (3.20) into (3.19) yields

$\begin{equation} {e^{i\alpha }}zf'(z)\left( {{\Upsilon ^ - } + p(z){\Upsilon ^ + }} \right) = g(z)\left[ {\left( {{e^{i\alpha }}{\Upsilon ^ - } - \psi } \right) + p(z)\left( {{e^{i\alpha }}{\Upsilon ^ + } + \psi } \right)} \right], \end{equation}$

(3.21)

where ${\Upsilon ^ - } = 1 - B$ and ${\Upsilon ^ + } = 1 + B$ .

Using the series expansions of $f\left(z \right), {\text{ }}h\left(z \right),$ and $p\left(z \right)$ , (3.21) becomes

${\Upsilon ^ - }{e^{i\alpha }}\left( {z + 2{a_2}{z^2} + 3{a_3}{z^3} + 4{a_4}{z^4} + 5{a_5}{z^5} + ...} \right) + {\Upsilon ^ + }{e^{i\alpha }}\\\left[ {z + \left( {{k_1} + 2{a_2}} \right){z^2} + \left( {{k_2} + 2{a_2}{k_1} + 3{a_3}} \right){z^3}} \right. \\ \left. { + \left( {{k_3} + 2{a_2}{k_2} + 3{a_3}{k_1} + 4{a_4}} \right){z^4} + \ldots } \right] \\ = \left( {{e^{i\alpha }}{\Upsilon ^ - } - \psi } \right)\left( {z + {a_3}{z^3} + {a_5}{z^5} + ...} \right) + \left( {{e^{i\alpha }}{\Upsilon ^ + } + \psi } \right)\\\left[ {z + {k_1}{z^2} + \left( {{k_2} + {a_3}} \right){z^3} + \left( {{k_3} + {a_3}{k_1}} \right){z^4} + \ldots } \right]. \\$

On comparing coefficients of ${z^2}, {\text{ }}{z^3},$ and ${\text{ }}{z^4}$ , respectively, we get

$\begin{equation} {a_2} = \frac{{{p_1}\psi {e^{ - i\alpha }}}}{4}, \end{equation}$

(3.22)

$\begin{equation} {a_3} = \frac{{\psi {e^{ - i\alpha }}\left( {2{p_2} - p_1^2{\Upsilon ^ + }} \right)}}{8}, \end{equation}$

(3.23)

and

$\begin{equation} {a_4} = \frac{{\psi {e^{ - 2i\alpha }}\left[ {8{p_3}{e^{i\alpha }} + 2{p_1}{p_2}\left( {\psi - 4{\Upsilon ^ + }{e^{i\alpha }}} \right) + {p_1}^3{\Upsilon ^ + }\left( { - \psi + 2{\Upsilon ^ + }{e^{i\alpha }}} \right)} \right]}}{{64}}. \end{equation}$

(3.24)

From (3.22)–(3.24), we obtain

$\begin{equation} {H_2}(2) = {a_2}{a_4} - {a_3}^2 \\ = \frac{{{\psi ^2}{e^{ - 2i\alpha }}}}{{256}}\left[ {8{p_1}{p_3} + 2{p_1}^2{p_2}\left( {\psi {e^{ - i\alpha }} + 4{\Upsilon ^ + }} \right) - 16{p_2}^2 + {p_1}^4{\Upsilon ^ + }\left( { - \psi {e^{ - i\alpha }} - 2{\Upsilon ^ + }} \right)} \right]. \\ \end{equation}$

(3.25)

By suitably rearranging the terms in (3.25) and using the triangle inequality, we get

$\begin{equation} \left| {{H_2}(2)} \right| \leqslant \frac{{{\psi ^2}}}{{256}}\left[ {8\left| {{p_1}} \right|\left| {{p_3} - {\eta _1}{p_1}{p_2}} \right| + \left| { - 16{p_2}} \right|\left| {{p_2} - {\eta _2}{p_1}^2} \right| + {{\left| {{p_1}} \right|}^4}\left| {{\Upsilon ^ + }} \right|\left| { - \psi {e^{ - i\alpha }} - 2{\Upsilon ^ + }} \right|} \right], \end{equation}$

(3.26)

where ${\eta _1} = - {\Upsilon ^ + }$ and ${\eta _2} = \frac{{\psi {e^{ - i\alpha }}}}{8}$ .

Further, by applying Lemma 2.2 and Lemma 2.3, we have

$8\left| {{p_1}} \right|\left| {{p_3} - {\eta _1}{p_1}{p_2}} \right| \leqslant 32\left| { 2{\Upsilon ^ + } + 1} \right|,$

$\left| { - 16{p_2}} \right|\left| {{p_2} - {\eta _2}{p_1}^2} \right| \leqslant 16\left| {\psi {e^{ - i\alpha }} - 4} \right|,$

and

${\left| {{p_1}} \right|^4}\left| {{\Upsilon ^ + }} \right|\left| { - \psi {e^{ - i\alpha }} - 2{\Upsilon ^ + }} \right| \leqslant 16{\Upsilon ^ + }\left| { \psi {e^{ - i\alpha }} + 2{\Upsilon ^ + }} \right|.$

Thus, from (3.26), we obtain the required inequality (3.18). This completes the proof of Theorem 3.2.

3.3. Fekete-Szegö inequality

Theorem 3.3. If $f\in S_{SC}^*\left({\alpha, \delta, A, B} \right),$ then for any complex number $\mu$ , we have

$\begin{equation} \left| {{a_3} - \mu {a_2}^2} \right| \leqslant \frac{\psi }{2} max \left\{ {1,\frac{{\left| {2B + \mu \psi {e^{ - i\alpha }}} \right|}}{2}} \right\}, \end{equation}$

(3.27)

where $\psi = \left({A - B} \right){\tau _{\alpha \delta }}$ and ${\mathit{\text{}}}{\tau _{\alpha \delta }} = \cos \alpha - \delta.$

Proof. In view of (3.22) and (3.23), we have

$\begin{equation} \left| {{a_3} - \mu {a_2}^2} \right| = \left| {\frac{{\psi {e^{ - i\alpha }}}}{4}\left( {{p_2} - \chi p_1^2} \right)} \right|, \end{equation}$

(3.28)

where $\chi = \frac{{2{\Upsilon ^ + } + \mu \psi {e^{ - i\alpha }}}}{4}.$

By the application of Lemma 2.2, our result follows.

3.4. Distortion and growth bounds

Theorem 3.4. If $f \in S_{SC}^*\left({\alpha, \delta, A, B} \right),$ then for $\left| z \right| = r,$ $0 < r < 1$ , we have

$\begin{equation} \frac{1}{{1 + {r^2}}}\left[ {\frac{{\left( {1 - Ar} \right){\tau _{\alpha \delta }}}}{{1 - Br}} + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} } \right] \leqslant \left| {f'\left( z \right)} \right| \leqslant \frac{1}{{1 - {r^2}}}\left[ {\frac{{\left( {1 + Ar} \right){\tau _{\alpha \delta }}}}{{1 + Br}} + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} } \right], \end{equation}$

(3.29)

where ${\mathit{\text{}}}{\tau _{\alpha \delta }} = \cos \alpha - \delta.$ The bound is sharp.

Proof. In view of (1.10), we have

$\begin{equation} \left| {{e^{i\alpha }}\frac{{zf'\left( z \right)}}{{h\left( z \right)}} - \left( {\delta + i\sin \alpha } \right)} \right| = \left| {{\tau _{\alpha \delta }}} \right|\left| {\frac{{1 + A\omega \left( z \right)}}{{1 + B\omega \left( z \right)}}} \right|. \end{equation}$

(3.30)

Since $h$ is an odd star-like function, it follows that ^[24]

$\begin{equation} \frac{r}{{1 + {r^2}}} \leqslant \left| {h\left( z \right)} \right| \leqslant \frac{r}{{1 - {r^2}}}. \end{equation}$

(3.31)

Furthermore, for $\omega \in H,$ it can be easily established that ^[1]

$\begin{equation} \frac{{1 - Ar}}{{1 - Br}} \leqslant \left| {\frac{{1 + A\omega \left( z \right)}}{{1 + B\omega \left( z \right)}}} \right| \leqslant \frac{{1 + Ar}}{{1 + Br}}. \end{equation}$

(3.32)

Applying (3.31) and (3.32), and for $\left| z \right| = r$ , we find that, after some simplification,

$\begin{equation} \left| {f'\left( z \right)} \right| \leqslant \frac{1}{{1 - {r^2}}}\left[ {\frac{{\left( {1 + Ar} \right){\tau _{\alpha \delta }}}}{{1 + Br}} + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} } \right] \end{equation}$

(3.33)

and

$\begin{equation} \left| {f'\left( z \right)} \right| \geqslant \frac{1}{{1 + {r^2}}}\left[ {\frac{{\left( {1 - Ar} \right){\tau _{\alpha \delta }}}}{{1 - Br}} + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} } \right], \end{equation}$

(3.34)

which yields the desired result (3.29). The result is sharp due to the extremal functions corresponding to the left and right sides of (3.29), respectively,

$\begin{equation*} f\left( z \right) = \int_0^z {\frac{1}{{1 + {t^2}}}\left[ {\frac{{\left( {1 - At} \right){\tau _{\alpha \delta }}}}{{1 - Bt}} + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} } \right]} dt \end{equation*}$

and

$\begin{equation*} f\left( z \right) = \int_0^z {\frac{1}{{1 - {t^2}}}\left[ {\frac{{\left( {1 + At} \right){\tau _{\alpha \delta }}}}{{1 + Bt}} + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} } \right]} dt. \end{equation*}$

Theorem 3.5. If $f \in S_{SC}^*\left({\alpha, \delta, A, B} \right),$ then for $\left| z \right| = r,$ $0 < r < 1$ , we have

$\begin{equation} \frac{1}{{1 + {B^2}}}\left[ {\psi \ln \left( {\frac{{1 - Br}}{{\sqrt {1 + {r^2}} }}} \right) + {\tau _{\alpha \delta }}\left( {1 + AB} \right){{\tan }^{ - 1}}r} \right] + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} {\tan ^{ - 1}}r \leqslant \left| {f\left( z \right)} \right| \\ \leqslant \frac{1}{{1 - {B^2}}}\left[ {\psi \ln \left( {\frac{{1 + Br}}{{\sqrt {1 - {r^2}} }}} \right) + {\tau _{\alpha \delta }}\left( {1 - AB} \right)\ln \left( {\sqrt {\frac{{1 + r}}{{1 - r}}} } \right)} \right] + \sqrt {{{\sin }^2}\alpha + {\delta ^2}} \ln \left( {\sqrt {\frac{{1 + r}}{{1 - r}}} } \right), \\ \end{equation}$

(3.35)

where $\psi = \left({A - B} \right){\tau _{\alpha \delta }}$ and ${\mathit{\text{}}}{\tau _{\alpha \delta }} = \cos \alpha - \delta.$ The bound is sharp.

Proof. Upon elementary integration of (3.29) yields (3.35). The result is sharp due to the extremal functions corresponding to the left and right sides of (3.35), respectively,

and

4. Consequences and corollaries

In this section, we apply our main results in Section 3 to deduce each of the following consequences and corollaries as shown in Tables 1-4.

Table 1. General Taylor-Maclaurin coefficients.

$\mathrm{Corollary}$ $\mathrm{4.1}$	$\mathrm{Class}$	$\left\| {{a_n}} \right\|, \; n \geqslant 2$
$\mathrm{(a)}$	$S_{SC}^*\left({0, 0, 1, -1} \right)$	$\left\| {{a_{2n}}} \right\| \leqslant \frac{{{2^{2 - n}}}}{{n!}}\prod\limits_{j = 1}^{n - 1} {\left({1 + j} \right)}, \; n \geqslant 1$
		$\left\| {{a_{2n + 1}}} \right\| \leqslant \frac{{{2^{2 - n}}}}{{n!}}\prod\limits_{j = 1}^{n - 1} {\left({1 + j} \right)}, \; n \geqslant 1$
$\mathrm{(b)}$	$S_{SC}^*\left({0, \delta, 1, -1} \right)$	$\left\| {{a_{2n}}} \right\| \leqslant \frac{{{2^{2 - n}}\left({1 - \delta } \right)}}{{n!}}\prod\limits_{j = 1}^{n - 1} {\left({\left({1 - \delta } \right) + j} \right)}, \; n \geqslant 1$
		$\left\| {{a_{2n + 1}}} \right\| \leqslant \frac{{{2^{2 - n}}\left({1 - \delta } \right)}}{{n!}}\prod\limits_{j = 1}^{n - 1} {\left({\left({1 - \delta } \right) + j} \right)}, \; n \geqslant 1$
$\mathrm{(c)}$	$S_{SC}^*\left({0, 0, A, B} \right)$	$\left\| {{a_{2n}}} \right\| \leqslant \frac{{\left({A - B} \right)}}{{n!{2^n}}}\prod\limits_{j = 1}^{n - 1} {\left({\left({A - B} \right) + 2j} \right)}, \; n \geqslant 1$
		$\left\| {{a_{2n + 1}}} \right\| \leqslant \frac{{\left({A - B} \right)}}{{n!{2^n}}}\prod\limits_{j = 1}^{n - 1} {\left({\left({A - B} \right) + 2j} \right)}, \; n \geqslant 1$

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Table 2. Second Hankel determinant.

$\mathrm{Corollary}$ $\mathrm{4.2}$	$\mathrm{Class}$	$\left\| {{H_2}(2)} \right\|$
$\mathrm{(a)}$	$S_{SC}^*\left({0, 0, 1, -1} \right)$	$\left\| {{H_2}(2)} \right\| \leqslant 1\$
$\mathrm{(b)}$	$S_{SC}^*\left({0, \delta, 1, -1} \right)$	$\left\| {{H_2}(2)} \right\| \leqslant \frac{{{{\left({1 - \delta } \right)}^2}\left({2 + \delta } \right)}}{2}\$
$\mathrm{(c)}$	$S_{SC}^*\left({0, 0, A, B} \right)$	$\left\| {{H_2}(2)} \right\| \leqslant \frac{{{{\left({A - B} \right)}^2}}}{{16}}\left[{2\left\| {2{\Upsilon ^ + } + 1} \right\| + \left\| {\left({A - B} \right) - 4} \right\| + {\Upsilon ^ + }\left\| {\left({A - B} \right) + 2{\Upsilon ^ + }} \right\|} \right]\$

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Table 3. Fekete-Szegö inequality.

$\mathrm{Corollary}$ $\mathrm{4.3}$	$\mathrm{Class}$	$\left\| {{a_3} - \mu {a_2}^2} \right\|$
$\mathrm{(a)}$	$S_{SC}^*\left({0, 0, 1, -1} \right)$	$\left\|a_3-\mu a_2^2\right\| \leqslant max \{1, \|-1+\mu\|\}$
$\mathrm{(b)}$	$S_{SC}^*\left({0, \delta, 1, -1} \right)$	$\left\|a_3-\mu a_2^2\right\| \leqslant(1-\delta) max \{1, \|-1+\mu(1-\delta)\|\}$
$\mathrm{(c)}$	$S_{SC}^*\left({0, 0, A, B} \right)$	$\left\|a_3-\mu a_2^2\right\| \leqslant \frac{(A-B)}{2} max \left\{1, \frac{\|2 B+\mu(A-B)\|}{2}\right\}$

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Table 4. Distortion bound.

$\mathrm{Corollary}$ $\mathrm{4.4}$	$\mathrm{Class}$	Distortion bound
$\mathrm{(a)}$	$S_{SC}^*\left({0, 0, 1, -1} \right)$	$\frac{{1 - r}}{{\left({1 + {r^2}} \right)\left({1 + r} \right)}} \leqslant \left\| {f'\left(z \right)} \right\| \leqslant \frac{{1 + r}}{{\left({1 - {r^2}} \right)\left({1 - r} \right)}}$
$\mathrm{(b)}$	$S_{SC}^*\left({0, \delta, 1, -1} \right)$	$\frac{1}{{1 + {r^2}}}\left[ {\frac{{\left({1 - r} \right)\left({1 - \delta } \right)}}{{1 + r}} + \delta } \right] \leqslant \left\| {f'\left(z \right)} \right\| \leqslant \frac{1}{{1 - {r^2}}}\left[ {\frac{{\left({1 + r} \right)\left({1 - \delta } \right)}}{{1 - r}} + \delta } \right]$
$\mathrm{(c)}$	$S_{SC}^*\left({0, 0, A, B} \right)$	$\frac{{1 - Ar}}{{\left({1 + {r^2}} \right)\left({1 - Br} \right)}} \leqslant \left\| {f'\left(z \right)} \right\| \leqslant \frac{{1 + Ar}}{{\left({1 - {r^2}} \right)\left({1 + Br} \right)}}$

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Remark 4.1. The results obtained in Corollary 4.1(c) coincide with the results obtained in ^[4].

Remark 4.2. The result in Corollary 4.2(a) coincides with the findings of Singh ^[10].

Remark 4.3. The result obtained in Corollary 4.4(c) coincides with the result obtained in ^[4].

Remark 4.4. Setting $\alpha = 0{\rm{ and }}\delta = 0,$ Theorem 3.5 reduces to the result of Ping and Janteng ^[4].

5. Conclusions

In this paper, we considered a new subclass of tilted star-like functions with respect to symmetric conjugate points. Various interesting properties of these functions were investigated, such as coefficient bounds, the second Hankel determinant, Fekete-Szegö inequality, distortion bound, and growth bound. The results presented in this paper not only generalize some results obtained by Ping and Janteng ^[4] and Singh ^[10], but also give new results as special cases based on the various special choices of the involved parameters. Other interesting properties for functions belonging to the class $S_{SC}^*\left({\alpha, \delta, A, B} \right)$ could be estimated in future work, such as the upper bounds of the Toeplitz determinant, the Hankel determinant of logarithmic coefficients, the Zalcman coefficient functional, the radius of star-likeness, partial sums, etc.

Acknowledgments

This research was funded by Universiti Teknologi MARA grant number 600-RMC/GPM LPHD 5/3 (060/2021). The authors wish to thank the referees for their valuable comments and suggestions.

Conflict of interest

The authors declare that they have no competing interests.

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