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# Application of a subordination theorem associated with certain new generalized subclasses of analytic and univalent functions

## Abstract

The prime focus of the present work is to investigate some fascinating relations of some analytic and univalent functions using a subordination theorem.

## Introduction

Let H denote the class of normalized analytic functions f(z) having the form:

$$f(z)=z+a_{2}z^{2}+a_{3}z^{3}+...$$
(1)

in the unit disk $$U=\left \{ z\in \mathbb {C}:|z|<1\right \}$$. Also, let S denote the subclass of H univalent in U. Suppose that Sâˆ— denote the subclass of S consisting of the functions f(z) which are starlike in U. A function f(z)âˆˆK is said to be convex in U if f(z)âˆˆS satisfies the condition that zfâ€²(z)âˆˆSâˆ—. If f(z)âˆˆH satisfies the geometric condition:

$$\Re \left(\frac{zf^{\prime }(z)}{f(z)}\right)>\beta,\;\;z\in U$$

for some real Î²(0â‰¤Î²<1), then we say that f(z) belongs to the class Sâˆ—(Î²) starlike of order Î², and if f(z)âˆˆH satisfies the geometric condition:

$$\Re \left(1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\right)>\beta,\;\;z\in U$$

for some real Î²(0â‰¤Î²<1), then we say that f(z) belongs to the class K(Î²) convex of order Î² (see [1, 2]). Let the function g(z) of the form:

$$g(z)=z+z^{3}+z^{5}+...\;\;\;z\in U$$
(2)

be in the class Sâˆ— while the function g(z) of the form:

$$g(z)=z+z^{2}+z^{3}+...\;\;\;z\in U$$
(3)

be in the class K. With reference to (2) and (3), we can write that:

$$g_{\alpha }(z)=\frac{z}{1-z^{\alpha }}=z+\sum\limits_{k=1}^{\infty }z^{1+k\alpha }\;\;\;\;z\in U,$$
(4)

where we consider the principal value of z kÎ± for some real Î±(0<Î±â‰¤2). See Darus and Owa [3] for some properties of functions fÎ±(z) of the form (4). Here, we present a more generalized form of (4) such that:

$$g_{\alpha,n}(z)=\frac{A^{n}z}{\left(A+Bz^{\alpha }\right)^{n}} =z+\sum\limits_{k=1}^{\infty }(-1)^{k}\frac{B^{k}}{A^{k}}n_{k}z^{1+k\alpha }\;\;\;\;z\in U$$
(5)

for some real Î±(0<Î±â‰¤2),âˆ’1â‰¤B<Aâ‰¤1,nâ‰¥0 and nk is given by $$n_{k}=\Pi _{j=1}^{k}\left (\frac {n+j-1}{j!}\right)$$.

In view of (1) and (5), we introduce a class HÎ±,n of analytic function fÎ±,n(z) which is a convolution (or Hadamard product) of f(z) and gÎ±,n (f(z)âˆ—gÎ±,n(z)) such that:

$$f_{\alpha,n}(z)=z+\sum\limits_{k=1}^{\infty }(-1)^{k}\frac{B^{k}}{A^{k}} n_{k}a_{k+1}z^{1+k\alpha }\;\;\;\;z\in U$$
(6)

In addition, if fÎ±,n(z)âˆˆHÎ±,n satisfies the following condition:

$$\Re \left(\frac{zf_{\alpha,n}^{\prime }(z)}{f_{\alpha,n}(z)}\right)>\gamma \;\;\;z\in U$$
(7)

for some real Î±(0<Î±â‰¤2),n>0, and Î³(0â‰¤Î³<1), then fÎ±,n belong to the starlike class $$S_{\alpha,n}^{\ast }(A,B,\gamma)$$ (of order Î³). Also, if fÎ±,n(z)âˆˆHÎ±,n satisfies the following condition:

$$\Re \left(1+\frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right)>\gamma \;\;\;z\in U$$
(8)

for some real Î±(0<Î±â‰¤2),n>0, and Î³(0â‰¤Î³<1), then fÎ±,n belong to the convex class $$K_{\alpha,n}^{\ast }(A,B,\gamma)$$ (of order Î³). Here, it is noted that fÎ±,n(z)âˆˆHÎ±,n(z) belong to the convex class $$K_{\alpha,n}(A,B,\gamma)\Leftrightarrow zf_{\alpha,n}^{\prime }(z)$$ belong to the starlike class $$S_{\alpha,n}^{\ast }(A,B,\gamma)$$. For the purpose of the present investigation, we shall call to mind the following definitions and lemmas.

### Definition 1

(Subordination principle) For two functions f and g analytic in U, we say that f is subordinate to g, and write fâ‰ºg in U or f(z)â‰ºg(z), if there exists a Schwarz function w(z), which is analytic in U with w(0)=0 and |w(z)|<1 (zâˆˆU), such that f(z)=g(w(z)). It is known that:

$$f(z)\prec g(z)\,\,\Rightarrow f(0)=g(0)\ {and}\ f(U)\subset g(U).$$

Furthermore, if the function g is univalent in U:

$$f(z)\prec g(z)\,\,\,\Leftrightarrow f(0)=g(0)\ { and }\ f(U)\subset g(U).$$
(9)

Also, we say that g(z)is superordinate to f(z) in U (see [4â€“6]).

### Definition 2

(Subordinating factor sequence) A sequence $$\left \{ b_{k}\right \}_{k=1}^{\infty }$$ of complex numbers is called subordinating factor sequence if for every univalent function f(z) in K, we have the subordination given by:

$$\sum\limits_{k=1}^{\infty }a_{k}b_{k}z^{k}\prec f(z)\;\;\;\left(z\in U,\;\;a_{1}=1\right) \ (see {[4-6]}).$$
(10)

### Lemma 1

The sequence $$\left \{ b_{k}\right \}_{k=1}^{\infty }$$ is a subordinating factor sequence if and only if:

$$\Re \left\{ 1+2\sum\limits_{k=1}^{\infty }b_{k}z^{k}\right\} >0\;\;\;(z\in U).$$
(11)

The lemma above is due to Wilf [7]. Interested reader can also refer to [4â€“6].

### Lemma 2

Let s(z)(s(z)â‰ 0)be a univalent function in U. Also, let Î¼â‰ 0 be a complex number, then we have that:

$$\Re \left\{ 1+z\frac{s^{\prime \prime }(z)}{s^{\prime }(z)}-z\frac{s^{\prime }(z)}{s(z)}\right\} >{\text{max}}\left\{ 0,\Re \Big(\frac{\mu -1}{\mu }s(z)\Big) \right\}.$$
(12)

Suppose that r (r(z)â‰ 0) satisfies the differential equation:

$$(1-\mu)(r(z)-1)+\mu \frac{zr^{\prime }(z)}{r(z)}\prec (1-\mu)(s(z)-1)+\mu \frac{zs\prime (z)}{s(z)},\;z\in U$$
(13)

then râ‰ºs and s is the best dominant (see [8] among others).

### Lemma 3

Let Ï‰be regular in H with Ï‰(0)=0. Also, suppose that |Ï‰(z)| attains its maximum value on the circle |z|<1 at a point z0, then:

$$z_{0}\omega^{\prime }(z_{0})=\sigma \omega (z_{0}),$$
(14)

where Ïƒ is any real number and Ïƒâ‰¥1 (see [8] among others).

## Coefficient inequality

In this section, we consider the coefficient inequalities for function fÎ±,n(z) given by (6) belonging to both classes $$S_{\alpha,n}^{\ast }(A,B,\gamma)$$ and KÎ±,n(A,B,Î³) in the unit disk U.

### Theorem 1

Let the function fÎ±,n(z) of the form (6) satisfy the inequality:

$$\sum_{k=1}^{\infty }\left(k\alpha -\gamma +1\right)n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|\leq 1-\gamma.$$
(15)

Then, $$f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$$ for 0â‰¤Î³<1s,0<Î±â‰¤2,âˆ’1â‰¤B<Aâ‰¤1,0<Aâ‰¤1 and n>0. The equality holds true for fÎ±,n(z)given by:

$$f_{\alpha,n}(z)=z+\frac{\left(1-\gamma \right) e^{i\pi }}{\big(k\alpha -\gamma +1\big)n_{k}\frac{|B|^{k}}{A^{k}}}z^{1+k\alpha }\ \ \ (k\geq 1).$$

### Proof

Suppose that the function fÎ±,n(z) given by (6) satisfies (15), then:

$$\left|\frac{zf_{\alpha,n}^{\prime }(z)}{f_{\alpha,n}(z)}-1\right|=\left| \frac{\sum_{k=1}^{\infty }(-1)^{k}k\alpha n_{k}\frac{B^{k}}{A^{k}} a_{k+1}z^{k\alpha }}{1+\sum_{k=1}^{\infty }(-1)^{k}n_{k}\frac{B^{k}}{A^{k}} a_{k+1}z^{k\alpha }}\right|$$
$$\leq \frac{\sum_{k=1}^{\infty }k\alpha n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}||z|^{k\alpha }}{1-\sum_{k=1}^{\infty }n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}||z|^{k\alpha }}|<\frac{\sum_{k=1}^{\infty }k\alpha n_{k}\frac{ |B|^{k}}{A^{k}}|a_{k+1}|}{1-\sum_{k=1}^{\infty }n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|}\leq 1-\gamma.$$

This shows that $$f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$$, and this ends the proof. â–¡

### Corollary 1

Let the function fÎ±,n(z)of the form (6) satisfy the inequality:

$$\sum_{k=1}^{\infty }\big(k\alpha +1\big)n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|\leq 1.$$

Then, $$f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,0)$$.

### Theorem 2

Let the function fÎ±,n(z) of the form (6) satisfy the inequality:

$$\sum\limits_{k=1}^{\infty }\left(k\alpha +1\right)\left(k\alpha -\gamma +1\right)n_{k} \frac{|B|^{k}}{A^{k}}|a_{k+1}|\leq 1-\gamma.$$
(16)

Then, fÎ±,n(z)âˆˆKÎ±,n(A,B,Î³) for 0â‰¤Î³<1, 0<Î±â‰¤2,âˆ’1â‰¤B<Aâ‰¤1,0<Aâ‰¤1 and n>0. The equality holds true for fÎ±,n(z)given by:

$$f_{\alpha,n}(z)=z+\frac{\left(1-\gamma \right) e^{i\pi }}{\left(k\alpha +1 \right)\left(k\alpha -\gamma +1\right)n_{k}\frac{|B|^{k}}{A^{k}}}z^{1+k\alpha }\ \ \ \ \ (k\geq 1).$$

### Proof

The proof is similar to that of Theorem 1. â–¡

### Corollary 2

Let the function fÎ±,n(z) of the form (6) satisfy the inequality:

$$\sum\limits_{k=1}^{\infty }\left(k\alpha +1\right)^{2}n_{k}\frac{|B|^{k}}{A^{k}} |a_{k+1}|\leq 1.$$

Then, fÎ±,n(z)âˆˆKÎ±,n(A,B,0).

### Remark 1

Putting A=n=1 and B=âˆ’1 in Theorems 1 and 2, we obtain the results obtained by Darus and Owa [[3], Theorems 3 and 4].

Next, we present some subordination results.

## Some subordination results

Our prime objective here is to establish sufficient conditions for functions belonging to the analytic class $$S_{\alpha,n}^{\ast }(A,B,\gamma)$$.

### Theorem 3

Suppose that the function fÎ±,n(z) is as defined in (6). Let 0<Î±â‰¤2,n>0,Ïƒâ‰ âˆ’1 and Î¼be a non-zero complex number in U such that:

$$\Re \left\{ 1+\frac{z[1-\sigma (1-2z)]}{(1-z)(1+\sigma z)}\right\} >{\text{max}}\left\{ 0,\Re \left(\frac{\mu -1}{\mu }\right)\left(\frac{1+\sigma z}{1-z} \right)\right\}.$$

If

$${}(1-\mu)\left(f_{\alpha,n}^{\prime }(z)-1\right)+\mu \left(\frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right)\prec (1-\mu)\left(\left(\frac{1+\sigma z}{1-z}\right)-1\right)+\mu \left(\frac{(1+\sigma)z}{ (1+\sigma z)(1-z)}\right)$$

holds true, then $$f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$$.

### Proof

Suppose that we let:

$$r(z)=f_{\alpha,n}^{\prime }(z)\;\;\;{\text{and}}\;\;\;s(z)=\frac{1+\sigma z}{1-z}.$$
(17)

Then,

$${}\Re \left\{ 1+\frac{zs^{\prime \prime }(z)}{s^{\prime }(z)}-\frac{zs^{\prime }(z)}{s(z)}\right\} >{\text{max}}\left\{ 0,\Re \left(\frac{\mu -1}{\mu }\right)\left(\frac{1+\sigma z}{1-z}\right)\right\} ={\text{max}}\left\{ 0,\Re \left(\frac{\mu -1}{\mu }s(z)\right)\right\}$$

and

$$(1-\mu)(r(z)-1)+\mu \frac{zr^{\prime }(z)}{r(z)}=(1-\mu)\left(f_{\alpha,n}^{\prime }(z)-1\right)+\mu \left(\frac{zf_{\alpha,n}^{\prime \prime }(z)}{ f_{\alpha,n}^{\prime }(z)}\right)$$
$$\prec (1-\mu)\left(\left(\frac{1+\sigma z}{1-z}\right)-1\right)+\mu \left(\frac{ (1+\sigma)z}{(1+\sigma z)(1-z)}\right)=(1-\mu)(s(z)-1)+\mu \frac{zs^{\prime }(z)}{s(z)}.$$
(18)

Using Lemma 2 in (18), then we obtain the desired result. â–¡

### Theorem 4

Let the analytic function fÎ±,n(z)be defined as in (6). Suppose that fÎ±,n(z) satisfies the condition that:

$$\Re \left\{ \frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right\} <-\frac{1+\sigma }{2(1-\sigma)}.$$
(19)

Then, for 0<Î±â‰¤2,n>0 and $$\sigma >1, f_{\alpha,n}(z)\in S_{\alpha,n}^{\ast }(A,B,\gamma)$$.

### Proof

Setting:

$$f_{\alpha,n}^{\prime }(z)=\left(\frac{1+\sigma \omega (z)}{1-\omega (z)} \right),\;\;\;\omega (z)\neq 1.$$

Then, Ï‰ is regular in U, and since Ïƒâ‰ âˆ’1, then Ï‰(0)=0. Also, it follows that:

$$\Re \left\{ \frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right\} =\Re \left\{ \frac{(1+\sigma)z\omega^{\prime }(z)}{(1-\omega (z))(1+\sigma \omega (z))}\right\} <\frac{\sigma +1}{2(\sigma -1)},\;\;\sigma \neq 1.$$

Next, we show that |Ï‰(z)|<1. So, let there exists a point z0âˆˆU such that for |z|â‰¤|z0|:

$${\text{max}}|\omega (z)|=|\omega (z)|=1.$$

Then, appealing to Lemma 3 and setting Ï‰(z0)=eiÎ¸,z0Ï‰â€²(z0)=Î´eiÎ¸ and for Î´â‰¥1,Ïƒ>1, we have that:

\begin{aligned} \Re \left\{\frac{zf_{\alpha,n}^{\prime \prime}(z)}{f_{\alpha,n}^{\prime}(z)}\right\} &=\Re \left\{\frac{(1+\sigma)z_{0}\omega^{\prime}(z_{0})}{(1-\omega (z_{0}))(1+\sigma \omega (z_{0}))}\right\} =\Re \left\{\frac{\delta e^{i\theta}(1+\sigma)}{(1-e^{i\theta})(1+\sigma e^{i\theta})} \right\} \\&=\frac{\delta(\sigma +1)}{2(\sigma -1)}\geq \frac{(\sigma +1)}{2(\sigma -1)}. \end{aligned}

Therefore,

$$\Re \left\{ \frac{zf_{\alpha,n}^{\prime \prime }(z)}{f_{\alpha,n}^{\prime }(z)}\right\} \geq -\frac{1+\sigma }{2(1-\sigma)}\;\;\;z\in U$$

which negates the hypothesis (19).

Hence, we conclude that |Ï‰(z)|<1 for all zâˆˆU and:

$$f_{\alpha,n}^{\prime }(z)\prec \left(\frac{1+\sigma z}{1-z}\right),\;\;\sigma \neq 1,\;z\in U$$

and this obviously ends the proof. â–¡

## Application of a subordination theorem

Let $$\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)$$ and $$\overline {K} _{\alpha,n}(A,B,\gamma)$$ denote the classes of functions fa,nâˆˆHa,n whose coefficients satisfy conditions (15) and (16), respectively. We note that $$\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)\subseteq S_{\alpha,n}^{\ast }(A,B,\gamma)\$$and $$\overline {K}_{\alpha,n}(A,B,\gamma)\subseteq K_{\alpha,n}(A,B,\gamma).$$ Here, we consider an application of the subordination result given in Lemma 1 to both classes $$\overline {S} _{\alpha,n}^{\ast }(A,B,\gamma)$$ and $$\overline {K}_{\alpha,n}(A,B,\gamma)$$.

### Theorem 5

Let $$f_{\alpha,n}(z)\in \overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)$$. If 0â‰¤Î³<1,0<Î±â‰¤2,âˆ’1â‰¤B<Aâ‰¤1,0<Aâ‰¤1 and n>0, then:

$$\frac{n(\alpha -\gamma +1)|B|}{2\big[n\alpha |B|+(1-\gamma)(A+n|B|)\big]} \left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z)$$
(20)

for every function gÎ± in KÎ± and:

$$\Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha |B|+(1-\gamma)(A+n|B|) \right]}{n(\alpha -\gamma +1)|B|}.$$
(21)

The constant factor:

$$\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]}$$

in the subordination result (20) is sharp.

### Proof

Let $$f_{\alpha,n}\in \overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)$$ and let gÎ± be any function in KÎ±. Then:

$$\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} \left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z)$$
$$=\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} \left(z+\sum\limits_{k=1}^{\infty }a_{k+1}b_{k+1}z^{k\alpha +1}\right).$$

Thus, by Definition 2, the subordination result (20) will hold true if:

$$\left\{ \frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|) \right]}a_{k}\right\}_{k=1}^{\infty }$$

is a subordinating factor sequence, with a1=1, appealing to Lemma 1, this is equivalent to:

$$\Re \left\{ 1+\sum\limits_{k=1}^{\infty }\frac{n(\alpha -\gamma +1)|B|}{\left[ n\alpha |B|+(1-\gamma)(A+n|B|)\right]}a_{k}z^{(k-1)\alpha +1}\right\} >0\;\;\;(z\in U).$$
(22)

Since $$n_{k}\left (k\alpha -\gamma +1\right)\frac {|B|^{k}}{A^{k}}$$ is an increasing function of k (kâ‰¥1), we have that:

$$\Re \left\{ 1+\sum\limits_{k=1}^{\infty }\frac{n(\alpha -\gamma +1)|B|}{\left[ n\alpha |B|+(1-\gamma)(A+n|B|)\right]}a_{k}z^{(k-1)\alpha +1}\right\}$$
$$=\Re \left\{ 1+\frac{n(\alpha -\gamma +1)|B|}{M}z+\frac{A}{M} \sum\limits_{k=2}^{\infty }n(\alpha -\gamma +1)\frac{|B|}{A}a_{k}z^{(k-1)\alpha +1}\right\}$$
$$\geq 1-\frac{n(\alpha -\gamma +1)|B|}{M}r-\frac{A}{M}\sum\limits_{k=2}^{\infty }n_{k-1}\Big((k-1)\alpha -\gamma +1\Big)\frac{|B|^{k-1}}{A^{k-1}} a_{k}r^{(k-1)\alpha +1}$$
$$>1-\frac{n(\alpha -\gamma +1)|B|}{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r-\frac{A\left(1-\gamma \right) }{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r^{\alpha +1}$$
$$>1-\frac{n(\alpha -\gamma +1)|B|}{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r-\frac{A\left(1-\gamma \right) }{\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]} r=1-r>0$$
(23)
$$\left(|z|=r<1\right),$$

where M=[nÎ±|B|+(1âˆ’Î³)(A+n|B|)]. Therefore, (22) holds true in U and this obviously proves the inequality (20) while (21) follows by taking:

$$g_{\alpha }(z)=\frac{z}{1-z^{\alpha }}\in K_{\alpha }$$

in (20). Now, suppose that we consider the function qÎ±,n(z) of the form:

$$q_{\alpha,n}(z)=z-\frac{1-\gamma }{n(\alpha -\gamma +1)\frac{|B|}{A}} z^{\alpha +1}$$

which belongs to the class $$\overline {S}_{\alpha,n}^{\ast }(A,B,\gamma)$$. Then, using (20), we have that:

$$\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]}.q_{\alpha,n}(z)\prec \frac{z}{1-z^{\alpha }}\;\;\;(z\in U)$$

which can easily be verified that for 0â‰¤Î³<1,0<Î±â‰¤2,âˆ’1â‰¤B<Aâ‰¤1,0<Aâ‰¤1,nâ‰¥0 and |z|â‰¤r:

$${\text{min}}\left\{ \Re \left(\frac{n(\alpha -\gamma +1)|B|}{2\left[n\alpha |B|+(1-\gamma)(A+n|B|)\right]}.q_{\alpha,n}(z)\right)\right\} =-\frac{1}{2} \;\;(z\in U)$$

and this evidently completes the proof of Theorem 5. For various choices of the parameters involved, several interesting results are obtained. Given below are few instances. â–¡

### Corollary 3

Let $$f_{\alpha,n}(z)\in \overline {S}_{\alpha,n}^{\ast }(1,-1,\gamma)$$. Then:

$$\frac{n(\alpha -\gamma +1)}{2\left[n\alpha +(1-\gamma)(1+n)\right]}\left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z)$$

for every function gÎ± in KÎ± and:

$$\Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha +(1-\gamma)(1+n)\right]}{ n(\alpha -\gamma +1)}.$$

The constant factor:

$$\frac{n(\alpha -\gamma +1)}{2\left[n\alpha +(1-\gamma)(1+n)\right]}$$

is sharp.

### Corollary 4

Let $$f_{\alpha,1}(z)\in \overline {S}_{\alpha,1}^{\ast }(1,-1,\gamma)$$. Then:

$$\frac{(\alpha -\gamma +1)}{2(\alpha -2\gamma +2)}\left(f_{\alpha,1}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z)$$

for every function gÎ± in KÎ± and:

$$\Re \left(f_{\alpha,1}(z)\right)>-\frac{\left(\alpha -2\gamma +2\right)}{(\alpha -\gamma +1)}.$$

The constant factor:

$$\frac{(\alpha -\gamma +1)}{2\left(\alpha -2\gamma +2\right)}$$

is sharp.

### Corollary 5

[9,10] Let $$f_{1,1}(z)\in \overline {S} _{1,1}^{\ast }(1,-1,\gamma)$$. Then:

$$\frac{(2-\gamma)}{2\left(3-2\gamma \right)}\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z)$$

for every function g1 in K1 and:

$$\Re \left(f_{1,1}(z)\right)>-\frac{\left(3-2\gamma \right)}{(2-\gamma)}.$$

The constant factor:

$$\frac{(2-\gamma)}{2\left(3-2\gamma \right)}$$

is sharp.

### Corollary 6

[9â€“11] Let $$f_{1,1}(z)\in \overline {S} _{1,1}^{\ast }(1,-1,0)$$. Then:

$$\frac{1}{3}\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z)$$

for every function g1 in K1 and:

$$\Re \left(f_{1,1}(z)\right)>-\frac{3}{2}.$$

### Theorem 6

Let $$f_{\alpha,n}(z)\in \overline {K}_{\alpha,n}(A,B,\gamma)$$. If 0â‰¤Î³<1,0<Î±â‰¤2,âˆ’1â‰¤B<Aâ‰¤1 and n>0, then:

$$\frac{n(\alpha +1)(\alpha -\gamma +1)|B|}{2\left[n\alpha (\alpha +1)|B|+(1-\gamma)(A+n(\alpha +1)|B|)\right]}\left(f_{\alpha,n}\ast g_{\alpha } \right)(z)\prec g_{\alpha }(z)$$
(24)

for every function gÎ± in KÎ± and:

$$\Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha (\alpha +1)|B|+(1-\gamma)(A+n(\alpha +1)|B|)\right]}{n(\alpha +1)(\alpha -\gamma +1)|B|}.$$
(25)

The constant factor:

$$\frac{n(\alpha +1)(\alpha -\gamma +1)|B|}{2\left[n\alpha (\alpha +1)|B|+(1-\gamma)(A+n(\alpha +1)|B|)\right]}$$

in the subordination result (24) cannot be replaced by a larger one, and the proof of which is similar to that of Theorem 3.

### Corollary 7

Let $$f_{\alpha,n}(z)\in \overline {K}_{\alpha,n}(1,-1,\gamma)$$. Then:

$$\frac{n(\alpha +1)(\alpha -\gamma +1)}{2\left[n\alpha (\alpha +1)+(1-\gamma)(1+n(\alpha +1))\right]}\left(f_{\alpha,n}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z)$$
(26)

for every function gÎ± in KÎ± and:

$$\Re \left(f_{\alpha,n}(z)\right)>-\frac{\left[n\alpha (\alpha +1)+(1-\gamma)(1+n(\alpha +1))\right]}{n(\alpha +1)(\alpha -\gamma +1)}.$$
(27)

The constant factor:

$$\frac{n(\alpha +1)(\alpha -\gamma +1)}{2\left[n\alpha (\alpha +1)+(1-\gamma)(1+n(\alpha +1))\right]}$$

cannot be replaced by a larger one.

### Corollary 8

Let $$f_{\alpha,1}(z)\in \overline {K}_{\alpha,1}(1,-1,\gamma)$$. Then:

$$\frac{(\alpha +1)(\alpha -\gamma +1)}{2\left[\alpha (\alpha +1)+(1-\gamma)(\alpha +2)\right]}\left(f_{\alpha,1}\ast g_{\alpha }\right)(z)\prec g_{\alpha }(z)$$
(28)

for every function gÎ± in KÎ± and:

$$\Re \left(f_{\alpha,1}(z)\right)>-\frac{\left[\alpha (\alpha +1)+(1-\gamma)(\alpha +2)\right]}{(\alpha +1)(\alpha -\gamma +1)}.$$
(29)

The constant factor:

$$\frac{(\alpha +1)(\alpha -\gamma +1)}{2\left[\alpha (\alpha +1)+(1-\gamma)(\alpha +2)\right]}$$

cannot be replaced by a larger one.

### Corollary 9

[9,10] Let $$f_{1,1}(z)\in \overline {K} _{1,1}(1,-1,\gamma)$$. Then:

$$\frac{2-\gamma }{5-3\gamma }\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z)$$
(30)

for every function g1 in K1 and:

$$\Re \left(f_{1,1}(z)\right)>-\frac{5-3\gamma }{2(2-\gamma)}.$$
(31)

The constant factor:

$$\frac{2-\gamma }{5-3\gamma }$$

cannot be replaced by a larger one.

### Corollary 10

[9,10] Let $$f_{1,1}(z)\in \overline {K} _{1,1}(1,-1,0)$$. Then:

$$\frac{2}{5}\left(f_{1,1}\ast g_{1}\right)(z)\prec g_{1}(z)$$
(32)

for every function g1 in K1 and:

$$\Re \left(f_{1,1}(z)\right)>-\frac{5}{4}.$$
(33)

The constant factor:

$$\frac{2}{5}$$

cannot be replaced by a larger one.

For further illustrations on the applications of the subordination result stated in Lemma 1, interested reader can see [4,6,8â€“11].

## Availability of data and materials

All data generated or analyzed during this study are included in this published article.

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## Acknowledgements

The authors would like to express their sincerest thanks to the referees for a careful reading and various suggestions made for the improvement of the paper.

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Hamzat, J.O., El-Ashwah, R.M. Application of a subordination theorem associated with certain new generalized subclasses of analytic and univalent functions. J Egypt Math Soc 28, 33 (2020). https://doi.org/10.1186/s42787-020-00094-4