On Harmonic Univalent Spirallike Mappings

Lemma 1.1. [9] Let $\lambda \in (-\pi/2, \pi/2)$ . Suppose that a function $f \in C^1(\mathbb{D})$ satisfies the condition that f(z) = 0 if and only if z = 0, and that $J_f = |f_z|^2 - |f_{\overline{z}}|^2 > 0$ on $\mathbb{D}$ . Then f is one-one in $\mathbb{D}$ and $f(\mathbb{D}_r)$ is $\lambda$ -spirallike for each 0 < r < 1 if and only if Re $$\left(e^{-i\lambda}\frac{Df(z)}{f(z)}\right) > 0, \quad z \in \mathbb{D} \setminus \{0\}.$$ Let $\mathcal{SP}_H(\lambda)$ and $\mathcal{SP}_{H'}(\lambda)$ denote the classes of harmonic hereditarily $\lambda$ -spirallike functions in $\mathcal{H}$ and $\mathcal{H}'$ , respectively. In this article, we provide some necessary and sufficient conditions for a harmonic mapping to be hereditarily spirallike function. Also, we give growth estimate for certain harmonic hereditarily spirallike mappings. Moreover, we connect this concept of harmonic hereditarily spirallike mapping with analytic spirallike function and develop some examples in support of our results. In Section 2, we present our results and examples. In Section 3, we provide detail proof of the results.

Theorem 2.1. Let $f = h + \overline{g}$ be a sense-preserving harmonic mapping on $\mathbb{D}$ such that f(0) = 0 only for z = 0. Then $f \in \mathcal{SP}_H(\lambda)$ if and only if <span id="page-2-2"></span> $$(2.1) |h(z)|^2 \operatorname{Re} \left( e^{-i\lambda} \frac{zh'(z)}{h(z)} \right) > |g(z)|^2 \operatorname{Re} \left( e^{i\lambda} \frac{zg'(z)}{g(z)} \right)$$ $$+ \operatorname{Re} \left( ze^{i\lambda} (h(z)g'(z) - e^{-2i\lambda}g(z)h'(z)) \right).$$ Theorem 2.1 can be used to identify whether a harmonic mapping is hereditarily spirallike or not. In this context, we present the following example. <span id="page-2-3"></span>Example 2.1. For $\alpha \in \mathbb{D}$ , the harmonic mapping $f_1(z) = z + \alpha \overline{z}$ is not hereditarily $\pi/4$ -spirallike. Because for $\alpha = -1/2$ and z = 1/2(1+i) the inequality (2.1) becomes 3/4 > 1. Here we note that, for $\alpha \in \mathbb{D}$ and an analytic univalent mapping $h \in \mathcal{S}$ , the harmonic mapping $f = h + \alpha \overline{h}$ is sense-preserving fully starlike in $\mathbb{D}$ if and only if h is starlike in $\mathbb{D}$ , because $$\operatorname{Re}\left(\frac{Df(z)}{f(z)}\right) = \frac{(1-|\alpha|^2)|h(z)|^2}{|h(z)+\alpha\overline{h(z)}|^2}\operatorname{Re}\left(\frac{zh'(z)}{h(z)}\right).$$ But the Example 2.1 shows that this does not happen in the case of spirallike mappings, and so the class $\mathcal{SP}_H(\lambda)$ is not affine invariant. In 1998, Silverman [14] considered the classes $\mathcal{H}$ and $\mathcal{H}'$ of harmonic mappings and obtained sufficient conditions for functions in $\mathcal{H}$ and $\mathcal{H}'$ to be a member of the class $\mathcal{ST}_H$ and $\mathcal{ST}_{H'}$ , respectively, in term of coefficients. Furthermore, the author proved that the obtained condition is also necessary for the the later case. For convenient of the reader, we state the result here.

Lemma 2.1. [14] Let $f = h + \overline{g} \in \mathcal{H}$ be of the form (1.2) such that <span id="page-3-0"></span>(2.2) $$\sum_{n=1}^{\infty} n(|a_n| + |b_n|) \le 2.$$ Then $f \in \mathcal{ST}_H$ . Moreover, the condition (2.2) is necessary and sufficient for a harmonic mapping $f = h + \overline{g} \in \mathcal{H}'$ of the form (1.3) to be a member of the class $\mathcal{ST}'_H$ . We present a few similar results for harmonic hereditarily spirallike mappings in the unit disk. Before that we note down a few notations which we will use throughout the paper. For $\lambda \in (-\pi/2, \pi/2)$ and $n \geq 1$ , let <span id="page-3-2"></span><span id="page-3-1"></span>(2.3) $$A_n = |1 + ne^{-i\lambda}| + |1 - ne^{-i\lambda}| \text{ and } B = |1 + e^{-i\lambda}| - |1 - e^{-i\lambda}|.$$

Theorem 2.2. Let $\lambda \in (-\pi/2, \pi/2)$ and $f = h + \overline{g}$ be of the form (1.2) such that <span id="page-3-4"></span>(2.4) $$\sum_{n=2}^{\infty} \frac{A_n}{B} |a_n| + \sum_{n=1}^{\infty} \frac{A_n}{B} |b_n| \le 1$$ where $A_n$ and B are given by (2.3). Then f is harmonic hereditarily $\lambda$ -spirallike function. For $\lambda=0$ , Theorem 2.2 reduces to Lemma 2.1, which is expected because $\mathcal{SP}_H(0)=\mathcal{ST}_H$ . There are harmonic hereditarily $\lambda$ -spirallike mappings for which the value of the left hand side of (2.4) is 1. For example, the harmonic hereditarily $\lambda$ -spirallike mappings $$f(z) = h(z) + \overline{g(z)} = z + \sum_{n=2}^{\infty} \frac{B}{A_n} x_n z^n + \sum_{n=1}^{\infty} \frac{B}{A_n} \overline{y_n z^n}$$ where $\sum_{n=2} |x_n| + \sum_{n=1} |y_n| = 1$ have this property. The inequality (2.4) is useful to construct harmonic univalent spirallike functions in the unit disk. Using the inequality (2.4), one can easily check that $f_2(z) =$ $z + \alpha B/A_2\overline{z^2}$ and $f_3(z) = z + B\alpha/A_1\overline{z} + B/A_3(1-|\alpha|)\overline{z^3}$ where $\alpha \in \mathbb{D}$ are sense-preserving harmonic hereditarily $\lambda$ -spirallike mappings for every $\lambda \in (-\pi/2, \pi/2)$ . The images of $\mathbb{D}$ under $f_2$ for certain values of $\alpha$ and $\lambda$ are shown in Figure 1. <span id="page-4-0"></span> FIGURE 1. $f_2(\mathbb{D})$ for certain values of $\alpha$ and $\lambda$ . The condition (2.4) in the Theorem 2.2 is not a necessary condition. This can be seen by an example of a hereditarily $\lambda$ -spirallike function which does not satisfy the condition (2.4). We consider the function $f_4(z) = z(1-z)^{i-1} = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{H}$ . It is known that $f_4$ is hereditarily $\pi/4$ -spirallike function (see [13]). Clearly, $$a_2 = \frac{f_3''(0)}{2} = 1 - i.$$ But for $\lambda = \pi/4$ , we have $$\sum_{n=2}^{\infty} \frac{A_n}{B} |a_n| + \sum_{n=1}^{\infty} \frac{A_n}{B} |b_n| \ge \frac{A_2}{B} |a_2| = \sqrt{2} \frac{A_2}{B} > 2.$$ Our next result gives necessary and sufficient condition for harmonic mappings in $\mathcal{H}'$ to be a member of the class $\mathcal{SP}_{H'}(\lambda)$ .

Theorem 2.3. Let $\lambda \in (-\pi/2, \pi/2)$ and $f = h + \overline{g} \in \mathcal{H}'$ be of the form (1.3). If f satisfies (2.4) then $f \in \mathcal{SP}_{H'}(\lambda)$ . Conversely, every harmonic function $f \in \mathcal{SP}_{H'}(\lambda)$ satisfies the coefficient inequality <span id="page-5-2"></span>(2.5) $$\sum_{n=2}^{\infty} \frac{B}{A_n} |a_n| + \sum_{n=1}^{\infty} \frac{B}{A_n} |b_n| \le 1$$ where $A_n$ and B are given by (2.3). <span id="page-5-3"></span>The proof of the converse part of the Theorem 2.3 provides the following result. Corollary 2.1. Let $$\lambda \in (-\pi/2, \pi/2)$$ and $f = h + \overline{g} \in \mathcal{SP}_{H'}(\lambda)$ be of the form (1.3). Then $\sum_{n=1}^{\infty} (n|a_n| + n|b_n|) \leq 2$ , and $f \in \mathcal{ST}'_H$ . By Theorem 2.2 and Theorem 2.3, we have constructed harmonic hereditarily $\lambda$ -spirallike mapping in infinite series form. <span id="page-5-1"></span>Example 2.2. Let $h_1(z) = z$ , $h_n(z) = z + B/A_n z^n$ , $n \ge 2$ and $g_n(z) = z + B/A_n \overline{z^n}$ , $n \ge 1$ and $X_n, Y_n \ge 0$ are such that $\sum_{n=1}^{\infty} (X_n + Y_n) = 1$ . Then $f(z) = \sum_{n=1}^{\infty} (X_n h_n(z) + Y_n g_n(z))$ is hereditarily $\lambda$ -spirallike. Clearly, f can be written as $$f(z) = \sum_{n=1}^{\infty} (X_n h_n(z) + Y_n g_n(z))$$ $$= \sum_{n=1}^{\infty} (X_n + Y_n) z + \sum_{n=2}^{\infty} \left( \frac{X_n B}{A_n} z^n \right) + \sum_{n=1}^{\infty} \left( \frac{Y_n B}{A_n} \overline{z^n} \right)$$ $$= z + \sum_{n=2}^{\infty} \left( \frac{X_n B}{A_n} z^n \right) + \sum_{n=1}^{\infty} \left( \frac{Y_n B}{A_n} \overline{z^n} \right).$$ Thus the function f is of the form (1.2) and satisfies $$\sum_{n=2}^{\infty} \frac{A_n}{B} \left( \frac{X_n B}{A_n} \right) + \sum_{n=1}^{\infty} \frac{A_n}{B} \left( \frac{Y_n B}{A_n} \right) = \sum_{n=2}^{\infty} X_n + \sum_{n=1}^{\infty} Y_n = 1 - X_1 \le 1.$$ Hence by the Theorem 2.2, f is hereditarily $\lambda$ -spirallike. Remark 2.1. If we take $h_n(z) = z - B/A_n z^n$ , $n \ge 2$ in the Example 2.2, then the infinite series $\sum_{n=1}^{\infty} (X_n h_n(z) + Y_n g_n(z))$ ultimately belong to the class $\mathcal{SP}_{H'}(\lambda)$ . Example 2.3. Let $f = h + \overline{g} \in \mathcal{SP}_{H'}(\lambda)$ . Then f can be written as $f(z) = \sum_{n=1}^{\infty} (X_n h_n(z) + Y_n g_n(z))$ , where $h_1(z) = z$ , $h_n(z) = z - B/A_n z^n$ , $n \ge 2$ and $g_n(z) = z + B/A_n \overline{z^n}$ , $n \ge 1$ with some $X_n, Y_n \ge 0$ are such that $\sum_{n=1}^{\infty} (X_n + Y_n) = 1$ . Since $f = h + \overline{g} \in \mathcal{SP}_{H'}(\lambda)$ is of the form (1.3), by Theorem 2.3 the condition (2.5) holds. Let $$h_1(z) = z, \ h_n(z) = z - \frac{A_n}{B} z^n, \ n \ge 2$$ and $g_n(z) = z + \frac{A_n}{B} \overline{z^n}, \ n \ge 1$ and set $$X_n = \frac{B|a_n|}{A_n}$$ , $n \ge 2$ and $Y_n = \frac{B|b_n|}{A_n}$ , $n \ge 1$ with $X_1 = 1 - \sum_{n=2}^{\infty} X_n - \sum_{n=1}^{\infty} Y_n$ . Then from the condition (2.5), it follows that $0 \le X_n$ , $Y_n \le 1$ for $n \ge 1$ . Therefore, $$\sum_{n=1}^{\infty} (X_n h_n(z) + Y_n g_n(z)) = \sum_{n=1}^{\infty} (X_n + Y_n) z - \sum_{n=2}^{\infty} X_n \frac{A_n}{B} z^n + \sum_{n=1}^{\infty} Y_n \frac{A_n}{B} \overline{z^n}$$ $$= z - \sum_{n=2}^{\infty} |a_n| z^n + \sum_{n=1}^{\infty} |b_n| \overline{z^n} = f(z).$$ The growth estimate for sense-preserving harmonic $\lambda$ -spirallike functions is given below.

Theorem 2.4. Let $f = h + \overline{g} \in \mathcal{H}$ be a harmonic mapping of the form (1.2) which satisfies the condition (2.4). Then the sharp inequality $$\left(1 - \frac{B}{A_1}\right)r \le |f(z)| \le \left(1 + \frac{B}{A_1}\right)r \quad \text{for } |z| = r$$ holds and $$\left\{ w \in \mathbb{C} : |w| < 1 - \frac{B}{A_1} \right\} \subset f(\mathbb{D}).$$ There is an interesting relation between analytic starlike and spirallike functions which was obtained by Başgöze and Keogh [1]. For convenient, we state the result below.

Lemma 2.2. [1] Let $\lambda \in (-\pi/2, \pi/2)$ . For each analytic $\lambda$ -spirallike function $h \in \mathcal{A}$ there exist an unique analytic starlike function $g \in \mathcal{A}$ such that <span id="page-6-1"></span>(2.6) $$\frac{h(z)}{z} = \left(\frac{g(z)}{z}\right)^{e^{-i\lambda}\cos\lambda}.$$ <span id="page-6-3"></span>We extend the Lemma 2.2 for functions in the class $C^1(\mathbb{D})$ .

Theorem 2.5. Let $\lambda$ be a real number with $|\lambda| < \pi/2$ . For a hereditarily $\lambda$ -spirallike function $h \in C^1(\mathbb{D})$ satisfying h(z) = 0 only for z = 0 and $J_h = |h_z|^2 - |h_{\overline{z}}|^2 > 0$ for $z \in \mathbb{D}$ there exists a fully starlike function $g \in C^1(\mathbb{D})$ satisfying g(z) = 0 only for z = 0 and $J_q = |g_z|^2 - |g_{\overline{z}}|^2 > 0$ for $z \in \mathbb{D}$ such that relation (2.6) is satisfied. It is important to note that, the relation (2.6) hold for analytic functions, but not for complex-valued harmonic functions as the exponential power of harmonic function may not be harmonic. For example, for the harmonic function $g(z) = z + \alpha \overline{z}$ , $\alpha \in \mathbb{D}$ , the corresponding mapping h define by the relation (2.6), is not harmonic. Now, we give a relation between harmonic spirallike and starlike functions in some different way. <span id="page-7-1"></span>Theorem 2.6. Let $F(z) = z - \sum_{n=2}^{\infty} |a_n| z^n + \sum_{n=1}^{\infty} |b_n| z^n$ be harmonic hereditarily starlike mapping. Then $f(z) = z + \sum_{n=2}^{\infty} d_n a_n z^n + \sum_{n=1}^{\infty} d_n b_n z^n$ , where $\{d_n\}$ is a sequence such that $|d_n| \le nB/A_n$ for $n \ge 1$ , is harmonic hereditarily $\lambda$ -spirallike mapping. Conversely, if $f(z) = z - \sum_{n=2}^{\infty} |a_n| z^n + \sum_{n=1}^{\infty} |b_n| z^n$ is harmonic hereditarily $\lambda$ -spirallike mapping then $F(z) = z + \sum_{n=2}^{\infty} a_n z^n + \sum_{n=1}^{\infty} b_n z^n$ is hereditarily starlike mapping. For $0 < \alpha < 1$ , the harmonic mapping $f_5(z) = z + B\alpha/A_1\overline{z} + B/A_2(1-\alpha)\overline{z^2}$ is hereditarily $\lambda$ -spirallike as from Lemma 2.2, the function $F(z) = z + \alpha\overline{z} + 1/2(1-\alpha)\overline{z^2}$ is hereditarily starlike. Here, we note that $d_n = nB/A_n$ . The images of $\mathbb D$ under $f_5$ for certain values of $\alpha$ and $\lambda$ are shown in Figure 2. <span id="page-7-0"></span> FIGURE 2. $f_5(\mathbb{D})$ for certain values of $\lambda$ and $\alpha$ . Thus we see that one can construct harmonic spirallike functions from given starlike mapping, and vice-versa. As many geometric properties of analytic univalent functions are already known, it provide more geometric exposures. So, it is quite interesting to relate harmonic function theory to analytic function theory. In this regards, Nagpal and Ravichandran [11] established a relation between harmonic fully starlike function and analytic starlike function.

Lemma 2.3. [11] A sense-preserving harmonic function $f = h + \overline{g}$ is fully starlike in $\mathbb{D}$ if the analytic functions $h + \epsilon g$ are starlike in $\mathbb{D}$ for each $|\epsilon| = 1$ . <span id="page-8-1"></span>Our next result for spirallike function is based on similar concept.

Theorem 2.7. Let $\lambda \in (-\pi/2, \pi/2)$ and $\{d_n\}$ be sequence of complex numbers with $|d_n| \leq nB/A_n$ for $n \geq 1$ . Let $H(z) = z - \sum_{n=2}^{\infty} |a_n| z^n$ and $G(z) = \sum_{n=1}^{\infty} |b_n| z^n$ be two analytic functions. If $$F_{\epsilon}(z) = z \left(\frac{H(z) + \epsilon G(z)}{z}\right)^{e^{i\lambda}\cos\lambda}$$ is analytic and $\lambda$ -spirallike for each $|\epsilon|=1$ then the harmonic mapping $$f(z) = h(z) + \overline{g(z)} = z + \sum_{n=2}^{\infty} d_n a_n z^n + \sum_{n=1}^{\infty} d_n b_n z^n$$ <span id="page-8-0"></span>is hereditarily $\lambda$ -spiralike.