**1. Introduction**

We denote by A the class of analytic functions on the unit disc U = {*z* ∈ C : |*z*| < 1} having the following taylor series representation:

$$f(z) = z + \sum\_{n=2}^{\infty} a\_n z^n. \tag{1}$$

The analytic function *f* will be subordinate to an analytic function *g*, if there exists an analytic function *w*, known as a Schwarz function, with *w* (0) = 0 and |*w*(*z*)| < |*z*|, such that *f*(*z*) = *g*(*w*(*z*)). Moreover, if the function *g* is univalent in U, then we have the following (see [1,2]):

> *f*(*z*) ≺ *g*(*z*), *z* ∈ U ⇐⇒ *f*(0) = *g*(0) and *f*(U) ⊂ *g*(U).

Uralegaddi et al. [3] introduced the reciprocal classes M(*γ*) of starlike and N (*γ*) of convex functions for 1 ≤ *γ* ≤ 43, which were further studied by Owa et al. [4–6] for the values *γ* ≥ 1.

The classes M(*γ*) of starlike functions and N (*γ*) of reciprocal order convex functions *γ*, (*γ* > 1) are defined as follows:

$$\mathcal{M}\left(\gamma\right) \;= \; \left\{ f \in \mathcal{A} : \mathfrak{Re} \frac{zf'(z)}{f(z)} < \gamma, \; z \in \mathbb{U} \right\},$$

$$\mathcal{N}\left(\gamma\right) \;= \; \left\{ f \in \mathcal{A} : \mathfrak{Re} \left\{ 1 + \frac{zf''(z)}{f'(z)} \right\} < \gamma, \; z \in \mathbb{U} \right\}.$$

Using the same concept, together with the idea of *k*-uniformly starlike and *γ* ordered convex functions, Nishiwaki and Owa [7] defined the reciprocal classes of uniformly starlike MD (*k*, *γ*) and convex functions N D (*k*, *<sup>γ</sup>*). The class MD (*k*, *γ*) denotes the subclass of A consisting of functions *f* satisfying the inequality

$$\Re \text{Re} \frac{zf'(z)}{f(z)} < k \left| \frac{zf'(z)}{f(z)} - 1 \right| + \gamma\_{\prime\prime} \text{ ( $z \in \mathbb{U}$ )} \,,$$

for some *γ* (*γ* > 1) and *k* (*k* ≤ 0) and the class N D (*k*, *γ*) denotes the subclass of A consisting of functions *f*(*z*) satisfying the inequality

$$\Re \mathfrak{Re} \frac{\left(zf'(z)\right)'}{f'(z)} < \gamma + k \left|\frac{\left(zf'(z)\right)'}{f'(z)} - 1\right|, \ (z \in \mathcal{U})\ \rho$$

for some *γ* (*γ* > 1) and *k* (*k* ≤ <sup>0</sup>). They also proved that the well-known Alexander relation holds between MD (*k*, *γ*) and N D (*k*, *<sup>γ</sup>*). This means that

$$f \in \mathcal{ND}\left(k, \gamma\right) \quad \Leftrightarrow \quad zf' \in \mathcal{MD}\left(k, \gamma\right) .$$

For a more detailed and recent study on uniformly convex and starlike functions, we refer the reader to [8–12].

Considering the above defined classes, we introduce the following classes.

**Definition 1.** *Let f belong to* A*. Then, it will belong to the class* KD (*β*, *γ*) *if there exists g* ∈ MD (*γ*) *such that* 

$$\Re \mathfrak{Re} \left\{ \frac{z f'(z)}{g(z)} \right\} < \beta, \quad (z \in \mathbb{U}), \tag{2}$$

.

*for some β*, *γ* > 1*.*

**Definition 2.** *Let f belong to* A*. Then, it will belong to the class* QD (*β*, *γ*) *if there exists g* ∈ ND (*γ*) *such that* 

$$\Re \mathfrak{Re} \left\{ \frac{\left( z f'(z) \right)'}{g'(z)} \right\} < \beta, \quad (z \in \mathbb{U}), \tag{3}$$

*for some β*, *γ* > 1*.*

> It is clear, from (2) and (3), that

$$f\left(z\right) \in \mathcal{QD}\left(\beta, \gamma\right) \quad \Leftrightarrow \quad zf'\left(z\right) \in \mathcal{KD}\left(\beta, \gamma\right) \dots$$

**Definition 3.** *Let f belong to* A*. Then, it will belong to the class* KD (*k*, *β*, *γ*) *if there exists g* ∈ MD (*k*, *γ*) *such that* 

$$\Re \text{Re}\left\{\frac{zf'(z)}{g(z)}\right\} < k \left|\frac{zf'(z)}{g(z)} - 1\right| + \beta, \quad (z \in \mathbb{U}), \tag{4}$$

*for some k* ≤ 0 *and β*, *γ* > 1*.* **Definition 4.** *Let f belong to* A*. Then, it is said to be in the class* QD (*k*, *β*, *γ*) *if there exists g* ∈ND (*k*, *γ*) *such that* 

$$\mathfrak{Re}\left\{\frac{\left(zf'(z)\right)'}{\mathfrak{g}'(z)}\right\} < k \left|\frac{\left(zf'(z)\right)'}{\mathfrak{g}'(z)} - 1\right| + \mathfrak{\beta}, \quad (z \in \mathbb{U}), \tag{5}$$

*for some k* ≤ 0 *and β*, *γ* > 1*.*

We can see, from (4) and (5), that the well-known relation of Alexander type holds between the classes KD (*k*, *β*, *γ*) and QD (*k*, *β*, *<sup>γ</sup>*), which means that

$$f\left(z\right) \in \mathcal{QD}\left(k, \beta, \gamma\right) \quad \Leftrightarrow \quad zf'\left(z\right) \in \mathbb{X}\mathcal{D}\left(k, \beta, \gamma\right) \dots$$

#### **2. Preliminary Lemmas**

**Lemma 1.** *For positive integers t and σ, we have*

$$
\sigma \sum\_{j=1}^{t} \frac{(\sigma)\_{j-1}}{(j-1)!} = \frac{(\sigma)\_t}{(t-1)!},\tag{6}
$$

*where* (*σ*)*t is the Pochhammer symbol, defined by*

$$\sigma(\sigma)\_t = \frac{\Gamma\left(\sigma + t\right)}{\Gamma\left(\sigma\right)} = \sigma(\sigma + 1)(\sigma + 2)(\sigma + 3)\cdots(\sigma + t - 1).$$

**Proof.** Consider

$$\begin{array}{l} & \sigma \sum\_{j=1}^{t} \frac{(\sigma)\_{j-1}}{(j-1)!} \\ &= \ & \sigma \left(1 + \frac{\sigma}{1} + \frac{(\sigma)\_{2}}{2!} + \frac{(\sigma)\_{3}}{3!} + \frac{(\sigma)\_{4}}{4!} + \dots + \frac{(\sigma)\_{t-1}}{(t-1)!} \right) \\ &= & \sigma(1+\sigma) \left(1 + \frac{\sigma}{2} + \frac{\sigma(\sigma+2)}{2\times3} + \dots + \frac{\sigma(\sigma+2)\dots(\sigma+t-2)}{2\times\cdots\times(t-1)} \right) \\ &= & \sigma(1+\sigma) \frac{(\sigma+2)}{2} \left(1 + \frac{\sigma}{3} + \dots + \frac{\sigma(\sigma+3)\dots(\sigma+t-2)}{3\times4\times\cdots\times(t-1)} \right) \\ &= & \sigma(1+\sigma) \frac{(\sigma+2)}{2} \frac{(\sigma+3)}{3} \left(1 + \frac{\sigma}{4} + \dots + \frac{\sigma(\sigma+4)\dots(\sigma+t-2)}{4\times\cdots\times(t-1)} \right) \\ &= & \sigma(1+\sigma) \frac{(\sigma+2)}{2} \frac{(\sigma+3)}{3} \frac{(\sigma+4)}{4} \left(1 + \frac{\sigma}{5} + \dots + \frac{\sigma\cdots(\sigma+t-2)}{5\times6\times\cdots\times(t-1)} \right) \\ &= & \sigma(1+\sigma) \frac{(\sigma+2)}{2} \frac{(\sigma+3)}{3} \frac{(\sigma+4)}{4} \dots \left(1 + \frac{\sigma}{t-1} \right) \\ &= & \frac{(\sigma)\_t}{(t-1)!} . \end{array}$$

**Lemma 2.** *If f*(*z*) ∈ MD (*k*, *<sup>γ</sup>*)*, then*

$$f(z) \in \mathcal{MD}\left(\frac{\gamma - k}{1 - k}\right).$$

**Proof.** Using the definition , we write

$$\begin{aligned} \mathfrak{Re}\frac{zf'(z)}{f(z)} &< |k| \frac{zf'(z)}{f(z)} - 1 \Big| + \gamma \\ &\le |k \mathfrak{Re}\frac{zf'(z)}{f(z)} + \gamma - k| \end{aligned}$$

which implies that

$$(1 - k)\Re\overline{e}\frac{zf'(z)}{f(z)} < \gamma - k\_{-}$$

After simplification, we obtain

$$\Re \mathfrak{a} \frac{z f'(z)}{f(z)} < \frac{\gamma - k}{1 - k}, \ (k \le 0, \ \gamma > 1) \dots$$

$$\text{As } \frac{\gamma - k}{1 - k} > 1 \text{, we have } f(z) \in \mathcal{MD}\left(\frac{\gamma - k}{1 - k}\right). \text{ With this, we obtain the required result.} \quad \square$$

**Lemma 3.** *If f belongs to the class* MD (*k*, *<sup>γ</sup>*)*, then*

$$|a\_n| \le \frac{\left(\delta\_{k,\gamma}\right)\_{n-1}}{(n-1)!},\tag{7}$$

*where*

$$
\delta\_{k,\gamma} = \frac{2(\gamma - 1)}{1 - k}. \tag{8}
$$

**Proof.** Let us define a function

$$p(z) = \frac{(\gamma - k) - (1 - k)\left(\frac{zf'(z)}{f(z)}\right)}{\gamma - 1},\tag{9}$$

where *p* ∈ P, the class of Caratheodory functions (see [1]). One may write

$$\frac{z f'(z)}{f(z)} = \frac{(\gamma - k) - (\gamma - 1)}{1 - k} {}\_{\prime}\tag{10}$$

or

$$z f'(z) = \left(\frac{\gamma - k}{1 - k} - \frac{\gamma - 1}{1 - k} p(z)\right) f(z). \tag{11}$$

.

Let us write *p*(*z*) as *p*(*z*) = 1 + ∞ ∑ *<sup>n</sup>*=1 *pnz<sup>n</sup>* and let *f* have the series form, as in (1). Then, (11) can be written as 

$$\sum\_{n=1}^{\infty} n a\_n z^n = \left( \sum\_{n=1}^{\infty} a\_n z^n \right) \left( \frac{\gamma - k}{1 - k} - \frac{\gamma - 1}{1 - k} \left( 1 + \sum\_{n=1}^{\infty} p\_n z^n \right) \right), \quad a\_1 = 1, 2$$

which reduces to

$$\begin{aligned} \sum\_{n=1}^{\infty} n a\_n z^n &= \quad \left( \sum\_{n=1}^{\infty} a\_n z^n \right) \left( 1 - \frac{\gamma - 1}{1 - k} \sum\_{n=1}^{\infty} p\_n z^n \right) \\ &= \quad \sum\_{n=1}^{\infty} a\_n z^n - \frac{\gamma - 1}{1 - k} \left( \sum\_{n=1}^{\infty} a\_n z^n \right) \left( \sum\_{n=1}^{\infty} p\_n z^n \right) . \end{aligned}$$

> This implies that

$$\sum\_{n=1}^{\infty} \left( n - 1 \right) a\_n z^n = -\frac{\gamma - 1}{1 - k} \sum\_{n=1}^{\infty} \left( \sum\_{j=0}^{n-1} a\_j p\_{n-j} \right) z^n.$$

After comparing the *nth* term's coefficients, appearing on both sides, combined with the fact that *a*0 = 0, we obtain

$$a\_n = \frac{-\left(\gamma - 1\right)}{\left(n - 1\right)\left(1 - k\right)} \sum\_{j = 1}^{n - 1} a\_j p\_{n - j}.$$

Now, we take the absolute value and then apply the triangle inequality to ge<sup>t</sup>

$$|a\_n| \le \frac{\gamma - 1}{(n - 1)\left(1 - k\right)} \sum\_{j = 1}^{n - 1} |a\_j| \left| |p\_{n - j}| \right| \cdot $$

Applying the coefficient estimates, such that |*pn*| ≤ 2 (*n* ≥ 1) for Caratheodory functions [1], we obtain

$$|a\_n| \le \frac{2\left(\gamma - 1\right)}{\left(n - 1\right)\left(1 - k\right)} \sum\_{j=1}^{n-1} |a\_j|\,.$$

$$|a\_n| \le \frac{\delta\_{k,\gamma}}{n - 1} \sum\_{j=1}^{n-1} |a\_j|\,. \tag{12}$$

where *<sup>δ</sup>k*,*<sup>γ</sup>* = <sup>2</sup>(*γ* − 1) 1 − *k* . We prove (7) by induction on *n*. Thus, first for *n* = 2, we obtain the following from (12):

$$|a\_2| \le \frac{\delta\_{k,\gamma}}{1} = \frac{\left(\delta\_{k,\gamma}\right)\_{2-1}}{(2-1)!}.\tag{13}$$

This proves that, for *n* = 2, (7) is true. For *n* = 3, we obtain

$$|a\_3| \le \frac{\delta\_{k,\gamma}}{2} \left( 1 + |a\_2| \right) = \frac{\delta\_{k,\gamma} \left( 1 + \delta\_{k,\gamma} \right)}{2} = \frac{\left( \delta\_{k,\gamma} \right)\_{3-1}}{(3-1)!}.$$

This proves that when *n* = 3, (7) holds true. Now, we assume that for *t* ≤ *n*, (7) is true, that means

$$|a\_t| \le \frac{\left(\delta\_{k,\gamma}\right)\_{t-1}}{(t-1)!} \quad t = 1,2,\ldots,n. \tag{14}$$

Using (12) and (14), we have

$$|a\_{t+1}| \le \frac{\delta\_{k,\gamma}}{t} \sum\_{j=1}^t |a\_j| \le \frac{\delta\_{k,\gamma}}{t} \sum\_{j=1}^t \frac{\left(\delta\_{k,\gamma}\right)\_{j-1}}{(j-1)!}.$$

After applying (6), we obtain

$$|a\_{t+1}| \le \frac{1}{t} \frac{(\delta\_{k,\gamma})\_t}{(t-1)!} = \frac{(\delta\_{k,\gamma})\_t}{t!}.$$

As a result of mathematical induction, it is shown that (7) is true for all *n* ≥ 2. Hence, the required bound is obtained.

**Lemma 4** ([13])**.** *Let w be analytic in* U *with w*(0) = 0. *If there exists z*0 ∈ U *such that*

$$\max\_{|z| \le |z\_0|} |w(z)| = |w(z\_0)| \text{ .}$$

*then*

$$z\_0 w'(z\_0) = c w(z\_0)\_r$$

*where c is real and c* ≥ 1*.*

#### **3. Main Results**

**Theorem 1.** *If f*(*z*) ∈ KD (*k*, *β*, *<sup>γ</sup>*)*, then*

$$f(z) \in \mathcal{KD}\left(\frac{\beta - k}{1 - k}, \gamma\right).$$

**Proof.** If *f*(*z*) ∈ KD (*k*, *β*, *<sup>γ</sup>*), then *k* ≤ 0, *β* > 1, and so we obtain

$$\begin{aligned} \left\| \mathfrak{Re} \left\{ \frac{zf'(z)}{g(z)} \right\} \right\| &< \left| k \left| \frac{zf'(z)}{g(z)} - 1 \right| + \beta \\ &\le \left| \beta + k \mathfrak{Re} \left\{ \frac{zf'(z)}{g(z)} - 1 \right\} \right| \end{aligned}$$

which leads to

$$\Re \mathfrak{Re}\left\{\frac{zf'(z)}{g(z)}\right\} - k \Re \mathfrak{e}\left\{\frac{zf'(z)}{g(z)}\right\} < -k + \beta.$$

After simplification, we obtain

$$\Re \text{Re} \left\{ \frac{zf'(z)}{g(z)} \right\} < \frac{\beta - k}{1 - k}, \quad (k \le 0, \ \beta > 1) \,. \tag{15}$$

This completes the proof.

In a similar way, one can easily prove the following important result.

**Theorem 2.** *If f* ∈ QD (*k*, *β*, *<sup>γ</sup>*)*, then*

$$f \in \mathcal{QD}\left(\frac{\beta - k}{1 - k}, \gamma\right).$$

**Theorem 3.** *If f*(*z*) ∈ KD (*k*, *β*, *<sup>γ</sup>*)*, then*

$$|a\_n| \le \frac{\left(\delta\_{k,\gamma}\right)\_{n-1}}{n!} + \frac{\left|\delta\_{k,\emptyset}\right|}{n} \sum\_{j=1}^{n-1} \frac{\left(\delta\_{k,\gamma}\right)\_{j-1}}{(j-1)!} \gamma$$

*where <sup>δ</sup>k*,*<sup>γ</sup> is given by* (8) *and*

$$
\delta\_{k,\emptyset} = \frac{2\left(\beta - 1\right)}{1 - k}.\tag{16}
$$

**Proof.** If *f* is in the class KD(*k*, *β*, *<sup>γ</sup>*), then there exists *g*(*z*) ∈ MD (*k*, *γ*) such that the function

$$p(z) = \frac{(\beta - k) - (1 - k)\left(\frac{zf'(z)}{g(z)}\right)}{\beta - 1} \tag{17}$$

belongs to P. Therefore, we write

$$z f'(z) = \frac{\beta - k}{1 - k} g(z) - \frac{\beta - 1}{1 - k} g(z) p(z). \tag{18}$$

Let us write *p*(*z*) as *p*(*z*) = 1 + ∑∞*<sup>n</sup>*=<sup>1</sup> *pnz<sup>n</sup>*, *g*(*z*) as *g*(*z*) = *z* + ∑∞*<sup>n</sup>*=<sup>2</sup> *bnzn*, and let *f*(*z*) have the series form as in (1). Then, (18) can be written as

$$z + \sum\_{n=2}^{\infty} n a\_n z^n = \frac{\beta - k}{1 - k} \left( z + \sum\_{n=2}^{\infty} b\_n z^n \right) - \frac{\beta - 1}{1 - k} \left( 1 + \sum\_{n=1}^{\infty} p\_n z^n \right) \left( z + \sum\_{n=2}^{\infty} b\_n z^n \right) \dots$$

Comparing the *nth* term's coefficients on both sides, we obtain

$$ma\_{\mathbb{H}} = b\_{\mathbb{H}} - \frac{\beta - 1}{1 - k} \left[ p\_{n-1} + p\_{n-2}b\_{\mathbb{2}} + p\_{n-3}b\_{\mathbb{3}} + \dots + p\_1 b\_{n-1} \right]$$

By taking the absolute value, we ge<sup>t</sup>

$$\begin{aligned} |n|a\_n| &= \left| b\_n - \frac{\beta - 1}{1 - k} \left[ p\_{n-1} + p\_{n-2}b\_2 + p\_{n-3}b\_3 + \dots + p\_1 b\_{n-1} \right] \right| \\ &\le \left| b\_n \right| + \frac{\beta - 1}{1 - k} \left| p\_{n-1} + p\_{n-2}b\_2 + p\_{n-3}b\_3 + \dots + p\_1 b\_{n-1} \right|. \end{aligned}$$

Applying the triangle inequality, we obtain

$$n|a\_{\boldsymbol{n}}| \le |b\_{\boldsymbol{n}}| + \frac{\beta - 1}{1 - k} \left\{ |p\_{n-1}| + |p\_{n-2}b\_2| + |p\_{n-3}b\_3| + \dots + |p\_1b\_{n-1}| \right\}.\tag{19}$$

.

As Re {*p*(*z*)} > 0 in U, we have |*pn*| ≤ 2 (*n* ≥ 1) (see [1]). Then, from (19), we have

$$n|a\_n| \le |b\_n| + \frac{2\left(\beta - 1\right)}{1 - k} \sum\_{j=1}^{n-1} |b\_j| \,\_{\beta}$$

where *b*1 = 1. Using Lemma (3), we obtain

$$n|a\_{\mathfrak{l}}| \le \frac{\left(\delta\_{k,\gamma}\right)\_{n-1}}{(n-1)!} + \delta\_{k,\emptyset} \sum\_{j=1}^{n-1} \frac{\left(\delta\_{k,\gamma}\right)\_{j-1}}{(j-1)!}.$$

where *<sup>δ</sup>k*,*<sup>β</sup>* = 2 (*β* − 1) 1 − *k* and *<sup>δ</sup>k*,*<sup>γ</sup>* is defined by (8). This can be written as

$$|a\_n| \le \frac{\left(\delta\_{k,\gamma}\right)\_{n-1}}{n!} + \frac{\delta\_{k,\emptyset}}{n} \sum\_{j=1}^{n-1} \frac{\left(\delta\_{k,\gamma}\right)\_{j-1}}{(j-1)!}.$$

This completes the proof.

From Definition 4 and Theorem 2, we immediately ge<sup>t</sup> the following corollary.

**Corollary 1.** *If f*(*z*) ∈ QD(*k*, *β*, *<sup>γ</sup>*)*, then*

$$|a\_n| \le \frac{1}{n} \left[ \frac{\left(\delta\_{k,\gamma}\right)\_{n-1}}{n!} + \frac{\delta\_{k,\emptyset}}{n} \sum\_{j=1}^{n-1} \frac{\left(\delta\_{k,\gamma}\right)\_{j-1}}{(j-1)!} \right].$$

*where <sup>δ</sup>k*,*<sup>β</sup> and <sup>δ</sup>k*,*<sup>γ</sup> are given by* (16) *and* (8)*, respectively.*

By taking *k* = 0 in the above results, we obtain the coefficient inequality for the classes KD(*β*, *γ*) and QD(*β*, *<sup>γ</sup>*).

**Theorem 4.** *If a function f* ∈ KD(*k*, *β*, *<sup>γ</sup>*)*, then there exists g* ∈ MD(*k*, *γ*) *such that*

$$\frac{z f'(z)}{g(z)} \prec 1 + 2\left(\beta\_1 - 1\right) - \frac{2\left(\beta\_1 - 1\right)}{1 - z}, \ (z \in \mathbb{U}), \tag{20}$$

*where*

$$
\beta\_1 = \frac{\beta - k}{1 - k}.\tag{21}
$$

**Proof.** Let *f*(*z*) ∈ KD(*k*, *β*, *<sup>γ</sup>*). Then, there exists *g*(*z*) in MD(*k*, *γ*) and a Schwarz function *w*(*z*) such that 

$$\frac{\beta\_1 - \left(\frac{zf'(z)}{g(z)}\right)}{\beta\_1 - 1} = \frac{1 + w(z)}{1 - w(z)}\tag{22}$$

as *w*(*z*) is analytic U with *w*(0) = 0 and

$$\Re e\left(\frac{1+w(z)}{1-w(z)}\right) > 0, \ (z \in \mathbb{U})\ .$$

So, from (22), we obtain

$$\begin{split} \frac{zf'(z)}{g(z)} &= \begin{array}{c} \beta\_1 - (\beta\_1 - 1) \left( \frac{1 + w(z)}{1 - w(z)} \right) \\ &= \frac{\beta\_1 \left( 1 - w(z) \right) - (\beta\_1 - 1) \left( 1 + w(z) \right)}{1 - w(z)} \end{array} \\ &= \frac{1 + w(z) - 2 \beta\_1 w(z)}{1 - w(z)} \\ &= \frac{1 - w(z) - 2 \left( \beta\_1 - 1 \right) w(z)}{1 - w(z)} \\ &= \frac{1 - w(z) + 2 \left( \beta\_1 - 1 \right) - 2 \left( \beta\_1 - 1 \right) w(z) - 2 \left( \beta\_1 - 1 \right)}{1 - w(z)} \\ &= \frac{1 - w(z) + 2 \left( \beta\_1 - 1 \right) \left( 1 - w(z) \right) - 2 \left( \beta\_1 - 1 \right)}{1 - w(z)}. \end{split}$$

This implies that

$$\frac{z f'(z)}{g(z)} = 1 + 2\left(\beta\_1 - 1\right) - \frac{2\left(\beta\_1 - 1\right)}{1 - w(z)}\prime$$

and hence

$$\frac{z f'(z)}{g(z)} \prec 1 + 2 \left(\beta\_1 - 1\right) - \frac{2 \left(\beta\_1 - 1\right)}{1 - z}, \quad (z \in \mathbb{U}),$$

which is as required in (20).

**Corollary 2.** *If f* ∈ QD(*k*, *β*, *<sup>γ</sup>*)*, then there exists g* ∈ND (*k*, *γ*) *such that*

$$\frac{\left(zf'(z)\right)'}{g'(z)} \prec 1 + 2\left(\beta\_1 - 1\right) - \frac{2\left(\beta\_1 - 1\right)}{\left(1 - z\right)}, \ (z \in \mathbb{U}), \tag{23}$$

*where β*1 *is given by* (21)*.*

**Theorem 5.** *If f* ∈ KD(*k*, *β*, *<sup>γ</sup>*)*, then there exists a function g* ∈ MD (*k*, *γ*) *such that*

$$\frac{1 - (2\beta\_1 - 1)r}{1 - r} \le \Re \epsilon \frac{zf'(z)}{g(z)} \le \frac{1 + (2\beta\_1 - 1)r}{1 + r},\tag{24}$$

*where* |*z*| = *r* < 1 *and β*1 *is given by* (21)*.*

**Proof.** Using Theorem 4, we define the function *φ* as follows

$$\phi(z) = 1 + 2\left(\beta\_1 - 1\right) + \frac{2\left(1 - \beta\_1\right)}{1 - z}, \left(z \in \mathbb{U}\right).$$

Letting *z* = *rei<sup>θ</sup>* (0 ≤ *r* < <sup>1</sup>), we observe that

$$\Re \mathfrak{e}\phi(z) = 1 + 2\left(\beta\_1 - 1\right) + \frac{2\left(1 - \beta\_1\right)\left(1 - r\cos\theta\right)}{1 + r^2 - 2r\cos\theta}.$$

Let us define

$$\psi(t) = \frac{1 - rt}{1 + r^2 - 2rt}, (t = \cos \theta) \dots$$

$$\text{As } \psi'(t) = \frac{r\left(1 - r^2\right)}{\left(1 + r^2 - 2rt\right)^2} \ge 0 \text{ (since } r < 1\text{), we get}$$

$$1 + 2\left(\beta\_1 - 1\right) + \frac{2\left(1 - \beta\_1\right)}{1 - r} \le \Re\phi(z) \le 1 + 2\left(\beta\_1 - 1\right) + \frac{2\left(1 - \beta\_1\right)}{1 + r}.$$

After simplification, we have

$$\frac{1 - (2\beta\_1 - 1)r}{1 - r} \le \Re \mathfrak{e}\phi(z) \le \frac{1 + (2\beta\_1 - 1)r}{1 + r}.$$

With the fact that *z f* (*z*) *g*(*z*) ≺ *φ*(*z*), (*z* ∈ U) and as *φ* is univalent in U, by using (22), we ge<sup>t</sup> the required result.

**Corollary 3.** *If f* ∈ QD(*k*, *β*, *<sup>γ</sup>*)*, then there exists g* ∈ND (*k*, *γ*) *such that*

$$\frac{1 - \left(2\beta\_1 - 1\right)r}{1 - r} \le \Re \mathfrak{e} \frac{\left(zf'(z)\right)'}{g'(z)} \le \frac{1 + (2\beta\_1 - 1)r}{1 + r},\tag{25}$$

.

*where* |*z*| = *r* < 1 *and β*1 *is given by* (21)*.*

**Theorem 6.** *Assume that a function f* ∈ A *satisfies*

$$\Re \mathfrak{Re} \left( \frac{zg'(z)}{g(z)} - \frac{zf''(z)}{f'(z)} \right) > \frac{\beta\_1 + 1}{2\beta\_1}, \ (z \in \mathbb{U}), \tag{26}$$

*for some g*(*z*) ∈ MD (*k*, *γ*) *and for real β*1 *given by* (21)*. If*

$$\phi(z) = \frac{zf'(z)}{g(z)}$$

*is analytic in* U *and φ*(*z*) = 0 *and φ*(*z*) = 2*β*1 − 1 *in* U*, then f* ∈ KD(*k*, *β*1)*.*

**Proof.** Let us define a function *w*(*z*) by

$$w(z) = \frac{\phi(z) - 1}{\phi(z) + (1 - 2\beta\_1)}, \ z \in \mathbb{U}.$$

Then, *w*(*z*) is analytic in U as *φ*(*z*) = 2*β*1 − 1 and

$$\phi(z) = \frac{zf'(z)}{g(z)} = \frac{1 + (1 - 2\beta\_1) \, w(z)}{1 - w(z)}.\tag{27}$$

Because *φ*(*z*) = 0, we use logarithmic differentiation to ge<sup>t</sup>

$$\frac{1}{z} + \frac{f'''(z)}{f'(z)} - \frac{g'(z)}{g(z)} = \frac{(1 - 2\beta\_1) \, w'(z)}{1 + (1 - 2\beta\_1) \, w(z)} + \frac{w'(z)}{1 - w(z)}.$$

which further yields

$$\frac{zg'(z)}{g(z)} - \frac{zf''(z)}{f'(z)} = 1 - \frac{(1 - 2\beta\_1)}{1 + (1 - 2\beta\_1)} \frac{zw'(z)}{w(z)} - \frac{zw'(z)}{1 - w(z)}.\tag{28}$$

Then, we note that *w* is analytic in open unit disk and *w*(0) = 0. Therefore, from (28), we obtain

$$\begin{split} \mathfrak{Re}\left(\frac{zg'(z)}{g(z)} - \frac{zf''(z)}{f'(z)}\right) &= \mathfrak{Re}\left(1 - \frac{(1 - 2\beta\_1)zw'(z)}{1 + (1 - 2\beta\_1)w(z)} - \frac{zw'(z)}{1 - w(z)}\right) \\ &> \frac{\beta\_1 + 1}{2\beta\_1} .\end{split}$$

Suppose there exists a point *z*0 ∈ U such that

$$\max\_{|z| \le |z\_0|} |w(z)| = |w(z\_0)| = 1,$$

then, by Lemma 4, we can write *<sup>w</sup>*(*<sup>z</sup>*0) = *ei<sup>θ</sup>* and *<sup>z</sup>*0*w*(*<sup>z</sup>*0) = *cei<sup>θ</sup>* for a point *z*0, and we have

$$\begin{split} &\quad \mathfrak{Re}\left(\frac{z\_0 g'(z\_0)}{g(z\_0)} - \frac{z\_0 g^{\partial}(z\_0)}{f'(z\_0)}\right) \\ &= \quad \mathfrak{Re}\left(1 - \frac{(1 - 2\beta\_1)c\epsilon^{\partial}}{1 + (1 - 2\beta\_1)c\epsilon^{\partial}} - \frac{cc^{\partial}}{1 - \epsilon^{\partial}}\right) \\ &= \quad \mathfrak{Re}\left(1 - \frac{c\left(1 - 2\beta\_1\right)\left(c^{\partial} + (1 - 2\beta\_1)\right)}{1 + \left(1 - 2\beta\_1\right)^2 + 2\left(1 - 2\beta\_1\right)\cos\theta} + \frac{c\left(1 - \epsilon^{\partial}\right)}{2\left(1 - \cos\theta\right)}\right) \\ &= \quad 1 + \frac{c\left(2\beta\_1 - 1\right)\left[\cos\theta + \left(1 - 2\beta\_1\right)\right]}{1 + \left(1 - 2\beta\_1\right)^2 + 2\left(1 - 2\beta\_1\right)\cos\theta} + \frac{c}{2} \\ &\leq \quad 1 - \frac{c\left(2\beta\_1 - 1\right)}{2\beta\_1} + \frac{c}{2} \\ &= \quad 1 - \frac{c\left(\beta\_1 - 1\right)}{2\beta\_1} \\ &\leq \quad 1 - \frac{\beta\_1 - 1}{2\beta\_1}, \text{ as } c < 1 \\ &= \quad \frac{\beta\_1 + 1}{2\beta\_1}, \end{split}$$

which gives that

$$\Re \mathfrak{Re} \left\{ \frac{z\_0 g'(z\_0)}{g(z\_0)} - \frac{z\_0 f''(z\_0)}{f'(z\_0)} \right\} \le \frac{\beta\_1 + 1}{2\beta\_1}.$$

which is the contradiction to the supposed condition (26). Hence, there is no *z*0 ∈ U such that |*w*(*<sup>z</sup>*0)| = 1. This implies that |*w*(*z*)| < 1,(*z* ∈ U) and, therefore, by (27), we have

$$\frac{zf'(z)}{g(z)} \prec \frac{1 - (2\beta\_1 - 1)z}{1 - z}$$

or

$$\mathfrak{Re}\left\{\frac{zf'(z)}{g(z)}\right\} < \beta\_1, \ z \in \mathbb{U}.$$

Hence, we conclude that *f*(*z*) ∈ KD(*k*, *β*1).

**Theorem 7.** *Assume that k* ≤ 0 *and β* > 1*. If f* ∈ A *and if there exists g* ∈ MD (*k*, *γ*) *such that*

$$\left|\frac{zf'(z)}{g(z)} - 1\right| < \frac{\beta - 1}{1 - k} \quad z \in \mathbb{U},\tag{29}$$

*then f* ∈ KD(*k*, *β*, *<sup>γ</sup>*)*.*

**Proof.** We have

$$\begin{aligned} &\left|\frac{zf'(z)}{g(z)} - 1\right| < \frac{\beta - 1}{1 - k} \\ \Rightarrow & (1 - k) \left|\frac{zf'(z)}{g(z)} - 1\right| + 1 < \beta \\ \Rightarrow & \left|\frac{zf'(z)}{g(z)} - 1\right| + 1 < k \left|\frac{zf'(z)}{g(z)} - 1\right| + \beta \\ \Rightarrow & \Re \frac{zf'(z)}{g(z)} < k \left|\frac{zf'(z)}{g(z)} - 1\right| + \beta \\ \Rightarrow & f \in \mathcal{K}\mathcal{D}(k, \beta, \gamma). \end{aligned}$$

**Corollary 4.** *Let f* ∈ A *have the form* (1)*. Assume that g* = *z* + *<sup>b</sup>*2*z*<sup>2</sup> + ··· *belongs to the class* MD (*k*, *γ*) *and satisfies*

$$\left| \frac{\sum\_{n=2}^{\infty} (na\_n - b\_n)z^{n-1}}{1 + \sum\_{n=2}^{\infty} b\_n z^{n-1}} \right| < \frac{\beta - 1}{1 - k} \quad z \in \mathbb{U},\tag{30}$$

*for some k* (*k* ≤ <sup>0</sup>)*, β* (*β* > 1)*. Then, f*(*z*) ∈ KD(*k*, *β*, *<sup>γ</sup>*)*.*

**Proof.** We have

$$\begin{aligned} & \left| \frac{z f'(z)}{g(z)} - 1 \right| \\ &= \left| \frac{z + \sum\_{n=2}^{\infty} n a\_n z^n}{z + \sum\_{n=2}^{\infty} b\_n z^n} - 1 \right| \\ &= \left| \frac{\sum\_{n=2}^{\infty} (n a\_n - b\_n) z^{n-1}}{1 + \sum\_{n=2}^{\infty} b\_n z^{n-1}} \right| \\ &< \left| \frac{\beta - 1}{1 - k} \right| \end{aligned}$$

and hence (29) follows immediately from (30). **Theorem 8.** *Let f* ∈ A *have the form* (1) *and let g* = *z* + ∑∞*<sup>n</sup>*=<sup>2</sup> *bnzn, belonging to the class* MD (*k*, *<sup>γ</sup>*)*, satisfy*

$$1 + \sum\_{n=2}^{\infty} \left( n|a\_n| + y|b\_n| \right) < y \quad z \in \mathbb{U},\tag{31}$$

*for some k* (*k* ≤ <sup>0</sup>)*, β* (*β* > 1) *and where*

$$y = \frac{(\beta - 1)}{(1 - k)} > 0.$$

*Then, f*(*z*) ∈ KD(*k*, *β*, *<sup>γ</sup>*)*.*

**Proof.** Consider

$$1 + \sum\_{n=2}^{\infty} \left( n \left| a\_n \right| + y \left| b\_n \right| \right) < y \tag{32}$$

$$\implies \quad 1 + \sum\_{n=2}^{\infty} n \left| a\_n \right| < y - y \sum\_{n=2}^{\infty} \left| b\_n \right| $$

$$\implies \quad 0 < y - y \sum\_{n=2}^{\infty} \left| b\_n \right| $$

$$\implies \quad 0 < y - y \sum\_{n=2}^{\infty} \left| b\_n \right| \left| z^{n-1} \right| $$

$$\implies \quad 0 < y \left| 1 + \sum\_{n=2}^{\infty} b\_n z^{n-1} \right| . \tag{33}$$

We have

$$\begin{aligned} &1 + \sum\_{n=2}^{\infty} \left( n|a\_{\mathbb{n}}| + y|b\_{\mathbb{n}}| \right) < y \\ &\Rightarrow \quad 1 + \sum\_{n=2}^{\infty} n|a\_{\mathbb{n}}| < y - y \sum\_{n=2}^{\infty} |b\_{\mathbb{n}}| \\ &\Rightarrow \quad 1 + \sum\_{n=2}^{\infty} n|a\_{\mathbb{n}}| \left| z^{n-1} \right| < y - y \sum\_{n=2}^{\infty} |b\_{\mathbb{n}}| \left| z^{n-1} \right| \\ &\Rightarrow \quad \left| 1 + \sum\_{n=2}^{\infty} n a\_{\mathbb{n}} z^{n-1} \right| < y \left| 1 + \sum\_{n=2}^{\infty} b\_{n} z^{n-1} \right| \\ &\Rightarrow \quad \left| \frac{1 + \sum\_{n=2}^{\infty} n a\_{n} z^{n-1}}{1 + \sum\_{n=2}^{\infty} b\_{n} z^{n-1}} \right| < y, \end{aligned}$$

from (33). By (30), it follows that *f* ∈ KD(*k*, *β*, *<sup>γ</sup>*).

**Author Contributions:** Conceptualization, S.M.; Formal analysis, S.N.M. and J.S.; Funding acquisition, S.M.; Investigation, S.M.; Methodology, S.M. and S.N.M.; Supervision, H.M.S. and J.S.; Validation, H.M.S.; Visualization, S.M.;Writing—original draft, S.M.;Writing—review and editing, S.M. and S.N.M.

**Funding:** This research is supported by Sarhad University of Science & I.T, Peshawar 25000, Pakistan.

**Acknowledgments:** The authors are grateful to referees for their valuable comments which improved the quality of work and presentation of paper.

**Conflicts of Interest:** The authors declare no conflict of interest.
