Univalent Functions involving Generalized Hypergeometric Series

Lemma 2. [9] For some $\alpha$ (1 < $\alpha \leq \frac{4}{3}$ ) and $\lambda$ (0 $\leq \lambda$ < 1), and if $f \in \mathcal{V}$ , then $f \in \mathcal{M}^*(\lambda, \alpha)$ if and only if (3) $$\sum_{n=2}^{\infty} [n - (1 + n\lambda - \lambda)\alpha] a_n \leq \alpha - 1.$$

Lemma 4. [9] For some $\alpha$ (1 < $\alpha \leq \frac{4}{3}$ ) and $\lambda$ (0 $\leq \lambda$ < 1), and if $f \in \mathcal{V}$ , then $f \in \mathcal{N}^*(\lambda, \alpha)$ if and only if (5) $$\sum_{n=2}^{\infty} n \left[ n - (1 + n\lambda - \lambda)\alpha \right] a_n \leq \alpha - 1.$$

Lemma 11. [12] For some $\alpha (1 < \alpha \leq \frac{4}{3})$ and $\lambda (0 \leq \lambda < 1)$ , and if $f \in \mathcal{V}$ , then $f \in \mathcal{M}^*(\lambda, \alpha)$ if and only if (12) $$\sum_{n=2}^{\infty} [n - (1 + n\lambda - \lambda)\alpha] a_n \leq \alpha - 1.$$

Lemma 13. [12] For some $\alpha$ (1 < $\alpha \leq \frac{4}{3}$ ) and $\lambda$ (0 $\leq \lambda$ < 1), and if $f \in \mathcal{V}$ , then $f \in \mathcal{N}^*(\lambda, \alpha)$ if and only if (14) $$\sum_{n=2}^{\infty} n \left[ n - (1 + n\lambda - \lambda)\alpha \right] a_n \leq \alpha - 1.$$ The following summation formula for Clausen's hypergeometric function due to Driver and Johnston [9] is given by:

Theorem 15. [9] If Re(c) > Re(b) > 0 and Re(c - a - b) > 0, then <span id="page-6-2"></span> $$(16) _{4}F_{3}\left(\begin{array}{ccc} a, & \frac{b}{3}, & \frac{b+1}{3}, & \frac{b+2}{3} \\ \frac{c}{3}, & \frac{c+1}{3}, & \frac{c+2}{3} \end{array}; 1\right) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \left(\sum_{n=0}^{\infty} \frac{(a)_{n}(-1)^{n}(b)_{n}}{n!(c-a)_{n}}\right) \times {}_{2}F_{1}(-n, b+n; c-a+n; -1).$$ Now, we state the following lemma due to Chandrasekran and Prabhakaran [4] which is useful to prove our main results.

Lemma 17. [4] Let a, b, c > 0. (1) For c > a + b + 1. $$\sum_{n=0}^{\infty} \frac{(n+1)(a)_n \left(\frac{b}{3}\right)_n \left(\frac{b+1}{3}\right)_n \left(\frac{b+2}{3}\right)_n}{\left(\frac{c}{3}\right)_n \left(\frac{c+2}{3}\right)_n \left(\frac{c+2}{3}\right)_n (1)_n}$$ $$= \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)} \left(\sum_{n=0}^{\infty} \left(\frac{(a)_{n+1} (-1)^n (b)_{n+3}}{n! (c-a)_{n+2} (c-a-b-1)}\right) \times {}_{2}F_{1}(-n,b+3+n;c-a+2+n;-1) + \sum_{n=0}^{\infty} \left(\frac{(a)_n (-1)^n (b)_n}{n! (c-a)_n}\right) {}_{2}F_{1}(-n,b+n;c-a+n;-1)\right).$$ (2) For c > a + b + 2 $$\begin{split} \sum_{n=0}^{\infty} \frac{(n+1)^2 \, (a)_n \, \left(\frac{b}{3}\right)_n \, \left(\frac{b+1}{3}\right)_n \, \left(\frac{b+2}{3}\right)_n}{\left(\frac{c}{3}\right)_n \, \left(\frac{c+1}{3}\right)_n \, \left(\frac{c+2}{3}\right)_n \, (1)_n} \\ &= \frac{\Gamma(c) \, \Gamma(c-a-b)}{\Gamma(c-a) \, \Gamma(c-b)} \left(\sum_{n=0}^{\infty} \left(\frac{(a)_{n+2} \, (-1)^n \, (b)_{n+6}}{n! \, (c-a)_{n+4} \, (c-a-b-2)_2}\right) \\ &\qquad \times {}_2F_1(-n,b+6+n;c-a+4+n;-1) \end{split}$$ $$+3\sum_{n=0}^{\infty} \left(\frac{(a)_{n+1} (-1)^n (b)_{n+3}}{n! (c-a)_{n+2} (c-a-b-1)}\right) \times {}_{2}F_{1}(-n,b+3+n;c-a+2+n;-1) + \sum_{n=0}^{\infty} \left(\frac{(a)_{n} (-1)^n (b)_{n}}{n! (c-a)_{n}}\right) {}_{2}F_{1}(-n,b+n;c-a+n;-1)\right).$$ (3) For c > a + b + 3 $$\sum_{n=0}^{\infty} \frac{(n+1)^3 (a)_n \left(\frac{b}{3}\right)_n \left(\frac{b+1}{3}\right)_n \left(\frac{b+2}{3}\right)_n}{\left(\frac{c}{3}\right)_n \left(\frac{c+1}{3}\right)_n \left(\frac{c+2}{3}\right)_n (1)_n}$$ $$= \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)} \left(\sum_{n=0}^{\infty} \left(\frac{(a)_{n+3} (-1)^n (b)_{n+9}}{n! (c-a)_{n+6} (c-a-b-3)_3}\right) \times {}_2F_1(-n,b+9+n;c-a+6+n;-1) \right)$$ $$+6 \sum_{n=0}^{\infty} \left(\frac{(a)_{n+2} (-1)^n (b)_{n+6}}{n! (c-a)_{n+4} (c-a-b-2)_2}\right) \times {}_2F_1(-n,b+6+n;c-a+4+n;-1)$$ $$+7 \sum_{n=0}^{\infty} \left(\frac{(a)_{n+1} (-1)^n (b)_{n+3}}{n! (c-a)_{n+2} (c-a-b-1)}\right) \times {}_2F_1(-n,b+3+n;c-a+2+n;-1)$$ $$+\sum_{n=0}^{\infty} \left(\frac{(a)_n (-1)^n (b)_n}{n! (c-a)_n}\right) {}_2F_1(-n,b+n;c-a+n;-1)\right).$$ (4) For $a \neq 1$ , $b \neq 1$ , 2, 3 and $c > \max\{a+2, a+b-1\}$ , $$\sum_{n=0}^{\infty} \frac{(a)_n \left(\frac{b}{3}\right)_n \left(\frac{b+1}{3}\right)_n \left(\frac{b+2}{3}\right)_n}{\left(\frac{c}{3}\right)_n \left(\frac{c+1}{3}\right)_n \left(\frac{c+2}{3}\right)_n (1)_{n+1}}$$ $$= \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}$$ $$\times \left(\sum_{n=0}^{\infty} \frac{(c-a-1) (c-a-b) (a-1)_n (-1)^n (b-3)_n}{n! (a-1) (b-3)_3 (c-a-2)_n} \right)$$ $$\times {}_2F_1(-n,b-3+n;c-a-2+n;-1) - \left(\frac{(c-3)_3}{(a-1) (b-3)_3}\right).$$

Theorem 18. Let $a, b \in \mathbb{C}\setminus\{0\}$ , c > 0 and c > |a| + |b| + 1. A sufficient condition for the function z F(z) to belong to the class $\mathcal{M}^*(\lambda, \alpha)$ , $1 < \alpha \leq \frac{4}{3}$ and $0 \leq \lambda < 1$ is that <span id="page-7-0"></span>(19) $$\sum_{n=0}^{\infty} \left( \frac{(1-\alpha\lambda)(|a|)_{n+1}(-1)^n (|b|)_{n+3}}{n! (c-|a|)_{n+2} (c-|a|-|b|-1)} \right)_2 F_1(-n,|b|+3+n;c-|a|+2+n;-1)$$ $$\leq (\alpha-1) \sum_{n=0}^{\infty} \left( \frac{(|a|)_n (-1)^n (|b|)_n}{n! (c-|a|)_n} \right)_2 F_1(-n,|b|+n;c-|a|+n;-1).$$

Corollary 20. Let $a, b \in \mathbb{C}\setminus\{0\}$ , c > 0 and c > |a| + |b| + 1. A sufficient condition for the function z F(z) to belong to the class $\mathcal{M}^*(\alpha)$ , $1 < \alpha \leq \frac{4}{3}$ is that (21) $$\sum_{n=0}^{\infty} \left( \frac{(|a|)_{n+1} (-1)^n (|b|)_{n+3}}{n! (c-|a|)_{n+2} (c-|a|-|b|-1)} \right)_2 F_1(-n,|b|+3+n;c-|a|+2+n;-1)$$ $$\leq (\alpha-1) \sum_{n=0}^{\infty} \left( \frac{(|a|)_n (-1)^n (|b|)_n}{n! (c-|a|)_n} \right)_2 F_1(-n,|b|+n;c-|a|+n;-1).$$

Theorem 22. Let $a, b \in \mathbb{C}\setminus\{0\}$ , c > 0 and c > |a| + |b| + 1. A sufficient condition for the function z F(z) to belong to the class $\mathcal{N}^*(\lambda, \alpha)$ , $1 < \alpha \leq \frac{4}{3}$ and $0 \leq \lambda < 1$ is that <span id="page-9-0"></span> $$(23) \sum_{n=0}^{\infty} \left( \frac{(1-\alpha\lambda)(|a|)_{n+2}(-1)^n(|b|)_{n+6}}{n!(c-|a|)_{n+4}(c-|a|-|b|-2)_2} \right) {}_{2}F_{1}(-n,b+|6|+n;c-|a|+4+n;-1)$$ $$-\sum_{n=0}^{\infty} \left( \frac{(3-2\alpha\lambda-\alpha)(|a|)_{n+1}(-1)^n(|b|)_{n+3}}{n!(c-|a|)_{n+2}(c-|a|-|b|-1)} \right) \times {}_{2}F_{1}(-n,|b|+3+n;c-|a|+2+n;-1)$$ $$\leq (\alpha-1)\sum_{n=0}^{\infty} \left( \frac{(|a|)_{n}(-1)^n(|b|)_{n}}{n!(c-|a|)_{n}} \right) {}_{2}F_{1}(-n,|b|+n;c-|a|+n;-1)$$

Lemma 25. [8] If $f \in \mathcal{R}^{\tau}(A, B)$ is of the form (7), then (26) $$|a_n| \leq (A-B)\frac{|\tau|}{n}, n \in \mathbb{N} \setminus \{1\}.$$ The result is sharp. <span id="page-12-3"></span>Using the Lemma 25, we prove the following results:

Theorem 27. Let $a, b \in \mathbb{C} \setminus \{0\}, \ c > 0, \ c > |a| + |b| + 1.$ and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}_{\frac{c}{2}, \frac{c+1}{2}, \frac{c+2}{3}}^{a, \frac{b+1}{3}, \frac{b+2}{3}}(f)(z) \in \mathcal{N}^*(\alpha, \lambda), \ 1 < \alpha \leq \frac{4}{3} \ and \ 0 \leq \lambda < 1 \ if$ <span id="page-12-1"></span>(28) $$\left( \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \right) \left( \sum_{n=0}^{\infty} \left( \frac{(1 - \alpha \lambda) (|a|)_{n+1} (-1)^n (|b|)_{n+3}}{n! (c - |a|)_{n+2} (c - |a| - |b| - 1)} \right) \times_{2} F_{1}(-n, |b| + 3 + n; c - |a| + 2 + n; -1)$$ $$-(\alpha - 1) \sum_{n=0}^{\infty} \left( \frac{(|a|)_n (-1)^n (|b|)_n}{n! (c - |a|)_n} \right) {}_{2}F_{1}(-n, |b| + n; c - |a| + n; -1)$$ $$\leq (\alpha - 1) \left( \frac{(1 - (A - B)|\tau|)}{(A - B)|\tau|} \right).$$ Proof. Let f be of the form [\(7\)](#page-4-0) belong to the class R^τ (A, B) ∩ V. Because of Lemma [13,](#page-6-3) it is enough to show that $$\sum_{n=2}^{\infty} n \left[ n(1 - \alpha \lambda) - \alpha (1 - \lambda) \right] \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+1}{3} \right)_{n-1} \left( \frac{c+2}{3} \right)_{n-1} (1)_{n-1}} \right) |a_n| \le \alpha - 1$$ since f ∈ R^τ (A, B) ∩ V, then by Lemma [25,](#page-12-0) we have $$|a_n| \le (A-B)\frac{|\tau|}{n}, n \in \mathbb{N} \setminus \{1\}.$$ Letting $$\mathcal{T}_{3}(\alpha,\lambda) = \sum_{n=2}^{\infty} n \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right] \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+1}{3} \right)_{n-1} \left( \frac{c+2}{3} \right)_{n-1} (1)_{n-1}} \right) |a_{n}|$$ we derived that $$\begin{split} \mathcal{T}_{3}(\alpha,\lambda) &= (A-B) \, |\tau| \, \sum_{n=2}^{\infty} \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right] \\ &\times \left( \frac{\left( |a| \right)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \, \left( \frac{|b|+1}{3} \right)_{n-1} \, \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \, \left( \frac{c+1}{3} \right)_{n-1} \, \left( \frac{1}{3} \right)_{n-1} \, \left( \frac{|b|+1}{3} \right)_{n-1}} \right) \\ &= (A-B) \, |\tau| \, \left( (1-\alpha\lambda) \, \sum_{n=2}^{\infty} \, n \, \left( \frac{\left( |a| \right)_{n-1} \, \left( \frac{|b|}{3} \right)_{n-1} \, \left( \frac{|b|+1}{3} \right)_{n-1} \, \left( \frac{|b|+2}{3} \right)_{n-1} \, \left( 1 \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \, \left( \frac{c+1}{3} \right)_{n-1} \, \left( \frac{|b|+2}{3} \right)_{n-1} \, \left( 1 \right)_{n-1}} \right) \\ &- \alpha \, (1-\lambda) \, \sum_{n=2}^{\infty} \, \left( \frac{\left( |a| \right)_{n-1} \, \left( \frac{|b|}{3} \right)_{n-1} \, \left( \frac{|b|+1}{3} \right)_{n-1} \, \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \, \left( \frac{c+1}{3} \right)_{n-1} \, \left( 1 \right)_{n-1}} \right) \right) \\ &= (A-B) \, |\tau| \, \left( (1-\alpha\lambda) \, \sum_{n=0}^{\infty} \, \left( \frac{\left( n+1 \right) \, \left( |a| \right)_{n} \, \left( \frac{|b|}{3} \right)_{n} \, \left( \frac{|b|+1}{3} \right)_{n} \, \left( \frac{|b|+2}{3} \right)_{n}}{\left( \frac{c}{3} \right)_{n} \, \left( \frac{c+1}{3} \right)_{n} \, \left( \frac{c+2}{3} \right)_{n} \, \left( 1 \right)_{n}} \right) \\ &- (1-\alpha\lambda) - \alpha \, (1-\lambda) \, \sum_{n=0}^{\infty} \, \left( \frac{\left( |a| \right)_{n} \, \left( \frac{|b|}{3} \right)_{n} \, \left( \frac{|b|+1}{3} \right)_{n} \, \left( \frac{|b|+2}{3} \right)_{n}}{\left( \frac{c}{3} \right)_{n} \, \left( \frac{c+1}{3} \right)_{n} \, \left( \frac{c+2}{3} \right)_{n} \, \left( 1 \right)_{n}} \right) \\ &+ \alpha \, (1-\lambda) \right) \end{split}$$ Using the result (1) of Lemma 17 and the formula (16) in above mentioned equation, we derived that $$= (A - B) |\tau| \left( (1 - \alpha \lambda) \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \left( \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n+1} (-1)^n (|b|)_{n+3}}{n! (c - |a| - |b| - 1)} \right) \right) \times {}_{2}F_{1}(-n, |b| + 3 + n; c - |a| + 2 + n; -1)$$ $$+ \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n} (-1)^{n} (|b|)_{n}}{n! (c - |a|)_{n}} \right) {}_{2}F_{1}(-n, |b| + n; c - |a| + n; -1) \right)$$ $$-\alpha (1 - \lambda) \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \left( \sum_{n=0}^{\infty} \frac{(|a|)_{n} (-1)^{n} (|b|)_{n}}{n! (c - |a|)_{n}} \right)$$ $$\times {}_{2}F_{1}(-n, |b| + n; c - |a| + n; -1) + \alpha - 1 \right)$$ $$= (A - B) |\tau| \left( \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \left( \sum_{n=0}^{\infty} \left( \frac{(1 - \alpha \lambda) (|a|)_{n+1} (-1)^{n} (|b|)_{n+3}}{n! (c - |a|)_{n+2} (c - |a| - |b| - 1)} \right) \right)$$ $$\times {}_{2}F_{1}(-n, |b| + 3 + n; c - |a| + 2 + n; -1)$$ $$-(\alpha - 1) \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n} (-1)^{n} (|b|)_{n}}{n! (c - |a|)_{n}} \right) {}_{2}F_{1}(-n, |b| + n; c - |a| + n; -1) + \alpha - 1 \right)$$ The above expression is bounded above by $\alpha - 1$ if and only if the equation (28) holds, which completes proof. By taking $\lambda = 0$ in Theorem 27, we have the following corollary: Corollary 29. Let $a, b \in \mathbb{C} \setminus \{0\}$ , c > 0, c > |a| + |b| + 1 and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}_{\frac{c}{2}, \frac{c+1}{2}, \frac{c+2}{2}}^{a, \frac{b}{3}, \frac{b+1}{3}, \frac{b+2}{3}}(f)(z) \in \mathcal{N}^*(\alpha), 1 < \alpha \leq \frac{4}{3}$ if $$\left(\frac{\Gamma(c)\,\Gamma(c-|a|-|b|)}{\Gamma(c-|a|)\,\Gamma(c-|b|)}\right) \left(\sum_{n=0}^{\infty} \left(\frac{(|a|)_{n+1}\,(-1)^n\,(|b|)_{n+3}}{n!\,(c-|a|)_{n+2}\,(c-|a|-|b|-1)}\right) \times {}_{2}F_{1}(-n,|b|+3+n;c-|a|+2+n;-1) - (\alpha-1)\sum_{n=0}^{\infty} \left(\frac{(|a|)_{n}\,(-1)^n\,(|b|)_{n}}{n!\,(c-|a|)_{n}}\right) {}_{2}F_{1}(-n,|b|+n;c-|a|+n;-1)\right) \\ \leq (\alpha-1)\left(\frac{(1-(A-B)|\tau|)}{(A-B)\,|\tau|}\right).$$

Theorem 30. Let $a, b \in \mathbb{C} \setminus \{0\}$ , c > 0, c > |a| + |b| + 1 and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}^{a, \frac{b}{3}, \frac{b+1}{3}, \frac{b+2}{3}}_{\frac{c+1}{3}, \frac{c+1}{3}, \frac{c+2}{3}}(f)(z) \in \mathcal{M}^*(\alpha, \lambda), 1 < \alpha \leq \frac{4}{3} \text{ and } 0 \leq \lambda < 1 \text{ if}$ <span id="page-14-0"></span>(31) $$\frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \left( \sum_{n=0}^{\infty} \left( \frac{(1 - \alpha \lambda) (|a|)_n (-1)^n (|b|)_n}{n! (c - |a|)_n} \right) \times {}_{2}F_{1}(-n, |b| + n; c - |a| + n; -1) \right)$$ $$-\alpha (1 - \lambda) \left( \sum_{n=0}^{\infty} \frac{(c - |a| - 1) (c - |a| - |b|) (|a| - 1)_n (-1)^n (|b| - 3)_n}{n! (|a| - 1) (|b| - 3)_3 (c - |a| - 2)_n} \right) \times {}_{2}F_{1}(-n, |b| - 3 + n; c - |a| - 2 + n; -1) - \left( \frac{(c - 3)_3}{(|a| - 1) (|b| - 3)_3} \right) \right)$$ $$\leq (\alpha - 1) \left( \frac{(1 - (A - B) |\tau|)}{(A - B) |\tau|} \right).$$ Proof. Let f be of the form [\(7\)](#page-4-0) belong to the class R^τ (A, B) ∩ V. Because of Lemma [11,](#page-6-0) it is enough to show that $$\sum_{n=2}^{\infty} \left[ n(1 - \alpha \lambda) - \alpha (1 - \lambda) \right] \left( \frac{\left( |a| \right)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+1}{3} \right)_{n-1} \left( \frac{c+2}{3} \right)_{n-1} (1)_{n-1}} \right) |a_n| \le \alpha - 1$$ since f ∈ R^τ (A, B) ∩ V, then by Lemma [25](#page-12-0) the inequality [\(26\)](#page-12-3) holds. Letting $$\mathcal{T}_{4}(\alpha,\lambda) = \sum_{n=2}^{\infty} \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right] \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+1}{3} \right)_{n-1} \left( \frac{c+2}{3} \right)_{n-1} (1)_{n-1}} \right) |a_{n}|$$ We get $$\mathcal{T}_{4}(\alpha,\lambda) = (A-B) |\tau| \sum_{n=2}^{\infty} \frac{1}{n} [n(1-\alpha\lambda) - \alpha(1-\lambda)]$$ $$\times \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+1}{3} \right)_{n-1} \left( \frac{1}{3} \right)_{n-1} \left( \frac{1}{3} \right)_{n-1}} \right)$$ $$= (A-B) |\tau| \left( (1-\alpha\lambda) \sum_{n=2}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+1}{3} \right)_{n-1} \left( \frac{c+2}{3} \right)_{n-1} \left( 1 \right)_{n-1}} \right)$$ $$-\alpha (1-\lambda) \sum_{n=2}^{\infty} \frac{1}{n} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{3} \right)_{n-1} \left( \frac{|b|+1}{3} \right)_{n-1} \left( \frac{|b|+2}{3} \right)_{n-1}}{\left( \frac{c}{3} \right)_{n-1} \left( \frac{c+2}{3} \right)_{n-1} \left( 1 \right)_{n-1}} \right)$$ $$= (A-B) |\tau| \left( (1-\alpha\lambda) \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n} \left( \frac{|b|}{3} \right)_{n} \left( \frac{|b|+1}{3} \right)_{n} \left( \frac{|b|+2}{3} \right)_{n}}{\left( \frac{c}{3} \right)_{n} \left( \frac{c+1}{3} \right)_{n} \left( \frac{c+2}{3} \right)_{n} \left( 1 \right)_{n}} \right) - (1-\alpha\lambda)$$ $$-\alpha (1-\lambda) \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n} \left( \frac{|b|}{3} \right)_{n} \left( \frac{|b|+1}{3} \right)_{n} \left( \frac{|b|+2}{3} \right)_{n}}{\left( \frac{c}{3} \right)_{n} \left( \frac{c+1}{3} \right)_{n} \left( \frac{|b|+2}{3} \right)_{n}} \right) + \alpha (1-\lambda) \right)$$ Using the formula (16) and the result (4) of Lemma 17 in above mentioned equation, we have $$= (A - B) |\tau| \left( (1 - \alpha \lambda) \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \left( \sum_{n=0}^{\infty} \frac{(|a|)_n (-1)^n (|b|)_n}{n! (c - |a|)_n} \right) \right. \\ \left. \times {}_2F_1(-n, |b| + n; c - |a| + n; -1) \right. \\ \left. - \alpha (1 - \lambda) \left( \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \right. \right. \\ \left. \times \left( \sum_{n=0}^{\infty} \frac{(c - |a| - 1) (c - |a| - |b|) (|a| - 1)_n (-1)^n (|b| - 3)_n}{n! (|a| - 1) (|b| - 3)_3 (c - |a| - 2)_n} \right. \\ \left. \times {}_2F_1(-n, |b| - 3 + n; c - |a| - 2 + n; -1) - \left( \frac{(c - 3)_3}{(|a| - 1) (|b| - 3)_3} \right) \right) + \alpha - 1 \right) \right. \\ \left. = (A - B) |\tau| \left( \frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)} \left( (1 - \alpha \lambda) \sum_{n=0}^{\infty} \frac{(|a|)_n (-1)^n (|b|)_n}{n! (c - |a|)_n} \right. \\ \left. \times {}_2F_1(-n, |b| + n; c - |a| + n; -1) \right. \\ \left. - \alpha (1 - \lambda) \left( \sum_{n=0}^{\infty} \frac{(c - |a| - 1) (c - |a| - |b|) (|a| - 1)_n (-1)^n (|b| - 3)_n}{n! (|a| - 1) (|b| - 3)_3 (c - |a| - 2)_n} \right. \\ \left. \times {}_2F_1(-n, |b| - 3 + n; c - |a| - 2 + n; -1) - \left( \frac{(c - 3)_3}{(|a| - 1) (|b| - 3)_3} \right) \right) + \alpha - 1 \right) \right.$$ The above expression is bounded above by $\alpha - 1$ if and only if the equation (31) holds, which completes proof. By taking $\lambda = 0$ in Theorem 30, we have the following corollary: Corollary 32. Let $a, b \in \mathbb{C} \setminus \{0\}$ , c > 0, c > |a| + |b| + 1 and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}_{\frac{c}{2}, \frac{c+1}{2}, \frac{c+2}{2}}^{a, \frac{b}{3}, \frac{b+1}{3}, \frac{b+2}{3}}(f)(z) \in \mathcal{M}^*(\alpha), 1 < \alpha \leq \frac{4}{3}$ if $$\begin{split} \frac{\Gamma(c)\,\Gamma(c-|a|-|b|)}{\Gamma(c-|a|)\,\Gamma(c-|b|)} \left( \, \sum_{n=0}^{\infty} \left( \frac{(|a|)_n\,(-1)^n\,(|b|)_n}{n!\,(c-|a|)_n} \right) \\ & \times {}_2F_1(-n,|b|+n;c-|a|+n;-1) \\ -\alpha \left( \, \sum_{n=0}^{\infty} \frac{(c-|a|-1)\,(c-|a|-|b|)\,(|a|-1)_n\,(-1)^n\,(|b|-3)_n}{n!\,(|a|-1)\,(|b|-3)_3\,(c-|a|-2)_n} \right. \\ & \times {}_2F_1(-n,|b|-3+n;c-|a|-2+n;-1) - \left( \frac{(c-3)_3}{(|a|-1)\,(|b|-3)_3} \right) \right) \right) \\ & \leq (\alpha-1) \left( \frac{(1-(A-B)\,|\tau|)}{(A-B)\,|\tau|} \right). \end{split}$$

Lemma 38. [14] For some $\alpha (1 < \alpha \leq \frac{4}{3})$ and $\lambda (0 \leq \lambda < 1)$ , and if $f \in \mathcal{V}$ , then $f \in \mathcal{M}^*(\lambda, \alpha)$ if and only if (39) $$\sum_{n=2}^{\infty} [n - (1 + n\lambda - \lambda)\alpha] a_n \leq \alpha - 1.$$

Lemma 40. [14] For some $\alpha$ (1 < $\alpha \leq \frac{4}{3}$ ) and $\lambda$ (0 $\leq \lambda$ < 1), and if $f \in \mathcal{V}$ , then $f \in \mathcal{N}^*(\lambda, \alpha)$ if and only if (41) $$\sum_{n=2}^{\infty} n \left[ n - (1 + n\lambda - \lambda)\alpha \right] a_n \leq \alpha - 1.$$ In 2009, Coffey and Johnston [10] derived a summation formula for ${}_{5}F_{4}$ (1) hypergeometric function in terms of Gaussian hypergeometric function. We recall their summation formula as follows:

Theorem 42. [10] For Re(c) > Re(b) > 0 and Re(c-a-b) > 0, <span id="page-20-1"></span> $$(43) \quad {}_{5}F_{4}\left(\frac{a,\frac{b}{4},\frac{b+1}{4},\frac{b+2}{4},\frac{b+3}{4}}{4,\frac{c+2}{4},\frac{c+3}{4}};1\right) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(b)\Gamma(c-b)} \sum_{n=0}^{\infty} {\binom{-a}{n}} \left(\frac{\Gamma(b+2n)}{\Gamma(c-a+2n)}\right) \times {}_{2}F_{1}(a,b+2n;c-a+2n;-1).$$ Now, we state the following lemma due to Chandrasekran and Prabhakaran [6], which is useful to prove our main results. <span id="page-21-0"></span>Lemma 44. [\[6\]](#page-31-7) Let a, b, c > 0. Then we have the following: (1) For c > a + b + 1, we have $$\begin{split} &\sum_{n=0}^{\infty} \frac{(n+1) \left(a\right)_n \, \left(\frac{b}{4}\right)_n \, \left(\frac{b+1}{4}\right)_n \, \left(\frac{b+2}{4}\right)_n \, \left(\frac{b+3}{4}\right)_n}{\left(\frac{c}{4}\right)_n \, \left(\frac{c+1}{4}\right)_n \, \left(\frac{c+2}{4}\right)_n \, \left(\frac{c+3}{4}\right)_n \, (1)_n} \\ &= \frac{\Gamma(c) \, \Gamma(c-a-b)}{\Gamma(b) \, \Gamma(c-b)} \left(\sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left(\frac{a}{c-a-b-1}\right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \right) \\ &\qquad \times {}_2F_1(a+1,b+4+2n;c-a+3+2n;-1) \\ &+ \sum_{n=0}^{\infty} \binom{-a}{n} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} \, {}_2F_1(a,b+2n;c-a+2n;-1) \right) \end{split}$$ (2) For c > a + b + 2, we have $$\sum_{n=0}^{\infty} \frac{(n+1)^{2} (a)_{n} \left(\frac{b}{4}\right)_{n} \left(\frac{b+1}{4}\right)_{n} \left(\frac{b+2}{4}\right)_{n} \left(\frac{b+3}{4}\right)_{n}}{\left(\frac{c}{4}\right)_{n} \left(\frac{c+1}{4}\right)_{n} \left(\frac{c+2}{4}\right)_{n} \left(\frac{c+3}{4}\right)_{n} (1)_{n}}$$ $$= \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(b) \Gamma(c-b)} \left(\sum_{n=0}^{\infty} \binom{-(a+2)}{n} \left(\frac{(a)_{2}}{(c-a-b-2)_{2}}\right) \frac{\Gamma(b+8+2n)}{\Gamma(c-a+6+2n)} \right)$$ $$\times {}_{2}F_{1}(a+2,b+8+2n;c-a+6+2n;-1)$$ $$+3 \sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left(\frac{a}{c-a-b-1}\right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \right)$$ $$\times {}_{2}F_{1}(a+1,b+4+2n;c-a+3+2n;-1)$$ $$+\sum_{n=0}^{\infty} \binom{-a}{n} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} {}_{2}F_{1}(a,b+2n;c-a+2n;-1)$$ (3) For c > a + b + 3, we have $$\sum_{n=0}^{\infty} \frac{(n+1)^3 (a)_n \left(\frac{b}{4}\right)_n \left(\frac{b+1}{4}\right)_n \left(\frac{b+2}{4}\right)_n \left(\frac{b+3}{4}\right)_n}{\left(\frac{c}{4}\right)_n \left(\frac{c+1}{4}\right)_n \left(\frac{c+2}{4}\right)_n \left(\frac{c+3}{4}\right)_n (1)_n}$$ $$= \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(b) \Gamma(c-b)} \left(\sum_{n=0}^{\infty} \binom{-(a+3)}{n} \left(\frac{(a)_3}{(c-a-b-3)_3}\right) \frac{\Gamma(b+12+2n)}{\Gamma(c-a+9+2n)} \right) \left(\frac{(a)_3}{\Gamma(c-a+9+2n)} + 6\sum_{n=0}^{\infty} \binom{-(a+2)}{n} \left(\frac{(a)_2}{(c-a-b-2)_2}\right) \frac{\Gamma(b+8+2n)}{\Gamma(c-a+6+2n)} \right) \left(\frac{(a)_2}{\Gamma(c-a+6+2n)} + 7\sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left(\frac{a}{c-a-b-1}\right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \right) \left(\frac{a}{\Gamma(c-a+3+2n)} + 7\sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left(\frac{a}{c-a-b-1}\right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \right) \left(\frac{a}{\Gamma(c-a+3)} + 2\pi (a+1,b+4+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2\pi (a+3+2n) + 2$$ (4) For $$a \neq 1$$ , $b \neq 1$ , 2, 3, 4 and $c > \max\{a+3, a+b-1\}$ , we have $$\sum_{n=0}^{\infty} \frac{(a)_n \left(\frac{b}{4}\right)_n \left(\frac{b+1}{4}\right)_n \left(\frac{b+2}{4}\right)_n \left(\frac{b+3}{4}\right)_n}{\left(\frac{c}{4}\right)_n \left(\frac{c+1}{4}\right)_n \left(\frac{c+2}{4}\right)_n \left(\frac{c+3}{4}\right)_n (1)_{n+1}}$$ $$= \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(b) \Gamma(c-b)} \left(\frac{c-a-b}{a-1}\right) \times \sum_{n=0}^{\infty} {\binom{-a}{n}} \frac{\Gamma(b-4+2n)}{\Gamma(c-a-3+2n)}$$ $$\times_2 F_1(a,b-4+2n;c-a-3+2n;-1) - \frac{(c-4)_4}{(a-1)(b-4)_4}$$ <span id="page-22-1"></span>Theorem 45. Let a, b ∈ C\{0}, and c > |a| + |b| + 1 > 0. A sufficient condition for the function z G(z) to belong to the class M^∗ (λ, α), 1 < α ≤ 4 3 and 0 ≤ λ < 1 is that <span id="page-22-0"></span>(46) $$(1 - \alpha \lambda) \sum_{n=0}^{\infty} {\binom{-(a+1)}{n}} \left( \frac{a}{c-a-b-1} \right) \left( \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \right)$$ $$\times {}_{2}F_{1}(a+1,b+4+2n;c-a+3+2n;-1)$$ $$\leq (\alpha - 1) \sum_{n=0}^{\infty} {\binom{-a}{n}} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} {}_{2}F_{1}(a,b+2n;c-a+2n;-1)$$ Proof. Let f(z) = z ⁵F⁴ ^a1, a2, a3, a4, a⁵ b1, b2, b3, b4 ; z . Then, by Lemma [38,](#page-20-0) it is enough to show that $$\mathcal{L}_1(\alpha, \lambda) = \sum_{n=2}^{\infty} [n - (1 + n\lambda - \lambda)\alpha] |A_n| \le \alpha - 1.$$ Using the fact |(a)n| ≤ (|a|)n, one can get $$\mathcal{L}_{1}(\alpha,\lambda) = \sum_{n=2}^{\infty} \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right]$$ $$\times \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$= (1-\alpha\lambda) \sum_{n=2}^{\infty} n \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$-\alpha (1-\lambda) \sum_{n=2}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$= (1-\alpha\lambda) \sum_{n=0}^{\infty} \left( \frac{(n+1) (|a|)_{n} \left( \frac{|b|}{4} \right)_{n} \left( \frac{|b|+1}{4} \right)_{n} \left( \frac{|b|+2}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}}{\left( \frac{c+3}{4} \right)_{n} \left( \frac{|b|+1}{4} \right)_{n} \left( \frac{|b|+2}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}}{\left( \frac{c+3}{4} \right)_{n} \left( \frac{|b|+1}{4} \right)_{n} \left( \frac{|b|+2}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}} \right) - (1-\alpha\lambda) \right)$$ $$-\alpha \left(1-\lambda\right) \sum_{n=0}^{\infty} \left(\frac{\left(|a|\right)_n \left(\frac{|b|}{4}\right)_n \left(\frac{|b|+1}{4}\right)_n \left(\frac{|b|+2}{4}\right)_n \left(\frac{|b|+3}{4}\right)_n}{\left(\frac{c}{4}\right)_n \left(\frac{c+1}{4}\right)_n \left(\frac{c+2}{4}\right)_n \left(\frac{c+3}{4}\right)_n \left(1\right)_n}\right) + \alpha \left(1-\lambda\right)$$ Using the result (1) of Lemma 44 and the formula (43) in above mentioned equation, we derived that $$= (1 - \alpha\lambda) \frac{\Gamma(c) \Gamma(c - a - b)}{\Gamma(b) \Gamma(c - b)} \left( \sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left( \frac{a}{c - a - b - 1} \right) \frac{\Gamma(b + 4 + 2n)}{\Gamma(c - a + 3 + 2n)} \right) \times {}_{2}F_{1}(a + 1, b + 4 + 2n; c - a + 3 + 2n; -1) + \sum_{n=0}^{\infty} \binom{-a}{n} \frac{\Gamma(b + 2n)}{\Gamma(c - a + 2n)} {}_{2}F_{1}(a, b + 2n; c - a + 2n; -1) \right) - \alpha (1 - \lambda) \frac{\Gamma(c) \Gamma(c - a - b)}{\Gamma(b) \Gamma(c - b)} \sum_{n=0}^{\infty} \binom{-a}{n} \left( \frac{\Gamma(b + 2n)}{\Gamma(c - a + 2n)} \right) \times {}_{2}F_{1}(a, b + 2n; c - a + 2n; -1) + (\alpha - 1) + (\alpha - 1) \right) + \sum_{n=0}^{\infty} \binom{-a}{\Gamma(b) \Gamma(c - b)} \left( (1 - \alpha\lambda) \sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left( \frac{a}{c - a - b - 1} \right) \frac{\Gamma(b + 4 + 2n)}{\Gamma(c - a + 3 + 2n)} \right) \times {}_{2}F_{1}(a + 1, b + 4 + 2n; c - a + 3 + 2n; -1) + (\alpha - 1) - (\alpha - 1) \sum_{n=0}^{\infty} \binom{-a}{n} \frac{\Gamma(b + 2n)}{\Gamma(c - a + 2n)} {}_{2}F_{1}(a, b + 2n; c - a + 2n; -1) + (\alpha - 1) \right)$$ The above expression is bounded above by $\alpha - 1$ if and only if the equation (46) holds, which completes proof. By taking $\lambda = 0$ in Theorem 45, we have the following corollary: Corollary 47. Let $a, b \in \mathbb{C} \setminus \{0\}$ , and c > |a| + |b| + 1 > 0. A sufficient condition for the function z G(z) to belong to the class $\mathcal{M}^*(\alpha)$ , $1 < \alpha \leq \frac{4}{3}$ is that $$\sum_{n=0}^{\infty} {\binom{-(a+1)}{n}} \left(\frac{a}{c-a-b-1}\right) \left(\frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)}\right) \times {}_{2}F_{1}(a+1,b+4+2n;c-a+3+2n;-1) \\ \leq (\alpha-1) \sum_{n=0}^{\infty} {\binom{-a}{n}} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} {}_{2}F_{1}(a,b+2n;c-a+2n;-1)$$

Theorem 48. Let $a, b \in \mathbb{C} \setminus \{0\}$ , and c > |a| + |b| + 1 > 0. A sufficient condition for the function z G(z) to belong to the class $\mathcal{N}^*(\lambda, \alpha)$ , $1 < \alpha \leq \frac{4}{3}$ and $0 \leq \lambda < 1$ is that <span id="page-23-0"></span> $$(49) \qquad (1-\alpha\lambda) \sum_{n=0}^{\infty} {\binom{-(a+2)}{n}} \left(\frac{(a)_2}{(c-a-b-2)_2}\right) \frac{\Gamma(b+8+2n)}{\Gamma(c-a+6+2n)} \times {}_2F_1(a+2,b+8+2n;c-a+6+2n;-1) + (3-2\alpha\lambda-\alpha) \sum_{n=0}^{\infty} {\binom{-(a+1)}{n}} \left(\frac{a}{c-a-b-1}\right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \times {}_2F_1(a+1,b+4+2n;c-a+3+2n;-1)$$ $$\leq (\alpha - 1) \sum_{n=0}^{\infty} {\binom{-a}{n}} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} {}_{2}F_{1}(a,b+2n;c-a+2n;-1)$$ Proof. Let f(z) = z G(z). Then, by Lemma [40,](#page-20-2) it is enough to show that $$\mathcal{L}_2(\alpha, \lambda) = \sum_{n=2}^{\infty} [n - (1 + n\lambda - \lambda)\alpha] |A_n| \le \alpha - 1$$ Using the fact |(a)n| ≤ (|a|)n, one can get $$\mathcal{L}_{2}(\alpha,\lambda) = \sum_{n=2}^{\infty} n \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right]$$ $$\times \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$= \sum_{n=2}^{\infty} \left[ n^{2} \left( 1 - \alpha\lambda \right) \right] \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}} \right)$$ $$- \sum_{n=2}^{\infty} \left[ \alpha(1-\lambda) n \right] \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ Replace n = (n − 1) + 1 and n ² = (n − 1)(n − 2) + 3(n − 1) + 1 in above, we find that $$= \sum_{n=2}^{\infty} \left[ \left( (n-1)(n-2) + 3(n-1) + 1 \right) (1-\alpha\lambda) \right]$$ $$\times \left( \frac{\left( |a| \right)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$- \sum_{n=2}^{\infty} \left[ \alpha(1-\lambda) \left( (n-1)+1 \right) \right] \left( \frac{\left( |a| \right)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$= (1-\alpha\lambda) \sum_{n=2}^{\infty} \left( \frac{(n-1)(n-2) \left( |a| \right)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$+ (3-2\alpha\lambda-\alpha) \sum_{n=2}^{\infty} \left( \frac{(n-1) \left( |a| \right)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{(b+1)}{4} \right)_{n-1} \left( \frac{(b+1)}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}} \right)$$ $$+ (1 - \alpha) \sum_{n=2}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$= (1 - \alpha\lambda) \sum_{n=3}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}} \right)$$ $$+ (3 - 2\alpha\lambda - \alpha) \sum_{n=2}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{(c+1)}{4} \right)_{n-1} \left( \frac{(b+1)}{4} \right)_{n-1} \left( \frac{(b+1)}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \right)$$ $$+ (1 - \alpha) \sum_{n=2}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{(c+3)}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} \left($$ Using the formula (43) and using the fact that $\Gamma(a+1) = a\Gamma(a)$ in above mentioned equation, we find that $$\mathcal{L}_{2}(\lambda,\alpha) = (1-\alpha\lambda) \sum_{n=0}^{\infty} \binom{-(a+2)}{n} \left( \frac{(a)_{2}}{(c-a-b-2)_{2}} \right) \frac{\Gamma(b+8+2n)}{\Gamma(c-a+6+2n)} \times {}_{2}F_{1}(a+2,b+8+2n;c-a+6+2n;-1) + (3-2\alpha\lambda-\alpha) \sum_{n=0}^{\infty} \binom{-(a+1)}{n} \left( \frac{a}{c-a-b-1} \right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \times {}_{2}F_{1}(a+1,b+4+2n;c-a+3+2n;-1) + (1-\alpha) \sum_{n=0}^{\infty} \binom{-a}{n} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} {}_{2}F_{1}(a,b+2n;c-a+2n;-1) + \alpha-1.$$ The above expression is bounded above by $\alpha - 1$ if and only if the equation (49) holds, which completes proof. By taking $\lambda = 0$ in Theorem 48, we have the following corollary:

Corollary 50. Let $a, b \in \mathbb{C} \setminus \{0\}$ , and c > |a| + |b| + 1 > 0. A sufficient condition for the function z G(z) to belong to the class $\mathcal{N}^*(\alpha)$ , $1 < \alpha \leq \frac{4}{3}$ is that $$\sum_{n=0}^{\infty} {\binom{-(a+2)}{n}} \left( \frac{(a)_2}{(c-a-b-2)_2} \right) \frac{\Gamma(b+8+2n)}{\Gamma(c-a+6+2n)} \times {}_{2}F_{1}(a+2,b+8+2n;c-a+6+2n;-1)$$ $$+(3-\alpha) \sum_{n=0}^{\infty} {\binom{-(a+1)}{n}} \left( \frac{a}{c-a-b-1} \right) \frac{\Gamma(b+4+2n)}{\Gamma(c-a+3+2n)} \times {}_{2}F_{1}(a+1,b+4+2n;c-a+3+2n;-1)$$ $$\leq (\alpha-1) \sum_{n=0}^{\infty} {\binom{-a}{n}} \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} {}_{2}F_{1}(a,b+2n;c-a+2n;-1)$$

Lemma 51. If $f \in \mathcal{R}^{\tau}(A, B)$ is of the form (7), then (52) $$|a_n| \leq (A-B)\frac{|\tau|}{n}, n \in \mathbb{N} \setminus \{1\}.$$ The result is sharp. <span id="page-26-3"></span><span id="page-26-2"></span>Using the Lemma 51, we prove the following results:

Theorem 53. Let $a, b \in \mathbb{C} \setminus \{0\}$ , c > |a| + |b| + 1 > 0 and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}^{a, \frac{b}{4}, \frac{b+1}{4}, \frac{b+2}{4}, \frac{b+3}{4}}_{\frac{c}{4}, \frac{c+1}{4}, \frac{c+3}{4}, \frac{c+3}{4}}(f)(z) \in \mathcal{N}^*(\alpha, \lambda), 1 < \alpha \leq \frac{4}{3} \text{ and } 0 \leq \lambda < 1 \text{ if}$ <span id="page-26-1"></span> $$(54) \quad \left(\frac{\Gamma(c) \Gamma(c - |a| - |b|)}{\Gamma(c - |a|) \Gamma(c - |b|)}\right) \left(\sum_{n=0}^{\infty} {\binom{-(a+1)}{n}}\right) \\ \times \left(\frac{(1 - \alpha \lambda) a}{c - a - b - 1}\right) \frac{\Gamma(b + 4 + 2n)}{\Gamma(c - a + 3 + 2n)} {}_{2}F_{1}(a + 1, b + 4 + 2n; c - a + 3 + 2n; -1) \\ - (\alpha - 1) \sum_{n=0}^{\infty} {\binom{-a}{n}} \frac{\Gamma(b + 2n)}{\Gamma(c - a + 2n)} {}_{2}F_{1}(a, b + 2n; c - a + 2n; -1)\right) \\ \leq (\alpha - 1) \left(\frac{(1 - (A - B)|\tau|)}{(A - B)|\tau|}\right).$$

Theorem 56. Let $a, b \in \mathbb{C} \setminus \{0\}$ , c > |a| + |b| + 1 > 0 and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}_{\frac{a}{4}, \frac{b+1}{4}, \frac{b+2}{4}, \frac{c+3}{4}}^{\frac{b+1}{4}, \frac{b+3}{4}}(f)(z) \in \mathcal{M}^*(\alpha, \lambda), 1 < \alpha \leq \frac{4}{3} \text{ and } 0 \leq \lambda < 1 \text{ if}$ <span id="page-28-0"></span>(57) $$\left( \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(b) \Gamma(c-b)} \right) \left( (1-\alpha\lambda) \sum_{n=0}^{\infty} {\binom{-a}{n}} \left( \frac{\Gamma(b+2n)}{\Gamma(c-a+2n)} \right) \right. \\ \left. \times {}_{2}F_{1}(a,b+2n;c-a+2n;-1) \right. \\ \left. -\alpha \left( 1-\lambda \right) \left( \left( \frac{c-a-b}{a-1} \right) \sum_{n=0}^{\infty} {\binom{-a}{n}} \left( \frac{\Gamma(b-4+2n)}{\Gamma(c-a-3+2n)} \right) \right. \\ \left. \times {}_{2}F_{1}(a,b-4+2n;c-a-3+2n;-1) - \frac{(c-4)_{4}}{(a-1)(b-4)_{4}} \right) \right)$$ $$\leq (\alpha - 1) \left( \frac{(1 - (A - B) |\tau|)}{(A - B) |\tau|} \right).$$ Proof. Let f be of the form [\(33\)](#page-18-0) belong to the class R^τ (A, B) ∩ V. Because of Lemma [38,](#page-20-0) it is enough to show that $$\sum_{n=2}^{\infty} \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right] \times \left( \frac{\left(|a|\right)_{n-1} \left(\frac{|b|}{4}\right)_{n-1} \left(\frac{|b|+1}{4}\right)_{n-1} \left(\frac{|b|+2}{4}\right)_{n-1} \left(\frac{|b|+3}{4}\right)_{n-1}}{\left(\frac{c}{4}\right)_{n-1} \left(\frac{c+1}{4}\right)_{n-1} \left(\frac{c+2}{4}\right)_{n-1} \left(\frac{c+3}{4}\right)_{n-1} (1)_{n-1}} \right) |a_n| \le \alpha - 1$$ since f ∈ R^τ (A, B) ∩ V, then by Lemma [51](#page-26-0) the inequality [\(52\)](#page-26-3) holds. Letting $$\mathcal{L}_{4}(\alpha,\lambda) = \sum_{n=2}^{\infty} \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right] \times \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right) |a_{n}|$$ We get $$\mathcal{L}_{4}(\alpha,\lambda) = (A-B) |\tau| \sum_{n=2}^{\infty} \frac{1}{n} \left[ n(1-\alpha\lambda) - \alpha(1-\lambda) \right]$$ $$\times \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+2}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$= (A-B) |\tau|$$ $$\times \left( (1-\alpha\lambda) \sum_{n=2}^{\infty} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+1}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1} (1)_{n-1}} \right)$$ $$-\alpha (1-\lambda) \sum_{n=2}^{\infty} \frac{1}{n} \left( \frac{(|a|)_{n-1} \left( \frac{|b|}{4} \right)_{n-1} \left( \frac{|b|+1}{4} \right)_{n-1} \left( \frac{|b|+2}{4} \right)_{n-1} \left( \frac{|b|+3}{4} \right)_{n-1}}{\left( \frac{c}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} \left( \frac{c+3}{4} \right)_{n-1} \left( 1 \right)_{n-1}} \right) \right)$$ $$= (A-B) |\tau| \left( (1-\alpha\lambda) \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n} \left( \frac{|b|}{4} \right)_{n} \left( \frac{|b|+1}{4} \right)_{n} \left( \frac{|b|+2}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}}{\left( \frac{c}{4} \right)_{n} \left( \frac{c+1}{4} \right)_{n} \left( \frac{c+2}{4} \right)_{n} \left( \frac{c+3}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}} \right)$$ $$-\alpha (1-\lambda) \sum_{n=0}^{\infty} \left( \frac{(|a|)_{n} \left( \frac{|b|}{4} \right)_{n} \left( \frac{|b|+1}{4} \right)_{n} \left( \frac{|b|+2}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}}{\left( \frac{c+1}{4} \right)_{n} \left( \frac{c+3}{4} \right)_{n} \left( \frac{|b|+3}{4} \right)_{n}} \right) + (\alpha-1) \right)$$ Using the formula (43) and the result (4) of Lemma 44 in above mentioned equation, we have $$= (A - B) |\tau| \left( (1 - \alpha \lambda) \frac{\Gamma(c) \Gamma(c - a - b)}{\Gamma(b) \Gamma(c - b)} \sum_{n=0}^{\infty} {\binom{-a}{n}} \left( \frac{\Gamma(b + 2n)}{\Gamma(c - a + 2n)} \right) \right.$$ $$\times {}_{2}F_{1}(a, b + 2n; c - a + 2n; -1)$$ $$-\alpha (1 - \lambda) \left( \frac{\Gamma(c) \Gamma(c - a - b)}{\Gamma(b) \Gamma(c - b)} \left( \frac{c - a - b}{a - 1} \right) \sum_{n=0}^{\infty} {\binom{-a}{n}} \left( \frac{\Gamma(b - 4 + 2n)}{\Gamma(c - a - 3 + 2n)} \right) \right.$$ $$\times {}_{2}F_{1}(a, b - 4 + 2n; c - a - 3 + 2n; -1) - \frac{(c - 4)_{4}}{(a - 1)(b - 4)_{4}} + \alpha - 1 \right)$$ $$= (A - B) |\tau| \left( \frac{\Gamma(c) \Gamma(c - a - b)}{\Gamma(b) \Gamma(c - b)} \left( (1 - \alpha \lambda) \sum_{n=0}^{\infty} {\binom{-a}{n}} \left( \frac{\Gamma(b + 2n)}{\Gamma(c - a + 2n)} \right) \right.$$ $$\times {}_{2}F_{1}(a, b + 2n; c - a + 2n; -1)$$ $$-\alpha (1 - \lambda) \left( \left( \frac{c - a - b}{a - 1} \right) \sum_{n=0}^{\infty} {\binom{-a}{n}} \left( \frac{\Gamma(b - 4 + 2n)}{\Gamma(c - a - 3 + 2n)} \right) \right.$$ $$\times {}_{2}F_{1}(a, b - 4 + 2n; c - a - 3 + 2n; -1) - \frac{(c - 4)_{4}}{(a - 1)(b - 4)_{4}} \right) + \alpha - 1 \right)$$ The above expression is bounded above by $\alpha-1$ if and only if the equation (57) holds, which completes proof. By taking $\lambda = 0$ in Theorem 56, we have the following corollary: Corollary 58. Let $a, b \in \mathbb{C} \setminus \{0\}$ , c > |a| + |b| + 1 > 0 and $f \in \mathcal{R}^{\tau}(A, B) \cap \mathcal{V}$ . Then $\mathcal{I}_{\underline{c}, \underline{c+1}, \underline{c+2}, \underline{c+3}}^{a, \underline{b}, \underline{b+1}, \underline{b+2}, \underline{b+3}, \underline{d}}(f)(z) \in \mathcal{M}^*(\alpha), 1 < \alpha \leq \frac{4}{3}, if$ $$\left(\frac{\Gamma(c)\,\Gamma(c-a-b)}{\Gamma(b)\,\Gamma(c-b)}\right) \left(\sum_{n=0}^{\infty} \binom{-a}{n} \left(\frac{\Gamma(b+2n)}{\Gamma(c-a+2n)}\right) \,_{2}F_{1}(a,b+2n;c-a+2n;-1) \right) \\ -\alpha \left(\left(\frac{c-a-b}{a-1}\right)\sum_{n=0}^{\infty} \binom{-a}{n} \left(\frac{\Gamma(b-4+2n)}{\Gamma(c-a-3+2n)}\right) \\ \times_{2}F_{1}(a,b-4+2n;c-a-3+2n;-1) - \frac{(c-4)_{4}}{(a-1)(b-4)_{4}}\right) \\ \leq (\alpha-1) \left(\frac{(1-(A-B)|\tau|)}{(A-B)|\tau|}\right).$$

Definition 1.1. [3] For some $\alpha$ $\left(1 < \alpha \leq \frac{4}{3}\right)$ and $\lambda$ $\left(0 \leq \lambda < 1\right)$ , the functions of the form (1) be in the subclass $\mathcal{M}(\lambda, \alpha)$ of $\mathcal{S}$ is $$\mathcal{M}(\lambda, \alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{zf'(z)}{(1-\lambda)f(z) + \lambda z f'(z)}\right) < \alpha, z \in \mathbb{D} \right\}$$

Definition 1.2. [3] For some $\alpha$ $\left(1 < \alpha \leq \frac{4}{3}\right)$ and $\lambda$ $\left(0 \leq \lambda < 1\right)$ , the functions of the form (1) be in the subclass $\mathcal{N}(\lambda, \alpha)$ of $\mathcal{S}$ is $$\mathcal{N}(\lambda, \alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{f'(z) + zf''(z)}{f'(z) + \lambda z f''(z)}\right) < \alpha, z \in \mathbb{D} \right\}$$ <span id="page-1-0"></span>Also, let $\mathcal{M}^(\lambda, \alpha) \equiv \mathcal{M}(\lambda, \alpha) \cap \mathcal{V}$ and $\mathcal{N}^(\lambda, \alpha) \equiv \mathcal{N}(\lambda, \alpha) \cap \mathcal{V}$ .

Definition 1.3. [8] A function $f \in \mathcal{A}$ is said to be in the class $\mathcal{R}^{\tau}(A, B)$ , with $\tau \in \mathbb{C} \setminus \{0\}$ and $-1 \leq B \leq A \leq 1$ , if it satisfies the inequality $$\left| \frac{f'(z) - 1}{(A - B)\tau - B[f'(z) - 1]} \right| < 1, z \in \mathbb{D}$$ Dixit and Pal [8] introduced the Class $\mathcal{R}^{\tau}(A, B)$ . Which is stated as in the definition 1.3. If we substitute $\tau = 1$ , $A = \beta$ and $B = -\beta$ , $(0 < \beta \le 1)$ in the definition 1.3, then we obtain the class of functions $f \in \mathcal{A}$ satisfying the inequality $$\left| \frac{f'(z) - 1}{f'(z) + 1} \right| < \beta, \ z \in \mathbb{D}$$ which was studied by Padmanabhan [10] and others subsequently.

Definition 1.4. [1] The generalized hypergeometric series is defined by (6) $${}_{p}F_{q}(z) = {}_{p}F_{q}\begin{pmatrix} a_{1}, & a_{2}, \cdots & a_{p} \\ b_{1}, & b_{2}, \cdots & b_{q} \end{pmatrix}; z = \sum_{n=0}^{\infty} \left( \frac{(a_{1})_{n} \cdots (a_{p})_{n}}{(b_{1})_{n} \cdots (b_{q})_{n}(1)_{n}} \right) z^{n}.$$ This series converges absolutely for all z if $p \leq q$ and for |z| < 1 if p = q + 1, and it diverges for all $z \neq 0$ if p > q + 1. For |z| = 1 and p = q + 1, the series ${}_pF_q(z)$ converges absolutely if $Re(\sum b_i - \sum a_i) > 0$ . The series converges conditionally if $z = e^{i\theta} \neq 1$ and $-1 < Re(\sum b_i - \sum a_i) \leq 0$ and diverges if $Re(\sum b_i - \sum a_i) \leq -1$ . We consider the linear operator $\mathcal{I}_q^p(f): \mathcal{A} \to \mathcal{A}$ is defined by the convolution product $$\mathcal{I}_{q}^{p}(f)(z) = z_{p}F_{q}(z) * f(z) = z + \sum_{n=2}^{\infty} A_{n} z^{n}, z \in \mathbb{D}$$ with $A_1 = 1$ and for n > 1, $$A_n = \left(\frac{(a_1)_{n-1}\cdots(a_p)_{n-1}}{(b_1)_{n-1}\cdots(b_q)_{n-1}(1)_{n-1}}\right) a_n.$$ Motivated by the results in connections between various subclasses of analytic univalent functions, by using hypergeometric functions [4, 5, 6, 7, 11], and Poisson distributions [3] we obtain the necessary and sufficient conditions for ${}_{p}F_{q}(z)$ hypergeometric function to be in the classes $\mathcal{M}^{}(\lambda, \alpha)$ and $\mathcal{N}^{}(\lambda, \alpha)$ and information regarding the image of function ${}_{p}F_{q}(z)$ hypergeometric function belonging to $\mathcal{R}^{\tau}(A, B)$ by applying the convolution operator. Now, we list out the work done by us in details. In section 2, we obtain the necessary and sufficient conditions on the parameters for the function F(z) to be in the classes $\mathcal{M}^(\lambda, \alpha)$ and $\mathcal{N}^(\lambda, \alpha)$ and information regarding the image of functions F(z) belonging to $\mathcal{R}^{\tau}(A, B)$ by applying the convolution operator in open unit disc $\mathbb{D}$ . We find the necessary and sufficient conditions for the function G(z) to be in the classes $\mathcal{M}^(\lambda, \alpha)$ and $\mathcal{N}^(\lambda, \alpha)$ and information regarding the image of functions G(z) belonging to $\mathcal{R}^{\tau}(A, B)$ by applying the convolution operator in open unit disc $\mathbb{D}$ in section 3. At the end of the each section, we have listed out only the books and the research articles which are used directly to prove our main results in the present research work.

Definition 2.1. [15] The class $\mathcal{M}(\alpha)$ of starlike functions of order $\alpha$ , with $1 < \alpha \leq \frac{4}{3}$ , defined by $$\mathcal{M}(\alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{zf'(z)}{f(z)}\right) < \alpha, z \in \mathbb{D} \right\}$$

Definition 2.2. [15] The class $\mathcal{N}(\alpha)$ of convex functions of order $\alpha$ , with $1 < \alpha \leq \frac{4}{3}$ , defined by $$\mathcal{N}(\alpha) = \left\{ f \in \mathcal{A} : \Re\left(1 + \frac{zf''(z)}{f'(z)}\right) < \alpha, \ z \in \mathbb{D} \right\} = \left\{ f \in \mathcal{A} : zf'(z) \in \mathcal{M}(\alpha) \right\}$$ In this paper, we consider the two subclasses $\mathcal{M}(\lambda, \alpha)$ and $\mathcal{N}(\lambda, \alpha)$ of $\mathcal{S}$ to discuss some inclusion properties based on F(z) hypergeometric function. These two subclasses was introduced by Bulboaca and Murugusundaramoorthy [1], which are stated as follows:

Definition 2.3. [1] For some $\alpha$ $\left(1 < \alpha \leq \frac{4}{3}\right)$ and $\alpha$ $\left(0 \leq \lambda < 1\right)$ , the functions of the form (7) be in the subclass $\mathcal{M}(\lambda, \alpha)$ of $\mathcal{S}$ is $$\mathcal{M}(\lambda, \alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{zf'(z)}{(1-\lambda)f(z) + \lambda z f'(z)}\right) < \alpha, \ z \in \mathbb{D} \right\}$$

Definition 2.4. [1] For some $\alpha$ $\left(1 < \alpha \leq \frac{4}{3}\right)$ and $\alpha$ $\left(0 \leq \lambda < 1\right)$ , the functions of the form (7) be in the subclass $\mathcal{N}(\lambda, \alpha)$ of $\mathcal{S}$ is $$\mathcal{N}(\lambda, \alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{f'(z) + zf''(z)}{f'(z) + \lambda z f''(z)}\right) < \alpha, z \in \mathbb{D} \right\}$$ Also, let $\mathcal{M}^(\lambda, \alpha) \equiv \mathcal{M}(\lambda, \alpha) \cap \mathcal{V}$ and $\mathcal{N}^(\lambda, \alpha) \equiv \mathcal{N}(\lambda, \alpha) \cap \mathcal{V}$ .

Definition 2.5. [8] An function $f \in \mathcal{A}$ is said to be in the class $\mathcal{R}^{\tau}(A, B)$ with $\tau \in \mathbb{C} \setminus \{0\}$ and $-1 \leq B < A \leq 1$ , if it satisfies the inequality $$\left| \frac{f'(z) - 1}{(A - B)\tau - B(f'(z) - 1)} \right| < 1, z \in \mathbb{D}$$ The class $\mathcal{R}^{\tau}(A,B)$ was introduced earlier by Dixit and Pal [8]. If we replace $$\tau = 1, A = \beta, \text{ and } B = -\beta \ (0 < \beta < 1),$$ then, we obtain the class of functions $f \in \mathcal{A}$ satisfying the inequality $$\left| \frac{f'(z) - 1}{f'(z) + 1} \right| < 1, z \in \mathbb{D}.$$ Which was studied by Padmanabhan [13] and many others.

Definition 2.6. [14] The hypergeometric function is defined as (8) $${}_{4}F_{3}\left(\begin{array}{ccc} a_{1}, & a_{2}, & a_{3}, & a_{4} \\ b_{1}, & b_{2}, & b_{3} \end{array}; z\right) = \sum_{n=0}^{\infty} \frac{(a_{1})_{n}(a_{2})_{n}(a_{3})_{n}(a_{4})_{n}}{(b_{1})_{n}(b_{2})_{n}(b_{3})_{n}(1)_{n}} z^{n}, \quad |z| < 1,$$ We consider the linear operator $\mathcal{I}^{a,\frac{b}{3},\frac{b+1}{3},\frac{b+2}{3}}_{\frac{c}{3},\frac{c+1}{3},\frac{c+2}{3}}(f):\mathcal{A}\to\mathcal{A}$ is defined by convolution product (9) $$\mathcal{I}_{\frac{c}{3}, \frac{b+1}{3}, \frac{b+2}{3}}^{a, \frac{b}{3}, \frac{b+2}{3}}(f)(z) = z F(z) * f(z)$$ $$= z + \sum_{n=2}^{\infty} A_n z^n,$$ with $A_1 = 1$ and for n > 1, (10) $$A_n = \frac{(a)_{n-1} \left(\frac{b}{3}\right)_{n-1} \left(\frac{b+1}{3}\right)_{n-1} \left(\frac{b+2}{3}\right)_{n-1}}{\left(\frac{c}{3}\right)_{n-1} \left(\frac{c+1}{3}\right)_{n-1} \left(\frac{c+2}{3}\right)_{n-1} (1)_{n-1}} a_n.$$ Due to the various interesting results in connections between various subclasses of analytic univalent functions using hypergeometric functions [2, 3, 4, 5, 6, 7, 15] and Poisson distributions [1], we try to find the necessary and sufficient conditions on a, b, c, $\lambda$ and $\alpha$ for F(z) hypergeometric series to be in the classes $\mathcal{M}^(\lambda, \alpha)$ and $\mathcal{N}^(\lambda, \alpha)$ and information regarding the image of functions F(z) hypergeometric series belonging to $\mathcal{R}^{\tau}(A, B)$ by applying the convolution operator. 2.2. Main Results and Proofs. First, we recall the following results to prove our main theorems.

Definition 3.1. [17] The class $\mathcal{M}(\alpha)$ of starlike functions of order $\alpha$ , with $1 < \alpha \leq \frac{4}{3}$ , defined by $$\mathcal{M}(\alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{zf'(z)}{f(z)}\right) < \alpha, \ z \in \mathbb{D} \right\}$$

Definition 3.2. [17] The class $\mathcal{N}(\alpha)$ of convex functions of order $\alpha$ , with $1 < \alpha \leq \frac{4}{3}$ , defined by $$\mathcal{N}(\alpha) = \left\{ f \in \mathcal{A} : \Re\left(1 + \frac{zf''(z)}{f'(z)}\right) < \alpha, \ z \in \mathbb{D} \right\} = \left\{ f \in \mathcal{A} : zf'(z) \in \mathcal{M}(\alpha) \right\}$$ The above two subclass were introduced by Uralegaddi et al [17]. Also, let $\mathcal{M}^(\alpha) \equiv \mathcal{M}(\alpha) \cap \mathcal{V}$ and $\mathcal{N}^(\alpha) \equiv \mathcal{N}(\alpha) \cap \mathcal{V}$ . In this study, we consider the two subclasses $\mathcal{M}(\lambda, \alpha)$ and $\mathcal{N}(\lambda, \alpha)$ of $\mathcal{S}$ was introduced by Bulboaca and Murugusundaramoorthy [2] to discuss some inclusion properties based on generalized hypergeometric function. Which are stated as follows:

Definition 3.3. [2] For some $\alpha$ $\left(1 < \alpha \leq \frac{4}{3}\right)$ and $\lambda$ $\left(0 \leq \lambda < 1\right)$ , the functions of the form (33) be in the subclass $\mathcal{M}(\lambda, \alpha)$ of $\mathcal{S}$ is $$\mathcal{M}(\lambda, \alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{zf'(z)}{(1-\lambda)f(z) + \lambda z f'(z)}\right) < \alpha, z \in \mathbb{D} \right\}$$

Definition 3.4. [2] For some $\alpha$ $\left(1 < \alpha \leq \frac{4}{3}\right)$ and $\alpha$ $\left(0 \leq \lambda < 1\right)$ , the functions of the form (33) be in the subclass $\mathcal{N}(\lambda, \alpha)$ of $\mathcal{S}$ is $$\mathcal{N}(\lambda, \alpha) = \left\{ f \in \mathcal{A} : \Re\left(\frac{f'(z) + zf''(z)}{f'(z) + \lambda z f''(z)}\right) < \alpha, z \in \mathbb{D} \right\}$$ Also, let $\mathcal{M}^(\lambda, \alpha) \equiv \mathcal{M}(\lambda, \alpha) \cap \mathcal{V}$ and $\mathcal{N}^(\lambda, \alpha) \equiv \mathcal{N}(\lambda, \alpha) \cap \mathcal{V}$ .

Definition 3.5. [11] A function $f \in \mathcal{A}$ is said to be in the class $\mathcal{R}^{\tau}(A, B)$ , with $\tau \in \mathbb{C} \setminus \{0\}$ and $-1 \leq B \leq A \leq 1$ , if it satisfies the inequality $$\left| \frac{f'(z) - 1}{(A - B)\tau - B[f'(z) - 1]} \right| < 1, z \in \mathbb{D}$$ Dixit and Pal [11] introduced the Class $\mathcal{R}^{\tau}(A, B)$ . Which is stated as the above definition. If we substitute $\tau = 1$ , $A = \beta$ and $B = -\beta$ , $(0 < \beta \le 1)$ in the definition 3.5, then we obtain the class of functions $f \in \mathcal{A}$ satisfying the inequality $$\left| \frac{f'(z) - 1}{f'(z) + 1} \right| < \beta, \ z \in \mathbb{D}$$ which was studied by Padmanabhan [10] and others subsequently. The Special functions [1, 15] plays an important role in to characterize, various subclasses of univalent functions in geometric function theory [12]. The important integral operator is the Hohlov convolution operator [13], which is none other than convolution of the normalized analytic univalent function with Gaussian hypergeometric function. Recently, Chandrasekran and Prabhakaran introduced an integral operator involving the Generalized hypergeometric function and they derived geometric properties of various subclasses of univalent function [3, 4, 5, 6]. Now, we are in the position to recall the generalized hypergeometric series and its convergence.

Definition 3.6. [16] The generalized hypergeometric series is defined by <span id="page-19-1"></span>(34) $${}_{p}F_{q}\begin{pmatrix} a_{1}, & a_{2}, \cdots & a_{p} \\ b_{1}, & b_{2}, \cdots & b_{q} \end{pmatrix} = \sum_{n=0}^{\infty} \frac{(a_{1})_{n} \cdots (a_{p})_{n}}{(b_{1})_{n} \cdots (b_{q})_{n}(1)_{n}} z^{n}.$$ This series converges absolutely for all z if $p \leq q$ and for |z| < 1 if p = q + 1, and it diverges for all $z \neq 0$ if p > q + 1. For |z| = 1 and p = q + 1, the series ${}_{p}F_{q}(a_{1} \ldots, a_{p}; b_{1} \ldots b_{q}; z)$ converges absolutely if $Re(\sum b_{i} - \sum a_{i}) > 0$ . The series converges conditionally if $z = e^{i\theta} \neq 1$ and $-1 < Re(\sum b_{i} - \sum a_{i}) \leq 0$ and diverges if $Re(\sum b_{i} - \sum a_{i}) \leq -1$ . If we put p = 5 and q = 4 in the equation (34), the generalized hypergeometric function becomes the ${}_{5}F_{4}\left({}_{b_{1},b_{2},b_{3},b_{4}}^{a_{1},a_{2},a_{3},a_{4},a_{5}};z\right)$ hypergeometric function. Further, the formal definition of ${}_{5}F_{4}\left({}_{b_{1},b_{2},b_{3},b_{4}}^{a_{1},a_{2},a_{3},a_{4},a_{5}};z\right)$ hypergeometric function is stated as follows.

Definition 3.7. The hypergeometric function ${}_{5}F_{4}(z)$ is defined as $$(35) _{5}F_{4}\left(\substack{a_{1},a_{2},a_{3},a_{4},a_{5} \\ b_{1},b_{2},b_{3},b_{4}};z\right) = \sum_{n=0}^{\infty} \frac{(a_{1})_{n}(a_{2})_{n}(a_{3})_{n}(a_{4})_{n}(a_{5})_{n}}{(b_{1})_{n}(b_{2})_{n}(b_{3})_{n}(b_{4})_{n}(1)_{n}} z^{n}; |z| < 1$$ with $a_1, a_2, a_3, a_4, a_5, b_1, b_2, b_3, b_4 \in \mathbb{C}$ provided $b_1, b_2, b_3, b_4 \neq 0, -1, -2, -3 \cdots$ , which is an analytic function in unit disc $\mathbb{D}$ . We consider the linear operator $\mathcal{I}^{a,\frac{b}{4},\frac{b+1}{4},\frac{b+2}{4},\frac{b+3}{4}}_{\frac{a}{4},\frac{c+1}{4},\frac{c+2}{4},\frac{c+3}{4}}(f):\mathcal{A}\to\mathcal{A}$ is defined by convolution product (36) $$\mathcal{I}_{\frac{c}{4}, \frac{c+1}{4}, \frac{c+2}{4}, \frac{c+3}{4}}^{a, \frac{b+1}{4}, \frac{b+2}{4}, \frac{b+3}{4}}(f)(z) = z G(z) * f(z)$$ $$= z + \sum_{n=2}^{\infty} A_n z^n,$$ with $A_1 = 1$ and for n > 1, (37) $$A_n = \frac{(a)_{n-1} \left(\frac{b}{4}\right)_{n-1} \left(\frac{b+1}{4}\right)_{n-1} \left(\frac{b+2}{4}\right)_{n-1} \left(\frac{b+3}{4}\right)_{n-1}}{\left(\frac{c}{4}\right)_{n-1} \left(\frac{c+1}{4}\right)_{n-1} \left(\frac{c+2}{4}\right)_{n-1} \left(\frac{c+3}{4}\right)_{n-1} (1)_{n-1}} a_n.$$ Motivated by the results in connections between various subclasses of analytic univalent functions using hypergeometric functions [7, 8, 9] and Poisson distributions [2], we try to find the necessary and sufficient conditions on a, b, c, $\lambda$ and $\alpha$ for z $_5F_4$ $\binom{a_1,a_2,a_3,a_4,a_5}{b_1,b_2,b_3,b_4}$ ; z) hypergeometric series to be in the classes $\mathcal{M}^(\lambda,\alpha)$ and $\mathcal{N}^(\lambda,\alpha)$ and information regarding the image of functions z $_5F_4$ $\binom{a_1,a_2,a_3,a_4,a_5}{b_1,b_2,b_3,b_4}$ ; z) hypergeometric series belonging to $\mathcal{R}^{\tau}(A,B)$ by applying the convolution operator. 3.2. Main Results and Proofs. First, we recall the following results to prove our main theorems: