Sufficient Conditions For Janowski Starlike Functions With Fixed Second Coefficient

Lemma 1.1. [2] Let $p \in \mathcal{H}_{\mu,n}$ with $0 < \mu \le 2$ . Let $\psi \in \Psi_{\mu,n}$ with associated domain D. If $(p(z), zp'(z)) \in D$ and $\operatorname{Re} \psi(p(z), zp'(z)) > 0$ , then $\operatorname{Re} p(z) > 0$ for $z \in \mathbb{D}$ .

Lemma 2.1. Suppose that $-1 \le B < A \le 1$ , $-1 \le E < D \le 1$ , and $\alpha E < 0$ . Assume that $$G = n (2 + \mu') + (2 - \mu'),$$ $$H = (n - 1) (2 + \mu') + (2 - \mu'),$$ where $0 < \mu' = 2\mu/(A - B) \le 2$ . In addition for all n > 1, let $$((D-E)(1-B)^{2} + \alpha E(A-B))^{2} G^{2} + (D-E)^{2}(1+B)^{4}H^{2}$$ $$(2.2) \leq \alpha^2 (A-B)^2 G^2 + 2(D-E)(1+B)^2 ((D-E)(1-B)^2 - \alpha E(A-B))GH.$$ If $p \in \mathcal{H}_{\mu,n}$ and $$(2.3) 1 + \alpha z p'(z) \prec \frac{1 + Dz}{1 + Ez},$$ then (2.1) $$p(z) \prec \frac{1 + Az}{1 + Bz}.$$

Theorem 2.1. Let the conditions of Lemma 2.1 hold. If $f \in A_{n,b}$ satisfies $$1 + \alpha \frac{zf'(z)}{f(z)} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \prec \frac{1 + Dz}{1 + Ez},$$ then $f \in S^*[A, B]$ . For $0 \le \lambda < 1$ and $\delta \le 1$ , by taking $\alpha = 1$ , $A = -B = \lambda$ , and $D = -E = \delta$ in Theorem 2.1, we get the following result.

Corollary 2.1. Let $0 \le \lambda < 1$ . If $f \in \mathcal{A}_{n,h}$ satisfies $$\left| \frac{zf'(z)}{f(z)} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \right| < \delta \left| 2 + \frac{zf'(z)}{f(z)} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \right|,$$ where $\delta = (\lambda G) / \sqrt{((1+\lambda)^2 - \lambda)^2 G^2 + (1-\lambda)^4 H^2 - 2(1-\lambda)^2 ((1+\lambda)^2 + \lambda) GH}$ . Then $f \in S^*[\lambda]$ . If we take n = 1 and $\mu = A = -B$ , then Corollary 2.1 reduces to [3, Corollary 2.4] for $\alpha = \lambda$ .

Corollary 2.2. If $f \in A_{n,b}$ satisfies $$\left|\frac{zf'(z)}{f(z)}\left(1+\frac{zf''(z)}{f'(z)}-\frac{zf'(z)}{f(z)}\right)\right|<\frac{1-\lambda}{2},$$ then $f \in S_{\lambda}^*$ .

Corollary 2.3. If $f \in A_{n,b}$ satisfies $$\left| \frac{zf'(z)}{f(z)} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \right| < \delta,$$ where $\delta = (1 - \lambda)G/(G - H)$ , and $0 \le \lambda < 1$ . Then $f \in S^*(\lambda)$ .

Corollary 2.4. If $f \in A$ satisfies $$\left| \frac{zf'(z)}{f(z)} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \right| < 1 - \lambda.$$ Then $f \in S^*(\lambda)$ .

Lemma 2.2. Suppose that $-1 \le B < A \le 1$ , $0 < E < D \le 1$ , and $\alpha E < 0$ . Also let G and H as in (2.1). In addition for all n > 1, let $$((D-E)(1+A)^2 - E(A-B))^2 G^2 + (D-E)^2 (1-A)^4 H^2$$ $$(2.8) \leq (A-B)^2 G^2 + 2(D-E)(1-A)^2 ((D-E)(1+A)^2 + E(A-B))GH.$$ If $p \in \mathcal{H}_{\mu,n}$ and $$1 + \frac{zp'(z)}{p^2(z)} \prec \frac{1 + Dz}{1 + Ez},$$ then $$p(z) \prec \frac{1 + Az}{1 + Bz}$$ .

Theorem 2.2. Let the conditions of Lemma 2.2 hold. If $f \in A_{n,b}$ satisfies $$\frac{1+zf''(z)/f'(z)}{zf'(z)/f(z)} \prec \frac{1+Dz}{1+Ez},$$ then $f \in S^*[A, B]$ .

Corollary 2.5. If $f \in F_c$ , where $c = (1-\lambda)G/|(2-\lambda)^2G - \lambda^2H|$ , where $0 \le \lambda < 1$ , then $f \in S^*(\lambda)$ .

Corollary 2.6. If $f \in A_{n,b}$ satisfies (2.9) $$\left| 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right| < \delta \left| 1 + \frac{zf''(z)}{f'(z)} + \frac{zf'(z)}{f(z)} \right|,$$ where $\delta = (\lambda G)/\sqrt{((1+\lambda)^2 + \lambda)^2 G^2 + (1-\lambda)^4 H^2 - 2(1-\lambda)^2 ((1+\lambda)^2 - \lambda) GH}$ , and $0 \le \lambda < 1$ , then $f \in S^*[\lambda]$ .

Corollary 2.7. If (2.9) holds with $$\delta = ((1 - \lambda)G) / \sqrt{(1 - \lambda)^2 (5 - 4\lambda)^2 G^2 + 16\lambda^2 H^2 - 8\lambda^2 (1 - \lambda)(3 - 4\lambda)GH},$$ where $0 \le \lambda < 1$ , then $f \in S_{\lambda}^*$ .

Lemma 2.3. Suppose that $-1 \le B < A \le 1$ , $-1 \le E < D \le 1$ , and $\alpha E < 0$ . Also let G and H as in (2.1). In addition, for all n > 1, let $$\left( \left( (D-E)(1-A)(1-B) + \alpha E(A-B) \right)^2 - \alpha^2 (A-B)^2 \right) G^2 + (D-E)^2 (1+A)^2 (1+B)^2 H^2$$ $$\leq 2(D-E) \left( (D-E) \left( (1-AB)^2 + (A-B)^2 \right) + \alpha E(A-B)(1+A)(1+B) \right) GH.$$ If $p \in \mathcal{H}_{\mu,n}$ and $$(2.10) 1 + \alpha \frac{zp'(z)}{p(z)} \prec \frac{1 + Dz}{1 + Ez},$$ then $$p(z) \prec \frac{1 + Az}{1 + Bz}.$$ Proof. Define the function $q: \mathbb{D} \to \mathbb{C}$ as (2.4). Then q is analytic on $\mathbb{D}$ and $q(z) = 1 + 2\mu'z^n + q_{n+1}z^{n+1} + \cdots \in \mathcal{H}_{\mu',n}$ , where $0 < \mu' = 2\mu/(A-B) \le 2$ . It follows from (2.4) and (2.10) that $$\operatorname{Re}\left\{\frac{(D-E)\big((1-B)+(1+B)q(z)\big)\big((1-A)+(1+A)q(z)\big)+2\alpha(1-E)(A-B)zq'(z)}{(D-E)\big((1-B)+(1+B)q(z)\big)\big((1-A)+(1+A)q(z)\big)-2\alpha(1+E)(A-B)zq'(z)}\right\}>0.$$ Define $\psi: \mathbb{C}^2 \to \mathbb{C}$ by $$\psi(r,s) = \frac{(D-E)((1-B)+(1+B)r)((1-A)+(1+A)r)+2\alpha(1-E)(A-B)s}{(D-E)((1-B)+(1+B)r)((1-A)+(1+A)r)-2\alpha(1+E)(A-B)s}.$$ Then $\psi$ is continuous of r and s on $D =: \mathbb{C}^2 - \{(r,s) : N'(r,s) = 0\}$ , where $N'(r,s) =: (D - E)((1-B) + (1+B)r)((1-A) + (1+A)r) - 2\alpha(1+E)(A-B)s$ . Note that $(1,0) \in D$ and $\text{Re } \{\psi(1,0)\} > 0$ . It also follows that for all $(i\rho,\sigma) \in D$ , $$\operatorname{Re}\left\{\psi(i\rho,\sigma)\right\} = \operatorname{Re}\left\{\frac{(D-E)[(1-B)+(1+B)i\rho][(1-A)+(1+A)i\rho]+2\alpha(1-E)(A-B)\sigma}{(D-E)[(1-B)+(1+B)i\rho][(1-A)+(1+A)i\rho]-2\alpha(1+E)(A-B)\sigma}\right\}.$$ Let $$a = (D - E)(1 - A)(1 - B),$$ $$b = 2\alpha(1 - E)(A - B),$$ $$c = -(D - E)(1 + A)(1 + B),$$ $$d = 2(D - E)(1 - AB),$$ $$e = -2\alpha(1+E)(A-B).$$ Then Re $\psi(i\rho,\sigma)$ = Re $\{(a+b\sigma+c\rho^2+di\rho)/(a+e\sigma+c\rho^2+di\rho)\}$ . For Re $\psi(i\rho,\sigma)<0$ , we need to prove $$a^{2} + a(b+e)\sigma + be\sigma^{2} + (2ac + d^{2} + c(b+e)\sigma)\rho^{2} + c^{2}\rho^{4} < 0.$$ Since $\sigma < -1/2$ , it follows that $$2ac + d^{2} + (b+e)c\sigma \geqslant 2ac + d^{2} - \frac{1}{2}(b+e)c$$ $$= 2(D-E)((D-E)((1-AB)^{2} + (A-B)^{2})$$ $$-\alpha E(A-B)(1+A)(1+B)) \le 0,$$ (2.11) provided $\alpha E < 0$ . Also, $\rho^2 \le -\left(\left(2(2+\mu')\sigma\right)/\left(n(2+\mu')\right) + (2-\mu')\right) + 1\right)$ where $\mu' = 2\mu/(A-B)$ , then the proof follows on lines similar to Lemma 2.1.

Theorem 2.3. Let $(0 < A \le 1)$ . If $p \in \mathcal{H}_{\mu,n}$ , and $$|zp'(z)/p(z)| < (AG)/\sqrt{(1-A)^2G^2 + (1+A)^2H^2 - 2(1+A^2)GH},$$ then $p(z) \prec 1 + Az$ . With n = 1, by taking p(z) = zf'(z)/f(z) where $f \in A_b$ , and b = A - B, then for $A = 1 - \lambda$ , theorem 2.3 reduces to the following result

Corollary 2.8. If $f \in A$ satisfies $$\left|1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)}\right| < \frac{1-\lambda}{\lambda}, \quad (0 \le \lambda < 1),$$ then $f(z) \in S^*(\lambda)$ .

Lemma 2.4. Suppose that $-1 \le B < A \le 1, -1 \le E < D \le 1,$ $$\begin{split} K &= \left( (D-1)(1+B)^2 - \alpha E(1+A)^2 \right)^2 - \alpha^2 (1+A)^4 \ge 0, \\ L &= \lambda (A-B) \left( E(D-1)(1+B)^2 + \alpha (1-E^2)(1+A)^2 \right) < 0, \\ M &= \left( (D-1)(1-B^2) - \alpha E(1-A^2) \right)^2 + 4\alpha E(D-1)(A-B)^2 - \alpha^2 (1-A^2)^2 \\ &- \lambda (A-B) \left( \alpha (1-E^2)(1+A)^2 + E(D-1)(1+B)^2 \right) > 0, \\ N &= \left( (D-1)(1-B)^2 - \alpha E(1-A)^2 \right)^2 - \left( \alpha (1-A)^2 - \lambda (A-B) \right)^2 \\ &- \lambda (A-B) \left( E^2 \left( 2\alpha (1-A)^2 - \lambda (A-B) \right) - 2E(D-1)(1-B)^2 \right). \end{split}$$ Also let G and H as in (2.1). In addition, for all n > 1, let $$(2.12) NG^2 - 2MGH + KH^2 < 0$$ If $p \in \mathcal{H}_{u.n}$ and (2.13) $$\alpha p^2(z) + \lambda z p'(z) \prec \frac{1 + Dz}{1 + Ez},$$ then $$p(z) \prec \frac{1 + Az}{1 + Bz}.$$

Theorem 2.4. Let the conditions of Lemma 2.4 hold. If $f \in A_{n,b}$ satisfies $$\alpha \left( \frac{zf'(z)}{f(z)} + \frac{z^2f''(z)}{f(z)} \right) \prec \frac{1 + Dz}{1 + Ez},$$ then $f \in S^*[A, B]$ .

Corollary 2.9. If $f \in A$ satisfies Re $$\left(\frac{zf'(z)}{f(z)} + \frac{z^2f''(z)}{f(z)}\right) > \delta$$ , then $f \in S^*$ .

Lemma 2.5. Suppose that $-1 \le B < A \le 1, -1 \le E < D \le 1,$ $$K = ((D-1)(1+B)^2 - (1-\alpha)E(1+A)^2)^2 - (1-\alpha)^2(1+A)^4 - \alpha(1+A)(1+B)$$ $$\times (2E(1+B)^2 + \alpha(1-E^2)(1+A)(1+B) + 2(1-\alpha)(1-E^2)(1+A)^2) \ge 0,$$ $$L = \lambda(A-B)(E(D-1)(1+B)^2 + \alpha(1-E^2)(1+A)(1+B)$$ $$+ (1 - \alpha)(1 - E^{2})(1 + A)^{2}) < 0$$ $$M = \left( \left( (D - 1)(1 - B^{2}) - E(1 - \alpha)(1 - A^{2}) \right)^{2} - (1 - \alpha)^{2}(1 - A^{2})^{2} \right.$$ $$- E(D - 1) \left( 2(1 - B^{2}) \left( (1 - \alpha)(1 - A^{2}) + \alpha(1 - AB) \right) \right.$$ $$- (1 - \alpha) \left( (A - B)^{2} + (1 - AB)^{2} \right) + \lambda(A - B)(1 + B)^{2} \right)$$ $$- (1 - E^{2}) \left( 2\alpha \left( (1 - \alpha)(1 - A^{2})(1 - AB) \right) + \alpha^{2} \left( (A - B)^{2} + (1 - AB)^{2} \right) \right.$$ $$+ \lambda(A - B)(1 + A) \left( (1 - \alpha)(1 + A) + \alpha(1 + B) \right) \right) > 0,$$ $$N = \left( \left( (D - 1)(1 - B)^{2} - (1 - \alpha)E(1 - A)^{2} \right)^{2} - (1 - \alpha)^{2}(1 - A)^{4} \right)$$ $$- (1 - E^{2}) \left( \left( \alpha(1 - A)(1 - B) - \lambda(A - B) \right)^{2} - 2(1 - \alpha)(1 - A)^{2} \right.$$ $$\times \left( \lambda(A - B) - \alpha(1 - A)(1 - B) \right) \right) + 2E(D - 1)(1 - B)^{2}$$ Also let G and H as in (2.1). In addition, for all n > 1, let $$NG^2 - 2MGH + KH^2 \le 0$$ If $p \in \mathcal{H}_{u,n}$ and (2.16) $$\alpha p(z) + (1 - \alpha)p^{2}(z) + \lambda z p'(z) \prec \frac{1 + Dz}{1 + Ez},$$ then $$p(z) \prec \frac{1 + Az}{1 + Bz}.$$

Theorem 2.5. Let the conditions of Lemma 2.5 hold. If $f \in A_{n,b}$ satisfies $$\frac{zf'(z)}{f(z)} + \beta \frac{z^2 f''(z)}{f(z)} \prec \frac{1 + Dz}{1 + Ez},$$ then $f \in S^*[A, B]$ .

Theorem 2.6. Let the conditions of Lemma 2.5 hold. If $f \in A_{n,b}$ satisfies $$f'(z) + \beta z f''(z) \prec \frac{1 + Dz}{1 + Ez},$$ then $f'(z) \prec \frac{1+Az}{1+Bz}$ .

Lemma 2.6. Suppose that $-1 \le B < A \le 1, -1 \le E < D \le 1,$ $$K = (1+A)^{2}(1+B)^{2} ((D-1) - \alpha E)^{2} - \alpha^{2}) - (1-\alpha)(1+A)^{3}(1-E^{2})$$ $$\times (2\alpha(1+B) + (1-\alpha)(1+A)) \ge 0,$$ $$L = \alpha \Big( E(D-1)(1+B) + (1-E^{2}) (\alpha(1+B) + (1-\alpha)(1+A)) \Big) < 0,$$ $$M = \Big( (((D-1) - \alpha E)^{2} - \alpha^{2}) ((A-B)^{2} + (1-AB)^{2}) - (1+A)(E(D-1) + \alpha(1-E^{2})) \Big)$$ $$\times (2(1-\alpha)(1-A)(1-AB) + \alpha(1+B)(A-B)) - (1-\alpha)(1-E^{2})(1+A)$$ $$\times ((1-\alpha)(1-A)(1-A^{2}) + \alpha(1+A)(A-B)) \Big) > 0,$$ $$N = (1-A)^{2}(1-B)^{2} \Big( ((D-1) - \alpha E)^{2} - \alpha^{2} \Big) + (E^{2}-1) \Big( (1-\alpha)(1-A)^{2} - \alpha(A-B) \Big)^{2} + 2(1-A)(1-B)(E(D-1) + \alpha(1-E^{2})) (\alpha(A-B) - (1-\alpha)(1-A)^{2}).$$ Also let G and H as in (2.1). In addition, for all n > 1, let $$NG^2 - 2MGH + KH^2 \le 0$$ If $p \in \mathcal{H}_{u,n}$ and $$(2.18) (1-\alpha)p(z) + \alpha \left(1 + \frac{zp'(z)}{p(z)}\right) \prec \frac{1+Dz}{1+Ez},$$ then $$p(z) \prec \frac{1+Az}{1+Bz}$$

Theorem 2.7. Let the conditions of Lemma 2.6 hold. If $p \in \mathcal{H}_{\mu,n}$ satisfies (2.20) $$1 + \frac{zp'(z)}{p(z)} < \frac{1+Dz}{1+Ez},$$ then $$p(z) \prec \frac{1 + Az}{1 + Bz}.$$

Theorem 2.8. Let the conditions of Lemma [2.6](#page-11-0) hold. If f ∈ An,b satisfies $$(1 - 2\alpha)\frac{zf'(z)}{f(z)} + \alpha\left(2 + \frac{zf''(z)}{f'(z)}\right) \prec \frac{1 + Dz}{1 + Ez},$$ then f ∈ S ∗ [A, B]. Proof. The proof of this theorem is derived from Lemma [2.6](#page-11-0) by letting p(z) = zf′ (z)/f(z), where f ∈ An,b and 0 < µ = nb ≤ A − B.