Janowski type close-to-convex functions associated with conic regions

Lemma 2.1. ([23]) Let $p(z) = 1 + \sum_{n=1}^{\infty} p_n z^n \prec F(z) = 1 + \sum_{n=1}^{\infty} d_n z^n$ in E. If F(z) is univalent in E and F(E) is convex, then $$|p_n| < |d_1|, \quad n > 1.$$

Lemma 2.2. ([1]) Let $p(z) = 1 + \sum_{n=1}^{\infty} c_n z^n \in k - \mathcal{P}[A, B]$ . Then $$|c_n| \leq |\delta(A, B, k)|,$$ where $$\delta(A, B, k) = \frac{(A - B)\delta_k}{2},\tag{2.1}$$ and $$\delta_{k} = \begin{cases} \frac{8(\cos^{-1}k)^{2}}{\pi^{2}(1-k^{2})}, & 0 \leq k < 1, \\ \frac{8}{\pi^{2}}, & k = 1, \\ \frac{\pi^{2}}{4\sqrt{t}(k^{2}-1)R^{2}(t)(1+t)}, & k > 1. \end{cases}$$ (2.2)

Lemma 2.3. ([2]) Let f and g be in the class C and $S^*$ , respectively. Then, for every function F(z) analytic in E with F(0) = 1, we have $$\frac{f(z) g(z)F(z)}{f(z) g(z)} \in \overline{\operatorname{co}}\big(F(E)\big), \quad z \in E,$$ where "\*" denotes the well-known convolution of two analytic functions and $\overline{\text{co}}F(E)$ denotes the closed convex hull F(E).

Lemma . ([\[](#page-13-3)]) Let g ∈ k–ST [C,D] with k ≥ and be given by $$g(z)=z+\sum_{n=2}^{\infty}b_nz^n.$$ Then $$|b_n| \le \prod_{j=0}^{n-2} \frac{|(C-D)\delta_k - 2jD|}{2(j+1)},$$ where δ^k is defined by ([.\)](#page-4-0).

Theorem . Let f ∈ k–UK[A,B,C,D], and let it be of the form given by [\(.\)](#page-0-1). Then, for n ≥ , we have $$|a_n| \leq \frac{1}{n} \prod_{i=0}^{n-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)} + \frac{(A-B)|\delta_k|}{2n} \sum_{j=1}^{n-1} \prod_{i=0}^{j-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)},$$ where δ^k is defined by ([.\)](#page-4-0). This result is not sharp. Proof Let us take $$zf'(z) = g(z)p(z), (3.1)$$ where p ∈ k–P[A,B] and g ∈ k–ST [C,D]. Let zf (z) = z+ ^∞ ⁿ= nanzⁿ, g(z) = z+ ^∞ ⁿ= bnzⁿ and p(z)=+ ^∞ ⁿ= cnzⁿ. Then ([.\)](#page-5-0) becomes $$z + \sum_{n=2}^{\infty} n a_n z^n = \left(z + \sum_{n=2}^{\infty} b_n z^n\right) \left(1 + \sum_{n=1}^{\infty} c_n z^n\right).$$ Equating the coefficients of zⁿ on both sides, we have $$na_n = b_n + \sum_{j=1}^{n-1} b_j c_{n-j}.$$ This implies that $$n|a_n| \le |b_n| + \sum_{j=1}^{n-1} |b_j| |c_{n-j}|.$$ (3.2) Since p ∈ k – P[A,B] and g ∈ k – ST [C,D], therefore by Lemma [.](#page-4-1) and Lemma [.](#page-5-1) we have $$|b_n| \le \prod_{i=0}^{n-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)}$$ and $$|c_n| \leq \frac{1}{2}(A-B)|\delta_k|.$$ Hence [\(.](#page-5-2)) becomes $$n|a_n| \leq \prod_{i=0}^{n-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)} + \sum_{j=1}^{n-1} \frac{(A-B)|\delta_k|}{2} \prod_{i=0}^{j-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)},$$ which implies that $$|a_n| \leq \frac{1}{n} \prod_{i=0}^{n-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)} + \frac{(A-B)|\delta_k|}{2n} \sum_{j=1}^{n-1} \prod_{i=0}^{j-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)}.$$ This completes the proof. -

Corollary . ([\[](#page-13-11)]) Let f ∈ k – UK[, –, , –] which has the form ([.](#page-0-1)). Then, for n ≥ , $$|a_n| \le \frac{(|\delta_k|)_{n-1}}{n!} + \frac{|\delta_k|}{n} \sum_{j=0}^{n-1} \frac{(|\delta_k|)_{j-1}}{(j-1)!}.$$

Corollary . Let f ∈ k –UK[ – β, –, – γ , –] = f ∈ k –UK[β, γ ] which has the form ([.](#page-0-1)). Then, for n ≥ , $$|a_n| \le \frac{(|\delta_k|)_{n-1}}{n!} + \frac{|\delta_k|}{n} \sum_{j=0}^{n-1} \frac{(|\delta_k|)_{j-1}}{(j-1)!}.$$

Corollary . ([\[](#page-13-20)]) Let f ∈ – UK[, –, , –] = K which has the form ([.](#page-0-1)). Then, for n ≥ , $$|a_n| \leq n$$ . Using relation [\(.\)](#page-3-0) and Theorem [.,](#page-5-3) we obtain immediately the following result.

Theorem . Let f ∈ k – UQ[A,B,C,D] which has the form [\(.\)](#page-0-1). Then, for n ≥ , $$|a_n| \le \frac{1}{n^2} \prod_{i=0}^{n-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)} + \frac{(A-B)|\delta_k|}{2n^2} \sum_{j=1}^{n-1} \prod_{i=0}^{j-2} \frac{|(C-D)\delta_k - 2iD|}{2(i+1)}.$$ By assigning different permissible values to the parameters, we obtain several known results, see [17, 21, 22]. 2. Sufficient conditions

Theorem 3.6. Let $f \in A$ and be given by (1.1). Then $f \in k-\mathcal{UK}[A,B,C,D]$ if $$\sum_{n=2}^{\infty} \left[ 2(k+1)|b_n - na_n| + \left| (B+1)na_n - (A+1)b_n \right| \right] < |B-A|.$$ (3.3) Proof Let us assume that equation (3.3) holds true. It is sufficient to show that $$k \left| \frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1 \right| - \text{Re}\left(\frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1\right) < 1.$$ Now consider $$\begin{split} \left| \frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1 \right| &= \left| \frac{(B-1)zf'(z) - (A-1)g(z)}{(B+1)zf'(z) - (A+1)g(z)} - 1 \right| \\ &= 2 \left| \frac{g(z) - zf'(z)}{(B+1)zf'(z) - (A+1)g(z)} \right| \\ &= 2 \left| \frac{\sum_{n=2}^{\infty} (b_n - na_n)z^n}{(B-A)z + \sum_{n=2}^{\infty} [(B+1)na_n - (A+1)b_n]z^n} \right| \\ &\leq \frac{2\sum_{n=2}^{\infty} |b_n - na_n|}{(B-A) - \sum_{n=2}^{\infty} |(B+1)na_n - (A+1)b_n|}. \end{split}$$ Since $$k \left| \frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1 \right| - \operatorname{Re}\left(\frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1\right)$$ $$\leq (k+1) \left| \frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1 \right| \leq \frac{2(k+1)\sum_{n=2}^{\infty} |b_n - na_n|}{(B-A) - \sum_{n=2}^{\infty} |(B+1)na_n - (A+1)b_n|}.$$ The last inequality is bounded by 1 if $$2(k+1)\sum_{n=2}^{\infty}|b_n-na_n|\leq |B-A|-\sum_{n=2}^{\infty}|(B+1)na_n-(A+1)b_n|.$$ Hence we have $$\sum_{n=2}^{\infty} \left[ 2(k+1)|b_n - na_n| + \left| (B+1)na_n - (A+1)b_n \right| \right] \le |B-A|.$$ This completes the

Theorem 3.7. Let $f \in A$ which has the form (1.1). Then $f \in k - UQ[A, B, C, D]$ if $$\sum_{n=2}^{\infty} n [2(k+1)|b_n - na_n| + |(B+1)na_n - (A+1)b_n|] < |B-A|.$$ The proof follows immediately by using Theorem 3.6 and relation (1.5).

Corollary 3.8. ([24]) A function is said to be in the class $1 - \mathcal{UQ}[1 - 2\beta, -1, 1, -1] = \mathcal{UQ}(\beta)$ for g(z) = z if $$\sum_{n=2}^{\infty} n^2 |a_n| < \frac{1-\beta}{2}.$$ 3. Necessary condition

Theorem 3.9. Let $f \in k - \mathcal{UK}[A, B, C, D]$ . Then, for $\theta_1 < \theta_2, z \in E$ , $$\int_{\theta_1}^{\theta_2} \operatorname{Re}\left\{\frac{(zf'(z))'}{f'(z)}\right\} d\theta > -\left[\frac{C-D}{2k+1-D} + \lambda\right]\pi,$$ where $\lambda$ is defined by (1.4). Proof Since $f \in k - \mathcal{UK}[A, B, C, D]$ , there exists $g \in k - \mathcal{CV}[C, D] \subset \mathcal{C}(\beta_1)$ , $$\beta_1 = \frac{2k+1-C}{2k+1-D} \tag{3.4}$$ such that $$f'(z) = g'(z)p(z),$$ where $p \in k - \mathcal{P}[A, B]$ . We can write $$f'(z) = (g_1'(z))^{1-\beta_1} h^{\lambda}(z), \quad h \in \mathcal{P}[A,B] \subset \mathcal{P}, g_1 \in \mathcal{C}.$$ For $z = re^{i\theta}$ , $0 \le r < 1$ , $0 \le \theta_1 \le \theta_2 \le 2\pi$ , we have $$\int_{\theta_{1}}^{\theta_{2}} \operatorname{Re}\left\{\frac{(zf'(z))'}{f'(z)}\right\} d\theta = (1 - \beta_{1}) \int_{\theta_{1}}^{\theta_{2}} \operatorname{Re}\left\{\frac{(zg'_{1}(z))'}{g'_{1}(z)}\right\} d\theta + \beta_{1}(\theta_{2} - \theta_{1}) + \lambda \int_{\theta_{1}}^{\theta_{2}} \operatorname{Re}\left\{\frac{zh'(z)}{h(z)}\right\} d\theta.$$ $$(3.5)$$ Also, we observe that, for $h \in \mathcal{P}[A, B]$ , $$\frac{\partial}{\partial \theta} \arg h(re^{i\theta}) = \frac{\partial}{\partial \theta} \operatorname{Re} \left\{ -i \ln h(re^{i\theta}) \right\}$$ $$= \operatorname{Re} \left\{ \frac{re^{i\theta} h'(re^{i\theta})}{h(re^{i\theta})} \right\}.$$ Therefore $$\int_{\theta_1}^{\theta_2} \operatorname{Re} \left\{ \frac{r e^{i\theta} h'(r e^{i\theta})}{h(r e^{i\theta})} \right\} d\theta = \arg h(r e^{i\theta_2}) - \arg h(r e^{i\theta_1}),$$ this implies that $$\max_{h \in P[A,B]} \left| \int_{\theta_1}^{\theta_2} \operatorname{Re} \left\{ \frac{re^{i\theta} h'(re^{i\theta})}{h(re^{i\theta})} \right\} d\theta \right| = \max_{h \in P[A,B]} \left| \arg h(re^{i\theta_2}) - \arg h(re^{i\theta_1}) \right|. \tag{3.6}$$ Since $h \in \mathcal{P}[A, B]$ , so $$\left| h(z) - \frac{1 - ABr^2}{1 - B^2r^2} \right| \le \frac{(A - B)r}{1 - B^2r^2}.$$ From (3.6), we observe that $$\max_{h \in P[A,B]} \left| \int_{\theta_1}^{\theta_2} \operatorname{Re} \left\{ \frac{r e^{i\theta} h'(r e^{i\theta})}{h(r e^{i\theta})} \right\} d\theta \right| \leq 2 \sin^{-1} \left( \frac{(A-B)r}{1-ABr^2} \right) \\ \leq \pi - 2 \cos^{-1} \left( \frac{(A-B)r}{1-ABr^2} \right).$$ (3.7) Also, for $g_1 \in \mathcal{C}$ , we have $$\int_{\theta_1}^{\theta_2} \operatorname{Re} \left\{ \frac{(zg_1'(z))'}{g_1'(z)} \right\} d\theta > -\pi.$$ Using (3.6) and (3.7) in (3.5), we obtain $$\int_{\theta_{1}}^{\theta_{2}} \operatorname{Re}\left\{\frac{(zf'(z))'}{f'(z)}\right\} d\theta > -(1 - \beta_{1})\pi + \beta_{1}(\theta_{2} - \theta_{1}) - \lambda\pi + 2\lambda \cos^{-1}\left(\frac{(A - B)r}{1 - ABr^{2}}\right)$$ $$> -\left[\frac{(C - D)}{2k + 1 - D} + \lambda\right]\pi(r \to 1).$$ This completes the required result. Remark 3.10 Let $f \in k - \mathcal{UK}[A, B, C, D]$ . Then, for $\frac{C-D}{2k+1-D} < 1 - \lambda$ , $$\int_{\theta_1}^{\theta_2} \operatorname{Re}\left\{\frac{(zf'(z))'}{f'(z)}\right\} d\theta > -\pi,$$ and hence f is univalent in E, see [21].

Corollary 3.11. ([21]) Let $f \in 0 - \mathcal{UK}[1, -1, 1, -1]$ . Then, for $\theta_1 < \theta_2, z \in E$ , $$\int_{\theta_1}^{\theta_2} \operatorname{Re} \left\{ \frac{(zf'(z))'}{f'(z)} \right\} d\theta > -\pi.$$ 4. Arc length problem

Theorem 3.12. Let $f \in k - \mathcal{UK}[A, B, C, D]$ which has the form (1.1). Then $$\mathcal{L}_r(f) \le C(\lambda, C, D) \left(\frac{1}{1-r}\right)^{2(1-\beta_1)-1} \quad (r \to 1),$$ where $\beta_1$ is defined by (3.4) and $C(\lambda, A, B)$ is a constant depending upon $\lambda$ , A and B. Proof Let $$zf'(z) = g(z)h^{\lambda}(z), \tag{3.8}$$ where $g \in k - \mathcal{ST}[C,D]$ and $h \in \mathcal{P}[A,B] \subset \mathcal{P}$ . Since $k - \mathcal{ST}[C,D] \subseteq \mathcal{S}^*(\beta_1)$ , see [1]. We can write $$g(z) = z \left(\frac{g_1(z)}{z}\right)^{1-\beta_1}, \quad g_1 \in \mathcal{S}^*.$$ Equation (3.8) gives $$zf'(z) = z^{\beta_1} (g_1(z))^{1-\beta_1} h^{\lambda}(z).$$ Now, for $z = re^{i\theta}$ , $$\mathcal{L}_r(f) = \int_0^{2\pi} \left| zf'(z) \right| d\theta = \int_0^{2\pi} \left| z^{\beta_1} (g_1(z))^{1-\beta_1} h^{\lambda}(z) \right| d\theta.$$ Using Holder's inequality, we have $$\mathcal{L}_{r}(f) \leq 2\pi \left(\frac{1}{2\pi} \int_{0}^{2\pi} \left| \left( g_{1}(z) \right)^{(1-\beta_{1})(\frac{2}{2-\lambda})} \right| d\theta \right)^{\frac{2-\lambda}{2}} \left( \frac{1}{2\pi} \int_{0}^{2\pi} \left| h(z) \right|^{2} d\theta \right)^{\frac{\lambda}{2}}. \tag{3.9}$$ Since $h \in \mathcal{P}[A, B] \subset \mathcal{P}$ , so $$\frac{1}{2\pi} \int_0^{2\pi} \left| h(z) \right|^2 d\theta \le \frac{1 + \{ (A - B)^2 - 1 \} r^2}{1 - r^2}.$$ (3.10) Using (3.10) and the distortion result for a starlike function in (3.9), we obtain $$\mathcal{L}_{r}(f) \leq 2\pi \left(\frac{1}{2\pi} \int_{0}^{2\pi} \frac{r^{(1-\beta_{1})(\frac{2}{2-\lambda})}}{|1-re^{i\theta}|^{\frac{4(1-\beta_{1})}{2-\lambda}}} d\theta\right)^{\frac{2-\lambda}{2}} \left(\frac{1+\{(A-B)^{2}-1\}r^{2}}{1-r^{2}}\right)^{\frac{\lambda}{2}}$$ $$\leq C(\lambda, A, B) \left(\frac{1}{1-r}\right)^{2(1-\beta_{1})+\lambda-1} \qquad (r \to 1),$$ where $C(\lambda, A, B) = \pi^{\frac{\lambda}{2}}(A - B)^{\lambda}$ and $2(1 - \beta_1) + \lambda > 1$ . This completes the

Corollary 3.13. ([25]) Let $f \in 0 - \mathcal{UK}[1, -1, 1, -1]$ . Then, for 0 < r < 1, $$\mathcal{L}_r(f) \le \mathcal{O}\left(\mathcal{M}(r)\log\frac{1}{1-r}\right) \quad as \ r \to 1,$$ where $\mathcal{O}$ -notation denotes that the constant is absolute and $\mathcal{M}(r) = \max_{|z|=r} |f(z)|$ . 5. Growth rate of coefficients

Theorem 3.14. Let $f \in k - \mathcal{UK}[A, B, C, D]$ which has the form (1.1). Then, for $k \geq 0$ , we have $$|a_n| \leq C(\lambda, A, B)n^{2(1-\beta_1)+\lambda-2}$$ , where $\beta_1$ is defined by (3.4). Proof From Cauchy's theorem with $z = re^{i\theta}$ , one can easily have $$na_n = \frac{1}{2\pi r^n} \int_0^{2\pi} z f'(z) e^{-in\theta} d\theta.$$ This implies that $$|na_n| \le \frac{1}{2\pi r^n} \int_0^{2\pi} |zf'(z)| d\theta$$ = $\frac{1}{2\pi r^n} \mathcal{L}_r(f)$ . Using Theorem 3.12 and putting $r = 1 - \frac{1}{n}$ , we obtain the required result. 6. Convolution properties

Theorem 3.15. If $f \in k - \mathcal{ST}[C,D]$ and $\varphi \in C$ , then $\varphi * f \in k - \mathcal{ST}[C,D]$ . Proof To prove the result, we need to prove $$\frac{z(\varphi(z)f(z))'}{\varphi(z)f(z)} \in k - \mathcal{P}[A,B].$$ Consider $$\frac{z(\varphi(z)f(z))'}{\varphi(z)f(z)} = \frac{\varphi(z)\frac{zf'(z)}{f(z)}f(z)}{\varphi(z)f(z)}$$ $$= \frac{\varphi(z)\Psi(z)f(z)}{\varphi(z)f(z)},$$ where $\frac{zf'(z)}{f(z)} = \Psi(z) \in k - \mathcal{P}[C,D]$ . Applying Lemma 2.3, we obtain the required result. $\Box$

Theorem 3.16. Let $f \in k - \mathcal{UK}[A, B, C, D]$ and $\varphi \in \mathcal{C}$ . Then $\varphi * f \in k - \mathcal{UK}[A, B, C, D]$ . Proof Since $f \in k - \mathcal{UK}[A, B, C, D]$ , there exists $g \in k - \mathcal{ST}[C, D]$ such that $\frac{zf'(z)}{g(z)} \in k - \mathcal{P}[A, B]$ . It follows from Lemma 2.3 that $\varphi * g \in k - \mathcal{ST}[C, D]$ . Now $$\begin{split} \frac{z(\varphi(z)f(z))'}{\varphi(z)g(z)} &= \frac{\varphi(z)zf'(z)}{\varphi(z)g(z)} = \frac{\varphi(z)\frac{zf'(z)}{g(z)}g(z)}{\varphi(z)g(z)} \\ &= \frac{\varphi(z)F(z)g(z)}{\varphi(z)g(z)}, \end{split}$$ where $F(z) \in k - \mathcal{P}[A, B]$ . Applying Lemma 2.3, we have $\frac{z(\varphi(z)f(z))'}{\varphi(z)g(z)} \in k - \mathcal{UK}[A, B, C, D]$ for $z \in E$ . 7. Radius of convexity problem

Theorem 3.17. Let $f \in k - \mathcal{UK}[A, B, C, D]$ in E. Then $f \in \mathcal{C}$ for $|z| < r_1$ , where $$r_1 = \frac{2}{2(1-\beta_1) + \lambda(A-B) + \sqrt{([2(1-\beta_1) + \lambda(A-B)]^2 + 4[2\beta_1 - 1])}},$$ (3.11) where $\lambda$ and $\beta_1$ are defined by (1.4) and (3.4), respectively. Proof Let $$zf'(z) = g(z)p(z),$$ where $g \in k - \mathcal{ST}[C,D]$ and $p \in k - \mathcal{P}[A,B]$ . Since $k - \mathcal{ST}[C,D] \subset \mathcal{S}^(\beta_1)$ , it is known [26] that there exists $g_1 \in \mathcal{S}$ such that $$g(z) = z \left(\frac{g_1(z)}{z}\right)^{1-\beta_1}.$$ We can write $$zf'(z) = z^{\beta_1} (g_1(z))^{1-\beta_1} h^{\lambda}(z), \quad h \in \mathcal{P}[A, B] \subset \mathcal{P}.$$ (3.12) The logarithmic differentiation of (3.12) yields $$\frac{(zf'(z))'}{f'(z)} = \beta_1 + (1-\beta_1) \frac{zg_1'(z)}{g_1(z)} + \lambda \frac{zh'(z)}{h(z)}.$$ Using the distortion result for the classes $S^*$ and $\mathcal{P}[A,B]$ , we obtain $$\operatorname{Re} \frac{(zf'(z))'}{f'(z)} \ge \beta_1 + (1 - \beta_1) \frac{1 - r}{1 + r} - \frac{\lambda(A - B)r}{1 - r^2}$$ $$\ge \frac{1 + (1 - 2\beta_1)r^2 - [2(1 - \beta_1) + \lambda(A - B)]r}{1 - r^2}.$$ (3.13) The right-hand side of (3.13) is positive for $|z| < r_1$ , where $r_1$ is given by (3.11). We note the following cases: (i). For A = 1, B = -1, C = 1 and D = -1, we obtain the radius of convexity problem for the class $k - \mathcal{UK}$ . - (ii). For k = , we have the radius of convexity for the class K[A,B,C,D]. - (iii). For A = , B = –, C = , D = and k = , we have the radius of convexity problem for the well-known class of close-to-convex functions studied by Kaplan [\[](#page-13-20)].

Definition 1.1. ([1]) A function p analytic in E belongs to the class $k - \mathcal{P}[A, B]$ if and only if $$p(z) \prec \frac{(A+1)p_k(z) - (A-1)}{(B+1)p_k(z) - (B-1)}, \quad k \ge 0,$$ where $p_k(z)$ is defined by (1.3) and $-1 \le B < A \le 1$ . Geometrically, the function $p(z) \in k - \mathcal{P}[A, B]$ takes all values in the domain $\Omega_k[A, B]$ , $-1 \le B < A \le 1$ , $k \ge 0$ , which is defined as follows: $$\Omega_k[A,B] = \left\{ w : \text{Re}\left(\frac{(B-1)w(z) - (A-1)}{(B+1)w(z) - (A+1)}\right) > k \left| \frac{(B-1)w(z) - (A-1)}{(B+1)w(z) - (A+1)} - 1 \right| \right\}$$ or equivalently $$\begin{split} \Omega_k[A,B] &= \left\{ u + iv : \left[ \left( B^2 - 1 \right) \left( u^2 + v^2 \right) - 2(AB - 1)u + \left( A^2 - 1 \right) \right]^2 \\ &> k^2 \left[ \left( -2(B+1) \left( u^2 + v^2 \right) + 2(A+B+2)u - 2(A+1) \right)^2 \\ &+ 4(A-B)^2 v^2 \right] \right\}. \end{split}$$ The domain $\Omega_k[A,B]$ retains the conic domain $\Omega_k$ inside the circular region defined by $\Omega_0[A,B] = \Omega[A,B]$ . The impact of $\Omega[A,B]$ on the conic domain $\Omega_k$ changes the original shape of the conic regions. The ends of hyperbola and parabola get closer to each other but never meet anywhere and the ellipse gets the shape of oval. When $A \longrightarrow 1$ , $B \longrightarrow -1$ , the radius of the circular disk defined by $\Omega[A,B]$ tends to infinity; consequently, the arms of hyperbola and parabola expand and the oval turns into ellipse.

Definition 1.2. ([1]) A function $f \in A$ is said to be in the class k - CV[C, D], $-1 \le D < C \le 1$ , if it satisfies the condition $$\operatorname{Re}\left(\frac{(D-1)\frac{(zf'(z))'}{f'(z)} - (C-1)}{(D+1)\frac{(zf'(z))'}{f'(z)} - (C+1)}\right) > k \left| \frac{(D-1)\frac{(zf'(z))'}{f'(z)} - (C-1)}{(D+1)\frac{(zf'(z))'}{f'(z)} - (C+1)} - 1 \right| \quad (k \ge 0; z \in E),$$ equivalently, we can write $$\frac{(zf'(z))'}{f'(z)} \in k - \mathcal{P}[C, D].$$

Definition 1.3. ([1]) The class $k - \mathcal{ST}[C,D]$ , $-1 \le D < C \le 1$ , is the family of all those functions $f \in \mathcal{A}$ such that $$\operatorname{Re}\left(\frac{(D-1)\frac{zf'(z)}{f(z)} - (C-1)}{(D+1)\frac{zf'(z)}{f(z)} - (C+1)}\right) > k \left| \frac{(D-1)\frac{zf'(z)}{f(z)} - (C-1)}{(D+1)\frac{zf'(z)}{f(z)} - (C+1)} - 1 \right| \quad (k \ge 0; z \in E),$$ or equivalently $$\frac{zf'(z)}{f(z)} \in k - \mathcal{P}[C, D].$$ These two classes were recently introduced by Noor and Malik [1]. Motivated by the recent work presented by Noor and Malik [1], we define some classes of analytic functions associated with conic domains as follows.

Definition 1.4. Let $f \in \mathcal{A}$ . Then $f \in k - \mathcal{UK}[A,B,C,D]$ if and only if there exists $g \in k - \mathcal{ST}[C,D]$ such that $$\operatorname{Re}\left(\frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)}\right) > k \left| \frac{(B-1)\frac{zf'(z)}{g(z)} - (A-1)}{(B+1)\frac{zf'(z)}{g(z)} - (A+1)} - 1 \right| \quad (k \ge 0),$$ or equivalently $$\frac{zf'(z)}{g(z)} \in k - \mathcal{P}[A, B],$$ where $-1 \le D \le C \le 1$ and $-1 \le B < A \le 1$ .

Definition 1.5. Let $f \in \mathcal{A}$ . Then $f \in k - \mathcal{UQ}[A, B, C, D]$ if and only if, for $-1 \le D < C \le 1$ , $-1 \le B < A \le 1$ and $k \ge 0$ , there exists $g \in k - \mathcal{CV}[C, D]$ such that $$\operatorname{Re}\left(\frac{(B-1)\frac{(zf'(z))'}{g'(z)} - (A-1)}{(B+1)\frac{(zf'(z))'}{g'(z)} - (A+1)}\right) > k \left| \frac{(B-1)\frac{(zf'(z))'}{g'(z)} - (A-1)}{(B+1)\frac{(zf'(z))'}{g'(z)} - (A+1)} - 1 \right|,$$ or equivalently $$\frac{(zf'(z))'}{g'(z)} \in k - \mathcal{P}[A, B].$$ It can easily be seen that $$f \in k-\mathcal{U}\mathcal{Q}[A,B,C,D] \Leftrightarrow zf' \in k-\mathcal{U}\mathcal{K}[A,B,C,D].$$ (1.5) Special cases: i. $0 - \mathcal{U}\mathcal{K}[A, B, C, D] = \mathcal{K}[A, B, C, D]$ and $0 - \mathcal{U}\mathcal{Q}[A, B, 1, -1] = \mathcal{Q}[A, B]$ , subclasses of close-to-convex and quasi-convex functions studied by Silvia and Noor, respectively, see [15, 16]. - ii. $k \mathcal{UK}[1, -1, 1, -1] = k \mathcal{UK}$ and $k \mathcal{UQ}[1, -1, 1, -1] = k \mathcal{UQ}$ , the class of k-uniformly close-to-convex and the class of k-uniformly quasi-convex functions studied by Acu [17]. - iii. $k \mathcal{UK}[1 2\beta, -1, 1 2\gamma, -1] = k \mathcal{UK}(\beta, \gamma)$ and $k \mathcal{UQ}[1 2\beta, -1, 1 2\gamma, -1] = k \mathcal{UQ}(\beta, \gamma)$ , the well-known class of k-uniformly close-to-convex and the class of k-uniformly quasi-convex functions of order $\beta$ and type $\gamma$ , see [18]. - iv. $0-\mathcal{U}\mathcal{K}[1-2\beta,-1,1-2\gamma,-1]=\mathcal{K}(\beta,\gamma)$ and $0-\mathcal{U}\mathcal{Q}[1-2\beta,-1,1-2\gamma,-1]=\mathcal{Q}(\beta,\gamma)$ , well-known classes of close-to-convex and quasi-convex functions of order $\beta$ type $\gamma$ , see [19, 20]. - v. $0 \mathcal{U}\mathcal{K}[1, -1, 1, -1] = \mathcal{K}$ and $0 \mathcal{U}\mathcal{Q}[1, -1, 1, -1] = \mathcal{Q}$ , the classes of close-to-convex and quasi-convex functions; for details, see [21, 22]. Throughout this paper, we assume that $-1 \le D < C \le 1$ , $-1 \le B < A \le 1$ and $k \ge 0$ unless otherwise specified.