Quantum approach on linear combination of harmonic univalent mappings

Theorem 1. [3] A locally univalent harmonic function $f = h + \overline{g}$ in $\mathbb{D}$ is a univalent mapping of real axis $\mathbb{D}$ onto a domain convex in the direction of real axis if and only if h - g is a analytic univalent mapping of $\mathbb{D}$ onto a domain convex in the direction of real axis. In this paper using quantum approach we find the new harmonic functions. Such that, if we have an analytic function of the form Example 1. $$h(z) - g(z) = z - \frac{1}{2}z^2$$ and $\omega_q(z) = \frac{\partial_q(g(z))}{\partial_q(h(z))} = \frac{[2]_q}{2}z^2$ $$\partial_q(h(z)) - \partial_q(g(z)) = 1 - \frac{[2]_q}{2}z \qquad \partial_q(g(z)) = \frac{[2]_q}{2}z\partial_q(h(z))$$ we get $$\partial_a(h(z)) = 1$$ which on q-integration and normalization gives $$h(z) = z$$ $g(z) = \frac{[2]_q}{4}z^2$ Thus the mapping $f = h + \bar{g}$ given by $$f = z - \frac{[2]_q}{4} \overline{z^2}$$ On $q \to 1^-$ , we get the original function $f = z - \frac{1}{2}\overline{z^2}$ Definition 2. Harmonic q right half-plane mappings: If we have an analytic function of the form $$h(z) + g(z) = \frac{z}{1 - z}$$ and $\omega_q(z) = \frac{z^2 - 2z + qz}{1 - qz}$ . Then $$\partial_q(h(z)) + \partial_q(g(z)) = \frac{1}{(1-z)(1-qz)}$$ we get $$\partial_q(h(z)) = \frac{1}{(1-z)^3}$$ which on q-integration and normalization gives $$h(z) = \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2[n+1]_q} z^{n+1}$$ Using this we get $$g(z) = \sum_{n=0}^{\infty} \frac{2[n+1]_q - (n+1)(n+2)}{2[n+1]_q} z^{n+1}$$ Thus h and g of the form $$h(z) = \sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2[n+1]_q} z^{n+1} \qquad g(z) = \sum_{n=0}^{\infty} \frac{2[n+1]_q - (n+1)(n+2)}{2[n+1]_q} z^{n+1}$$ On q → 1 ⁻, we get the mapping defined in [\(1.1\)](#page-2-0). <span id="page-3-3"></span>Lemma 1. [\[3\]](#page-6-16) Let Ω ⊂ C be a domain convex in the direction of the real axis. Also let p be a real-valued continuous function in Ω. Then the mapping ω 7→ ω + p(ω) is univalent in Ω if and only if it is locally univalent. If it is univalent, then its range is convex in the direction of the real axis. <span id="page-3-0"></span>Lemma 2. [\[18\]](#page-6-17)Let f be analytic function in D with f(0) = 0 and f ′ (0) 6= 0. Suppose also that $$\varphi(z) = \frac{z}{(1 + ze^{i\theta})(1 + ze^{-i\theta})} \quad (\theta \in \mathbb{R}; z \in \mathbb{D}).$$ If $$\Re\left(\frac{zf'(z)}{\varphi(z)}\right) > 0 \quad (z \in \mathbb{D}),$$ then f is convex in the direction of real axis. Dorff and Rolf [\[5\]](#page-6-15) applied another way of constructing a univalent harmonic map by taking two suitable harmonic maps f¹ and f² with same dilatations, whose linear combination f³ = tf¹ + (1 − t)f2, t ∈ [0, 1] is univalent and convex in the direction of the imaginary axis. Wang et al. [\[26\]](#page-6-18) derived several sufficient conditions on harmonic univalent functions f¹ and f² so that their linear combination f = tf¹ + (1 − t)f2, t ∈ [0, 1], is univalent and convex in the direction of the real axis. More results on the linear combination f of f¹ and f² may also be found in [\[6,](#page-6-19) [13,](#page-6-20) [21,](#page-6-21) [23,](#page-6-22) [24,](#page-6-23) [26\]](#page-6-18) etc. (also see the references cited in these). In this paper using quantum approach we find sufficient conditions on f¹ and f² so that their linear combination f = tf¹ + (1 − t)f2, t ∈ [0, 1], is univalent and convex in the direction of the real axis. Some examples are also illustrated to support our main results.

Theorem 2. Let for $j = 1, 2, f_j = h_j + \overline{g_j} \in S_{\mathcal{H}}$ with <span id="page-4-1"></span> $$F_j = h_j(z) - g_j(z).$$ (2.3) If $\omega_{q_1}(z) = \omega_{q_2}(z)$ and satisfy the condition $\Re\left(\frac{z\partial_q(F_j(z))}{\varphi_q(z)}\right) > 0$ for some function $\varphi_q(z)$ given by (2.2), then $f_3 = tf_1 + (1-t)f_2$ , $t \in [0,1]$ is univalent and convex in the direction of real axis.

Theorem 3. Let for $j=1,2, f_j=h_j+\overline{g_j}\in S_{\mathcal{H}}$ be convex in the direction of the real axis. If $\Re\left\{\left(1-\omega_{q_1}\overline{\omega_{q_2}}\right)\partial_q(h_1(z)\overline{\partial_q(h_2(z)}\right\}\geq 0$ , then $f_3=tf_1+(1-t)f_2,\ t\in[0,1]$ is convex in the direction of real axis.