Fekete-Szego and Zalcman Functional Estimates for Subclasses of Alpha-Convex Functions Related to Trigonometric Functions

1. Introduction and Preliminaries

Let $\mathcal{T}$ be the class consists of all analytic and normalized functions f, where f has the form

$$f\left(z\right)=z+a_2{z}^2+a_3{z}^3+\cdots ,z\in \mathbb{D},$$

(1)

and $\mathbb{D}:=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ is the open unit disc; also, the subclass of $\mathcal{T}$ consisting of univalent functions is denoted by $\mathcal{S}$.

Let us consider two analytic functions $\mathrm{P}$ and $\mathrm{Q}$ in $\mathbb{D}$. The function $\mathrm{P}$ is said to be subordinated to $\mathrm{Q}$, written symbolically as $\mathrm{P}\left(z\right)\prec \mathrm{Q}\left(z\right)$, if there exists an analytic function $\eta$ in $\mathbb{D}$, with $\eta \left(0\right)=0$ and $\left|\eta \right(z\left)\right|<1$ for all $z\in \mathbb{D}$, such that $\mathrm{P}=\mathrm{Q}\circ \eta$. Further, if $\mathrm{Q}$ is an univalent function in $\mathbb{D}$, then the following equivalence holds (see [1]):

$$\mathrm{P}\left(z\right)\prec \mathrm{Q}\left(z\right)\iff \mathrm{P}\left(0\right)=\mathrm{Q}\left(0\right)\mathrm{and}\mathrm{P}\left(\mathbb{D}\right)\subset \mathrm{Q}\left(\mathbb{D}\right).$$

The family of functions p analytic in $\mathbb{D}$ satisfying the condition $Rep\left(z\right)>0$, $z\in \mathbb{D}$, and of the form

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{t}_n{z}^n,z\in \mathbb{D},$$

(2)

is denoted by $\mathcal{P}$, that represents the well-known Carathéodory function class.

In [2] Mocanu introduced and studied the well-known class of α-convex functions, that is

$${\mathcal{M}}_{\alpha }:=\left\{f\in \mathcal{T}:Re\left[\left(1-\alpha \right){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)\right]>0,z\in \mathbb{D}\right\},\alpha \ge 0,$$

and the properties of this class of functions was extensively studied during a long period by many researchers (see, for example [3,4,5]). In [6] it was proved that all $\alpha$-convex functions are univalent and starlike, while the subclass ${\mathcal{S}}^{*}:={\mathcal{M}}_0$ is called the class of starlike (normalized) functions in $\mathbb{D}$ and ${\mathcal{S}}^c:={\mathcal{M}}_1$ represents the class of convex (normalized) functions in $\mathbb{D}$.

Definition 1.

Let us now define the new classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$, with $\alpha \ge 0$, connected with the sine and cosine functions, respectively, as follows:

$$\begin{array}{ccc}& & {\mathcal{M}}_{\alpha }(sin):=\left\{f\in \mathcal{T}:\left(1-\alpha \right){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)\prec 1+sinz=:\Phi \left(z\right)\right\},\hfill \end{array}$$

(3)

$$\begin{array}{ccc}& & {\mathcal{M}}_{\alpha }(cos):=\left\{f\in \mathcal{T}:\left(1-\alpha \right){\displaystyle \frac{z{f}^{\prime }z}{f\left(z\right)}}+\alpha \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)\prec cosz=:\Psi \left(z\right)\right\}.\hfill \end{array}$$

(4)

Remark 1. (i) Substituting the value of $\alpha =0$ and $\alpha =1$ in (3) we obtain the following subclasses which were studied in [7,8,9], respectively, that are

$${\mathcal{S}}_{sin}^{*}:={\mathcal{M}}_0\left(sin\right),{\mathcal{S}}_{sin}^c:={\mathcal{M}}_1\left(sin\right).$$

(ii) Taking $\alpha =0$ in Equation (4) we obtain the subclass ${\mathcal{S}}_{cos}^{*}:={\mathcal{M}}_0\left(cos\right)$ defined in [10], and by taking $\alpha =1$ in Equation (4) we obtain the subclass ${\mathcal{S}}_{cos}^c:={\mathcal{M}}_1\left(cos\right)$.

(iii) Since the functions Φ and Ψ defined above have real positive parts in $\mathbb{D}$, and moreover (see the Figure 1A,B made with MAPLE™ computer software)

$$Re\Phi \left(z\right)>\frac{1}{10},Re\Psi \left(z\right)>\frac{1}{2},z\in \mathbb{D},$$

it follows that the classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$ are subsets of the class ${\mathcal{M}}_{\alpha }$, that is ${\mathcal{M}}_{\alpha }(sin),{\mathcal{M}}_{\alpha }(cos)\subset {\mathcal{M}}_{\alpha }\subset {\mathcal{S}}^{*}\subset \mathcal{S}$, $\alpha \ge 0$.

The following lemmas are necessary to understand the proofs of our main results.

Lemma 1.

If $p\in \mathcal{P}$ has the form (2), then

$$\begin{array}{ccc}& & |t_n|\le 2,\mathrm{for}n\ge 1,\hfill \end{array}$$

(5)

$$\begin{array}{ccc}& & |t_{i+j}-\mu t_i{t}_j|\le 2max\{1;|1-2\mu |\},\mathrm{for}\mu \in \mathbb{C},,\hfill \end{array}$$

(6)

and for any complex number ζ we have

$$|t_2-\zeta t_1^2|\le 2max\left\{1;|2\zeta -1|\right\}.$$

(7)

The inequality (5) is the well-known Carathéodory’s result (see [11,12]), while (6) may be found in [1], and the inequality (7) is from [13] (see also [14], Lemma 2).

Lemma 2

([7](Lemma 2.2)). If $p\in \mathcal{P}$ has the form (2), then

$$|\alpha t_1^3-\beta t_1{t}_2+\gamma t_3|\le 2|\alpha |+2|\beta -2\alpha |+2|\alpha -\beta +\gamma |.$$

(8)

2. Initial Coefficients Estimates for the Classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$

In this section the coefficients of the functions of the classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$ are analysed, and the upper bounds for the first five coefficients is obtained.

Theorem 1.

If $f\in {\mathcal{M}}_{\alpha }(sin)$ has the form (1), then

$$\begin{array}{cc}\hfill |a_2|\le & \frac{1}{1+\alpha },\hfill \\ \hfill |a_3|\le & \frac{1}{2\left(1+2\alpha \right)}max\left\{1;\frac{3\alpha +1}{{\left(1+\alpha \right)}^2}\right\},\hfill \\ \hfill |a_4|\le & \frac{1}{6\left(1+3\alpha \right)}\left(max\left\{1;{\displaystyle \frac{6\alpha (\alpha -1)}{\left(1+2\alpha \right)\left(1+\alpha \right)}}\right\}+{\displaystyle \frac{-4{\alpha }^4+31{\alpha }^3+30{\alpha }^2+35\alpha +4}{3\left(1+2\alpha \right){\left(1+\alpha \right)}^3}}\right),\hfill \\ \hfill |a_5|\le & \frac{1}{4\left(1+4\alpha \right)}[\frac{1}{2}\left(max\left\{1;\frac{\left|27{\alpha }^2-20\alpha +1\right|}{9{\alpha }^2+12\alpha +3}\right\}+1\right)\hfill \\ \hfill & +\frac{|{\Phi }_1\left(\alpha \right)\left|\right|{\Phi }_2\left(\alpha \right)|}{18\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}+\mathrm{M}\left(\alpha \right)].\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\Phi }_1\left(\alpha \right)& :=-108{\alpha }^7+200{\alpha }^6+556{\alpha }^5+1042{\alpha }^4+631{\alpha }^3+409{\alpha }^2+145\alpha +5,\hfill \\ \hfill {\Phi }_2\left(\alpha \right)& :=-180{\alpha }^5-8{\alpha }^4+69{\alpha }^3+187{\alpha }^2+75\alpha +1,\hfill \end{array}$$

and

$$\mathrm{M}\left(\alpha \right):=\frac{8{\alpha }^2+1}{9{\left(1+2\alpha \right)}^2}max\left\{1;\left|1-\frac{{\Phi }_2\left(\alpha \right)}{\left(1+3\alpha \right){\left(1+\alpha \right)}^2\left(8{\alpha }^2+1\right)}\right|\right\}.$$

Proof. 

If $f\in {\mathcal{M}}_{\alpha }(sin)$, then there exists a function $\eta$ that is analytic in $\mathbb{D}$ and satisfy the conditions $\eta \left(0\right)=0$ and $\left|\eta \right(z\left)\right|<1$ for all $z\in \mathbb{D}$, such that

$$\left(1-\alpha \right){\displaystyle \frac{z{f}^{\prime }z}{f\left(z\right)}}+\alpha \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)=\Phi \left(\eta \left(z\right)\right)=1+sin\eta \left(z\right),z\in \mathbb{D}.$$

Since f is of the form (1), it follows that

$$\begin{array}{cc}\hfill \left(1-\alpha \right){\displaystyle \frac{z{f}^{\prime }z}{f\left(z\right)}}+\alpha \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)=& 1+\left(1+\alpha \right)a_{2}z+\left(-3\alpha a_2^2+4\alpha a_3-a_2^2+2a_3\right)z^2\hfill \\ \hfill & +\left(7\alpha a_2^3-15\alpha a_2{a}_3+a_2^3+9\alpha a_4-3a_2{a}_3+3a_4\right)z^3\hfill \\ \hfill & +\left(-15\alpha a_2^4+44\alpha a_2^2{a}_3-a_2^4-28\alpha a_2{a}_4-16\alpha a_3^2\right.\hfill \\ \hfill & \left.+4a_2^2{a}_3+16\alpha a_5-4a_2{a}_4-2a_3^2+4a_5\right)z^4+\cdots ,z\in \mathbb{D}.\hfill \end{array}$$

(9)

From the fact that $\eta \left(0\right)=0$ and $\left|\eta \right(z\left)\right|<1$ for all $z\in \mathbb{D}$, if we define the function p by

$$p\left(z\right):=\frac{1+\eta \left(z\right)}{1-\eta \left(z\right)}=1+t_{1}z+t_2{z}^2+\cdots ,z\in \mathbb{D},$$

we obtain that $p\in \mathcal{P}$ and

$$\eta \left(z\right)=\frac{p\left(z\right)-1}{p\left(z\right)+1}=\frac{t_{1}z+t_2{z}^2+\cdots }{2+t_{1}z+t_2{z}^2+\cdots },z\in \mathbb{D}.$$

According to the above relation we get

$$\begin{array}{cc}\hfill 1+sin\eta \left(z\right)=& 1+\frac{1}{2}{t}_{1}z+\left(\frac{t_2}{2}-\frac{t_1^2}{4}\right)z^2+\left(\frac{1}{2}{t}_3-\frac{1}{2}{t}_1{t}_2+\frac{5}{48}{t}_1^3\right)z^3\hfill \\ & +\left(\frac{5}{16}{t}_2{t}_1^2-\frac{1}{32}{t}_1^4+\frac{1}{2}{t}_4-\frac{1}{2}{t}_1{t}_3-\frac{1}{4}{t}_2^2\right)z^4+\cdots ,z\in \mathbb{D},\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill cos\eta \left(z\right)=& 1-\frac{t_1^2}{8}{z}^2+\left(-\frac{1}{4}{t}_1{t}_2+\frac{1}{8}{t}_1^3\right)z^3\hfill \\ \hfill & +\left(-\frac{35}{384}{t}_1^4-\frac{1}{8}{t}_2^2+\frac{3}{8}{t}_2{t}_1^2-\frac{1}{4}{t}_1{t}_3\right)z^4+\cdots ,z\in \mathbb{D},\hfill \end{array}$$

(11)

and equating the corresponding coefficients of (9) and (10) we obtain

$$\begin{array}{cc}\hfill a_2=& \frac{t_1}{2(1+\alpha )},\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill a_3=& \frac{2{\left(1+\alpha \right)}^2{t}_2+\alpha \left(1-\alpha \right)t_1^2}{8\left(1+2\alpha \right){\left(1+\alpha \right)}^2},\hfill \\ \hfill a_4=& \frac{5\left(1+2\alpha \right){\left(1+\alpha \right)}^3-6\left(1+7\alpha \right)\left(1+2\alpha \right)+9\left(1+5\alpha \right)\left(1-\alpha \right)\alpha }{144\left(1+3\alpha \right)\left(1+2\alpha \right){\left(1+\alpha \right)}^3}{t}_1^3\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill & +\frac{6\left(1+5\alpha \right)-8\left(1+2\alpha \right)\left(1+\alpha \right)}{48\left(1+3\alpha \right)\left(1+2\alpha \right)\left(1+\alpha \right)}{t}_1{t}_2+\frac{1}{6\left(1+3\alpha \right)}{t}_3,\hfill \\ \hfill a_5=& \frac{1}{4\left(1+4\alpha \right)}[\frac{{\Phi }_1\left(\alpha \right)}{288\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^4}{t}_1^4-\frac{{\Phi }_2\left(\alpha \right)}{48\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}{t}_1^2{t}_2\hfill \end{array}$$

(14)

$$\begin{array}{cc}\hfill & +\frac{2\left(1+7\alpha \right)-3\left(1+3\alpha \right)\left(1+\alpha \right)}{6\left(1+3\alpha \right)\left(1+\alpha \right)}{t}_1{t}_3+\frac{\left(1+8\alpha \right)-2{\left(1+2\alpha \right)}^2}{8{\left(1+2\alpha \right)}^2}{t}_2^2+\frac{1}{2}{t}_4].\hfill \end{array}$$

(15)

Using (12) we get

$$|a_2|=\frac{1}{2\left(1+\alpha \right)}\left|t_1\right|,$$

and from (5) we have $|t_1|\le 2$, hence

$$|a_2|\le \frac{1}{1+\alpha }.$$

The relation (13) leads to

$$|a_3|=\left|\frac{t_2}{4\left(1+2\alpha \right)}-\frac{\alpha \left(\alpha -1\right)t_1^2}{8\left(1+2\alpha \right){\left(1+\alpha \right)}^2}\right|,$$

using triangle inequality we get

$$|a_3|\le \frac{1}{4\left(1+2\alpha \right)}\left|t_2-\frac{\alpha \left(\alpha -1\right)t_1^2}{2{\left(1+\alpha \right)}^2}\right|,$$

and according to (7), since $\alpha \ge 0$, we obtain

$$\begin{array}{cc}\hfill |a_3|=\frac{1}{4\left(1+2\alpha \right)}\left|t_2-\frac{\alpha \left(\alpha -1\right)t_1^2}{2{\left(1+\alpha \right)}^2}\right|& \le \frac{1}{4\left(1+2\alpha \right)}2max\left\{1;\left|2·\frac{\alpha \left(\alpha -1\right)}{2{\left(1+\alpha \right)}^2}-1\right|\right\}\hfill \\ \hfill & =\frac{1}{2\left(1+2\alpha \right)}max\left\{1;\frac{3\alpha +1}{{\left(1+\alpha \right)}^2}\right\}.\hfill \end{array}$$

The equality (14) leads to

$$\begin{array}{cc}\hfill |a_4|=& \frac{1}{3\left(1+3\alpha \right)}\left|\frac{1}{4}\left(t_3-\frac{4\left(1+2\alpha \right)\left(1+\alpha \right)-3\left(1+5\alpha \right)}{2\left(1+2\alpha \right)\left(1+\alpha \right)}{t}_1{t}_2\right)\right.\hfill \\ \hfill & \left.+\frac{t_3}{4}-\frac{6\left(1+7\alpha \right)\left(1+2\alpha \right)-5\left(1+2\alpha \right){\left(1+\alpha \right)}^3-9\left(1+5\alpha \right)\left(1-\alpha \right)\alpha }{12\left(1+2\alpha \right){\left(1+\alpha \right)}^3}{t}_1^3\right|,\hfill \end{array}$$

and using the triangle inequality we get

$$\begin{array}{cc}\hfill |a_4|\le & \frac{1}{3\left(1+3\alpha \right)}[\frac{1}{4}\left|t_3-\frac{4\left(1+2\alpha \right)\left(1+\alpha \right)-3\left(1+5\alpha \right)}{2\left(1+2\alpha \right)\left(1+\alpha \right)}{t}_1{t}_2\right|\hfill \\ \hfill & +\frac{|t_3|}{4}+\frac{6\left(1+7\alpha \right)\left(1+2\alpha \right)-5\left(1+2\alpha \right){\left(1+\alpha \right)}^3-9\left(1+5\alpha \right)\left(1-\alpha \right)\alpha }{12\left(1+2\alpha \right){\left(1+\alpha \right)}^3}{\left|t_1\right|}^3].\hfill \end{array}$$

From (5), (6) and of Lemma 1, the above inequality implies that

$$\begin{array}{cc}\hfill |a_4|\le & \frac{1}{6\left(1+3\alpha \right)}\left(max\left\{1;{\displaystyle \frac{6\alpha (\alpha -1)}{\left(1+2\alpha \right)\left(1+\alpha \right)}}\right\}+{\displaystyle \frac{-4{\alpha }^4+31{\alpha }^3+30{\alpha }^2+35\alpha +4}{3\left(1+2\alpha \right){\left(1+\alpha \right)}^3}}\right).\hfill \end{array}$$

By rearranging (15) we get

$$\begin{array}{cc}\hfill |a_5|=& \frac{1}{4\left(1+4\alpha \right)}|\frac{1}{4}\left(t_4-\frac{6\left(1+3\alpha \right)\left(1+\alpha \right)-4\left(1+7\alpha \right)}{3\left(1+3\alpha \right)\left(1+\alpha \right)}{t}_1{t}_3\right)\hfill \\ \hfill & +\frac{1}{4}\left(t_4-\frac{-7\left(\left(1+8\alpha \right)-2{\left(1+2\alpha \right)}^2\right)}{18{\left(1+2\alpha \right)}^2}{t}_2^2\right)\hfill \\ \hfill & -\frac{5{\Phi }_2\left(\alpha \right)}{144\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}{t}_1^2{t}_2+\frac{{\Phi }_1\left(\alpha \right)}{288\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^4}{t}_1^4\hfill \\ \hfill & -\frac{2{\left(1+2\alpha \right)}^2-\left(1+8\alpha \right)}{36{\left(1+2\alpha \right)}^2}{t}_2\left(t_2-\frac{{\Phi }_2\left(\alpha \right)}{2\left(1+3\alpha \right){\left(1+\alpha \right)}^2\left(2{\left(1+2\alpha \right)}^2-\left(1+8\alpha \right)\right)}{t}_1^2\right)|,\hfill \end{array}$$

and using triangle inequality we get

$$\begin{array}{cc}\hfill |a_5|\le & \frac{1}{4\left(1+4\alpha \right)}[\frac{1}{4}\left|t_4-\frac{6\left(1+3\alpha \right)\left(1+\alpha \right)-4\left(1+7\alpha \right)}{3\left(1+3\alpha \right)\left(1+\alpha \right)}{t}_1{t}_3\right|\hfill \\ \hfill & +\frac{1}{4}\left|t_4-\frac{7\left(2{\left(1+2\alpha \right)}^2-\left(1+8\alpha \right)\right)}{18{\left(1+2\alpha \right)}^2}{t}_2^2\right|\hfill \\ \hfill & +\frac{5|{\Phi }_2\left(\alpha \right)|}{144\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}|t_1{|}^2|t_2|+\frac{|{\Phi }_1\left(\alpha \right)|}{288\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^4}{\left|t_1\right|}^4\hfill \\ \hfill & +\frac{8{\alpha }^2+1}{36{\left(1+2\alpha \right)}^2}\left|t_2\right||t_2-\frac{{\Phi }_2\left(\alpha \right)}{2\left(1+3\alpha \right){\left(1+\alpha \right)}^2\left(8{\alpha }^2+1)\right)}{t}_1^2|].\hfill \end{array}$$

(16)

Now we will find an upper bound for the each term of the right hand side of the above inequality, as follows.

(i) According to (6) we have

$$\begin{array}{cc}\hfill \left|t_4-\frac{6\left(1+3\alpha \right)\left(1+\alpha \right)-4\left(1+7\alpha \right)}{3\left(1+3\alpha \right)\left(1+\alpha \right)}{t}_1{t}_3\right|& \le 2max\left\{1;\left|1-\frac{12\left(1+3\alpha \right)\left(1+\alpha \right)-8\left(1+7\alpha \right)}{3\left(1+3\alpha \right)\left(1+\alpha \right)}\right|\right\}\hfill \\ \hfill & =2max\left\{1;\left|\frac{27{\alpha }^2-20\alpha +1}{9{\alpha }^2+12\alpha +3}\right|\right\},\hfill \end{array}$$

whenever $\alpha \ge 0$.

(ii) Using again the inequality (6) a simple computation shows that

$$\begin{array}{cc}\hfill \left|t_4-\frac{7\left(2{\left(1+2\alpha \right)}^2-\left(1+8\alpha \right)\right)}{18{\left(1+2\alpha \right)}^2}{t}_2^2\right|& \le 2max\left\{1;\left|1-\frac{7\left(2{\left(1+2\alpha \right)}^2-\left(1+8\alpha \right)\right)}{9{\left(1+2\alpha \right)}^2}\right|\right\}\hfill \\ \hfill & =2max\left\{1;\left|\frac{2(10{\alpha }^2-18\alpha -1)}{9{\left(1+2\alpha \right)}^2}\right|\right\}=2,\hfill \end{array}$$

whenever $\alpha \ge 0$.

(iii) For the sum of the third with the fourth term, using the inequality (5) we obtain

$$\begin{array}{c}\hfill \frac{5|{\Phi }_2\left(\alpha \right)|}{144\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}|t_1{|}^2|t_2|+\frac{|{\Phi }_1\left(\alpha \right)|}{288\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^4}{\left|t_1\right|}^4\\ \hfill \le \frac{|{\Phi }_1\left(\alpha \right)\left|\right|{\Phi }_2\left(\alpha \right)|}{18\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}.\end{array}$$

(v) To get a majorant for the last term of the sum, according to (5) and (7) we have

$$\begin{array}{c}\hfill \frac{8{\alpha }^2+1}{36{\left(1+2\alpha \right)}^2}\left|t_2\right||t_2-\frac{{\Phi }_2\left(\alpha \right)}{2\left(1+3\alpha \right){\left(1+\alpha \right)}^2\left(2{\left(1+2\alpha \right)}^2-\left(1+8\alpha \right)\right)}{t}_1^2|\}\\ \hfill \le \frac{8{\alpha }^2+1}{9{\left(1+2\alpha \right)}^2}max\left\{1;\left|1-\frac{{\Phi }_2\left(\alpha \right)}{\left(1+3\alpha \right){\left(1+\alpha \right)}^2\left(8{\alpha }^2+1\right)}\right|\right\}:=M\left(\alpha \right).\end{array}$$

Finally, using the upper bounds found to the items (i)–(v), from the inequality (16) we conclude that

$$\begin{array}{cc}\hfill |a_5|\le & \frac{1}{4\left(1+4\alpha \right)}\{\frac{1}{2}\left(max\left\{1;\frac{\left|27{\alpha }^2-20\alpha +1\right|}{9{\alpha }^2+12\alpha +3}\right\}+1\right)\hfill \\ \hfill & +\frac{|{\Phi }_1\left(\alpha \right)\left|\right|{\Phi }_2\left(\alpha \right)|}{18\left(1+3\alpha \right){\left(1+2\alpha \right)}^2{\left(1+\alpha \right)}^2}+\mathrm{M}\left(\alpha \right)\}.\hfill \end{array}$$

□

Remark 2.

A simple computation shows that the upper bounds obtained in the Theorem 1 could be written in the following forms:

$$|a_3|\le \left\{\begin{array}{ccc}{\displaystyle \frac{3\alpha +1}{2(1+2\alpha ){(1+\alpha )}^2}},\hfill & \mathrm{if}\hfill & 0\le \alpha \le 1,\hfill \\ {\displaystyle \frac{1}{2(1+2\alpha )}},\hfill & \mathrm{if}\hfill & \alpha \ge 1,\hfill \end{array}\right.$$

and

$$|a_4|\le \left\{\begin{array}{ccc}{\displaystyle \frac{2{\alpha }^4+52{\alpha }^3+57{\alpha }^2+50\alpha +7}{18\left(1+3\alpha \right)\left(1+2\alpha \right){\left(1+\alpha \right)}^3}},\hfill & \mathrm{if}\hfill & 0\le \alpha \le {\displaystyle \frac{9+\sqrt{97}}{8}},\hfill \\ {\displaystyle \frac{14{\alpha }^4+49{\alpha }^3+12{\alpha }^2+17\alpha +4}{18\left(1+3\alpha \right)\left(1+2\alpha \right){\left(1+\alpha \right)}^3}},\hfill & \mathrm{if}\hfill & \alpha \ge {\displaystyle \frac{9+\sqrt{97}}{8}}.\hfill \end{array}\right.$$

For $\alpha =0$ and $\alpha =1$, Theorem 1 reduces to the following corollary:

Corollary 1.(i) If $f\in {\mathcal{S}}_{sin}^{*}$ has the form (1), then

$$|a_2|\le 1,|a_3|\le \frac{1}{2},|a_4|\le \frac{7}{18},\left|a_5\right|\le \frac{25}{72}.$$

(ii) If $f\in {\mathcal{S}}_{sin}^c$ has the form (1), then

$$|a_2|\le \frac{1}{2},|a_3|\le \frac{1}{6},|a_4|\le \frac{7}{72},\left|a_5\right|\le \frac{145}{18}.$$

Remark 3.

The upper bounds given by Theorem 1 are not the best possible, excepting those for the first two coefficients.

(i) Thus, for the case $\alpha =0$, the function

$$\widehat{f}\left(z\right):=zexp\left({\int }_0^z\frac{sin\left(ct\right)}{t}\mathrm{d}t\right)=z+c{z}^2+\frac{c^2}{2}{z}^3+\frac{c^3}{9}{z}^4-\frac{c^4}{72}{z}^5-\cdots ,z\in \mathbb{D},\mathrm{with}\left|c\right|=1,$$

is the solution $\widehat{f}\in \mathcal{T}$ of the differential equation ${\displaystyle \frac{z{\widehat{f}}^{\prime }\left(z\right)}{\widehat{f}\left(z\right)}}=1+sin\left(cz\right)$, $\left|c\right|=1$, therefore $\widehat{f}\in {\mathcal{S}}_{sin}^{*}:={\mathcal{M}}_0(sin)$. For $\widehat{f}$ we have

$$|a_2|=1,|a_3|={\displaystyle \frac{1}{2}},|a_4|=\frac{1}{9}<\frac{7}{18},\left|a_5\right|={\displaystyle \frac{1}{72}}<{\displaystyle \frac{25}{72}},$$

hence the estimations given by Theorem 1 are not sharp for $|a_4|$ and $|a_5|$.

(ii) Similarly, for $\alpha =1$, the function

$$\begin{array}{cc}\hfill \tilde{f}\left(z\right)& :={\int }_0^z\left[exp\left({\int }_0^x\frac{sin\left(ct\right)}{t}\mathrm{d}t\right)\right]\mathrm{d}x\hfill \\ \hfill & =z+\frac{c}{2}{z}^2+\frac{c^2}{6}{z}^3+\frac{c^3}{36}{z}^4-\frac{c^4}{360}{z}^5--\cdots ,z\in \mathbb{D},\mathrm{with}\left|c\right|=1,\hfill \end{array}$$

is the solution $\tilde{f}\in \mathcal{T}$ of the differential equation $1+{\displaystyle \frac{z{\widehat{f}}^{\prime \prime }\left(z\right)}{{\widehat{f}}^{\prime }\left(z\right)}}=1+sin\left(cz\right)$, $\left|c\right|=1$, hence $\tilde{f}\in {\mathcal{S}}_{sin}^c:={\mathcal{M}}_1(sin)$. For this function

$$|a_2|={\displaystyle \frac{1}{2}},|a_3|={\displaystyle \frac{1}{6}},|a_4|=\frac{1}{36}<\frac{7}{72},\left|a_5\right|={\displaystyle \frac{1}{360}}<{\displaystyle \frac{145}{18}},$$

thus the estimations of Theorem 1 are not sharp for $|a_4|$ and $|a_5|$.

Theorem 2.

If $f\in {\mathcal{M}}_{\alpha }(cos)$ has the form (1), then

$$\begin{array}{cc}\hfill & |a_2|=0,|a_3|\le \frac{1}{4(1+2\alpha )},\hfill \\ \hfill & |a_4|\le \frac{1}{3\left(1+3\alpha \right)},\left|a_5\right|\le \frac{1}{8(1+4\alpha )}\left(\frac{3\alpha |1-\alpha |}{{(1+2\alpha )}^2}+\frac{11}{3}\right).\hfill \end{array}$$

Proof. 

If $f\in {\mathcal{M}}_{\alpha }(cos)$, then equating the corresponding coefficients of (9) and (11) we obtain

$$\begin{array}{cc}\hfill a_2=& 0,\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill a_3=& -\frac{t_1^2}{16(1+2\alpha )},\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill a_4=& -\frac{t_1}{12\left(1+3\alpha \right)}\left(t_2-\frac{t_1^2}{2}\right),\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill a_5=& \frac{4\alpha (1-\alpha )}{512\left(1+4\alpha \right){\left(1+2\alpha \right)}^2}{t}_1^4-\frac{t_1}{8(1+4\alpha )}\left(\frac{1}{2}{t}_3-\frac{3}{4}{t}_1{t}_2+\frac{1}{6}{t}_1^3\right)-\frac{t_2^2}{32\left(1+4\alpha \right)}.\hfill \end{array}$$

(20)

Using (18) we get

$$|a_3|=\frac{1}{16\left(1+2\alpha \right)}{\left|t_1\right|}^2,$$

and from (5) we have $|t_1|\le 2$, hence

$$|a_3|\le \frac{1}{4\left(1+2\alpha \right)}.$$

The relation (19) leads to

$$|a_4|=\left|\frac{t_1}{12\left(1+3\alpha \right)}\left(t_2-\frac{t_1^2}{2}\right)\right|,$$

and according to (5) and (7), we obtain

$$|a_4|\le \frac{1}{3\left(1+3\alpha \right)}.$$

From the equality (20) we have

$$|a_5|=\left|\frac{4\alpha (1-\alpha )}{512\left(1+4\alpha \right){\left(1+2\alpha \right)}^2}{t}_1^4-\frac{t_1}{8(1+4\alpha )}\left(\frac{1}{2}{t}_3-\frac{3}{4}{t}_1{t}_2+\frac{1}{6}{t}_1^3\right)-\frac{t_2^2}{32\left(1+4\alpha \right)}\right|,$$

and using the triangle inequality we get

$$|a_5|\le \frac{4\alpha \left|1-\alpha \right|}{512\left(1+4\alpha \right){\left(1+2\alpha \right)}^2}{\left|t_1\right|}^4+\frac{|t_1|}{8(1+4\alpha )}\left|\frac{1}{2}{t}_3-\frac{3}{4}{t}_1{t}_2+\frac{1}{6}{t}_1^3\right|+\frac{|t_2{|}^2}{32\left(1+4\alpha \right)}.$$

From (5) and Lemma 2 for the appropriate values $\alpha ={\displaystyle \frac{1}{6}}$, $\beta ={\displaystyle \frac{3}{4}}$, and $\gamma ={\displaystyle \frac{1}{2}}$, the above inequality implies that

$$|a_5|\le \frac{1}{8(1+4\alpha )}\left(\frac{3\alpha |1-\alpha |}{{(1+2\alpha )}^2}+\frac{11}{3}\right),$$

and all the estimations are proved. □

For $\alpha =0$ and $\alpha =1$ the Theorem 2 leads us to the following corollary.

Corollary 2.(i) If $f\in {\mathcal{S}}_{cos}^{*}$ has the form (1), then

$$|a_2|=0,|a_3|\le \frac{1}{4},|a_4|\le \frac{1}{3},\left|a_5\right|\le \frac{11}{24}.$$

(ii) If $f\in {\mathcal{S}}_{cos}^c$ has the form (1), then

$$|a_2|=0,|a_3|\le \frac{1}{12},|a_4|\le \frac{1}{12},\left|a_5\right|\le \frac{11}{120}.$$

Remark 4.

The estimations given by Theorem 2 are not the best possible, excepting those for the first two coefficients.

(i) Thus, for $\alpha =0$ and if $f\in {\mathcal{S}}_{cos}^{*}$, then the inequality $|a_3|\le {\displaystyle \frac{1}{4}}$ is sharp and it is attained for the function $f_{*}\in {\mathcal{S}}_{cos}^{*}$ that satisfies the differential equation ${\displaystyle \frac{z{f}_{*}^{\prime }\left(z\right)}{f_{*}\left(z\right)}}=cos\left(cz\right)$, $\left|c\right|=1$, that is

$$f_{*}\left(z\right):=zexp\left({\int }_0^z\frac{cos\left(ct\right)-1}{t}\mathrm{d}t\right)=z-\frac{c^2}{4}{z}^3+\frac{c^4}{24}{z}^5-\frac{47c^6}{8640}{z}^7+\cdots ,z\in \mathbb{D},\mathrm{with}\left|c\right|=1.$$

(ii) Also, for $\alpha =1$ and if $f\in {\mathcal{S}}_{cos}^c$, then the inequality $|a_3|\le {\displaystyle \frac{1}{12}}$ is sharp being attained for the function $f_{\circ }\in {\mathcal{S}}_{cos}^{*}$ that it is the solution of the differential equation $1+{\displaystyle \frac{z{f}_{\circ }^{\prime \prime }\left(z\right)}{f_{\circ }^{\prime }\left(z\right)}}=cos\left(cz\right)$, $\left|c\right|=1$, and

$$\begin{array}{cc}\hfill f_{\circ }\left(z\right)& :={\int }_0^z\left[exp\left({\int }_0^x\frac{cost-1}{t}\mathrm{d}t\right)\right]\mathrm{d}x\hfill \\ \hfill & =z-\frac{c^2}{12}{z}^3+\frac{c^4}{120}{z}^5-\frac{47c^6}{60480}{z}^7+\cdots ,z\in \mathbb{D},\mathrm{with}\left|c\right|=1.\hfill \end{array}$$

3. The Fekete-Szego Inequality for the Classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$

In this section we determine upper bounds for the Fekete-Szego functional for the new defined classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$.

Theorem 3.

If $f\in {\mathcal{M}}_{\alpha }(sin)$ has the form (1), then

$$|a_3-\rho a_2^2|\le \frac{1}{2\left(1+2\alpha \right)}max\left\{1;\frac{\left|2\rho \left(1+2\alpha \right)-\left(1+3\alpha \right)\right|}{{\left(1+\alpha \right)}^2}\right\},\rho \in \mathbb{C}.$$

Proof. 

If $f\in {\mathcal{M}}_{\alpha }(sin)$, then from (12) and (13) we get

$$\begin{array}{cc}\hfill a_3-\rho a_2^2=& \frac{1+3\alpha }{2\left(1+2\alpha \right)}\frac{t_1^2}{4{\left(1+\alpha \right)}^2}+\frac{1}{4\left(1+2\alpha \right)}\left(t_2-\frac{t_1^2}{2}\right)-\rho \frac{t_1^2}{4{\left(1+\alpha \right)}^2},\hfill \\ \hfill =& \frac{1}{4\left(1+2\alpha \right)}\left(t_2-\frac{2\rho \left(1+2\alpha \right)-\alpha \left(1-\alpha \right)}{2{\left(1+\alpha \right)}^2}{t}_1^2\right),\hfill \end{array}$$

and using (7) it follows that

$$\begin{array}{cc}\hfill |a_3-\rho a_2^2|\le & \frac{1}{4\left(1+2\alpha \right)}2max\left\{1;\left|\frac{2\rho \left(1+2\alpha \right)-\alpha \left(1-\alpha \right)}{{\left(1+\alpha \right)}^2}-1\right|\right\},\hfill \\ \hfill =& \frac{1}{2\left(1+2\alpha \right)}max\left\{1;\frac{\left|2\rho \left(1+2\alpha \right)-\left(1+3\alpha \right)\right|}{{\left(1+\alpha \right)}^2}\right\}.\hfill \end{array}$$

□

For $\alpha =0$ and $\alpha =1$, the following special are obtained.

Corollary 3.(i) If $f\in {\mathcal{S}}_{sin}^{*}$, then

$$|a_3-\rho a_2^2|\le \frac{1}{2}max\left\{1;|2\rho -1|\right\},\rho \in \mathbb{C}.$$

(ii) If $f\in {\mathcal{S}}_{sin}^c$, then

$$|a_3-\rho a_2^2|\le \frac{1}{6}max\left\{1;\frac{|3\rho -2|}{2}\right\},\rho \in \mathbb{C}.$$

Remark 5.

1. According to the Remark 3, the upper bounds given by Theorem 3 are the best possible for $\alpha =0$ and $\alpha =1$.

2. If $f\in {\mathcal{M}}_{\alpha }(cos)$ has the form (1), from (17), (18) and (5) we get

$$\left|a_3-\rho a_2^2\right|=\left|a_3\right|=\frac{{\left|t_1\right|}^2}{16(1+2\alpha )}\le \frac{1}{4\left(1+2\alpha \right)},\rho \in \mathbb{C},$$

hence to find the upper bound of the Fekete-Szego functional is obvious.

4. The Zalcman Functional Estimate for the Class ${\mathcal{M}}_{\alpha }(cos)$

Zalcman conjectured in 1960 that the coefficients of the functions $f\in \mathcal{S}$ having the form (1) satisfies the inequality

$$\left|a_n^2-a_{2n-1}\right|\le {(n-1)}^2,n\ge 2.$$

Further, the equality is obtained only for the Koebe function $k\left(z\right)={\displaystyle \frac{z}{{(1-z)}^2}}$ and its rotations. Like it was shown in [15,16] it implies the Bieberbach conjecture, that is $|a_n|\le n$, $n\ge 2$. It is noteworthy that for $n=2$ the above inequality is a well-known consequence of the Area Theorem and could be found in [1, Theorem 1.5]. In the recent years the Zalcman functional has been given a special interest by many researchers (see, for example, [17,18,19]).

In the next result, for $n=3$ we find the Zalcman functional upper bound for the class ${\mathcal{M}}_{\alpha }(cos)$ that allows us to prove that the Zalcman conjecture is holds in this case.

Theorem 4.

If $f\in {\mathcal{M}}_{\alpha }(cos)$ has the form (1), then

$$\left|a_3^2-a_5\right|\le \frac{170{\alpha }^2+170\alpha +41}{96\left(1+4\alpha \right){\left(1+2\alpha \right)}^2}.$$

(21)

Proof. 

For $f\in {\mathcal{M}}_{\alpha }(cos)$, using the equalities (18) and (20) it follows that

$$\begin{array}{cc}\hfill a_3^2-a_5=& \frac{70{\alpha }^2+70\alpha +19}{768{\left(2\alpha +1\right)}^2\left(4\alpha +1\right)}{t}_1^4+\frac{1}{8\left(1+4\alpha \right)}\left(\frac{1}{2}{t}_1{t}_3+\frac{1}{4}{t}_2^2-\frac{3}{4}{t}_1^2{t}_2\right),\hfill \\ \hfill =& \frac{t_1}{8\left(1+4\alpha \right)}\left(\frac{70{\alpha }^2+70\alpha +19}{96{\left(2\alpha +1\right)}^2}{t}_1^3-\frac{3}{4}{t}_1{t}_2+\frac{1}{2}{t}_3\right)+\frac{1}{32\left(1+4\alpha \right)}{t}_2^2,\hfill \end{array}$$

and from the triangle inequality we get

$$\begin{array}{cc}\hfill |a_3^2-a_5|\le & \frac{|t_1|}{8\left(1+4\alpha \right)}|\frac{70{\alpha }^2+70\alpha +19}{96{\left(2\alpha +1\right)}^2}{t}_1^3-\frac{3}{4}{t}_1{t}_2+\frac{1}{2}{t}_3|+\frac{1}{32\left(1+4\alpha \right)}{\left|t_2\right|}^2.\hfill \end{array}$$

Using the inequalities (5) of Lemma 1 and (8) of Lemma 2, the above relation leads easily to (21). □

Since

$$4-\frac{170{\alpha }^2+170\alpha +41}{\left(96+384\alpha \right){\left(1+2\alpha \right)}^2}=\frac{6144{\alpha }^3+7510{\alpha }^2+2902\alpha +343}{\left(96+384\alpha \right){\left(1+2\alpha \right)}^2}>0,\mathrm{for}\mathrm{all}\alpha \ge 0,$$

using the result of the Theorem 4 we deduce that:

Corollary 4.

If $f\in {\mathcal{M}}_{\alpha }(cos)$ has the form (1), then

$$\left|a_3^2-a_5\right|\le 4,$$

therefore the Zalcman conjecture hold for the class ${\mathcal{M}}_{\alpha }(cos)$ if $n=3$.

5. Conclusions

This paper mainly focuses on finding the upper bounds of the first five coefficients for the classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$ of $\alpha$-convex functions connected with the sine and cosine function. Also, we obtained the estimate for Fekete-Szego functional for these classes, we found the upper bound for Zalcman functional for these class ${\mathcal{M}}_{\alpha }(cos)$ for the case $n=3$, and this allows us to prove that the Zalcman inequality holds for this case.

Like we mentioned in the Remarks 3 and 4 the upper bounds we get for $\left|a_4\right|$ and $\left|a_5\right|$ for the functions that belong to the classes ${\mathcal{M}}_{\alpha }(sin)$ and ${\mathcal{M}}_{\alpha }(cos)$ are not the best possible, hence the estimation given in Theorem 4 is not sharp. The problem of finding the best bounds of the above mentioned coefficients and functionals for these classes remains an interesting open question.