Sub classes of Bi-Univalent Functions Defined by Salagean type $q-$ Difference Operator

Lemma 2.5. [13] If a function $p \in \mathcal{P}$ is given by $$p(z) = 1 + p_1 z + p_2 z^2 + \cdots$$ $(z \in \Delta),$ then $$|p_i| \leq 2 \qquad (i \in \mathbb{N}),$$ where $\mathcal{P}$ is the family of all functions p, analytic in $\Delta$ , for which $$p(0) = 1$$ and $\Re(p(z)) > 0$ $(z \in \Delta)$ .

Theorem 2.6. Let f given by (1.1) be in the class $\mathfrak{M}\Sigma_q^k(\lambda,\phi)$ . Then <span id="page-3-3"></span> $$|a_2| \le \frac{B_1 \sqrt{B_1}}{\sqrt{|(2(1+2\lambda)[3]_q^k - (1+3\lambda)[2]_q^{2k})B_1^2 + (1+\lambda)^2(B_1 - B_2)[2]_q^{2k}|}}$$ (2.5) and <span id="page-3-4"></span> $$|a_3| \le \frac{B_1}{2(1+2\lambda)[3]_q^k} + \left(\frac{B_1}{(1+\lambda)[2]_q^k}\right)^2$$ (2.6) where $0 \le \lambda \le 1$ .

Corollary 2.9. Let f given by (1.1) be in the class $\mathfrak{M}\Sigma_q(\lambda,\phi)$ . Then $$|a_2| \le \frac{B_1 \sqrt{B_1}}{\sqrt{|(2(1+2\lambda)[3]_q - (1+3\lambda)[2]_q)B_1^2 + (1+\lambda)^2(B_1 - B_2)[2]_q|}}$$ (2.23) and $$|a_3| \le \frac{B_1}{2(1+2\lambda)[3]_q} + \left(\frac{B_1}{(1+\lambda)[2]_q}\right)^2$$ (2.24) In the following section due to Frasin and Aouf [4] and Panigarhi and Murugusun-daramoorthy [14] we define the following new subclass involving the Sălăgean operator [15].

Theorem 3.4. Let f given by (1.1) be in the class $\mathfrak{F}\Sigma_q^k(\mu,\phi)$ . Then <span id="page-6-0"></span> $$|a_2| \le \frac{B_1 \sqrt{B_1}}{\sqrt{|(1+2\mu)[3]_q^k B_1^2 + (1+\mu)^2 [2]_q^{2k} (B_1 - B_2)|}}$$ (3.5) and <span id="page-7-4"></span> $$|a_3| \le B_1 \left( \frac{B_1}{(1+\mu)^2 [2]_q^{2k}} + \frac{1}{(1+2\mu)[3]_q^k} \right).$$ (3.6)

Theorem 4.1. Let the function $f(z) \in \mathcal{M}\Sigma_a^k(\lambda, \phi)$ and $\tau \in \mathbb{C}$ , then $$|a_3 - \tau a_2^2| \le \begin{cases} \frac{B_1}{2(1+2\lambda)[3]_q^k}, & 0 \le |\Theta(\tau)| < \frac{1}{8(1+2\lambda)[3]_q^k}, \\ 4B_1|\Theta(\tau)|, & |\Theta(\tau)| \ge \frac{1}{8(1+2\lambda)[3]_q^k}, \end{cases}$$ (4.1)

Theorem 4.2. Let the function $f(z) \in \mathcal{F}\Sigma_q^k(\mu, \phi)$ and $\tau \in \mathbb{C}$ , then $$|a_3 - \tau a_2^2| \le 2B_1 \left| \left( \Phi(\tau) + \frac{1}{4(1+2\mu)[3]_a^k} \right) + \left( \Phi(\tau) - \frac{1}{4(1+2\mu)[3]_a^k} \right) \right|$$ where $$\Phi(\tau) = \frac{B_1^2(1-\tau)}{4[(1+2\mu)[3]_q^k B_1^2 + (B_1 - B_2)(1+\mu)^2[2]_q^{2k}]}.$$ Concluding Remarks: Taking $\lambda=0$ (and 1) in Theorem 4.1, we can state the Fekete-Szegö inequality for the function class $S\Sigma_q^k(\phi).(and\ \mathcal{K}\Sigma_q^k(\phi)\ respectively$ . Putting $\mu=0$ (and 1) in Theorem 4.2, we can state the Fekete-Szegö inequality for the function class $\mathcal{H}\Sigma_q^k(\phi).(and\ \mathcal{P}\Sigma_q^k(\phi)\ respectively$ . Future Work: Making use of the values of $a_2$ and $a_3$ , and finding $a_4$ we can caluculate Hankel determinant coefficient for the function classes.

Definition 2.1. For $0 \le \lambda \le 1$ , a function $f \in \Sigma$ of the form (1.1) is said to be in the class $\mathcal{M}\Sigma_q^k(\lambda, \phi)$ if the following subordination hold: $$(1-\lambda)\frac{\mathcal{D}_q^{k+1}f(z)}{\mathcal{D}_q^kf(z)} + \lambda \frac{\mathcal{D}_q^{k+2}f(z)}{\mathcal{D}_q^{k+1}f(z)} \prec \phi(z)$$ (2.1) and $$(1-\lambda)\frac{\mathcal{D}_q^{k+1}g(w)}{\mathcal{D}_q^kg(w)} + \lambda \frac{\mathcal{D}_q^{k+2}g(w)}{\mathcal{D}_q^{k+1}g(w)} \prec \phi(w), \tag{2.2}$$ where $z, w \in \Delta$ and g is given by (1.5). Remark 2.2. Suppose $f \in \Sigma$ . If $\lambda = 0$ , then $\mathcal{M}\Sigma_q^k(\lambda, \phi) \equiv \mathbb{S}\Sigma_q^k(\phi)$ : thus $f \in \mathbb{S}\Sigma_q^k(\phi)$ if the following subordination holds: $$\frac{\mathcal{D}_q^{k+1} f(z)}{\mathcal{D}_q^k f(z)} \prec \phi(z) \qquad \text{and} \qquad \frac{\mathcal{D}_q^{k+1} g(w)}{\mathcal{D}_q^k g(w)} \prec \phi(w),$$ where $z, w \in \Delta$ and g is given by (1.5). Remark 2.3. Suppose $f \in \Sigma$ . If $\lambda = 1$ , then $\mathfrak{M}\Sigma_q^k(\lambda, \phi) \equiv \mathfrak{K}\Sigma_q^k(\phi)$ : thus $f \in \mathfrak{K}\Sigma_q^k(\phi)$ if the following subordination holds: $$\frac{\mathcal{D}_q^{k+2} f(z)}{\mathcal{D}_q^{k+1} f(z)} \prec \phi(z) \quad \text{and} \quad \frac{\mathcal{D}_q^{k+2} g(w)}{\mathcal{D}_q^{k+1} g(w)} \prec \phi(w),$$ where $z, w \in \Delta$ and g is given by (1.5). <span id="page-3-5"></span>Remark 2.4. For $0 \le \lambda \le 1$ and k = 0 a function $f \in \Sigma$ of the form (1.1) is said to be in the class $\mathcal{M}\Sigma_q^k(\lambda,\phi)$ if the following subordination hold: $$(1 - \lambda)\frac{z\mathcal{D}_q f(z)}{f(z)} + \lambda \frac{\mathcal{D}_q(z\mathcal{D}_q f(z))}{\mathcal{D}_q f(z))} \prec \phi(z)$$ (2.3) and $$(1 - \lambda) \frac{z \mathcal{D}_q g(w)}{g(w)} + \lambda \frac{\mathcal{D}_q(w \mathcal{D}_q g(w))}{\mathcal{D}_q g(w))} \prec \phi(w), \tag{2.4}$$ where $z, w \in \Delta$ and g is given by (1.5). It is of interest to note that $\mathcal{M}\Sigma_q^0(0,\phi) = \mathcal{S}\Sigma_q^*(\phi)$ , $\mathcal{M}\Sigma_q^0(1,\phi) = \mathcal{K}\Sigma_q(\phi)$ new subclasses of $\Sigma$ associated with q- difference operator not yet discussed sofar. In order to prove our main results, we require the following Lemma:

Definition 3.1. For $0 \le \mu \le 1$ , a function $f \in \Sigma$ of the form (1.1) is said to be in the class $\mathcal{F}\Sigma_q^k(\mu,\phi)$ if the following subordination hold: $$(1-\mu)\frac{\mathcal{D}_q^k f(z)}{z} + \mu(\mathcal{D}_q^k f(z))' \prec \phi(z)$$ (3.1) and $$(1-\mu)\frac{\mathcal{D}_q^k g(w)}{w} + \mu(\mathcal{D}_q^k g(w))' \prec \phi(w)$$ (3.2) where $z, w \in \Delta$ , g is given by (1.5) and $D_q^k f(z)$ is given by (1.9). Remark 3.2. Suppose $f(z) \in \Sigma$ . If $\mu = 0$ , then $\mathcal{F}\Sigma_q^k(0,\phi) \equiv \mathcal{H}\Sigma_q^k(\phi)$ : thus, $f \in \mathcal{H}\Sigma_q^k(\phi)$ if the following subordination holds: $$\frac{\mathcal{D}_q^k f(z)}{z} \prec \phi(z)$$ and $\frac{\mathcal{D}_q^k g(w)}{w} \prec \phi(w)$ where $z, w \in \Delta$ and g is given by (1.5). Remark 3.3. Suppose $f(z) \in \Sigma$ . If $\mu = 1$ , then $\mathcal{F}\Sigma_q^k(1, \phi) \equiv \mathcal{P}\Sigma_q^k(\phi)$ : thus, $f \in \mathcal{P}\Sigma_q^k(\phi)$ if the following subordination holds: $$(\mathcal{D}_a^k f(z))' \prec \phi(z)$$ and $(\mathcal{D}_a^k g(w))' \prec \phi(w)$ where $z, w \in \Delta$ and g is given by (1.5). It is of interest to note that $\mathcal{F}\Sigma_q^0(\mu,\phi) = \mathcal{F}\Sigma_q(\mu,\phi)$ if the following subordination hold: $$(1-\mu)\frac{f(z)}{z} + \mu(\mathcal{D}_q f(z)) \prec \phi(z) \tag{3.3}$$ and $$(1-\mu)\frac{g(w)}{w} + \mu(\mathcal{D}_q g(w)) \prec \phi(w) \tag{3.4}$$ where $z, w \in \Delta$ , g is given by (1.5) and $D_q^k f(z)$ is given by (1.9).