Univalent Harmonic Ring Mappings Vanishing on the Interior Boundary

Lemma 2.1. Let f be any harmonic function on $V_R$ (for some R > 1) satisfying (1.2). Then f has a harmonic continuation across $\partial U = \{z : |z| = 1\}$ which is of the form (2.1) $$f(z) = h(z) - h(\frac{1}{z}) + 2A \ln|z|, \quad A \in \mathbb{C},$$ where $h(z) = \sum_{j \in \mathbb{Z}} a_j z^j \in H(\frac{1}{R} < |z| < R)$ . Observe that $f(\frac{1}{z}) = -f(z)$ on $\{z : \frac{1}{R} < |z| < R\}$ . Conversely, each harmonic mapping on $\partial U$ satisfying (2.1) has the property (1.2). PROOF. Since f is harmonic on $V_R$ , f admits the representation $$(2.2) f(z) = h(z) + \overline{g(z)} + 2A \ln |z|,$$ where $A \in \mathbb{C}$ , $h(z) = \sum_{j \in \mathbb{Z}} a_j z^j \in H(V_R)$ , $g(z) = \sum_{j \in \mathbb{Z}} b_j z^j \in H(V_R)$ . This implies that $$\frac{1}{2\pi} \int_0^{2\pi} |f(e^{it})|^2 dt = \sum_{i \in \mathcal{I}} |a_i + \bar{b}_{-i}|^2 = 0,$$ and therefore h and g admit an analytic continuation onto $\{z: \frac{1}{R} < |z| < R\}$ satisfying (2.3) $$g(\frac{1}{\overline{z}}) = -\overline{h(z)} \text{ and } h(\frac{1}{\overline{z}}) = -\overline{g(z)}.$$ In the next Lemma we consider again harmonic mappings having the properties stated in Lemma 2.1. We give a necessary and sufficient condition in order that f is orientation-preserving on an exterior neighbourhood of $\bar{U}$ .

Lemma 2.2. Let $$f(z) = h(z) - h(\frac{1}{\overline{z}}) + 2A \ln|z|, \quad A \in \mathbb{C}$$ be harmonic on $\{z : |z| = 1\}$ , $f \not\equiv const.$ There is an exterior neighbourhood $V_R$ of $\bar{U}$ such that f is orientation-preserving on $V_R$ if and only if there exists a constant k > 0 such that $\psi(z) = zh'(z) + A$ satisfies $$(2.4) +\infty > \operatorname{Re} \frac{e^{it}\psi'(e^{it})}{\psi(e^{it})} \ge k > 0$$ whenever $\psi(e^{it}) \neq 0$ . PROOF. By Lemma 2.1. $h \in H(\frac{1}{R_1} < |z| < R_1)$ for some $R_1 > 1$ . Therefore, the dilatation function (2.5) $$a(z) = \frac{\overline{f_{\overline{z}}}(z)}{f_{\overline{z}}(z)} = \frac{\overline{\psi(\frac{1}{\overline{z}})}}{\psi(z)}$$ is meromorphic on $\{z:\frac{1}{R_1}<|z|< R_1\}$ . We may choose $R_1$ so close to one such that the only possible zeros of $\psi$ in $\{z:\frac{1}{R_1}<|z|< R_1\}$ are on $\partial U$ . Furthermore, we have $$|a(e^{it})| = \left|\frac{\overline{\psi(e^{it})}}{\psi(e^{it})}\right| = 1$$ whenever $\psi(e^{it}) \neq 0$ . Therefore, the zeros of $\psi$ on $\partial U$ are removable singularities of the function a(z) and we conclude that (2.6) $$a \in H(\frac{1}{R_1} < |z| < R_1) \text{ and } |a(e^{it})| \equiv 1 \text{ for all } t \in [0, 2\pi].$$ Moreover, by the reflection principle, we have $$a(z) = \frac{1}{\overline{a(\frac{1}{z})}}.$$ (a) Suppose now that f is orientation-preserving on $V_R$ . Then, by (2.7), |a(z)| < 1 on $V_R$ and |a(z)| > 1 on $\{z : \frac{1}{R} < |z| < 1\}$ , from which follows that $a'(e^{it}) \neq 0$ for all $t \in [0, 2\pi]$ and therefore $\frac{za'(z)}{a(z)}$ is a nonvanishing analytic function on $\partial U$ . We have, for all $z \in \partial U$ , $$\frac{za'(z)}{a(z)} = \operatorname{Re} \frac{za'(z)}{a(z)} = \frac{1}{2} \frac{\partial |a|^2}{\partial |z|} < 0,$$ which implies that there is a constant c > 0 such that (2.9) $$\frac{za'(z)}{a(z)} = \operatorname{Re} \frac{za'(z)}{a(z)} = \frac{1}{2} \frac{\partial |a|^2}{\partial |z|} \le -c < 0 \text{ for all } z \in \partial U.$$ Finally, we have from (2.5), whenever $\psi(z) \neq 0$ , (2.10) $$\frac{za'(z)}{a(z)} = -\left[\frac{z\psi'(z)}{\psi(z)} + \frac{1}{z} \cdot \frac{\overline{\psi'(\frac{1}{z})}}{\overline{\psi(\frac{1}{z})}}\right],$$ which implies that (2.11) $$\infty > \operatorname{Re} \frac{e^{it}\psi'(e^{it})}{\psi(e^{it})} = -\frac{1}{2}\frac{e^{it}a'(e^{it})}{a(e^{it})} \ge \frac{c}{2} = k > 0,$$ for all $e^{it}$ for which $\psi(e^{it}) \neq 0$ and (2.4) holds. (b) Now, suppose that the function $\psi$ satisfies (2.4). Then, by (2.11) and (2.9), we have in the case of $\psi(e^{it}) \neq 0$ (2.12) $$\frac{e^{it}a'(e^{it})}{a(e^{it})} = \frac{1}{2} \frac{\partial |a|^2}{\partial |z|} (e^{it}) \le -2k < 0.$$ Since, by (2.6), $a \in H(\frac{1}{R_1} < |z| < R_1)$ and $|a(e^{it})| \equiv 1$ , we conclude that (2.12) holds for all $t \in [0, 2\pi]$ . Put $R \in (1, R_1)$ such that $\operatorname{Re} \frac{za'(z)}{a(z)} \leq -k$ on $\{z : \frac{1}{R} < |z| < R\}$ . Then, the relation (2.13) $$\frac{1}{2} \frac{\partial |a|^2}{\partial |z|} = \frac{|a|^2}{|z|} \operatorname{Re} \frac{za'(z)}{a(z)}$$ shows that |a(z)| < 1 on $V_R$ , and Lemma 2.2 is proved. For the completeness we give a short proof of the following lemma.

Lemma 2.3. Let $\Phi$ be in $H(\frac{1}{R} < |z| < R)$ for some R > 1, such that $\Phi$ is real on $\partial U$ and $\frac{1}{2\pi} \oint_{|z|=\rho} d \arg z \Phi' = 1$ for all $\rho \in (1,R)$ . Then $\Phi'$ has exactly two zeros on $\partial U$ which are of order one. PROOF. By the reflection principle, we have $$\overline{\Phi(\frac{1}{\bar{z}})} = \Phi(z) \text{ and } -\frac{1}{z}\overline{\Phi'(\frac{1}{\bar{z}})} = z\Phi'(z).$$ Since $\Phi(\partial U)$ is a bounded real interval, there exists an $e^{i\beta}$ and an $e^{i\gamma}$ , $e^{i\beta} \neq e^{i\gamma}$ , such that $\Phi'(e^{i\beta}) = \Phi'(e^{i\gamma}) = 0$ . Applying the argument principle, we get for $\rho \in (1, R)$ $$\frac{1}{2\pi} \oint_{|z|=\rho} d \arg z \Phi' - \frac{1}{2\pi} \oint_{|z|=\frac{1}{\rho}} d \arg z \Phi' = 2 \cdot \frac{1}{2\pi} \oint_{|z|=\rho} d \arg z \Phi' = 2$$ and the result follows. The next lemma gives the important relation of some auxiliary analytic function $\Phi$ to a given harmonic mapping f.

Lemma 2.4. Let $$f(z) = h(z) - h(\frac{1}{\overline{z}}) + 2A \ln|z|, \quad A \in \mathbb{C}$$ be harmonic on $\{z: \frac{1}{R} < |z| < R\}$ and suppose that f is orientation-preserving on $V_R$ . Put $$e^{i\alpha} = \begin{cases} \bar{A}/|A| & \text{if } A \neq 0\\ 1 & \text{if } A = 0, \end{cases}$$ and define $$\Phi(z) = e^{i\alpha}h(z) + e^{-i\alpha}\overline{h(\frac{1}{z})}.$$ Then f is univalent on $V_{R_1}$ for some $R_1 \in (1, R)$ if and only if $\Phi$ is univalent on $V_{R_2}$ for some $R_2 \in (1, R)$ . PROOF. Observe that $\Phi \in H(\frac{1}{R} < |z| < R)$ and that $\Phi$ is real on $\partial U$ . Furthermore we have $$z\Phi'(z) = ze^{i\alpha} \left[ h'(z) - e^{-2i\alpha} \overline{h'(\frac{1}{z})} \frac{1}{z^2} \right] + |A| - |A|$$ $$= e^{i\alpha} \left[ zh'(z) + A \right] - e^{-i\alpha} \left[ \frac{1}{z} \overline{h'(\frac{1}{z})} + \overline{A} \right]$$ $$= e^{i\alpha} \psi(z) - e^{-i\alpha} \overline{\psi(\frac{1}{z})}$$ $$= e^{i\alpha} \psi(z) [1 - a(z)e^{-2i\alpha}],$$ where $\psi(z) = zf_z(z) = zh'(z) + A$ and $a(z) = \overline{\psi(\frac{1}{z})}/\psi(z)$ . Since f is orientation-preserving on $V_R$ , |a(z)| < 1 on $V_R$ . Therefore $\Phi'$ and $f_z$ vanishes simultaneously. In other words, $\Phi$ is locally univalent on $V_R$ if and only if f is locally univalent on $V_R$ . (a) Suppose first that $\Phi$ is univalent in $V_{R_2}$ for some $R_2 \in (1, R)$ . Put w = u + iv = f(z) and $\zeta = \xi + i\eta = \Phi(z)$ . Define (2.15) $$w = F(\zeta) = e^{i\alpha} f \circ \Phi^{-1}(\zeta)$$ $$= \zeta - 2 \operatorname{Re} \left\{ e^{i\alpha} h\left(\frac{1}{\bar{z}}\right) \right\} + 2|A| \ln|z|$$ $$= \zeta - 2 \operatorname{Re} \left\{ e^{i\alpha} h\left(\frac{1}{\Phi^{-1}(\zeta)}\right) \right\} + 2|A| \ln \Phi^{-1}(\zeta)|$$ $$= \zeta - q(\zeta),$$ where q is real on $\Phi(V_{R_2})$ . Put $J = \Phi(\partial U)$ and let $\gamma$ be a convex closed Jordan around $J, \gamma \in \Phi(V_{R_2})$ . Denote by G the doubly connected domain bounded by $\gamma$ and J. Then, F is locally univalent on G satisfying $$(2.16) v(\zeta) = \operatorname{Im} F(\zeta) = \operatorname{Im} \zeta = \eta,$$ and (2.17) $$\frac{\partial u}{\partial \xi} = \operatorname{Re}\left\{\frac{1 + a(\Phi^{-1}(\zeta))}{1 - a(\Phi^{-1}(\zeta))}\right\} > 0 \text{ on } G.$$ It follows that F is univalent on $G\setminus\{\zeta: \operatorname{Im}\zeta=0\}$ . Let $\xi_1$ and $\xi_2$ belong to $G\cap\{\zeta: \operatorname{Im}\zeta=0\}$ such that, $\xi_1<\xi_2$ . Note that $F(\xi_1)$ and $F(\xi_2)$ are real. We will show that $F(\xi_1)< F(\xi_2)$ , from which it follows that F is univalent in G. We can find closed intervals $\Gamma_n=[\xi_1+i\eta_n,\xi_2+i\eta_n]$ in G with $\eta_n\to 0$ as $n\to\infty$ . Since, by (2.17), Re F is strictly increasing on $\Gamma_n$ , we conclude that $F(\xi_1)\leq F(\xi_2)$ . On the other hand, $\frac{\partial \operatorname{Re} F}{\partial \xi}(\xi_i)>0$ , i=1,2, implies that $F(\xi_1)< F(\xi_2)$ . Choose now $R_1 \in (1, R)$ such that $V_{R_1} \subset \Phi^{-1}(G)$ . Then f is univalent on $V_{R_1}$ and in one direction Lemma 2.4 is proved. (b) Suppose now that f is univalent on $V_{R_1}$ . Then, by the same reasoning, we get that $\Phi$ is univalent on an exterior neighbourhood $V_{R_2}$ of $\bar{U}$ . Indeed, $$\zeta = F^{-1}(w) = \Phi \circ f^{-1}(e^{-i\alpha}w) = w + q_1(w),$$ where $q_1 = 2 \operatorname{Re} \left\{ e^{i\alpha} h \left( 1/\overline{f^{-1}(e^{-i\alpha}w)} \right) \right\} - 2|A| \ln |f^{-1}(e^{-i\alpha}w)|$ is real on $f(V_{R_1})$ . Define $G_1 = \left\{ w : 0 < |w| < \gamma \right\} \subset f(V_{R_2})$ . Then, from the equality $$\eta = \operatorname{Im} F^{-1}(w) \equiv \operatorname{Im} w = v$$ and the condition $\frac{\partial u}{\partial \xi} > 0$ , it follows that $\frac{\partial \xi}{\partial u} > 0$ . The rest of the proof goes as in (a) by reversing the role of w = u + iv and $\zeta = \xi + i\eta$ and by replacing F by $F^{-1}$ . The next lemma is known [7]. For completeness we give a short proof of it.

Lemma 2.5. Let F be analytic on $A = \{z : r_1 < |z| < r_2\}, 0 < r_1 < r_2$ , such that $$(2.18) 0 < \operatorname{Re} \frac{zF'(z)}{F(z)} < \infty \text{ on } A.$$ Then the following statements are equivalent: - (i) F is univalent on A; - (ii) $\frac{1}{2\pi} \oint_{|z|=\rho} d \arg F = 1$ , for some $\rho \in (r_1, r_2)$ ; - (iii) $p(z) = \frac{zF'(z)}{F(z)} = 1 + \sum_{j \in \mathbb{Z} \setminus \{0\}} p_j z^j \in H(A)$ . PROOF. (a) The fact (ii)⇔(iii) follows from the relation $$\frac{1}{2\pi} \oint_{|z|=\rho} d \arg F = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{zF'(z)}{F(z)} \cdot \frac{dz}{z} = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{p(z)}{z} dz.$$ - (b) The implication (i) $\Rightarrow$ (ii) is trivial since (2.18) excludes the case $\frac{1}{2\pi} \int_{|z|=\rho} d \arg F = -1$ . - (c) Going to prove (ii) $\Rightarrow$ (i), observe first that F and F' do not vanish on A. Indeed, if F has a zero of order m at $z_0 \in A$ , then, by (2.18), F' also has to have a zero of order m at $z_0$ which is impossible. We conclude therefore that F is locally univalent on A and that $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg z F' = \frac{1}{2\pi} \oint_{|z|=\rho} d\arg F = 1$$ for all $\rho \in (r_1, r_2)$ . Furthermore, using (2.18) and (ii), we conclude that for each $\rho \in (r_1, r_2)$ F is univalent on $\{z: |z| = \rho\}$ and that $\Gamma_{\rho} = \{F(\rho^{it}): 0 \le t \le 2\pi\}$ is a simple closed analytic and strictly starlike curve which winds once around the origin. Moreover, for each $r_1 < \rho_1 < \rho_2 < r_2$ we have $\Gamma_{\rho_1} \cap \Gamma_{\rho_2} = \emptyset$ . In fact, F is analytic and hence bounded on $\{z: \rho_1 \le |z| \le \rho_2\}$ and the image of it has to be a domain. Therefore, F is univalent on $\{z: \rho_1 \le |z| \le \rho_2\}$ . This holds for all $r_1 < \rho_1 < \rho_2 < r_2$ and hence F is univalent on A. We will close this section with the following lemma.

Lemma 2.6. Let F be in H(|z| = 1) such that - (i) $|F(e^{it})| \equiv 1 \text{ and } F'(e^{it}) \neq 0 \text{ for all } t \in [0, 2\pi];$ - (ii) $\frac{1}{2\pi} \int_{|z|=1}^{\infty} d \arg F = 1$ . Then there exists an $R_1 > 1$ such that F is univalent on $$\{z: \frac{1}{R_1} < |z| < R_1\}.$$ PROOF. The fact that $\frac{zF'(z)}{F(z)} \in \mathbb{R} \setminus \{0\}$ on $\partial U$ and that $$\frac{1}{2\pi} \oint_{|z|=1} d \arg z F' = \frac{1}{2\pi} \oint_{|z|=1} d \arg F = 1,$$ implies that there is a k > 0 such that y^- = Re y& > k > 0 and is finite on dU. Hence, there is an R\ > 1 such that F is analytic and 0 < Re ~^- < ooon {z : Y < \z\ < R\}. Applying Lemma 2.5, we conclude that Fis univalent on {z : £<N<\*i}-

Theorem 3.1. Let $$f(z) = h(z) + \overline{g(z)} + 2A \ln |z|, \quad A \in \mathbb{C},$$ be a harmonic mapping defined on the unit circle dU — {z : |z| = 1}. Put $$\psi(z) = zh'(z) + A = zf_z(z).$$ Then there exists an exterior neighbourhood VR of Û such thatf is univalent and positively oriented on VR and lim^^j f(z) = 0, if and only if the following conditions are z-V^f satisfied: - (a) h and g admit an analytic continuation across dU such that h(\) —g(z)for all z, )i < \z\ < R; - (b) if) has at most one zero ondU which is of order one; - (c) there exists an exterior neighbourhood V^ of ^Û and a constant k > 0 such that (3.1) $$\infty > \operatorname{Re} \frac{z\psi'(z)}{\psi(z)} \ge k > 0 \text{ on } V_{\hat{R}},$$ and (3.2) $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg \psi = 1 \text{ for all } \rho \in (1, \hat{R}).$$ Let us give some remarks about this Theorem. REMARK 3.1. The statement (c) says that xjj(z) = zfz(z) maps {z : |z| = r} , 1 < r < R, univalently onto an analytic strictly starlike Jordan curve with respect to the origin. For r — 1, xpidU) is still an analytic curve, but it may pass through the origin as the following example shows: the mapping $$f(z) = z - \frac{1}{\bar{z}} + 2\ln|z|$$ is univalent, harmonic and orientation-preserving onC\U and 1/; (z) = \+z vanishes at z = — 1. Observe that this zero is of order one. REMARK 3.2. The next example shows that (3.2) is essential. Let $\psi(z) = z + z^2$ . Then $\psi$ satisfies (b) and (3.1) with $k = \frac{3}{2}$ but not (3.2). The function $h(z) = z + z^2/2$ is analytic in $\mathbb{C}$ , but $$f(z) = z + \frac{z^2}{2} - \frac{1}{\bar{z}} - \frac{1}{2\bar{z}^2}$$ is not univalent on any circle $\{z:|z|=r\}$ , r>1. Indeed, we have $$f(re^{it}) = \frac{r^2 - 1}{r} [e^{it} + \frac{r^2 + 1}{2r} e^{2it}].$$ Putting $d = \frac{r^2+1}{r} > 1$ we see that the equation $\eta_1 + d\eta_1^2 = \eta_2 + d\eta_2^2$ has a solution for $|\eta_1| = |\eta_2| = 1$ , $\eta_1 \neq \eta_2$ . REMARK 3.3. The condition (3.1) cannot be replaced by $$\operatorname{Re} \frac{e^{it}\psi'(e^{it})}{\psi(e^{it})} \geq 0.$$ Indeed, consider the function $\psi(z) = z + \frac{1}{2}z^2$ . Then we have $$a(z) = \frac{1}{z^3} \cdot \frac{z + \frac{1}{2}}{1 + \frac{1}{2}z},$$ and, for $\varepsilon$ positive close to zero, $$a(-(1+\varepsilon)) = 1 + 2\varepsilon^3 + 0(\varepsilon^4) > 1.$$ Therefore f is not orientation-preserving on $V_R$ . REMARK 3.4. Put $p(z) = \frac{z\psi'(z)}{\psi(z)}$ . By Lemma 2.5 the statement (c) is equivalent to: (c') There exists an exterior neighbourhood $V_{\hat{R}}$ of $\bar{U}$ and a constant k > 0 such that $$p \in H(V_{\hat{R}}), \ p(z) = 1 + \sum_{j \in \mathbb{Z} \setminus \{0\}} p_j z^j \text{ and } \operatorname{Re} p(z) \ge k \text{ on } V_{\hat{R}}.$$ REMARK 3.5. The statements (b) can be replaced by (b') $\psi'(e^{it}) \neq 0$ for all $t \in [0, 2\pi]$ . In fact we have: $(\alpha)$ (a), (b) and (c) imply (a), (b') and (c). If $e^{it_0}$ is a zero of order 1 of $\psi$ then evidently $\psi'(e^{it_0}) \neq 0$ . Let $\psi(e^{it}) \neq 0$ . By (a), $\psi$ is analytic and hence bounded on $\partial U$ and by (c) we have $0 < k \le \operatorname{Re} \frac{e^{it}\psi'(e^{it})}{\psi(e^{it})} < \infty$ , which implies that $\psi'(e^{it}) \neq 0$ for all $t \in [0, 2\pi]$ . $(\beta)$ (a), (b'), (c) imply (a), (b), (c). Note that $\psi \in H(\partial U)$ and that all zeros of $\psi$ on $\partial U$ are of order one. On the other hand, if $\psi(e^{it}) \neq 0$ , then we conclude by (c) that $$(3.3) 0 < k \le \operatorname{Re} \frac{e^{it} \psi'(e^{it})}{\psi(e^{it})} < \infty.$$ Furthermore, for $\rho \in (1, \hat{R})$ , we have $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg z\psi' = \frac{1}{2\pi} \oint_{|z|=\rho} d\arg \psi = 1,$$ and, from (b'), we conclude that $\frac{1}{2\pi} \oint_{|z|=1} d \arg z \psi' = 1$ . Suppose that $\eta_k = e^{it_k}$ , $1 \le k \le N$ , are the zeros of $\psi$ on $\partial U$ . Then, by (3.3), $$\frac{1}{2\pi} \int_{(t_k,t_{k+1})} d\arg(z\psi')(e^{it}) = \frac{1}{2\pi} \int_{(\eta_k,\eta_{k+1})} d\arg z\psi' > \frac{1}{2}, \quad \eta_{N+1} = \eta_1,$$ and therefore $$1 = \frac{1}{2\pi} \oint_{|z|=1} d \arg z \psi'$$ = $\frac{1}{2\pi} \sum_{k=1}^{N} \oint_{(\eta_k, \eta_{k+1})} d \arg z \psi' + \frac{1}{2\pi} \sum_{k=1}^{N} \Delta [\arg(z \psi')(\eta_k)].$ Since the zeros of $\psi$ are of order one, we get $\Delta \arg(z\psi')(\eta_k) = 0$ . Therefore N = 0 or 1. Together with Remarks 4 and 5, Theorem 3.1 can be restated as follows.

Theorem 3.2. Let $$f(z) = h(z) + \overline{g(z)} + 2A \ln |z|, \quad A \in \mathbb{C},$$ be a harmonic mapping defined on the unit circle $\partial U = \{z : |z| = 1\}$ . Put $\psi(z) = zf_z(z) = zh'(z) + A$ and $p(z) = \frac{z\psi'(z)}{\psi(z)}$ . Then there exists an exterior neighbourhood $V_R$ of $\bar{U}$ such that f is univalent, positively oriented on $V_R$ and $\lim_{|z|>1} f(z) \equiv 0$ , if and only if the following conditions are satisfied: - (a) h and g admit an analytic continuation across $\partial U$ such that $h(\frac{1}{z}) = \overline{-g(z)}$ , $z \in \{z : \frac{1}{R} < |z| < R\}$ ; - $(b') \ \psi'(e^{it}) \neq 0 \ for \ all \ t \in [0, 2\pi];$ - (c') there is an exterior neighbourhood $V_{\hat{R}}$ and a constant k > 0 such that $p \in H(V_{\hat{R}})$ , is of the form $p(z) = 1 + \sum_{j \in \mathbb{Z} \setminus \{0\}} p_j z^j$ and satisfies the condition $\operatorname{Re} p(z) \ge k > 0$ on $V_{\hat{R}}$ . - 4. Proof of Theorem 3.1. Necessity: suppose that f is a univalent, positively oriented and harmonic mapping defined on an exterior neighbourhood $V_R$ of $\bar{U}$ such that $\lim_{|z|>1} f(z) \equiv 0$ . The statement (a) has been already proved in Lemma 2.1. Let us prove (3.2) of the statement (c). Since f is univalent and positively oriented we have for $\rho \in (1, R)$ : $$1 = \frac{1}{2\pi} \oint_{|z|=\rho} \arg[izf_z - i\bar{z}f_{\bar{z}}].$$ Recall that $f_z \neq 0$ on $V_R$ and, since f is orientation-preserving, we have $$a(z) = \frac{\overline{f_z}(z)}{f_z(z)} \in H(V_R) \text{ and } |a(z)| < 1, \quad z \in V_R.$$ Therefore, we get for p G (1,/?) (4.1) $$1 = \frac{1}{2\pi} \oint_{|z|=\rho} d \arg[izf_z - i\bar{z}f_{\bar{z}}] = \frac{1}{2\pi} \oint_{|z|=\rho} d \arg zf_z [1 - \bar{a}\frac{\bar{z}}{z}\frac{\bar{f}_z}{f_z}] \\ = \frac{1}{2\pi} \oint_{|z|=\rho} d \arg zf_z = \frac{1}{2\pi} \oint_{|z|=\rho} d \arg \psi,$$ and (3.2) is shown. Next we show (b). We apply Lemma 2.2. By (2.6), we have \a(elt)\ = 1 for all t G [0,27r], and (2.9) implies that (4.2) $$\frac{1}{2\pi} \oint_{|z|=1} d \arg a = \frac{1}{2\pi i} \oint_{|z|=1} \frac{z a'(z)}{a(z)} \cdot \frac{dz}{z} < 0.$$ Now, by Lemma 2.4, we know that O is univalent on VRx for some R\ G (1, R) and, from the formula (2.14) $$z\mathbf{\Phi}'(z) = e^{i\alpha}\psi(z)\left(1 - a(z)e^{-2i\alpha}\right) = e^{i\alpha}\psi(z) - e^{-i\alpha}\overline{\psi(\frac{1}{z})},$$ we obtain $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg z \Phi' = \frac{1}{2\pi} \oint_{|z|=\rho} d\arg \psi = 1 \text{ for all } \rho \in (1, R_1).$$ Lemma 2.3 implies that the function O' has exactly two zeros on dU which are of order one. This information together with (4.2) leads to the conclusion that either $$\frac{1}{2\pi} \oint_{|z|=1} d \arg a = -2 \text{ and } 0 \not\in \psi(\partial U)$$ or $$\frac{1}{2\pi} \oint_{|z|=1} d\arg a = -1$$ and then V> has exactly one zero of order one on dU. Therefore (b) has been established. It remains to show the statement (3.1 ). If xjj does not vanish on d U then, by Lemma 2.2, there is a k\ > 0 such that $$\operatorname{Re} \frac{z\psi'(z)}{\psi(z)} \ge k_1 \text{ on } \partial U.$$ Choose R such that Re ^ ^ > ifc = kx/ 2 on V^ and (3.1) is shown. Suppose now that ^ ( ^ ) = 0, ^ V⁷ \ ) ^ 0 and 1/; (e'⁷ ) ^ 0 on 3\7\ { }. By Lemma 2.2 we have $$0 < k \le \operatorname{Re} \frac{z\psi'(z)}{\psi(z)} < \infty \text{ on } \partial U \setminus \{e^{i\beta}\}$$ and therefore ijj'ie¹ \) =/=• 0 for all t G [0, 2TT]* (by the same reasoning as in the first part of Remark 3.5). Consider now the function $$m(z) = \frac{\psi(z)}{z\psi'(z)}.$$ Then m G H(dU) and satisfies the condition $$\left| m(e^{it}) - \frac{1}{2k} \right| \le \frac{1}{2k}.$$ Let K = { z : \z — e lf3\ < ft }, where ft > 0 is so small that ra(z) is analytic and univalent on K and such that either m(K Pi { z : | z| < 1} ) or m(^ Pi { z : | z\ > 1} ) is in the disk { w : | w — ^ | < } (this is possible since m'{e'^ ) ^ 0). The condition e ^l@ m'(e¹ ^ ) = 1 implies that the second case holds. Therefore, we have oo > Re ^ > k on [dU\ { e i(5} ] U [KH { z : | z\ > 1} ]. Hence there exists an exterior neighbourhood of Ù U K and hence an exterior neighbourhood VR2,R2 e (1,/?), such that Re ^ > k/ 2. Therefore (3.1) holds, and the proof of the necessity of the statements (a), (b), (c) is finished. Sufficiency: we now show the sufficiency of the statements (a), (b) and (c). Let h G H(dU) and suppose that ip (z) — zh'(z) + A, A G C, satisfies the statements (b) and (c) of Theorem 3.1, i.e. there is an R > 1 such that h G H(l < \z\ < R), oo > Re ^- > k > 0 on V^, 27§[z\=p dargil) = 1 for p G (1,\$) and ijj has at most one zero on dU which is of order 1. Put $$e^{i\alpha} = \left\{ \begin{array}{ll} \bar{A}/\left|A\right| & \text{if } A \neq 0 \ 1 & \text{if } A = 0 \end{array} \right.$$ and define: $$\Phi(z) = e^{i\alpha} h(z) + e^{-i\alpha} \overline{h(\frac{1}{\overline{z}})}, \quad z \in V_{\hat{R}},$$ $$f(z) = h(z) - h(\frac{1}{\overline{z}}) + 2A \ln|z|, \quad z \in V_{\hat{R}}.$$ Evidently,/ is harmonic on { z : A < \z\ < R} and/(z) = 0 on dU. First, we observe that/ is orientation preserving on an exterior neighbourhood of 0. This follows from the fact that Re ^0 p > k > 0 for all t for which ^{elt) ^ 0 and from Lemma 2.2. Call this exterior neighbourhood VRl ,R\ G (1, R). Next, we will show that/ is univalent on VR for some R G (l,R\). Indeed, since by (2.14) zO'(z) = e iai;(z)(l-a(z)e-2ia), the condition (3.2) implies that for all p G (l,/?i) $$\frac{1}{2\pi} \oint_{|z|=\rho} d \arg z \Phi' = \frac{1}{2\pi} \oint_{|z|=\rho} d \arg \psi = 1.$$ By Lemma 2.3, we conclude that <t>' has exactly two zeros of order one on dU; call thern^,^ ⁷ , ^ ^ j \ Denote by J the bounded real interval $$\Phi(\partial U) = [\Phi(e^{i\beta}), \Phi(e^{i\gamma})].$$ Next, define the function $$\zeta(s) = \frac{\Phi(e^{i\gamma}) - \Phi(e^{i\beta})}{4}(s + \frac{1}{s}) + \frac{\Phi(e^{i\gamma}) + \Phi(e^{i\beta})}{2},$$ and let $s = q(\zeta)$ be the univalent inverse function of $\zeta(s)$ which maps the exterior of $(\mathbb{C} \setminus J)$ onto the exterior of the unit disk. Put $Q = q \circ \Phi$ . Then Q satisfies the conditions of Lemma 2.6 and we conclude that there is an $R_2 \in (1, R_1)$ , such that Q is univalent. Hence $\Phi$ is univalent on $V_{R_2}$ . Finally, Lemma 2.4 shows that f is univalent on some $V_R$ , $R \in (1, R_2)$ . It remains to show that f is positively oriented. We already have seen that f is orientation-preserving. By (3.2), we have for $\rho \in (1, R)$ : $$\begin{split} \frac{1}{2\pi} \oint_{|z|=\rho} d \arg(izf_z - i\bar{z}f_{\bar{z}}) &= \frac{1}{2\pi} \oint_{|z|=\rho} d \arg \psi \left(1 - a(z) \frac{\bar{z}\overline{f_z}}{zf_z}\right) \\ &= \frac{1}{2\pi} \oint_{|z|=\rho} d \arg \psi = 1. \end{split}$$ Since f is univalent, $f(|z| = \rho)$ , $1 < \rho < R$ , are disjoint positively oriented Jordan curves. It remains to show, that they wind around the origin. Since f is harmonic on $\partial U$ and $f(\partial U) \equiv 0$ , $f_{\rho}(t) = f(\rho^{it})$ converges uniformly to $f_1 \equiv 0$ on $[0, 2\pi]$ as $\rho \downarrow 1$ . Using the fact that f is univalent on $V_R$ we conclude that $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg f = 1 \text{ for all } \rho \in (1,R)$$ and Theorem 3.1 is proved. 5. On the region of values of A. Let $h \in H(\partial U)$ and consider the harmonic mapping $$f(z) = h(z) - h(\frac{1}{z}) + 2A \ln |z|, \quad A \in \mathbb{C}.$$ Denote by $E_h$ the set of all $A \in \mathbb{C}$ for which f is univalent and positively oriented on $V_R$ for some R > 1 (which may depend on A). Evidently $E_h$ can be empty as the example $h(z) = z + \frac{1}{2}z^2$ or $h(z) = z^2$ shows. We have the following result.$

Theorem 5.1. Let $h \in H(\partial U)$ . If $E_h$ is nonempty, then we have the following properties: - (a) $E_h$ is a convex set. - (b) If, in addition, zh'(U) is a convex set, then the bounded component G of $\mathbb{C}\setminus[zh'(\partial U)]$ belongs to $-E_h$ . PROOF. (a) Let $A_1$ and $A_2$ be in $E_h$ . Since $E_h$ is non-empty, we conclude from Theorem 3.1 and Remark 3.5 that (a), (b') and (c) are satisfied. Observe that (a) and (b') are independent of A. Therefore we have only to consider the relation (c). There is a k > 0 and an R > 1 such that for i = 1, 2 $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg(zh' + A_i) = 1, \quad \rho \in (1, R)$$ and $$\operatorname{Re} p_i(z) = \operatorname{Re} \frac{\left(zh'(z)\right)'}{zh'(z) + A_i} \ge k > 0 \text{ on } V_R.$$ Put $A = \lambda A_1 + (1 - \lambda)A_2$ for some $\lambda \in (0, 1)$ . Then we have for all $z \in V_R$ . $$\frac{1}{p_i(z)} = \frac{zh'(z) + A_i}{\left(zh'(z)\right)'} \in D = \left\{w: \left|w - \frac{1}{2k}\right| \le \frac{1}{2k}\right\}.$$ Since D is convex $$\frac{1}{p(z)} = \frac{zh'(z) + A}{(zh'(z))'} \in D,$$ and therefore, $\operatorname{Re} p(z) \geq k$ . Furthermore, since $\Gamma_{\rho} = zh'(|z| = \rho)$ , $\rho \in (1, R)$ , is a positively oriented Jordan curve which is strictly starlike with respect to $-A_1$ and $-A_2$ , it is also strictly starlike with respect to -A. Hence, $$\frac{1}{2\pi} \oint_{|z|=\rho} d\arg(zh'+A) = 1,$$ which shows that $E_h$ is a convex set. (b) Put $\psi_0(z) = zh'(z)$ . Since $E_h \neq \emptyset$ , $\psi_0(\partial U)$ is an analytic closed Jordan curve. (This follows from (b), (c), (b') and (c') applied to $\psi_0 + A$ for an $A \in E_h$ ). Suppose, in addition, that $\psi_0(\partial U)$ is also a convex curve, i.e. that the bounded component G of $\mathbb{C} \setminus \psi_0(\partial U)$ is a convex domain. Let -A be in G and put $\psi = \psi_0 + A$ . Since $0 \notin \psi(\partial U)$ , $0 \notin \psi'(\partial U)$ and Re $\frac{z\psi'(z)}{\psi(z)} > 0$ on $\partial U$ , we conclude that (c) holds on $V_{R_1}$ , for some $R_1 > 1$ . The statements (a) and (b) hold already, since $E_h \neq \emptyset$ . Therefore $-G \subset E_h$ . We finish this section with the following remarks. REMARKS 5.2. (i) Define $\psi_0(z) = zh'(z)$ and suppose that $E_h \neq \emptyset$ . Let, as before, G be the bounded component of $\mathbb{C} \setminus [\psi_0(\partial U)]$ . Then $-E_h \subset \bar{G}$ . Therefore, $E_h$ is a bounded set. (ii) We cannot conclude that $E_h$ is closed (and hence compact) as the following example shows: Consider $h(z) = \frac{1}{1-z/2}$ . Then $\psi_0(z) = zh'(z) = \frac{1}{2} \cdot \frac{z}{(1-z/2)^2}$ and $\frac{z\psi_0'(z)}{\psi_0(z)} = \frac{1+z/2}{1-z/2}$ . It follows then that $0 \in E_h$ . There is even a disk $\{A : |A| < r\}, r > 0$ , which belongs to $E_h$ . On the other hand $A = \frac{4}{9} \notin E_h$ . Let $\hat{A} = \limsup\{A : A > 0, A \in E_h\}$ . Then, $\hat{A}$ violates (3.1) and is therefore not in $E_h$ .