**3. Main Results**

The first fuzzy subordination result obtained using the operator given by (2) is the following theorem:

**Theorem 1.** *Let the function q be analytic and univalent in U such that q*(*z*) = 0 *and* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H(*U*)*, for all z* ∈ *U, where a, c be complex numbers with c* = 0, −1, −2, ... *and λ*, *δ* > 0. *Suppose that zq*(*z*) *q*(*z*) *is starlike univalent in U. Let*

$$\operatorname{Re}\left(\frac{zq''(z)}{q'(z)} - \frac{zq'(z)}{q(z)} + \frac{2\mu}{\beta}(q(z))^2 + \frac{\xi}{\beta}q(z) + 1\right) > 0,\tag{6}$$

*for α*, *ξ*, *μ*, *β* ∈ C*, β* = 0*, z* ∈ *U and*

$$
\psi\_{\lambda}^{a,c}(\delta, a, \xi, \mu, \beta; z) := a + \beta \delta \left[ \frac{z \left( D\_z^{-\lambda} \phi(a, c; z) \right)^{\prime}}{D\_z^{-\lambda} \phi(a, c; z)} - 1 \right] + \tag{7}
$$

$$
\mu \left[ \frac{D\_z^{-\lambda} \phi(a, c; z)}{z} \right]^{2\delta} + \xi \left[ \frac{D\_z^{-\lambda} \phi(a, c; z)}{z} \right]^{\delta}.
$$

$$
\psi\_{\lambda}^{a, c} = \lambda \psi\_{\lambda}^{a, c} \psi\_{\lambda}^{a, c}.
$$

*If q satisfies the following fuzzy subordination*

$$F\_{\psi\_{\lambda}^{a,\varepsilon}(\mathrm{II})}\psi\_{\lambda}^{a,\varepsilon}(\delta,\mu,\underline{\mathfrak{x}},\mu,\beta;z) \le F\_{q(\mathrm{II})}\left(\mathfrak{a} + \beta \frac{zq'(z)}{q(z)} + \mu(q(z))^2 + \mathfrak{f}q(z)\right),\tag{8}$$

*for α*, *ξ*, *μ*, *β* ∈ C*, β* = 0, *then*

$$F\_{D\_z^{-\lambda}\phi(\mathcal{U})}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta} \le F\_{q(\mathcal{U})}q(z), \ z \in \mathcal{U},\tag{9}$$

*and q is the best dominant.*

**Proof.** Define *p*(*z*) := - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ*, *z* ∈ *U*, *z* = 0. Differentiating it we obtain *p*(*z*) = − *δ z p*(*z*) + *δ*- *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ*−1 (*<sup>D</sup>*−*<sup>λ</sup> z <sup>φ</sup>*(*<sup>a</sup>*,*c*;*<sup>z</sup>*)) *z* . Then *zp*(*z*) *p*(*z*) = *δ*/ *z*(*<sup>D</sup>*−*<sup>λ</sup> z <sup>φ</sup>*(*<sup>a</sup>*,*c*;*<sup>z</sup>*)) *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) − 10.

By setting *<sup>θ</sup>*(*w*) := *μw*<sup>2</sup> + *ξw* + *α* and *Q*(*w*) := *βw* , it is evident that *θ* is analytic in C, *φ* is analytic in C\{0} and *φ*(*w*) = 0, *w* ∈ C\{0}.

$$\text{Considering } Q(z) = zq'(z)\phi(q(z)) = \beta \frac{zq'(z)}{q(z)} \text{ and } h(z) = Q(z) + \theta(q(z)) = a + \beta \frac{zq'(z)}{q(z)} + \mu(q(z))^2 + \zeta q(z), \text{ we deduce that } Q \text{ is starlike interval in } \mathcal{U}.$$

Differentiating we obtain *<sup>h</sup>*(*z*) = *ξ* + *q*(*z*) + <sup>2</sup>*μq*(*z*)*q*(*z*) + *β* (*q*(*z*)+*zq*(*z*))*q*(*z*)−*<sup>z</sup>*(*q*(*z*))<sup>2</sup> (*q*(*z*))<sup>2</sup> and *zh*(*z*) *Q*(*z*) = *zh*(*z*) *β zq*(*z*) *q*(*z*) = 1+ *ξβ q*(*z*) + 2*μβ* (*q*(*z*))<sup>2</sup> − *zq*(*z*) *q*(*z*) + *zq*(*z*) *q*(*z*) , which imply that *Rezh*(*z*) *Q*(*z*) = *Re*-1 + *ξβ q*(*z*) + 2*μβ* (*q*(*z*))<sup>2</sup> − *zq*(*z*) *q*(*z*) + *zq*(*z*) *q*(*z*) > 0. We obtain *α* + *β zp*(*z*) *p*(*z*) + *μ*(*p*(*z*))<sup>2</sup> + *ξ p*(*z*) = *α* + *βδ*/ *z*(*<sup>D</sup>*−*<sup>λ</sup> z <sup>φ</sup>*(*<sup>a</sup>*,*c*;*<sup>z</sup>*)) *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) − 10+ *μ D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z* 2*δ* + *ξ D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ*. Using (8), we deduce *Fp*(*U*)*α* + *β zp*(*z*) *p*(*z*) + *μ*(*p*(*z*))<sup>2</sup> + *ξ p*(*z*) ≤ *Fq*(*U*)*α* + *βq*(*z*) + *μ*(*q*(*z*))<sup>2</sup> + *β zq*(*z*) *q*(*z*) . By an application of Lemma 1 we obtain *Fp*(*U*)*p*(*z*) ≤ *Fq*(*U*)*q*(*z*), *z* ∈ *U*, i.e.,

$$F\_{D\_z^{-\lambda}\phi(\mathcal{U})}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta} \le F\_{\phi(\mathcal{U})}q(z), z \in \mathcal{U} \text{ and } q \text{ is the fuzzy best dominant.} \quad \Box$$

**Corollary 1.** *Let c, a be complex numbers with c* = 0, −1, −2, . . . *and δ*, *λ* > 0. *Assume that (6) holds. If*

$$F\_{\boldsymbol{\Psi}\_{\lambda}^{a,c}(\boldsymbol{\mathcal{U}})} \boldsymbol{\Psi}\_{\lambda}^{a,c}(\boldsymbol{\delta}, \boldsymbol{a}, \boldsymbol{\xi}', \boldsymbol{\mu}, \boldsymbol{\beta}; z) \leq F\_{\boldsymbol{q}(\boldsymbol{\mathcal{U}})} \left( a + \beta \frac{(A - B)z}{(1 + Az)(1 + Bz)} + \mu \left( \frac{1 + Az}{1 + Bz} \right)^2 + \xi \frac{1 + Az}{1 + Bz} \right),$$

*for ξ*, *α*, *β*, *μ* ∈ C*, β* = 0, −1 ≤ *B* < *A* ≤ 1*, and ψ<sup>a</sup>*,*<sup>c</sup> λ is introduced in (7), then*

$$F\_{D\_z^{-\lambda}\phi(\mathcal{U})}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta} \le F\_{\eta(\mathcal{U})}\frac{1+Az}{1+Bz'}, \ z \in \mathcal{U},$$

*and* 1+*Az* 1+*Bzis the fuzzy best dominant.*

**Proof.** Consider in Theorem 1 *q*(*z*) = 1+*Az* 1+*Bz* , −1 ≤ *B* < *A* ≤ 1.

**Corollary 2.** *Let c, a be complex numbers with c* = 0, −1, −2, . . . *and δ*, *λ* > 0. *Assume that (6) holds. If*

$$F\_{\boldsymbol{\Psi}\_{\boldsymbol{\lambda}}^{a,c}(\boldsymbol{\Pi})}\boldsymbol{\Psi}\_{\boldsymbol{\lambda}}^{a,c}(\boldsymbol{\delta},a,\boldsymbol{\xi},\boldsymbol{\mu},\boldsymbol{\mu},\boldsymbol{\beta};z) \leq F\_{\boldsymbol{\eta}(\boldsymbol{\Pi})}\left(a + \boldsymbol{\xi}\left(\frac{1+z}{1-z}\right)^{\gamma} + \boldsymbol{\mu}\left(\frac{1+z}{1-z}\right)^{2\gamma} + \boldsymbol{\beta}\frac{2\gamma z}{1-z^{2}}\right),$$

*for ξ*, *α*, *β*, *μ* ∈ C*, β* = 0, 0 < *γ* ≤ 1*, where ψ<sup>a</sup>*,*<sup>c</sup> λ is introduced in (7), then*

$$F\_{D\_z^{-\lambda}\phi(\mathcal{U})}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta} \le F\_{q(\mathcal{U})}\left(\frac{1+z}{1-z}\right)^{\gamma}, z \in \mathcal{U}\_{\epsilon}$$

*and* -1+*z* <sup>1</sup>−*<sup>z</sup><sup>γ</sup> is the fuzzy best dominant.*

**Proof.** Theorem 1 give Corollary for *q*(*z*) = -1+*z* <sup>1</sup>−*<sup>z</sup><sup>γ</sup>*, 0 < *γ* ≤ 1.

**Theorem 2.** *Let q be analytic and univalent in U such that q*(*z*) = 0 *and zq*(*z*) *q*(*z*) *be starlike univalent in U. Assume that*

$$\operatorname{Re}\left(\frac{\zeta}{\beta}q(z) + \frac{2\mu}{\beta}(q(z))^2\right) > 0, \text{ for } \mu, \zeta, \beta \in \mathbb{C}, \ \beta \neq 0. \tag{10}$$

*Let c, a be complex numbers with c* = 0, −1, −2, ... *and δ*, *λ* > 0. *If ψ<sup>a</sup>*,*<sup>c</sup> λ* (*δ*, *α*, *ξ*, *μ*, *β*; *z*) *is univalent in U and* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H[0,(*λ* − 1)*δ*] ∩ *Q, where ψ<sup>a</sup>*,*<sup>c</sup> λ* (*δ*, *α*, *ξ*, *μ*, *β*; *z*) *is introduced in (7), then*

$$F\_{q(\mathcal{U})}\left(\mathfrak{a} + \beta \frac{zq'(z)}{q(z)} + \mu(q(z))^2 + \mathfrak{J}q(z)\right) \le F\_{\mathfrak{v}\_{\lambda}^{a,c}(\mathcal{U})} \mathfrak{v}\_{\lambda}^{a,c}(\delta, a, \mathfrak{z}, \mathfrak{v}\_{\lambda}\mu, \beta; z) \tag{11}$$

*implies*

$$F\_{q(\mathcal{U})}q(z) \le F\_{D\_z^{-\lambda}\phi(\mathcal{U})} \left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta}, z \in \mathcal{U},\tag{12}$$

*and q is the fuzzy best subordinant.*

**Proof.** Define *p*(*z*) := - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ*, *z* ∈ *U*, *z* = 0.

Consider *ν*(*w*) := *μw*<sup>2</sup> + *ξw* + *α* and *φ*(*w*) := *βw* it is evident that *ν* is analytic in C, *φ* is analytic in C\{0} and *φ*(*w*) = 0, *w* ∈ C\{0}. 

In this conditions *ν*(*q*(*z*)) *φ*(*q*(*z*)) = *q*(*z*)[*ξ*+2*μq*(*z*)]*q*(*z*) *β* , which imply *Reν*(*q*(*z*)) *φ*(*q*(*z*)) = *Reξβ q*(*z*) + 2*μβ* (*q*(*z*))<sup>2</sup>> 0, for *ξ*, *β*, *μ* ∈ C, *β* = 0.

We obtain

$$F\_{q(II)}\left(\mathfrak{a} + \beta \frac{zq'(z)}{q(z)} + \mu(q(z))^2 + \xi q(z)\right) \le F\_{p(II)}\left(\mathfrak{a} + \beta \frac{zp'(z)}{p(z)} + \mu(p(z))^2 + \xi p(z)\right).$$

Applying Lemma 2, we obtain

$$F\_{q(\mathcal{U})}q(z) \le F\_{D\_z^{-\lambda}\phi(\mathcal{U})} \left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta}, \quad z \in \mathcal{U},$$

and *q* is the fuzzy best subordinant.

**Corollary 3.** *Let c, a be complex numbers with c* = 0, −1, −2, ... *and δ*, *λ* > 0. *Assume that (10) holds. If* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H[0,(*λ* − 1)*δ*] ∩ *Q and*

$$F\_{q(\mathcal{U})}\left(a+\beta\frac{(A-B)z}{(1+Az)(1+Bz)}+\mu\left(\frac{1+Az}{1+Bz}\right)^2+\xi\frac{1+Az}{1+Bz}\right) \leq F\_{\psi\_{\lambda}^{\mathfrak{e}c}(\mathcal{U})}\psi\_{\lambda}^{\mathfrak{a}c}(\delta,a,\xi,\mu,\beta;z),$$

*for β*, *ξ*, *α*, *μ* ∈ C*, β* = 0, −1 ≤ *B* < *A* ≤ 1*, where ψ<sup>a</sup>*,*<sup>c</sup> λ is introduced in (7), then*

$$F\_{q(\mathcal{U})}\left(\frac{1+Az}{1+Bz}\right) \le F\_{D\_z^{-\lambda}\phi(\mathcal{U})}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta}, \ z \in \mathcal{U}\_{\epsilon}$$

*and* 1+*Az* 1+*Bz is the fuzzy best subordinant.*

**Proof.** Theorem 2 for *q*(*z*) = 1+*Az* 1+*Bz* , −1 ≤ *B* < *A* ≤ 1 give the corollary.

**Corollary 4.** *Let c, a be complex numbers with c* = 0, −1, −2, ... *and δ*, *λ* > 0. *Assume that (10) holds. If* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H[0,(*λ* − 1)*δ*] ∩ *Q and*

$$F\_{q(\mathcal{U})}\left(\alpha + \beta \frac{2\gamma z}{1-z^2} + \mu \left(\frac{1+z}{1-z}\right)^{2\gamma} + \mathfrak{z} \left(\frac{1+z}{1-z}\right)^{\gamma}\right) \le F\_{\mathfrak{P}\_{\lambda}^{\mathfrak{e}\varepsilon}(\mathcal{U})} \mathfrak{P}\_{\lambda}^{\mathfrak{e}\varepsilon}(\delta, \mathfrak{a}, \mathfrak{z}, \mathfrak{z}, \mu, \beta; z),$$

*for β*, *ξ*, *α*, *μ* ∈ C*,* 0 < *γ* ≤ 1*, β* = 0, *where ψ<sup>a</sup>*,*<sup>c</sup> λ is introduced in (7), then*

$$F\_{q(\mathcal{U})}\left(\frac{1+z}{1-z}\right)^{\gamma} \le F\_{D\_z^{-\lambda}\phi(\mathcal{U})}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^{\delta}, z \in \mathcal{U}\_{\epsilon}$$

*and* -1+*z* <sup>1</sup>−*<sup>z</sup><sup>γ</sup> is the fuzzy best subordinant.*

**Proof.** Theorem 2 for *q*(*z*) = -1+*z* <sup>1</sup>−*<sup>z</sup><sup>γ</sup>*, 0 < *γ* ≤ 1, give the corollary.

Theorems 1 and 2 combined give the following sandwich theorem.

**Theorem 3.** *Let q*1 *and q*2 *be analytic and univalent in U such that q*1(*z*) = 0 *and q*2(*z*) = 0*, for all z* ∈ *U, with zq*1(*z*) *q*1(*z*) *and zq*2(*z*) *q*2(*z*) *being starlike univalent. Suppose that q*1 *satisfies (6) and q*2 *satisfies (10). Let c, a be complex numbers with c* = 0, −1, −2, ... *and δ*, *λ* > 0. *If ψ<sup>a</sup>*,*<sup>c</sup> λ* (*δ*, *α*, *ξ*, *μ*, *β*; *z*) *is as introduced in (7) univalent in U and* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H[0,(*λ* − 1)*δ*] ∩ *Q, then*

$$\begin{aligned} F\_{q\_1(\mathcal{U})}\left(\boldsymbol{\alpha} + \beta \frac{zq\_1'(\boldsymbol{z})}{q\_1(\boldsymbol{z})} + \mu(q\_1(\boldsymbol{z}))^2 + \xi q\_1(\boldsymbol{z})\right) &\leq F\_{\boldsymbol{\varphi}\_{\lambda}^{ac}(\mathcal{U})} \boldsymbol{\upmu}\_{\lambda}^{a,c}(\delta, \boldsymbol{\alpha}, \boldsymbol{\xi}, \boldsymbol{\mu}, \beta; \boldsymbol{z}) \\ &\leq F\_{q\_2(\mathcal{U})}\left(\boldsymbol{\alpha} + \boldsymbol{\xi} q\_2(\boldsymbol{z}) + \mu(q\_2(\boldsymbol{z}))^2 + \beta \frac{zq\_2'(\boldsymbol{z})}{q\_2(\boldsymbol{z})}\right), \end{aligned}$$

*for β*, *ξ*, *α*, *μ* ∈ C*, β* = 0, *implies*

$$F\_{\mathfrak{q}\_1(\mathsf{U})}q\_1(z) \le F\_{D\_z^{-\lambda}\phi(\mathsf{U})} \left( \frac{D\_z^{-\lambda}\phi(a,c;z)}{z} \right)^{\delta} \le F\_{\mathfrak{q}\_2(\mathsf{U})}q\_2(z), \ z \in \mathsf{U},$$

*and q*1 *and q*2 *are respectively the fuzzy best subordinant and the fuzzy best dominant.*

For *q*1(*z*) = 1+*A*1*<sup>z</sup>* 1+*B*1*<sup>z</sup>* , *q*2(*z*) = 1+*A*2*<sup>z</sup>* 1+*B*2*<sup>z</sup>* , where −1 ≤ *B*2 < *B*1 < *A*1 < *A*2 ≤ 1, we obtain the following corollary.

**Corollary 5.** *Let c, a be complex numbers with c* = 0, −1, −2, ... *and δ*, *λ* > 0. *Assume that (6) and (10) hold. If* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H[0,(*λ* − 1)*δ*] ∩ *Q and*

$$\begin{aligned} F\_{q\_1(\mathsf{U})} \left( \mathfrak{a} + \beta \frac{(A\_1 - B\_1)z}{(1 + A\_1 z)(1 + B\_1 z)} + \mu \left( \frac{1 + A\_1 z}{1 + B\_1 z} \right)^2 + \xi \frac{1 + A\_1 z}{1 + B\_1 z} \right) &\leq F\_{\psi\_\lambda^{sc}(\mathsf{U})} \psi\_\lambda^{a, c} (\delta, a, \xi, \mu, \theta; z) \\ &\leq F\_{q\_2(\mathsf{U})} \left( \mathfrak{a} + \beta \frac{(A\_2 - B\_2)z}{(1 + A\_2 z)(1 + B\_2 z)} + \mu \left( \frac{1 + A\_2 z}{1 + B\_2 z} \right)^2 + \xi \frac{1 + A\_2 z}{1 + B\_2 z} \right) \end{aligned}$$

*for β*, *ξα*, , *μ*, ∈ C*, β* = 0, −1 ≤ *B*2 ≤ *B*1 < *A*1 ≤ *A*2 ≤ 1*, where ψ<sup>a</sup>*,*<sup>c</sup> λ is introduced in (7), then*

$${}\_{1}F\_{q\_{1}(\boldsymbol{\varPi})}\left(\frac{1+A\_{1}z}{1+B\_{1}z}\right)\leq F\_{D\_{z}^{-\lambda}}\phi\left(\frac{D\_{z}^{-\lambda}\phi(\boldsymbol{a},\boldsymbol{c};z)}{z}\right)^{\delta}\leq F\_{q\_{2}(\boldsymbol{\varPi})}\frac{1+A\_{2}z}{1+B\_{2}z}.$$

*hence* 1+*A*1*<sup>z</sup>* 1+*B*1*<sup>z</sup> and* 1+*A*2*<sup>z</sup>* 1+*B*2*<sup>z</sup> are the fuzzy best subordinant and the fuzzy best dominant, respectively.*

**Corollary 6.** *Let c, a be complex numbers with c* = 0, −1, −2, ... *and δ*, *λ* > 0. *Assume that (6) and (10) hold. If* - *D*−*<sup>λ</sup> z φ*(*<sup>a</sup>*,*c*;*<sup>z</sup>*) *z δ* ∈ H[0,(*λ* − 1)*δ*] ∩ *Q and*

$$\begin{aligned} F\_{q\_1(\mathcal{U})}\left(\mathfrak{a} + \beta \frac{2\gamma\_1 z}{1-z^2} + \mu \left(\frac{1+z}{1-z}\right)^{2\gamma\_1} + \xi \left(\frac{1+z}{1-z}\right)^{\gamma\_1}\right) &\leq F\_{\mathfrak{p}\_\lambda^{\mathfrak{a}c}(\mathcal{U})} \mathfrak{p}\_\lambda^{\mathfrak{a}c}(\delta, \mathfrak{a}, \mathfrak{z}, \mu, \beta; z) \\ &\leq F\_{q\_2(\mathcal{U})}\left(\mathfrak{a} + \xi \left(\frac{1+z}{1-z}\right)^{\gamma\_2} + \mu \left(\frac{1+z}{1-z}\right)^{2\gamma\_2} + \beta \frac{2\gamma\_2 z}{1-z^2}\right), \end{aligned}$$

*for β*, *ξα*, , *μ*, ∈ C*, β* = 0, 0 < *γ*1, *γ*2 ≤ 1*, where ψ<sup>a</sup>*,*<sup>c</sup> λ is introduced in (7), then*

$$F\_{q\_1(lI)}\left(\frac{1+z}{1-z}\right)^{\gamma\_1} \le F\_{D\_z^{-\lambda}\phi(lI)}\left(\frac{D\_z^{-\lambda}\phi(a,c;z)}{z}\right)^\delta \le F\_{q\_2(lI)}\left(\frac{1+z}{1-z}\right)^{\gamma\_2}$$

*hence* -1+*z* <sup>1</sup>−*<sup>z</sup><sup>γ</sup>*<sup>1</sup> *and* -1+*z* <sup>1</sup>−*<sup>z</sup><sup>γ</sup>*<sup>2</sup> *are the fuzzy best subordinant and the fuzzy best dominant, respectively.*
