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 \title{\bf New Subclass of Univalent Function involving the Modified Sigmoid Function using Subordination Principle}

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%\title{Coefficient condition for a subclass of harmonic univalent function.}

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\begin{abstract}
In this work, a new subclass of function, $G_\lambda(n, \mu, \lambda):n \in N_0, \mu \geq 1,
	\\0 \leq \lambda \leq 1$, was defined using the S\v{a}l\v{a}gean differential operator involving the modified sigmoid function and subordination principle. The initial coefficient bounds and the Fekete-Szego functional of this class were obtained.

\end{abstract}

%\begin{keyword}
%biharmonic univalent function, harmonic univalent function, sense-preserving, Salagean differential operator.

 \vspace{.2cm} {\bf
Keywords:} Analytic function, Sigmoid function, Subordination principle, S\v{a}l\v{a}gean differential operator.

 \vspace{.2cm}
 {\bf AMS Mathematics Subject Classification (2010):}  30C45.



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\section{Introduction}

Let the class of functions of the form 
$$f(z) = z + \sum_{k = 2}^{\infty}a_kz^k$$
that are analytic in the unit disk {\bf D} = $\{z: |z| < 1\}$ be denoted by {\bf A}.


 S\v{a}l\v{a}gean [8] introduced the differential operator $D^nf, n \in  N_0$ for functions $f(z)$ belonging to class ${\bf A}$ of analytic functions in the unit disk ${\bf D}$;
$$D^nf(z) = z + \sum_{k=2}^{\infty}k^n a_kz^k; \quad n \in N_0 $$
The sigmoid function
$$h(z) = \frac{1}{1 + e^{-z}}$$
is differentiable and has the following properties:
\begin{enumerate}
	\item It outputs real numbers between 0 and 1.
	\item It maps a very large input domain to a small range of outputs.
	\item It never loses information because it is a one-to-one function.
	\item It increases monotonically.
\end{enumerate}
Fadipe-Joseph et al [3] studied the modified sigmoid function 
$$G(z) = \frac{2}{1 + e^{-z}}$$
and obtained another series of the modified sigmoid function
$$G(z) = 1 + \sum_{m=1}^{\infty}\frac{(-1)^m}{2^m}\left(\frac{(-1)^n}{n!}z^n\right)^m$$\\
Ramachandran and Dhanalakshmi [6] established coefficient estimates for a class of spirallike functions in the space of sigmoid function. Furthermore, the Fekete- Szego functional for a subclass of analytic functions related to sigmoid function was obtained in [7]. 
\begin{defn}
Let $f(z) = a_0 + a_1z + a_2z^2 + \cdots$ be analytic and univalent in a domain $\textbf{D}$ and suppose $g(z) \in \textbf{D}$. If $g(z)$ is analytic in $\textbf{D}$, $f(0) = g(0)$ and $f(z) \subset g(z)$, then we say that $f(z)$ is subordinate to $g(z)$ in $\textbf{D}$ and it is denoted by
$$f(z) \prec g(z)$$
(Duren [2], Goodman [5])
\end{defn}
\begin{defn}
	A function $f \in {\bf A}$ is said to be in the class $G_\lambda (n, \mu, \lambda): n \in N_0,\\ \mu \geq 1, 0\leq \lambda \leq 1$, if the following subordination holds
	\begin{equation}
	\left(1 - \lambda\right)\left(\frac{D^nf_\gamma(z)}{z}\right)^\mu + \lambda f_\gamma'(z)\left(\frac{D^nf_\gamma(z)}{z}\right)^{\mu - 1} \prec G(z)
	\end{equation}
\end{defn}
Altinkaya and Yalcin [1] defined a subclass of univalent functions and obtained coefficients expansion using Chebyshev polynomial. The definition of this class was motivated by their work.
\begin{lem}
	Let $f(z)$ and $g(z)$ be analytic in {\bf D} and suppose that $g(z)$ is univalent in {\bf D}. Then $f(z) \prec g(z)$ if and only if there exists a Schwarz function $\omega(z)$ with the property $|\omega(z)| \leq |z|$ such that
	$$f(z) = g(\omega(z))$$
\end{lem}
\subsection{S\v{a}l\v{a}gean Differential Operator Involving Modified Sigmoid Function}
Consider the function
\begin{equation} 
f_\gamma(z) = z + \sum_{k = 2}^{\infty}\gamma(s)a_kz^k
\end{equation}
where
$$\gamma(s) = \frac{2}{1 + e^{-s}}$$
which are analytic and univalent in the unit disk. Then functions of the form (1.2) belong to the class ${\bf A_\gamma}$.\\
Let $D^nf_\gamma (z); n \in N_0 = \{0, 1, 2, \cdots \}$ denote the S\v{a}l\v{a}gean differential operator involving modified sigmoid function, then 
$$D^0f_\gamma(z) = f_\gamma(z)$$
$$D^1f_\gamma(z) = \gamma (s)zf'_\gamma (z)$$
$$\vdots$$
\begin{equation} 
D^nf_\gamma(z) = \gamma (s)z(D^{n - 1}f_\gamma (z))'
\end{equation}
Taking $\lim_{s \to 0} \gamma(s) = 1$, we have the S\v{a}l\v{a}gean differential operator. 
(Fadipe-Joseph et al [4])
\section{Main Results}
\begin{Theorem}
	If $f(z)$ belongs to the class $G_\lambda (n, \mu, \lambda): n \in N_0, \mu \geq 1, 0\leq \lambda \leq 1$, then
	$$|a_2| \leq \frac{1}{2(A + B)}$$
	$$|a_3| \leq \frac{1}{2(A' + B')} + \frac{D + E}{4(A + B)^2(A' + B')}$$
	$$|a_4| \leq \frac{11}{24(D' + E')} + \frac{F + G}{4(A + B)(A' + B')} + \frac{(D + E)(F + G)}{8(A + B)^3(A' + B')} + \frac{F' + G'}{8(A + B)^3}$$
	
	In particular, when
	$$G(z) = 1 + \left(\sum_{m = 1}^{\infty}\frac{(-1)^m}{2^m}\left(\sum_{n = 1}^{\infty}\frac{(-1)^n}{n!}z^n\right)^m\right)$$
	Taking $m = 1$, we have
	$$G(z) = 1 + \sum_{n = 1}^{\infty}c_nz^n; \hspace{0.5cm} c_n = \frac{(-1)^{n+1}}{2n!}$$
	If $\omega (z) = G(z) - 1$, then
	$$a_2 = \frac{1}{4(A + B)}$$
	$$a_3 = \frac{-1}{8(A' + B')} - \frac{D + E}{16(A + B)^2(A' + B')}$$
	$$a_4 = \frac{7}{192(D' + E')} + \frac{F + G}{32(A + B)(A' + B')(D' + E')} $$ $$~~~~~~~~~~~~~~~~~~+ \frac{(D + E)(F + G)}{64(A + B)^3(A' + B')(D' + E')} - \frac{F' + G'}{64(A + B)^3(D' + E')}$$
	where
	$$A = (1 - \lambda)2^n\mu \gamma^{n\mu + 1}(s)$$
	$$B = \lambda \gamma^{n(\mu - 1) + 1}(s)(2^n(\mu - 1) + 2)$$
	$$A' = (1 - \lambda)3^n\mu \gamma^{n\mu + 1}(s)$$
	$$B' = \lambda \gamma^{n(\mu - 1) + 1}(s)(3^n(\mu - 1) + 3)$$
	$$D = -(1 - \lambda)2^{2n}\frac{\mu (\mu - 1)}{2!}\gamma^{n\mu + 2}(s)$$
	$$E = -\lambda \gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{2n} + 2^{n + 1}(\mu - 1)\right)$$
	$$D' = (1 - \lambda)4^n\mu \gamma^{n\mu + 1}(s)$$
	$$E' = \lambda \gamma^{n(\mu - 1) + 1}(s)(4^n(\mu - 1) + 4)$$
	$$F = (1 - \lambda)2^{n + 1}\cdot 3^n \frac{\mu (\mu - 1)}{2!}\gamma^{n\mu + 2}(s)$$
	$$G = \lambda \gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{n + 1}\cdot 3^n + (\mu - 1)(2\cdot 3^n + 3 \cdot 2^n)\right)$$
	$$F' = (1 - \lambda)2^{3n}\frac{\mu (\mu - 1)(\mu - 2)}{3!}\gamma^{n\mu + 3}(s)$$
	$$G' = \lambda \gamma^{n(\mu - 1) + 3}(s)\left(\frac{(\mu - 1)(\mu - 2)(\mu - 3)}{3!}2^{3n} + \frac{(\mu - 1)(\mu - 2)}{2!}2^{2n + 1}\right)$$
\end{Theorem}
{\bf Proof:}
	If $f(z) \in G_\gamma (n, \mu, \lambda)$ , then from (1.1)
	$$\left(1 - \lambda\right)\left(\frac{D^nf_\gamma(z)}{z}\right)^\mu + \lambda f_\gamma'(z)\left(\frac{D^nf_\gamma(z)}{z}\right)^{\mu - 1} \prec G(z)
	$$
	\begin{equation}
	f_\gamma (z) = z + \sum_{k = 2}^{\infty}\gamma (s)a_kz^k 
	\end{equation}
	where
	$$\gamma (s) = \frac{2}{1 + e^{-s}}; \qquad s \geq 0$$
	\begin{equation}
	f_\gamma (z) = z + \gamma (s)(a_2z^2 + a_3z^3 + a_4z^4 + a_5z^5 + \cdots)
	\end{equation}
	From (1.3), we have
	$$\frac{D^nf_\gamma (z)}{z} = \frac{\gamma^n(s)z + \sum_{k = 2}^{\infty}k^n \gamma^{n+1}(s)a_k z^k}{z}$$
	\begin{equation}
	\frac{D^nf_\gamma (z)}{z} = \gamma^n(s) + \sum_{k = 2}^{\infty}k^n \gamma^{n+1}(s)a_k z^{k-1}
	\end{equation}
	$$\frac{D^nf_\gamma (z)}{z} = \gamma^n(s) + 2^n\gamma^{n+1}(s)a_2z + 3^n\gamma^{n+2}(s)a_3z^2 + 4^n\gamma^{n+3}(s)a_4z^3 + 5^n\gamma^{n+4}(s)a_5z^4 + \cdots $$
	$$\left(\frac{D^nf_\gamma (z)}{z}\right)^\mu = \gamma^{n\mu}(s)(1 + 2^n\gamma(s)a_2z + 3^n\gamma^{2}(s)a_3z^2 + 4^n\gamma^{3}(s)a_4z^3 + 5^n\gamma^{4}(s)a_5z^4 + \cdots )^\mu$$
	\begin{equation}
	\left(\frac{D^nf_\gamma (z)}{z}\right)^\mu = \gamma^{n\mu}(s)\left(\begin{array}{c}
	1 + \mu\gamma(s) \left(\begin{array}{cc} 2^na_2z + 3^na_3z^2 \\+ 4^na_4z^3  + 5^na_5z^4 + \cdots\end{array}\right)\\
	+ \frac{\mu(\mu - 1)}{2!}\gamma^2(s)\left(\begin{array}{cc} 2^na_2z + 3^na_3z^2 \\+ 4^na_4z^3 + 5^na_5z^4 + \cdots\end{array}\right)^2\\
	+ \frac{\mu(\mu - 1)(\mu - 2)}{3!}\gamma^{3}(s)\left(\begin{array}{cc} 2^na_2z + 3^na_3z^2 \\ + 4^na_4z^3 + 5^na_5z^4 + \cdots\end{array}\right)^3 + \cdots
	\end{array}\right)
	\end{equation}
	\begin{equation}
	\begin{array}{cc}
	\left(\frac{D^nf_\gamma (z)}{z}\right)^\mu = & \gamma^{n\mu}(s) + 2^n\mu\gamma^{n\mu + 1}(s)a_2z ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\
	&+ \left(3^n\mu\gamma^{n\mu + 1}(s)a_3 + 2^{2n}\frac{\mu(\mu - 1)}{2!}\gamma^{n\mu + 2}(s)a_2^2\right)z^2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\\ 
	& + \left(\begin{array}{c} 4^n\mu\gamma^{n\mu + 1}(s)a_4 + 2^n \cdot 3^n 2^{2n}\frac{2\mu(\mu - 1)}{2!}\gamma^{n\mu + 2}(s)a_2a_3 \\ + 2^{3n}\frac{\mu(\mu - 1)(\mu - 2)}{3!}\gamma^{n\mu + 3}(s)a_2^3\end{array}\right)z^3 ~~~~~~~~~~~~~~~~~~\\
	& + \cdots
	\end{array}
	\end{equation}
	Similarly,
	\begin{equation}
	\begin{array}{cc}
	\left(\frac{D^nf_\gamma (z)}{z}\right)^{\mu - 1} = & \gamma^{n(\mu - 1)}(s) + 2^n(\mu - 1)\gamma^{n(\mu - 1) + 1}(s)a_2z \\
	&+ \left(3^n(\mu - 1)\gamma^{n(\mu - 1) + 1}(s)a_3 + 2^{2n}\frac{(\mu - 1)(\mu - 2)}{2!}\gamma^{n(\mu - 1) + 2}(s)a_2^2\right)z^2 \\ 
	& + \left(\begin{array}{c} 4^n(\mu - 1)\gamma^{n(\mu - 1) + 1}(s)a_4 \\ + 2^n \cdot 3^n 2^{2n}\frac{2(\mu - 1)(\mu - 2)}{2!}\gamma^{n(\mu - 1) + 2}(s)a_2a_3 \\ + 2^{3n}\frac{(\mu - 1)(\mu - 2)(\mu - 3)}{3!}\gamma^{n(\mu - 1) + 3}(s)a_2^3\end{array}\right)z^3 \\
	& + \cdots
	\end{array}
	\end{equation}
	Differentiating (2.2) with respect to $z$, we have
	\begin{equation}
	f'_\gamma(z) = 1 + \gamma(s)(2a_2z + 3a_3z^2 + 4a_4z^3 + 5a_5z^4 + \cdots)
	\end{equation}
	Multiplying (2.6) and (2.7), we have\\
	$f'_\gamma(z)\left(\frac{D^nf_\gamma(z)}{z}\right)^{\mu - 1} = \gamma^{n(\mu - 1)}(s) + \gamma^{n(\mu - 1) + 1}(s)(2^n(\mu - 1) + 2)a_2z \\ + \left(\begin{array}{c}\gamma^{n(\mu - 1) + 1}(s)(3^n(\mu - 1) + 3)a_3 + \gamma^{n(\mu - 1) + 2}\left(2^{2n}\frac{(\mu - 1)(\mu - 2)}{2!} + 2^{n+1}(\mu - 1)\right)a_2^2\end{array}\right)z^2\\ + \left(\begin{array}{c}\gamma^{n(\mu - 1) + 1}(s)(4^n(\mu - 1) + 4)a_4 \\ + \gamma^{n(\mu - 1) + 2}(s) \left(2^{n+1}\cdot 3^n\frac{(\mu - 1)(\mu - 2)}{2!} + (\mu - 1)(2^n\cdot 3 + 2\cdot 3^n) \right)a_2a_3 \\ + \gamma^{n(\mu - 1) + 3}(s)\left(2^{3n}\frac{(\mu - 1)(\mu - 2)(\mu  - 3)}{3!} + 2^{2n+1}\frac{(\mu - 1)(\mu - 2)}{2!}\right)a_2^3\end{array}\right)z^3  + \cdots$
	\begin{equation}
	\begin{array}{c} 
	\left(1 - \lambda\right)\left(\frac{D^nf_\gamma(z)}{z}\right)^\mu + \lambda f_\gamma'(z)\left(\frac{D^nf_\gamma(z)}{z}\right)^{\mu - 1} = (1 + \lambda)\gamma^{n\mu}(s) + \lambda\gamma^{n(\mu - 1)}(s)\\ \begin{array}{c}
	+ ((1 - \lambda)2^n\mu\gamma^{n\mu + 1}(s) + \lambda\gamma^{n(\mu - 1) + 1}(s)(2^n(\mu - 1) + 2))a_2z \\
	+ \left(\begin{array}{c}((1 - \lambda)3^n\mu \gamma^{n\mu + 1}(s) + \lambda\gamma^{n(\mu - 1) + 1}(s)(3^n(\mu - 1) + 3))a_3\\ \left(\begin{array}{c}(1 - \lambda)2^{2n} \frac{\mu(\mu - 1)}{2!}\gamma^{n\mu + 2}(s) \\ + \lambda\gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{2n} + 2^{n+1}(\mu - 1)\right)\end{array}\right)a_2^2\end{array}\right)z^2\\ 
	+ \left(\begin{array}{c}((1 - \lambda)4^n\mu \gamma^{n\mu + 1}(s) + \lambda\gamma^{n(\mu - 1) + 1}(s)(4^n(\mu - 1) + 4))a_4 \\+ \left(\begin{array}{c}(1 - \lambda)2^{n+1}\cdot 3^n\frac{\mu(\mu - 1)}{2!}\gamma^{n\mu + 2}(s) \\+ \lambda\gamma^{n(\mu - 1) + 2}(s)\left(\begin{array}{c}\frac{(\mu - 1)(\mu - 2)}{2!}2^{n+1}\cdot 3^n \\+ (\mu - 1)(2\cdot 3^n + 3\cdot 2^n)\end{array}\right)\end{array}\right)a_2a_3\\ + \left(\begin{array}{c} (1 - \lambda)2^{3n}\frac{\mu(\mu - 1)(\mu - 2)}{3!}\gamma^{n\mu + 3}(s) \\+ \lambda\gamma^{n(\mu - 1) + 3}(s)\left(\frac{(\mu - 1)(\mu - 2)(\mu - 3)}{3!}2^{3n} + \frac{(\mu - 1)(\mu - 2)}{2!}2^{2n+ 1}\right)\end{array}\right)a_2^3\end{array}\right)z^3 \\ + \cdots
	\end{array}
	\end{array}
	\end{equation}
	\begin{equation}
	G(z) = 1 + \frac{z}{2} - \frac{z^3}{24} + \frac{z^5}{240} - \cdots
	\end{equation}
	\begin{equation}
	G(\omega(z)) = 1 + \frac{1}{2}\omega(z) - \frac{1}{24}\omega(z)^3 + \frac{1}{240}\omega(z)^5 - \cdots
	\end{equation}
	\begin{equation}
	\omega(z) = c_1z + c_2z^2 + c_3z^3 + c_4z^4 + \cdots
	\end{equation}
	\begin{equation}
	\omega^3(z) = c_1^3z^3 + 3c_1^2z^4 + (3c_1^2c_3 + 3c_1c_2^2)z^5 + \cdots
	\end{equation}
	\begin{equation}
	\omega^5(z) = c_1^5z^5 + \cdots 
	\end{equation}
	Substituting (2.11), (2.12) and (2.13) into (2.10), we have
	\begin{equation}
	G(\omega(z)) = 1 + \frac{c_1}{2}z + \frac{c_2}{2}z^2 + \left(\frac{c_3}{2} - \frac{c_1^3}{24}\right)z^3 + \left(\frac{c_4}{2} - \frac{c_1^2c_2}{8}\right)z^4 + \cdots
	\end{equation}
	It is well known that if $|\omega(z)| = |c_1z + c_2z^ + \cdots| < 1$, then
	\begin{equation}
	|c_j| \leq 1
	\end{equation}
	for all $j \in {\bf N}$ and
	\begin{equation}
	|c_2 - \rho c_1^2| \leq max\{1, |\rho|\}
	\end{equation}
	Equating (2.8) and (2.14) and comparing the coefficients, we have \\
	$ ((1 - \lambda)2^n\mu\gamma^{n\mu + 1}(s) + \lambda\gamma^{n(\mu - 1) + 1}(s)(2^n(\mu - 1) + 2))a_2z = \frac{c_1}{2}$\\
	$((1 - \lambda)3^n\mu \gamma^{n\mu + 1}(s) + \lambda\gamma^{n(\mu - 1) + 1}(s)(3^n(\mu - 1) + 3))a_3\\ \left((1 - \lambda)2^{2n} \frac{\mu(\mu - 1)}{2!}\gamma^{n\mu + 2}(s) + \lambda\gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{2n} + 2^{n+1}(\mu - 1)\right)\right)a_2^2 = \frac{c_2}{2}$\\
	$((1 - \lambda)4^n\mu \gamma^{n\mu + 1}(s) + \lambda\gamma^{n(\mu - 1) + 1}(s)(4^n(\mu - 1) + 4))a_4 \\+ \left(\begin{array}{c}(1 - \lambda)2^{n+1}\cdot 3^n\frac{\mu(\mu - 1)}{2!}\gamma^{n\mu + 2}(s) \\+ \lambda\gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{n+1}\cdot 3^n + (\mu - 1)(2\cdot 3^n + 3\cdot 2^n)\right)\end{array}\right)a_2a_3\\ + \left(\begin{array}{c} (1 - \lambda)2^{3n}\frac{\mu(\mu - 1)(\mu - 2)}{3!}\gamma^{n\mu + 3}(s) \\+ \lambda\gamma^{n(\mu - 1) + 3}(s)\left(\frac{(\mu - 1)(\mu - 2)(\mu - 3)}{3!}2^{3n} + \frac{(\mu - 1)(\mu - 2)}{2!}2^{2n+ 1}\right)\end{array}\right)a_2^3 = \frac{c_3}{2} - \frac{c_1^3}{24}$\\
	Simplifying the above, we have
	\begin{equation}
	\begin{array}{c}
	a_2 = \frac{c_1}{2(A + B)}\\
	a_3 = \frac{c_2}{2(A' + B')} - \frac{c_1^2(D + E)}{4(A + B)^2(A' + B')}\\
	a_4 = \begin{array}{c}
	\frac{1}{D' + E'}\left(\frac{c_1}{2} - \frac{c_1^3}{24}\right) - \frac{c_1c_2(F + G)}{4(A + B)(A' + B')}\\ + \frac{c_1^3(D + E)(F + G)}{8(A + B)^3(A' + B')} - \frac{c_1^3(F' + G')}{8(A + B)^3}
	\end{array}
	\end{array}
	\end{equation}
	Then, from the inequality (2.15), we have
	\begin{equation}
	\begin{array}{c}
	|a_2| = \left| \frac{c_1}{2(A + B)} \right| \leq \frac{1}{2(A + B)}\\
	|a_3| = \left| \frac{c_2}{2(A' + B')} - \frac{c_1^2(D + E)}{4(A + B)^2(A' + B')}\right| \leq \frac{1}{2(A' + B')} + \frac{(D + E)}{4(A + B)^2(A' + B')}\\
	|a_4| = \left| \begin{array}{c}
	\frac{1}{D' + E'}\left(\frac{c_1}{2} - \frac{c_1^3}{24}\right) - \frac{c_1c_2(F + G)}{4(A + B)(A' + B')}\\ + \frac{c_1^3(D + E)(F + G)}{8(A + B)^3(A' + B')} - \frac{c_1^3(F' + G')}{8(A + B)^3}
	\end{array} \right|\leq \begin{array}{c}
	\frac{11}{24(D' + E')} + \frac{(F + G)}{4(A + B)(A' + B')}\\ + \frac{(D + E)(F + G)}{8(A + B)^3(A' + B')} + \frac{(F' + G')}{8(A + B)^3}
	\end{array}
	\end{array}
	\end{equation}
	
	Also, when
	$$G(z) = 1 + \left(\sum_{m = 1}^{\infty}\frac{(-1)^m}{2^m}\left(\sum_{n = 1}^{\infty}\frac{(-1)^n}{n!}z^n\right)^m\right)$$
	For $m = 1$,
	$$G(z) - 1 = \frac{1}{2}z - \frac{1}{4}z^2 + \frac{1}{12}z^3 - \cdots$$
	Comparing with (2.11), we have
	\begin{equation}
	\begin{array}{cccc}
	c_1 = \frac{1}{2}; & c_2 = - \frac{1}{4}; & c_3 = \frac{1}{12}; & \cdots
	\end{array}
	\end{equation}
	Substituting (2.19) into (2.17), we have
	$$a_2 = \frac{1}{4(A + B)}$$
	$$a_3 = \frac{-1}{8(A' + B')} - \frac{D + E}{16(A + B)^2(A' + B')}$$
	$$a_4 = \frac{7}{192(D' + E')} + \frac{F + G}{32(A + B)(A' + B')(D' + E')} $$ $$~~~~~~~~~~~~~~~~~~+ \frac{(D + E)(F + G)}{64(A + B)^3(A' + B')(D' + E')} - \frac{F' + G'}{64(A + B)^3(D' + E')}$$
	where
	$$A = (1 - \lambda)2^n\mu \gamma^{n\mu + 1}(s)$$
	$$B = \lambda \gamma^{n(\mu - 1) + 1}(s)(2^n(\mu - 1) + 2)$$
	$$A' = (1 - \lambda)3^n\mu \gamma^{n\mu + 1}(s)$$
	$$B' = \lambda \gamma^{n(\mu - 1) + 1}(s)(3^n(\mu - 1) + 3)$$
	$$D = -(1 - \lambda)2^{2n}\frac{\mu (\mu - 1)}{2!}\gamma^{n\mu + 2}(s)$$
	$$E = -\lambda \gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{2n} + 2^{n + 1}(\mu - 1)\right)$$
	$$D' = (1 - \lambda)4^n\mu \gamma^{n\mu + 1}(s)$$
	$$E' = \lambda \gamma^{n(\mu - 1) + 1}(s)(4^n(\mu - 1) + 4)$$
	$$F = (1 - \lambda)2^{n + 1}\cdot 3^n \frac{\mu (\mu - 1)}{2!}\gamma^{n\mu + 2}(s)$$
	$$G = \lambda \gamma^{n(\mu - 1) + 2}(s)\left(\frac{(\mu - 1)(\mu - 2)}{2!}2^{n + 1}\cdot 3^n + (\mu - 1)(2\cdot 3^n + 3 \cdot 2^n)\right)$$
	$$F' = (1 - \lambda)2^{3n}\frac{\mu (\mu - 1)(\mu - 2)}{3!}\gamma^{n\mu + 3}(s)$$
	$$G' = \lambda \gamma^{n(\mu - 1) + 3}(s)\left(\frac{(\mu - 1)(\mu - 2)(\mu - 3)}{3!}2^{3n} + \frac{(\mu - 1)(\mu - 2)}{2!}2^{2n + 1}\right)$$
\begin{cor}
 If $f(z)$ belongs to the class $G_\gamma(0, 1, 0)$, then
 $$|a_2| \leq \frac{1}{2\gamma(s)}$$
 $$|a_3| \leq \frac{1}{2\gamma(s)}$$
 $$|a_4| \leq \frac{11}{24\gamma(s)}$$
\end{cor}
{\bf Proof:}
	Substituting $n = 0$, $\mu = 1$ and $\lambda = 0$ into the inequalities (2.18), we have the above result.
\begin{cor} 
If $f(z)$ belongs to the class $G_\gamma(0, 1, 1)$, then
$$|a_2| \leq \frac{1}{4\gamma(s)}$$
$$|a_3| \leq \frac{1}{6\gamma(s)}$$
$$|a_4| \leq \frac{11}{96\gamma(s)}$$
\end{cor}
{\bf Proof:}
	Substituting $n = 0$, $\mu = 1$ and $\lambda = 1$ into the inequalities (2.18), we have the above result.
\begin{cor} 
 If $f(z)$ belongs to the class $G_\gamma(1, 1, 0)$, then
$$|a_2| \leq \frac{1}{4\gamma^2(s)}$$
$$|a_3| \leq \frac{1}{6\gamma^2(s)}$$
$$|a_4| \leq \frac{11}{96\gamma^2(s)}$$
\end{cor} 
{\bf Proof:}
	Substituting $n = 1$, $\mu = 1$ and $\lambda = 0$ into the inequalities (2.18), we have the above result.
\begin{cor} 
If $f(z)$ belongs to the class $G_\gamma(1, 1, 1)$, then
$$|a_2| \leq \frac{1}{4\gamma(s)}$$
$$|a_3| \leq \frac{1}{6\gamma(s)}$$
$$|a_4| \leq \frac{11}{96\gamma(s)}$$
\end{cor} 
{\bf Proof:}
	Substituting $n = 1$, $\mu = 1$ and $\lambda = 1$ into the inequalities (2.18), we have the above result.
\subsection{Fekete-Szego Inequality}
The Fekete-Szego functional $|a_3 - \rho a_2^2|$ for functions of the form
$$f(z) = z + a_2z^2 + a_3z^3 + \cdots$$
that are normalized and univalent in the unit disk is given as 
$$|a_3 - \rho a_2^2| \leq 1 + 2exp(-2\rho/(1 - \rho)); \qquad 0 \leq \rho \leq 1$$
\begin{Theorem}
	If $f(z)$ belongs to the class $G_\lambda (n, \mu, \lambda): n \in N_0, \mu \geq 1, 0\leq \lambda \leq 1$, then
	$$|a_3 - \rho a_2^2| \leq \left\lbrace \begin{array}{cc}
	\frac{1}{2(A' + B')}\left(\frac{D + E}{2(A + B)^2}\right) ; & \rho = 0\\
	\frac{1}{2(A' + B')}\left(\frac{D + E}{2(A + B)^2} + \frac{A' + B'}{A + B}\right) ; & \rho = 1\\
	\frac{1}{2(A' + B')}\left(\frac{D + E}{2(A + B)^2} + \frac{\rho(A' + B')}{A + B}\right) ; & 0 < \rho < 1\end{array} \right\rbrace$$
\end{Theorem}
{\bf Proof:}
	From (2.17),
	$$a_2 = \frac{c_1}{2(A + B)}$$
	$$a_3 = \frac{c_2}{2(A' + B')} - \frac{c_1^2(D + E)}{4(A + B)^2(A' + B')}$$
	Substituting these into $|a_3 - \rho a_2^2|$, we have
	$$|a_3 - \rho a_2^2| = \left|\frac{c_2}{2(A' + B')} - \frac{c_1^2(D + E)}{4(A + B)^2(A' + B')} - \frac{\rho c_1}{2(A + B)} \right|$$
	$$= \left|\frac{c_2}{2(A' + B')} - c_1^2\left(\frac{D + E}{4(A + B)^2(A' + B')} + \frac{\rho}{A + B}\right)\right|$$
	$$= \frac{1}{2(A' + B')}\left|c_2 - c_1^2\left(\frac{D + E}{2(A + B)^2} + \frac{\rho (A' + B')}{A + B}\right)\right|$$
	Then, from the inequality (2.16), we have
	$$|a_3 - \rho a_2^2| \leq \frac{1}{2(A' + B')}max \left\{1, \left|\frac{D + E}{2(A + B)^2} + \frac{\rho (A' + B')}{A + B}\right|\right\}$$
	When $\rho = 0$, we have
	$$|a_3 - \rho a_2^2| \leq \frac{1}{2(A' + B')}\left(\frac{D + E}{2(A + B)^2}\right)$$
	When $\rho = 1$, we have
	$$|a_3 - \rho a_2^2| \leq \frac{1}{2(A' + B')}\left(\frac{D + E}{2(A + B)^2}+ \frac{A' + B'}{A + B}\right)$$
\bibliographystyle{model1b-num-names}
\bibliography{<your-bib-database>}


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\end{document}