**3. Main Results**

The new hypergeoemtric integral operator is defined using Definition 3 and the integral operator given by relation (6).

**Definition 4.** *Let β* > 1*, γ* > 0 *and the confluent (Kummer) hypergeometric function φ given by (5).*

*Let M* : H(*U*) → H(*U*) *be given by:*

$$M(z) = \frac{\beta}{z^{\beta - 1}} \int\_0^z \left[ \frac{\Gamma(\upsilon)}{\Gamma(\mu)} \sum\_{k=0}^\infty \frac{\Gamma(\upsilon + k)}{\Gamma(\upsilon + k)} \frac{t^k}{k!} \right] t^{\beta - 1} dt, \quad z \in \mathsf{U}. \tag{7}$$

**Remark 3.** *(a) For β* > 1*, γ* > 0 *and φ*(*<sup>u</sup>*, *v*; *z*) = <sup>Γ</sup>(*v*) <sup>Γ</sup>(*u*) ∞∑*k*=0 <sup>Γ</sup>(*u*+*k*) <sup>Γ</sup>(*v*+*k*) *zkk*! = 1 + *uv z*1! + *u*(*u*+<sup>1</sup>) *v*(*v*+<sup>1</sup>) *z*22! + ..., *u*, *v* ∈ C*, v* = 0, −1, −2, . . . *, we have*

$$M(z) = \frac{\beta}{z^{\beta - 1}} \int\_0^z \left[ 1 + \frac{u}{v} \frac{t}{1!} + \frac{u(u+1)}{v(v+1)} \frac{t^2}{2!} + \dots \right]^\gamma t^{\beta - 1} dt = \tag{8}$$

$$\frac{\beta}{z^{\beta - 1}} \int\_0^z \left( 1 + \gamma \frac{u}{v} t + p\_2 t^2 + \dots \right) t^{\beta - 1} dt = $$

$$\frac{\beta}{z^{\beta - 1}} \int\_0^z \left[ t^{\beta - 1} + \gamma \frac{u}{v} t^{\beta} + \dots \right] dt = z + \gamma \frac{u}{v} \frac{\beta}{\beta + 1} z^2 + p\_2 \frac{\beta}{\beta + 2} z^3 + \dots$$

*which is the analytic expression of the operator M*.

$$\text{(b) For } z \in \mathcal{U}, \mathcal{M}'(z) = 1 + 2\gamma \frac{u}{v} \frac{\mathcal{B}}{\mathcal{B} + \Gamma} z + \dots, \text{ with } \mathcal{M}'(0) = 1.$$

Using the method of differential subordination, next, a theorem is proved, giving the best dominant of a certain fuzzy differential subordination. Using specific functions as the fuzzy best dominant, conditions for starlikeness and convexity of the operator *M* are obtained as corollaries.

**Theorem 1.** *For β*, *γ* ∈ C*, β* > 1*, γ* > 0*, let the fuzzy function F* : C → [0, 1] *be given by*

$$F(z) = \frac{|z|}{1+|z|}, \ z \in \mathcal{U},\tag{9}$$

*and consider a holomorphic function in U given by the equation*

$$h(z) = q(z) + \frac{zq'(z)}{\beta - 1 + \gamma q(z)}, \ h(0) = q(0), \tag{10}$$

*when q is a univalent solution in U which satisfies the fuzzy differential subordination:*

$$\frac{|q(z)|}{1+|q(z)|} \le \frac{|h(z)|}{1+|h(z)|} \text{ i.e., } q(z) \preccurlyeq\_{\mathcal{F}} h(z), \; z \in \mathcal{U}. \tag{11}$$

*Consider φ*(*<sup>u</sup>*, *v*; *<sup>z</sup>*)*, a confluent (or Kummer) hypergeometric function given by (5) and the operator <sup>M</sup>*(*z*) *given by (7).*

*If* 1 + *γ<sup>z</sup>φ*(*<sup>u</sup>*,*v*;*<sup>z</sup>*) *φ*(*<sup>u</sup>*,*v*;*<sup>z</sup>*) *is analytic in U, and*

$$\frac{\left|1+\gamma \frac{z\phi'(u,v;z)}{\phi(u,v;z)}\right|}{1+\left|1+\gamma \frac{z\phi'(u,v;z)}{\phi(u,v;z)}\right|} \le \frac{|h(z)|}{1+|h(z)|} \quad i.e.,\ 1+\gamma \frac{z\phi'(u,v;z)}{\phi(u,v;z)} \prec\_{\mathcal{F}} h(z),\tag{12}$$

*then*

$$\frac{\left|\frac{zM'(z)}{M(z)}\right|}{1+\left|\frac{zM'(z)}{M(z)}\right|} \le \frac{|q(z)|}{1+|q(z)|} \quad \text{i.e.,} \ \frac{zM'(z)}{M(z)} \preccurlyeq\_{\mathcal{F}} q(z), \ z \in \mathcal{U},$$

*and q is the fuzzy best dominant.*

**Proof.** Relation (7) is equivalent to

$$z^{\beta -1}M(z) = \beta \int\_0^z \phi^\gamma(u, v; t) t^{\beta -1} dt. \tag{13}$$

Differentiating (13) and after short calculation we obtain

$$
\hat{\phi}\left(\beta - 1\right)M(z) + zM'(z) = \beta \phi^\gamma(u, v; z) \cdot z,\tag{14}
$$

which is equivalent to

$$M(z)\left[\beta - 1 + \frac{zM'(z)}{M(z)}\right] = \beta \phi^\gamma(u, v; z) \cdot z. \tag{15}$$

We let

$$p(z) = \frac{zM'(z)}{M(z)} = \frac{z\left(1 + 2\gamma \frac{\mu}{v} \frac{\beta}{\beta + 1} z + \dots \right)}{z\left(1 + \gamma \frac{\mu}{v} \frac{\beta}{\beta + 1} z + \dots \right)} = \frac{1 + 2\gamma \frac{\mu}{v} \frac{\beta}{\beta + 1} z + \dots}{1 + \gamma \frac{\mu}{v} \frac{\beta}{\beta + 1} z + \dots},\tag{16}$$

*p*(0) = 1.

> Using (16) in (15), we ge<sup>t</sup>

$$M(z)[\beta - 1 + p(z)] = \beta \phi^\gamma(\mu, v; z) \cdot z. \tag{17}$$

Differentiating (17), we obtain

$$\frac{zM'(z)}{M(z)} + \frac{zp'(z)}{\beta - 1 + p(z)} = \gamma \frac{z\phi'(u, v; z)}{\phi(u, v; z)} + 1.\tag{18}$$

Using (16) in (18), we have

$$p(z) + \frac{zp'(z)}{\beta - 1 + p(z)} = \gamma \frac{z\phi'(u, v; z)}{\phi(u, v; z)} + 1.\tag{19}$$

Using (19) in (12), we ge<sup>t</sup>

$$\frac{\left|p(z) + \frac{zp'(z)}{\beta - 1 + p(z)}\right|}{1 + \left|p(z) + \frac{zp'(z)}{\beta - 1 + p(z)}\right|} \le \frac{|h(z)|}{1 + |h(z)|}, z \in \mathcal{U}. \tag{20}$$

In order to obtain the desired relation, Lemma 1 will be applied. To apply the lemma, let the function *ψ* : C<sup>2</sup> × *U* → C,

$$
\psi(r, s; z) = r + \frac{s}{\beta - 1 + r}, \ r, s \in \mathbb{C}. \tag{21}
$$

For *r* = *p*(*z*), *s* = *zp*(*z*), relation (21) becomes

$$\psi(p(z), zp'(z)) = p(z) + \frac{zp'(z)}{\beta - 1 + p(z)}, \ z \in \mathcal{U}.\tag{22}$$

Using (22) in (20), we have

$$\frac{|\psi(p(z), zp'(z))|}{1 + |\psi(p(z), zp'(z))|} \le \frac{|h(z)|}{1 + |h(z)|}, z \in \mathcal{U}.\tag{23}$$

Applying Lemma 1, for *δ* = *β* − 1, *ω* = *γ* = 0, we obtain

$$
\psi\left(p(z), zp'(z)\right) \prec\_{\mathcal{F}} h(z), \; z \in \mathcal{U}.\tag{24}
$$

Using (22) in (24), we ge<sup>t</sup>

$$p(z) + \frac{zp'(z)}{\beta - 1 + p(z)} \prec\_{\mathcal{F}} h(z), \ z \in \mathcal{U}. \tag{25}$$

According to Lemma 1, relation (25) implies

$$p(z) \preccurlyeq q(z), \ z \in \mathsf{U}.\tag{26}$$

Using (16) in (26) we have

$$\frac{zM'(z)}{M(z)} \prec\_{\mathcal{F}} q(z), \; z \in \mathcal{U}. \tag{27}$$

Since *q* satisfies the differential Equation (10), *q* is the fuzzy best dominant.

**Remark 4.** *Using particular expressions for the fuzzy best dominant q, sufficient conditions for starlikeness of the operator <sup>M</sup>*(*z*) *given by (7) can be obtained.*

If in Theorem 1, function *q*(*z*) = 1−*<sup>z</sup>* 1+*z* is considered the following corollary is obtained.

**Corollary 1.** *For β*, *γ* ∈ C*, β* > 1*, γ* > 0*, let the fuzzy function F* : C → [0, 1] *given by (9) and consider a holomorphic function in U given by the equation*

$$h(z) = \frac{(\beta - 1)\left(1 - z^2\right) + 1 - 4z + z^2}{\left(\beta - 1\right)\left(1 + z\right)^2 + 1 - z^2} \lambda$$

*h*(0) = *q*(0) = 1*, where function q*(*z*) = 1−*<sup>z</sup>* 1+*z is a univalent solution in U, which satisfies the fuzzy differential subordination*

$$\frac{|1-z|}{|1+z|+|1-z|} \le \frac{| (\beta-1) \left( 1-z^2 \right) + 1 - 4z + z^2 |}{\left| (\beta-1)(1+z)^2 + 1 - z^2 \right| + | (\beta-1)(1-z^2) + 1 - 4z + z^2 |},$$

*i.e.,*

$$\frac{1-z}{1+z} \prec\_F \frac{\left(\beta -1\right)\left(1 - z^2\right) + 1 - 4z + z^2}{\left(\beta - 1\right)\left(1 + z\right)^2 + 1 - z^2}, \; z \in \mathcal{U}.$$

*Consider φ*(*<sup>u</sup>*, *v*; *z*) *to be a confluent (or Kummer) hypergeometric operator given by (5), and the operator <sup>M</sup>*(*z*) *given by (7).*

*If* 1 + *γ<sup>z</sup>φ*(*<sup>u</sup>*,*v*;*<sup>z</sup>*) *φ*(*<sup>u</sup>*,*v*;*<sup>z</sup>*) *is analytic in U, and*

$$\frac{\left|1+\gamma\frac{z\phi'(u,v;z)}{\phi(u,v;z)}\right|}{1+\left|1+\gamma\frac{z\phi'(u,v;z)}{\phi(u,v;z)}\right|} \le \frac{| (\beta-1)(1-z^2) + 1 - 4z + z^2 |}{\left| (\beta-1)(1+z)^2 + 1 - z^2 \right| + | (\beta-1)(1-z^2) + 1 - 4z + z^2 |},$$

$$\text{i.e., } 1+\gamma \frac{z\phi'(u,v;z)}{\phi(u,v;z)} \preccurlyeq \frac{(\beta-1)(1-z^2) + 1 - 4z + z^2}{(\beta-1)(1+z)^2 + 1 - z^2},$$

*then*

$$\frac{\left|\frac{zM'(z)}{M(z)}\right|}{1+\left|\frac{zM'(z)}{M(z)}\right|} \le \frac{\left|\frac{1-z}{1+z}\right|}{1+\left|\frac{1-z}{1+z}\right|} \text{ i.e., } \frac{zM'(z)}{M(z)} \precneq\_{\mathcal{F}} \frac{1-z}{1+z}, \; z \in \mathcal{U}, \text{ or } M \in \mathcal{S}^\*$$

*and q*(*z*) = 1−*<sup>z</sup>* 1+*z is the fuzzy best dominant.*

**Proof.** By using the function *q*(*z*) = 1−*<sup>z</sup>* 1+*z* in relation (27) from the proof of Theorem 1, the following fuzzy subordination is obtained:

$$\frac{zM'(z)}{M(z)} \preccurlyeq\_{\mathcal{F}} q(z) = \frac{1-z}{1+z}, \; z \in \mathcal{U}.\tag{28}$$

Since Re*zq*(*z*) *q*(*z*) + 1 = <sup>1</sup>−*ρ*<sup>2</sup> 1+2*ρ* cos *α*+*ρ*<sup>2</sup> > 0, 0 < *ρ* < 1, the *q* is convex, and Re 1−*<sup>z</sup>* 1+*z* > 0, *z* ∈ *U*, differential subordination (28) is equivalent to

> Re *zM*(*z*) *<sup>M</sup>*(*z*) > Re<sup>1</sup> − *z* 1 + *z* > 0, *z* ∈ *U*, hence *M* ∈ S∗. (29)

**Remark 5.** *Using the convex function q*(*z*) = 1+*z* 1−*<sup>z</sup> as the fuzzy best dominant in Theorem 1, sufficient conditions for the convexity of the operator <sup>M</sup>*(*z*) *given by (7) can be obtained as a corollary.*

**Corollary 2.** *For β*, *γ* ∈ C*, β* > 1*, γ* > 0*, let the fuzzy function F* : C → [0, 1] *given by (9) and consider a holomorphic function in U given by the equation*

$$h(z) = \frac{(\beta - 1)\left(1 - z^2\right) + \gamma(1 + z)^2}{\left(\beta - 1\right)\left(1 - z\right)^2 + \gamma(1 - z^2)},$$

*h*(0) = *q*(0) = 1*, where function q*(*z*) = 1+*z* 1−*<sup>z</sup> is a univalent solution in U which satisfies the fuzzy differential subordination*

$$\frac{|1+z|}{|1-z|+|1+z|} \le \frac{\left| (\beta-1)\left(1-z^2\right) + \gamma(1+z)^2 \right|}{\left| (\beta-1)(1-z)^2 + \gamma(1-z^2) \right| + \left| (\beta-1)(1-z^2) + \gamma(1+z)^2 \right|},$$

*i.e.,*

$$\frac{1+z}{1-z} \prec\_{\mathcal{F}} \frac{\left(\beta - 1\right)\left(1 - z^2\right) + \gamma\left(1 + z\right)^2}{\left(\beta - 1\right)\left(1 - z\right)^2 + \gamma\left(1 - z^2\right)}, \ z \in \mathcal{U}.$$

*Consider φ*(*<sup>u</sup>*, *v*; *z*) *the confluent (or Kummer) hypergeometric operator given by (5), and the operator <sup>M</sup>*(*z*) *given by (7).*

*If*

$$(1 + (\gamma - 1)\frac{z\phi'(u, v; z)}{\phi(u, v; z)} + \frac{(\gamma + 1)z\phi'(u, v; z) + z^2\gamma\phi''(u, v; z)}{\phi(u, v; z) + \gamma z\phi'(u, v; z)} = E(u, v; z)$$

*is analytic in U, and*

$$\frac{|E(u,v;z)|}{1+|E(u,v;z)|} \le \frac{\left| (\beta-1)\left(1-z^2\right) + \gamma(1+z)^2 \right|}{\left| (\beta-1)(1-z)^2 + \gamma(1-z^2) \right| + \left| (\beta-1)(1-z^2) + \gamma(1+z)^2 \right|} \tag{30}$$
 
$$\text{i.e., } E(u,v;z) \prec\_F \frac{\left(\beta-1\right)\left(1-z^2\right) + \gamma(1+z)^2}{\left(\beta-1\right)\left(1-z\right)^2 + \gamma(1-z^2)},$$

*then*

$$\frac{\left|1+\frac{zM''(z)}{M'(z)}\right|}{1+\left|1+\frac{zM''(z)}{M'(z)}\right|} \le \frac{\left|\frac{1+z}{1-z}\right|}{1+\left|\frac{1+z}{1-z}\right|} \text{ i.e., } 1+\frac{zM''(z)}{M'(z)} \precneq \frac{1+z}{1-z}, \; z \in \mathcal{U}, \text{ or } M \in \mathcal{K}$$

*and q*(*z*) = 1+*z* 1−*<sup>z</sup> is the fuzzy best dominant.*

**Proof.** Differentiating relation (14), from the proof of Theorem 1, we have

$$
\beta M'(z) + z M''(z) = \beta \phi^{\gamma - 1}(u, v; z) \left[ \phi(u, v; z) + \gamma z \phi'(u, v; z) \right],
$$

which is equivalent to

$$M'(z)\left[ (\beta - 1) + 1 + \frac{zM''(z)}{M'(z)} \right] = \beta \phi^{\gamma - 1}(u, v; z) \left[ \phi(u, v; z) + \gamma z \phi'(u, v; z) \right] \tag{31}$$

Let

$$1 + \frac{zM^{\prime\prime}(z)}{M^{\prime}(z)} = p(z), z \in \mathcal{U}, p(0) = 1. \tag{32}$$

By replacing (32) in (31) we obtain

$$[M'(z)](\beta - 1) + p(z)] = \beta \phi^{\gamma - 1}(u, v; z) \left[\phi(u, v; z) + \gamma z \phi'(u, v; z)\right].\tag{33}$$

Differentiating relation (33) we ge<sup>t</sup>

$$\frac{zM''(z)}{M'(z)} + \frac{zp'(z)}{\beta - 1 + p(z)} = (\gamma - 1)\frac{z\phi'(u, v; z)}{\phi(u, v; z)} + \frac{z(\gamma + 1)\phi'(u, v; z) + z^2\gamma\phi''(u, v; z)}{\phi(u, v; z) + \gamma z\phi'(u, v; z)}.$$

After some computations, we have

$$1 + \frac{zM''(z)}{M'(z)} + \frac{zp'(z)}{\beta - 1 + p(z)} = 1 + (\gamma - 1)\frac{z\phi'(u, v; z)}{\phi(u, v; z)} + \frac{z(\gamma + 1)\phi'(u, v; z) + z^2\gamma\phi''(u, v; z)}{\phi(u, v; z) + \gamma z\phi'(u, v; z)}.\tag{34}$$

Using relation (32) in (34) we can write

$$p(z) + \frac{zp'(z)}{\beta - 1 + p(z)} = E(u, v; z), \; z \in \mathcal{U}.\tag{35}$$

By considering (35) in (30), the following inequality emerges

$$\frac{\left|p(z) + \frac{zp'(z)}{\beta - 1 + p(z)}\right|}{1 + \left|p(z) + \frac{zp'(z)}{\beta - 1 + p(z)}\right|} \le \frac{\left|\left(\beta - 1\right)\left(1 - z^2\right) + \gamma(1 + z)^2\right|}{\left|\left(\beta - 1\right)(1 - z)^2 + \gamma(1 - z^2)\right| + \left|\left(\beta - 1\right)(1 - z^2) + \gamma(1 + z)^2\right|}. \tag{36}$$

In order to obtain the expected result, Lemma 1 will be used. For that, let *ψ* : C<sup>2</sup> × *U* → C, given by (21) and for *r* = *p*(*z*) and *s* = *zp*(*z*) from relation (35), we have

$$
\psi\left(p(z), zp'(z)\right) = E(u, v; z), \; z \in \mathcal{U}.\tag{37}
$$

Using (37) in (30), we ge<sup>t</sup>

$$\frac{|\psi(p(z), zp'(z))|}{1 + |\psi(p(z), zp'(z))|} \le \frac{\left| (\beta - 1) \left( 1 - z^2 \right) + \gamma (1 + z)^2 \right|}{\left| (\beta - 1) (1 - z)^2 + \gamma (1 - z^2) \right| + \left| (\beta - 1) (1 - z^2) + \gamma (1 + z)^2 \right|}. \tag{38}$$

Using Lemma 1, for *δ* = *β* − 1, *ω* = *γ* = 0, we have

$$\psi\left(p(z), zp'(z)\right) \prec\_{\mathcal{F}} h(z) = \frac{\left(\beta - 1\right)\left(1 - z^2\right) + \gamma\left(1 + z\right)^2}{\left(\beta - 1\right)\left(1 - z\right)^2 + \gamma\left(1 - z^2\right)},\tag{39}$$

which, according to Lemma 1, implies

$$p(z) \prec\_{\mathcal{F}} q(z) = \frac{1+z}{1-z}, \ z \in \mathsf{U}.\tag{40}$$

Using in (40) relation (32) we have

$$1 + \frac{zM^{\prime\prime}(z)}{M^{\prime}(z)} \preccurlyeq\_{\mathbb{Z}} q(z) = \frac{1+z}{1-z}, \; z \in \mathcal{U}.\tag{41}$$

Since *q* is convex, relation (41) is equivalent to

$$\operatorname{Re}\left(1+\frac{zM''(z)}{M'(z)}\right) > \operatorname{Re}\frac{1+z}{1-z} > 0, \; z \in \mathcal{U}, \text{hence } M \in \mathcal{K}.\tag{42}$$

**Remark 6.** *Using Lemma 3 and the convexity property proved for the operator <sup>M</sup>*(*z*), *the following corollary can be stated giving the property of the integral operator <sup>M</sup>*(*z*) *given by (7) to be starlike of order* 12.

**Corollary 3.** *Let the operator <sup>M</sup>*(*z*) *be given by (7). Then <sup>M</sup>*(*z*) ∈ K *implies <sup>M</sup>*(*z*) ∈ S∗-12.

**Proof.** Since Re1 + *zM*(*z*) *<sup>M</sup>*(*z*) > 0 using Lemma 3 we obtain Re *zM*(*z*) *<sup>M</sup>*(*z*) > 12 , hence *M* ∈ S∗-12.

**Remark 7.** *Using function q*(*z*) = 1−2*<sup>z</sup>* 1+*z as fuzzy best dominant in Theorem 1, we get the following corollary, which gives a sufficient condition for the operator <sup>M</sup>*(*z*) *given by (7) to be convex of order* -<sup>−</sup>12*.*

**Corollary 4.** *For β*, *γ* ∈ C*, β* > 1*, γ* > 0*, let the fuzzy function F* : C → [0, 1] *given by (9) and consider the holomorphic function in U given by the equation*

$$h(z) = \frac{(\beta - 1)(1 + z)(1 - 2z) + (1 - 2z)^2 - 3z}{\left(\beta - 1\right)(1 + z)^2 + (1 - 2z)(1 + z)}$$

,

*h*(0) = *q*(0) = 1*, where function q*(*z*) = 1−2*<sup>z</sup>* 1+*z is a univalent solution in U which satisfies the fuzzy differential subordination*

−

$$\frac{|1-2z|}{|1+z|+|1-2z|} \le$$

$$\left| \frac{\left(\beta - 1\right)\left(1+z\right)\left(1-2z\right) + \left(1-2z\right)^2 - 3z}{\left| \left(\beta - 1\right)\left(1+z\right)^2 + \left(1-2z\right)\left(1+z\right) \right| + \left| \left(\beta - 1\right)\left(1+z\right)\left(1-2z\right) + \left(1-2z\right)^2 - 3z \right|} \right|^2$$

*i.e.,*

$$\frac{1-2z}{1+z} \prec\_{\mathbb{F}} \frac{(\beta-1)(1+z)(1-2z)+(1-2z)^2-3z}{(\beta-1)(1+z)^2+(1-2z)(1+z)}, \; z \in \mathcal{U}.$$

*Consider the confluent (or Kummer) hypergeometric function φ*(*<sup>u</sup>*, *v*; *z*) *given by (5), and the operator <sup>M</sup>*(*z*) *given by (7).*

*If*

$$E(u,v;z) = 1 + (\gamma - 1) \frac{z\phi'(u,v;z)}{\phi(u,v;z)} + \frac{(\gamma + 1)z\phi'(u,v;z) + z^2\gamma\phi''(u,v;z)}{\phi(u,v;z) + \gamma z\phi'(u,v;z)}$$

*is analytic in U, and*

$$\frac{|\operatorname{E}(u,v;z)|}{1+|\operatorname{E}(u,v;z)|} \leq$$

$$\left| \frac{(\beta-1)(1+z)(1-2z)+(1-2z)^2-3z}{(\beta-1)(1+z)^2+(1-2z)(1+z)} \right| \leq \frac{1}{(\beta-1)(1+z)(1-2z)+(1-2z)^2-3z} \left| \frac{\beta-1}{\beta-1} \right| $$

$$\text{i.e., } \operatorname{E}(u,v;z) \prec\_{\mathcal{F}} \frac{(\beta-1)(1+z)(1-2z)+(1-2z)^2-3z}{(\beta-1)(1+z)^2+(1-2z)(1+z)},$$

*then*

$$\frac{\left|1+\frac{zM''(z)}{M'(z)}\right|}{1+\left|1+\frac{zM''(z)}{M(z)}\right|} \le \frac{\left|\frac{1-2z}{1+z}\right|}{1+\left|\frac{1-2z}{1+z}\right|} \text{ i.e., } 1+\frac{zM''(z)}{M'(z)} \precneq \frac{1-2z}{1+z}, \; z \in \mathcal{U}, \text{ or } M \in \mathcal{K}\left(-\frac{1}{2}\right),$$

*and q*(*z*) = 1−2*<sup>z</sup>* 1+*z is the fuzzy best dominant.*

**Proof.** Using in (41) *q*(*z*) = 1−2*<sup>z</sup>* 1+*z* , the relation becomes

$$1 + \frac{zM''(z)}{M'(z)} \prec\_{\mathcal{F}} \frac{1 - 2z}{1 + z}, \; z \in \mathcal{U}. \tag{43}$$

Since Re*zq*(*z*) *q*(*z*) + 1 = *Re* 1−*<sup>z</sup>* 1+*z* > 0, we have that *q* is convex and Re 1−2*<sup>z</sup>* 1+*z* = −12 . Then relation (43) is equivalent to

$$\operatorname{Re}\left(1+\frac{zM''(z)}{M'(z)}\right) > \operatorname{Re}\frac{1-2z}{1+z} > -\frac{1}{2}, \ z \in \mathcal{U}, \text{ hence } M \in \mathcal{K}\left(-\frac{1}{2}\right). \tag{44}$$

**Remark 8.** *Using Corollary 4 we next prove that the operator <sup>M</sup>*(*z*) *given by (7) is close-to-convex in U.*

,

**Theorem 2.** *Let <sup>M</sup>*(*z*) *be given by (7) satisfying the condition Re*-1 + *zM*(*z*) *<sup>M</sup>*(*z*) > −12 *, z* ∈ *U*. *Then <sup>M</sup>*(*z*) *is close-to-convex.*

**Proof.** For obtaining the desired result, Lemma 2 is applied. We calculate

$$\int\_{\theta\_1}^{\theta\_2} \text{Re}\left(1 + \frac{zM''(z)}{M'(z)}\right) d\theta = \int\_{\theta\_1}^{\theta\_2} -\frac{1}{2} d\theta = -\frac{1}{2}\theta|\_{\theta\_1}^{\theta\_2} = -\frac{1}{2}(\theta\_2 - \theta\_1) > -\frac{1}{2}2\pi = -\pi.$$

From Lemma 2, this means that *<sup>M</sup>*(*z*) ∈ C.

**Example 1.** *Let u* = <sup>−</sup>1*, v* = 13 *, φ*-−1, 13 ; *z* = 1 + 3*iz, and <sup>z</sup>φ*(−1, 13 ;*z*) *<sup>φ</sup>*(−1, 13 ;*z*) = 3*iz* 1+3*iz* = 1 − 1−3 sin *<sup>α</sup>*−3*i* cos *α* 10−6 sin *α* . *For β* = 2*, γ* = 1*, we calculate*

$$M(z) = \frac{2}{z} \int\_0^z (1 + 3it)t dt = \frac{2}{z} \left(\frac{z^2}{2} + iz^3\right) = z + 2iz^2$$

*and zM*(*z*) *<sup>M</sup>*(*z*) = 1 + 2*iz* 1+2*iz* .

*Using Corollary 1, we have:*

*Let β* = 2, *γ* = 1*, fuzzy function <sup>F</sup>*(*z*) = |*z*| <sup>1</sup>+|*z*| *, z* ∈ *U, and h*(*z*) = 1−2*<sup>z</sup>* 1+*z , h*(0) = 1*, with univalent solution q*(*z*) = 1−*<sup>z</sup>* 1+*z , z* ∈ *U, which satisfy the fuzzy differential subordination*

$$\frac{|1-z|}{|1+z|+|1-z|} \le \frac{|1-2z|}{|1+z|+|1-2z|}, \text{ i.e., } q(z) = \frac{1-z}{1+z} \preccurlyeq\_{\overline{\mathbb{F}}} \frac{1-2z}{1+z} = h(z),$$

*and φ*-−1, 13 ; *z* = 1 + 3*iz, Kummer hypergeometric function. If* 1+6*iz* 1+3*izis holomorphic in U, and*

$$\frac{|1+6iz|}{|1+3iz|+|1+6iz|} \le \frac{|1-2z|}{|1+z|+|1-2z|}, \text{ i.e., } \frac{1+6iz}{1+3iz} \precneq \frac{1-2z}{1+z}$$

*then*

$$\frac{|3iz|}{|1+3iz|+|3iz|} \le \frac{|1-z|}{|1+z|+|1-z|} \text{ i.e., } \frac{3iz}{1+3iz} \prec\_F \frac{1-z}{1+z}, z \in \mathcal{U}.$$

*In*

$$\operatorname{Re}\frac{z\phi'\left(-1,\frac{1}{3};z\right)}{\phi\left(-1,\frac{1}{3};z\right)} = \operatorname{Re}\left(1 - \frac{1 - 3\sin\alpha - 3i\cos\alpha}{10 - 6\sin\alpha}\right) = 1 - \frac{1 - 3\sin\alpha}{10 - 6\sin\alpha}$$

$$= \frac{10 - 6\sin\alpha - 1 + 3\sin\alpha}{4 + 6(1 - \sin\alpha)} = \frac{9 - 3\sin\alpha}{4 + 6(1 - \sin\alpha)} = \frac{6 + 3(1 - \sin\alpha)}{4 + 6(1 - \sin\alpha)} > 0.$$
