Bohr's Phenomenon for Some Univalent Harmonic Functions

Theorem 1.1. If $f(z) = \sum_{m=0}^{\infty} a_m z^m$ is a univalent function and $g(z) = \sum_{m=0}^{\infty} b_m z^m \in S(f)$ , then <span id="page-1-0"></span>(2) $$\sum_{m=1}^{\infty} |b_m| r^m \le d(f(0), \partial f(\mathbb{D}))$$ for all $|z| = r \le r_0 = 3 - \sqrt{8} = 0.17157...$ , where $d(f(0), \partial f(\mathbb{D}))$ is the Euclidean distance between f(0) and $\partial f(\mathbb{D})$ , the boundary of $f(\mathbb{D})$ . The value of $r_0$ is sharp for $f(z) = z/(1-z)^2$ , the Koebe function. Further, if f is convex univalent in $\mathbb{D}$ , then $r_0 = 1/3$ . In the recent years, a number of research articles (for example see [1], [7], [8]) are published and many hidden facts of this subject are brought into broad daylight. In particular, Bhowmik and Das [1] successfully extended the Bohr inequalities of type (2) for certain harmonic functions. A complex valued function f(z) = u(x,y) + iv(x,y) of $z = x + iy \in \mathbb{D}$ is said to be harmonic if both u(x,y) and v(x,y) are real harmonic in $\mathbb{D}$ . It is known that such an f can be uniquely represented as $f = h + \overline{g}$ , where h and g are analytic functions in $\mathbb{D}$ with f(0) = h(0). It immediately follows from this representation that f is locally univalent and sense preserving whenever its Jacobian $J_f$ , defined by $J_f(z) := |h'(z)| - |g'(z)|$ , satisfies $J_f(z) > 0$ for all $z \in \mathbb{D}$ ; or equivalently if $h' \neq 0$ in $\mathbb{D}$ and the (second complex) dilatation $w_f$ of f, defined by $w_f(z) = g'(z)/h'(z)$ , satisfies the condition $|w_f(z)| < 1$ in $\mathbb{D}$ . A harmonic function $f = h + \overline{q}$ defined in $\mathbb{D}$ is said to be K-quasiconformal if its dilatation $w_f$ satisfies $|w_f| \leq k, k = (K-1)/(K+1) \in [0,1)$ . In view of the work of Schaubroeck in [11], aforesaid definitions and notations for subordination of analytic functions can be extended to harmonic functions without any change. This lead Bhowmik and Das [1] to extend Theorem 1.1 as under:

Theorem 1.2. Let $f(z) = h(z) + \overline{g(z)} = \sum_{m=0}^{\infty} a_m z^m + \overline{\sum_{m=1}^{\infty} b_m z^m}$ be a sense preserving K-quasiconformal harmonic mapping defined in $\mathbb{D}$ such that h is univalent in $\mathbb{D}$ , and let $f_1(z) = h_1(z) + \overline{g_1(z)} = \sum_{m=0}^{\infty} c_m z^m + \overline{\sum_{m=0}^{\infty} d_m z^m} \in S(f)$ . Then <span id="page-1-1"></span>(3) $$\sum_{m=1}^{\infty} |c_m| r^m + \sum_{m=1}^{\infty} |d_m| r^m \le d(h(0), \partial h(\mathbb{D}))$$ for $|z|=r\leq r_0=(5K+1-\sqrt{8K(3K+1)}/(K+1))$ . This result is sharp for the function $p(z)=z/(1-z)^2+k\overline{z}/(1-z)^2$ , where k=(K-1)/(K+1). Moreover, if we take h to be convex univalent then the inequality in (3) holds for $|z|=r\leq r_0=(K+1)/(5K+1)$ . This result is again sharp for the function $q(z)=z/(1-z)+k\overline{z}/(1-z)$ . In this article, our aim is to establish the Bohr's phenomenon and compute Bohr radius for some subclasses of univalent harmonic functions. We also propose to improvise Theorem 1.1 and 1.2 stated above. We close this section by setting certain notations for subsequent use in this paper. We denote by $S_H$ , the class of univalent harmonic functions f normalized by the conditions f(0) = 0 and $f_z(0) = 1$ . In addition, if $f_{\overline{z}}(0) = 0$ also, then the class is denoted by $S_H^0$ . Further, $K_H^0$ is the usual subclass of $S_H^0$ consisting of convex functions. A domain $\Omega$ is said to be convex in the direction $\theta, 0 \le \theta < \pi$ , if the intersection of the straight line through the origin and the point $e^{i\theta}$ in the complex plane is connected or empty. A function f mapping the open unit disc $\mathbb{D}$ onto such a domain is called convex in direction $\theta$ .

Lemma 2.1. Let $f(z) = \sum_{m=0}^{\infty} a_m z^m$ and $g(z) = \sum_{m=0}^{\infty} b_m z^m$ be two analytic functions in $\mathbb{D}$ and $g \prec f$ . Then $$\sum_{m=0}^{\infty} |b_m| r^m \le \sum_{m=0}^{\infty} |a_m| r^m$$ for $|z| = r \le 1/3$ . Using this lemma, we now improvise Theorem 1.1 by taking univalent analytic function in $\mathbb{D}$ as $f(z) = z + \sum_{m=2}^{\infty} a_m z^m$ . Making use of well known De Brange's theorem: $|a_m| \leq m, m = 2, 3, ...$ , and after some simple calculations, we easily get:

Theorem 2.2. If $f(z) = z + \sum_{m=2}^{\infty} a_m z^m$ is a univalent analytic function in $\mathbb{D}$ and $g(z) = \sum_{m=1}^{\infty} b_m z^m \in S(f)$ , then $$(4) \sum_{m=1}^{\infty} |b_m| r^m \le 1$$ for all $|z| = r \le 1/3$ . In a similar manner, we restate Theorem 1.2 as under;

Theorem 2.3. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=1}^{\infty} b_m z^m}$ be a sense preserving K-quasiconformal harmonic mapping in $\mathbb{D}$ , such that h is analytic univalent in $\mathbb{D}$ . Then <span id="page-2-0"></span>(5) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le 1$$ for $|z|=r\leq r_0=(2K+1-\sqrt{K(3K+2)})/(K+1)$ and it is sharp for $p(z)=z/(1-z)^2+k\overline{z/(1-z)^2}$ . If we take h to be convex univalent then the inequality in (5) holds for $|z|=r\leq r_0=(K+1)/(3K+1)$ and it is sharp for $p(z)=z/(1-z)+k\overline{z/(1-z)}$ , where k=(K-1)/(K+1). Further, let $f_1(z)=h_1(z)+\overline{g_1(z)}=\sum_{m=1}^\infty c_mz^m+\overline{\sum_{m=0}^\infty d_mz^m}\in S(f)$ . Then <span id="page-3-0"></span>(6) $$\sum_{m=1}^{\infty} |c_m| r^m + \sum_{m=1}^{\infty} |d_m| r^m \le 1$$ for $|z| = r \le r_0 = min(1/3, (2K + 1 - \sqrt{K(3K + 2)})/(K + 1))$ . If we take h to be convex univalent then the inequality in (6) holds for $|z| = r \le r_0 = min(1/3, (K + 1)/(3K + 1))$ . In next theorem, we establish Bohr's phenomenon for univalent harmonic functions $f = h + \overline{g} \in S_H$ whose dilatation g'/h' is suitably chosen.

Theorem 2.4. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=1}^{\infty} b_m z^m}$ be a univalent and K-quasiconformal harmonic mapping in $\mathbb{D}$ , where h is analytic univalent in $\mathbb{D}$ and $g'(z)/h'(z) = ke^{i\theta}z^n, k = (K-1)/(K+1) \in (0,1), n \in \mathbb{N}, \theta \in \mathbb{R}$ . Then (7) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le 1$$ for $|z| = r \le r_0$ , where $r_0$ is the only root of the equation (8) $$\frac{(k+1)r}{(1-r)^2} - \frac{2nkr}{(1-r)} - kn^2 log(1-r) = 1$$ in (0,1) and this $r_0$ is best possible one <span id="page-3-1"></span>Letting $k \to 1$ (equivalently, $K \to \infty$ ) we obtain the following result. Corollary 2.5. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=1}^{\infty} b_m z^m}$ be a univalent harmonic mapping in $\mathbb{D}$ , where h is analytic univalent in $\mathbb{D}$ and $g'(z)/h'(z) = e^{i\theta} z^n, n \in \mathbb{N}, \theta \in \mathbb{R}$ . Then (9) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le 1, a_1 = 1$$ for $|z| = r \le r_0$ , where $r_0$ is the only root in (0,1) of the equation $\phi(r) = 0$ , where (10) $$\phi(r) = \frac{2r}{(1-r)^2} - \frac{2nr}{(1-r)} - n^2 \log(1-r) - 1.$$ This $r_0$ is the best possible one. By plotting the graph of $\phi(r)$ w.r.t r for different values of n, we observe that there is only one root of $\phi(r)$ in (0,1) which is the Bohr radius for that value of n in the dilatation function. Figure 1 illustrates the case when n=3  <span id="page-4-0"></span>FIGURE 1. $\phi(r)$ w.r.t r for n=3. and in the following table we have listed values of $r_0$ computed for n = 1, 2, 3 and 4. | n | $r_0$ | |---|--------| | 1 | 0.3485 | | 2 | 0.3121 | | 3 | 0.1794 | | 4 | 0.0959 | We observe that if $n \to \infty$ , then $r_0 \to 0$ . Lemma 2.1 and Theorem 2.4 together lead us to the following result for the subordination class S(f). Corollary 2.6. Let $f_1(z) = h_1(z) + \overline{g_1(z)} = \sum_{m=1}^{\infty} c_m z^m + \overline{\sum_{m=1}^{\infty} d_m z^m} \in S(f)$ where f is as defined in Theorem 2.4. Then (11) $$\sum_{m=1}^{\infty} |c_m| r^m + \sum_{m=1}^{\infty} |d_m| r^m \le 1$$ for $|z| = r \le r_1 = min(1/3, r_0)$ , where $r_0$ is same as obtained in Theorem 2.4. Next theorem shows the existence of Bohr's phenomenon for $f \in S_H$ with dilatation $w_f = (a+z)/(1+az), a \in (-1,1)$ .

Theorem 2.7. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=1}^{\infty} b_m z^m}$ be a univalent harmonic mapping in $\mathbb{D}$ , where h is univalent in $\mathbb{D}$ and $g'(z)/h'(z) = \frac{a+z}{1+az}, a \in (-1,1)$ Then (12) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le 1 + |a|$$ for $|z| = r \le r_0$ , where $r_0 = 0.2291...$ is a unique root lying in (0,1) of $r^3 - 3r^2 + 5r - 1 = 0$ . Remark 2.8. We observe that if we take $g'/h' = \frac{a-z}{1-az}$ , $a \in (-1,1)$ , in Theorem 2.7, then we obtain the same value of $r_0$ . In the following theorem we establish Bohr's phenomenon for univalent harmonic functions convex in one direction.

Theorem 2.9. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=1}^{\infty} b_m z^m}$ be a harmonic mapping in $\mathbb{D}$ , where h is analytic univalent in $\mathbb{D}$ and $h(z) + e^{i\theta}g(z)$ is convex univalent in $\mathbb{D}$ for some $\theta \in \mathbb{R}$ . Then (13) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le 1$$ for $|z| = r \le r_0 = 0.2192...$ We can drop the condition of univalency of h in Theorem 2.9 if we take $b_1 = 0$ .

Theorem 2.10. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=2}^{\infty} b_m z^m} \in S_H^0$ be a harmonic mapping in $\mathbb{D}$ , where $h(z) + e^{i\theta}g(z)$ is convex univalent in $\mathbb{D}$ for some $\theta \in \mathbb{R}$ . Then <span id="page-5-0"></span>(14) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=2}^{\infty} |b_m| r^m \le 1$$ for $|z| = r \le r_0 = 0.3134...$ , where $r_0$ is a unique root in (0,1) of $4r^3 - 9r^2 + 12r - 3 = 0$ . This result is sharp for Koebe function $K(z) = \frac{z - 1/2z^2 + 1/6z^3}{(1-z)^3} + \frac{1/2z^2 + 1/6z^3}{(1-z)^3}$ . Our last theorem gives Bohr radius for convex univalent harmonic functions in $S_H^0$ .

Theorem 2.11. Let $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=2}^{\infty} b_m z^m} \in K_H^0, z \in \mathbb{D}$ . Then (15) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=2}^{\infty} |b_m| r^m \le 1$$ for $|z| = r \le r_0 = (3 - \sqrt{5})/2 = 0.3819...$ This value of $r_0$ is sharp for $L(z) = \frac{1}{2} \left[ \frac{z}{1-z} + \frac{z}{(1-z)^2} + \frac{\overline{z}}{1-z} - \frac{\overline{z}}{(1-z)^2} \right]$ .

Lemma 3.1. Let $h(z) = \sum_{m=0}^{\infty} a_m z^m$ and $g(z) = \sum_{m=0}^{\infty} b_m z^m$ be two holomorphic functions in $\mathbb{D}$ such that h(z) = g(z). Then (16) $$\sum_{m=0}^{\infty} |a_m| r^m = \sum_{m=0}^{\infty} |b_m| r^m$$ for all |z| = r < 1. Proof of Theorem 2.4 From $g'(z) = ke^{i\theta}z^nh'(z)$ , we get $$\sum_{m=1}^{\infty} mb_m z^{m-1} = ke^{i\theta} \sum_{m=1}^{\infty} ma_m z^{n+m-1}, z \in \mathbb{D},$$ where $a_1 = 1$ and on integrating we obtain <span id="page-6-0"></span> $$\sum_{m=1}^{\infty} b_m z^m = k e^{i\theta} \sum_{m=1}^{\infty} \frac{m}{m+n} a_m z^{m+n}, z \in \mathbb{D}.$$ Now, applying Lemma 3.1, we get (17) $$\sum_{m=1}^{\infty} |b_m| r^m = k \sum_{m=1}^{\infty} \frac{m}{m+n} |a_m| r^{m+n}$$ for all |z| = r < 1. Since h is analytic univalent in $\mathbb{D}$ and according to De Brange's theorem, $|a_m| \le m, m = 2, 3, ...$ , therefore, from (17), we have $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m = \sum_{m=1}^{\infty} |a_m| r^m + k \sum_{m=1}^{\infty} \frac{m}{m+n} |a_m| r^{m+n}$$ $$\leq \sum_{m=1}^{\infty} m r^m + k \sum_{m=1}^{\infty} \frac{m^2}{m+n} r^{m+n}$$ $$= \sum_{m=1}^{\infty} m r^m + k \sum_{m=n+1}^{\infty} \frac{(m-n)^2}{m} r^m$$ $$\leq \sum_{m=1}^{\infty} m r^m + k \sum_{m=1}^{\infty} \frac{(m-n)^2}{m} r^m$$ $$= (k+1) \sum_{m=1}^{\infty} m r^m + k n^2 \sum_{m=1}^{\infty} \frac{1}{m} r^m - 2kn \sum_{m=1}^{\infty} r^m$$ $$= \frac{(k+1)r}{(1-r)^2} - kn^2 \log(1-r) - \frac{2knr}{1-r}.$$ Thus $\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \leq 1$ if (18) $$\frac{(k+1)r}{(1-r)^2} - kn^2 \log(1-r) - \frac{2knr}{1-r} \le 1.$$ Now, we need to verify that inequality (18) holds for $r \leq r_0$ , where $r_0$ is the unique root of the equation (8) lying in (0,1). For this let <span id="page-6-1"></span> $$\phi(r) = \frac{(k+1)r}{(1-r)^2} - kn^2 \log(1-r) - \frac{2knr}{1-r} - 1.$$ Then $\phi(r)$ is continuous in $(0,1), \phi(0) = -1 < 0$ and $\lim_{r \to 1^-} \phi(r) > 0$ implies that there is at least one root of $\phi(r) = 0$ in (0,1). But $\phi'(r) > 0$ for all $r \in (0,1), k \in (0,1)$ and for all $n \in \mathbb{N}$ shows that $\phi$ is strictly increasing in  <span id="page-7-0"></span>FIGURE 2. Image of |z| < 0.3485 under $f_0(z)$ . (0,1). Hence $\phi(r) = 0$ has a unique root $r_0$ in (0,1). To see that this $r_0$ is best possible one, we consider $f_0(z) = z/(1-z)^2 + \overline{z/(1-z)^2 - 2z/(1-z) - \log(1-z)}$ . $f_0$ maps |z| < 0.3485... onto the region given in the Figure 2 from which it is evident that $r_0$ is sharp and can not be improved further. Proof of Theorem 2.7 From $g'(z) = \left(\frac{a+z}{1+az}\right)h'(z)$ we obtain (19) $$\sum_{m=1}^{\infty} m b_m z^{m-1} = \left(\frac{a+z}{1+az}\right) \sum_{m=1}^{\infty} m a_m z^{m-1}, z \in \mathbb{D}$$ where $a_1 = 1$ and this gives $$\sum_{m=1}^{\infty}mb_mz^{m-1}+\sum_{m=1}^{\infty}mab_mz^m=\sum_{m=1}^{\infty}maa_mz^{m-1}+\sum_{m=1}^{\infty}ma_mz^m,z\in\mathbb{D}.$$ Thus we have $$\sum_{m=1}^{\infty} m|b_m||z|^{m-1} - \sum_{m=1}^{\infty} m|a||b_m||z|^m \leq \sum_{m=1}^{\infty} m|a||a_m||z|^{m-1} + \sum_{m=1}^{\infty} m|a_m||z|^m.$$ On integrating from 0 to r, we get $$\sum_{m=1}^{\infty}|b_m|r^m-\sum_{m=1}^{\infty}\frac{m}{m+1}|a||b_m|r^{m+1}\leq \sum_{m=1}^{\infty}|a||a_m|r^m+\sum_{m=1}^{\infty}\frac{m}{m+1}|a_m|r^{m+1},$$ and this implies (20) <span id="page-8-0"></span> $$\sum_{m=1}^{\infty} \left( |b_m| - \left(\frac{m-1}{m}\right)|a||b_{m-1}| \right) r^m \le \sum_{m=1}^{\infty} \left( |a||a_m| + \left(\frac{m-1}{m}\right)|a_{m-1}| \right) r^m.$$ Now, we have $$\begin{split} \sum_{m=1}^{\infty} (|a_m| + |b_m|) r^m &= \sum_{m=1}^{\infty} (|a_m| + |b_m|) r^m - \sum_{m=1}^{\infty} \left(\frac{m-1}{m}\right) |a| |b_{m-1}| r^{m-1} + \sum_{m=1}^{\infty} \left(\frac{m-1}{m}\right) |a| |b_{m-1}| r^{m-1} \\ &\leq \sum_{m=1}^{\infty} (|a_m| + |b_m|) r^m - \sum_{m=1}^{\infty} \left(\frac{m-1}{m}\right) |a| |b_{m-1}| r^m + \sum_{m=1}^{\infty} \left(\frac{m-1}{m}\right) |a| |b_{m-1}| r^{m-1} \\ &= \sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} \left(|b_m| - \left(\frac{m-1}{m}\right) |a| |b_{m-1}|\right) r^m + \sum_{m=1}^{\infty} \left(\frac{m-1}{m}\right) |a| |b_{m-1}| r^{m-1}. \end{split}$$ From (20), we get $$\begin{split} \sum_{m=1}^{\infty} (|a_m| + |b_m|) r^m &\leq \sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} \left( |a| |a_m| + \left( \frac{m-1}{m} \right) |a_{m-1}| \right) r^m + \sum_{m=1}^{\infty} \left( \frac{m-1}{m} \right) |a| |b_{m-1}| r^{m-1} \\ &\leq \sum_{m=1}^{\infty} (1+|a|) |a_m| r^m + \sum_{m=1}^{\infty} \left( \frac{m-1}{m} \right) |a_{m-1}| r^{m-1} + \sum_{m=1}^{\infty} \left( \frac{m-1}{m} \right) |b_{m-1}| r^{m-1} \\ &= \sum_{m=1}^{\infty} (1+|a|) |a_m| r^m + \sum_{m=1}^{\infty} \left( \frac{m}{m+1} \right) (|a_m+|b_m|) |r^m. \end{split}$$ Therefore, we get $$\sum_{m=1}^{\infty} \left( \frac{1}{m+1} \right) (|a_m| + |b_m|) r^m \le (1+|a|) \sum_{m=1}^{\infty} |a_m| r^m.$$ Multiplying both sides with r and then differentiating w.r.t r, we get (21) $$\sum_{m=1}^{\infty} (|a_m| + |b_m|) r^m \le (1 + |a|) \sum_{m=1}^{\infty} (m+1) |a_m| r^m.$$ As h is univalent, so $|a_m| \leq m$ by De Branges's Theorem. From (21), we have <span id="page-8-1"></span> $$\sum_{m=1}^{\infty} (|a_m| + |b_m|) r^m \le (1 + |a|) \sum_{m=1}^{\infty} (m+1) m r^m$$ $$= (1 + |a|) \left( \frac{r(1+r)}{(1-r)^3} + \frac{r}{(1-r)^2} \right) \le (1 + |a|)$$ for $r^3 - 3r^2 + 5r - 1 \le 0$ and this happens for $r \le 0.2291...$ Proof of Theorem 2.9 Let $h(z) + e^{i\theta}g(z) = \psi(z)$ , where $\psi(z) = \sum_{m=1}^{\infty} C_m z^m$ is a convex univalent function. So, we have $|a_m + e^{i\theta}b_m| = |C_m| \le 1$ for all $m \in \mathbb{N}$ . This implies $|b_m| \leq 1 + |a_m|, m \in \mathbb{N}$ . We have with $|a_1| = 1$ , $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le \sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} (1 + |a_m|) r^m$$ $$= 2 \sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} r^m.$$ Since h is univalent in $\mathbb{D}$ , so by De Brange's Theorem, we have $|a_m| \leq m$ and hence we get (22) $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=1}^{\infty} |b_m| r^m \le 1$$ for $\frac{2r}{(1-r)^2} + \frac{r}{1-r} \le 1$ i.e. for $2r^2 - 5r + 1 \ge 0$ . This is true for $r \le r_0 = \frac{5-\sqrt{17}}{4} = 0.2192...$ Now to prove Theorem 2.10, we first state the following result of Sheil-Small [12].

Lemma 3.2. If $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \overline{\sum_{m=2}^{\infty} b_m z^m} \in S_H^0$ is convex in one direction, then $$|a_m| \le \frac{(m+1)(2m+1)}{6}$$ $|b_m| \le \frac{(m-1)(2m-1)}{6}$ Proof of Theorem 2.10 $h + e^{i\theta}g$ is convex univalent implies that f is convex in the direction $-\theta/2$ , by the well known result of Clunie and Sheil-Small [4]. Therefore, from Lemma 3.2, we have $$|a_m| \le \frac{(m+1)(2m+1)}{6}$$ $|b_m| \le \frac{(m-1)(2m-1)}{6}$ and so $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=2}^{\infty} |b_m| r^m \le \sum_{m=1}^{\infty} \frac{(m+1)(2m+1)}{6} r^m + \sum_{m=2}^{\infty} \frac{(m-1)(2m-1)}{6} r^m$$ $$= \sum_{m=1}^{\infty} \frac{2m^2 + 1}{3} r^m$$ $$= \frac{2r(r+1)}{3(1-r)^3} + \frac{r}{3(1-r)}$$ $$\le 1$$ if $4r^3 - 9r^2 + 12r - 3 \le 0$ . This inequality holds for $r \le r_0 = 0.3134...$ , where $r_0$ is unique root of $4r^3 - 9r^2 + 12r - 3 = 0$ in (0,1). This result is sharp for $K(z) = \frac{z - 1/2z^2 + 1/6z^3}{(1-z)^3} + \frac{1/2z^2 + 1/6z^3}{(1-z)^3}$ , where K is harmonic mapping in $\mathbb{D}$ , which  FIGURE 3. Image of |z| < 0.3134 under K(z). maps |z| < 0.3134 onto region given in Figure 3. It is clear from Figure 3 that |K(z)| < 1 for |z| < 0.3134... and 0.3134... can not be improved. Hence this $r_0$ is sharp for inequality (14) also. To prove Theorem 2.11, we need following result of Duren [5]. Lemma 3.3. If a harmonic function $f(z) = h(z) + \overline{g(z)} = z + \sum_{m=2}^{\infty} a_m z^m + \sum_{m=2}^{\infty} b_m z^m \in K_H^0, z \in \mathbb{D}$ , then $$|a_m| \le \frac{m+1}{2} \qquad |b_m| \le \frac{m-1}{2}.$$ Proof of Theorem 2.11 In view of Lemma 3.3 $f(z) \in K_H^0$ implies that $$|a_m| \le \frac{m+1}{2} \qquad |b_m| \le \frac{m-1}{2}.$$ This gives for $a_1 = 1$ , $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=2}^{\infty} |b_m| r^m \le \sum_{m=1}^{\infty} \frac{m+1}{2} r^m + \sum_{m=2}^{\infty} \frac{m-1}{2} r^m$$ $$= 2 \sum_{m=1}^{\infty} m r^m$$ $$= \frac{r}{(1-r)^2}$$ < 1. for $r \le r_0 = 0.3819...$ This value of $r_0$ is best possible, as the result is sharp for $L(z) = \frac{1}{2} \left[ \frac{z}{1-z} + \frac{z}{(1-z)^2} + \frac{\overline{z}}{1-z} - \frac{\overline{z}}{(1-z)^2} \right]$ . For L(z) we have $$\sum_{m=1}^{\infty} |a_m| r^m + \sum_{m=2}^{\infty} |b_m| r^m = \sum_{m=1}^{\infty} \left| \frac{m+1}{2} \right| r^m + \sum_{m=2}^{\infty} \left| \frac{1-m}{2} \right| r^m$$ $$= \sum_{m=1}^{\infty} m r^m$$ $$= \frac{r}{(1-r)^2}$$ for $r \leq 0.3819...$ Thus $r_0$ is sharp.