Univalent Harmonic Ring Mappings Vanishing on the Interior Boundary

LEMMA 2.1. Letf be any harmonic function on VR (for some R > 1 ) satisfying (1.2). Thenf has a harmonic continuation across dU = { z : \z\ = 1} which is of the form (2.1) f(z) = h(z) - h(-) + 2AIn |z|, A G C, z where h(z) = E7GZ atf E H(± < \z\ < R). Observe thatf(\) = -f(z) on {z : {- < \z\ < R}. Conversely, each harmonic mapping on dU satisfying (2.1) has the property (1.2). PROOF. Since/ is harmonic onVR,f admits the representation (2.2) /(z) = A(z) + i ( 3 + 2Aln|z|,

LEMMA 2.2. Let f(z) = h(z) - h(-) + 2A In \z\, A G C z be harmonic on {z : \z\ — 1 } , / ^ const. There is an exterior neighbourhood VR oft) such thatf is orientation-preserving on VR if and only if there exists a constant k > 0 such that ijj (z) = zh!(z) + A satisfies (2.4) +oo > Re ;, \ > k > 0 yj (elt) whenever xjj (elt) ^ 0. PROOF. By Lemma 2.1. h G H(± < \z\ < R\) for some R{ > 1. Therefore, the

LEMMA 2.3. Let O be in H{j < \z\ < R)for some R > 1, such that O is real on dU and ^ jfz| =p d arg zO' = 1 for all p G ( 1, R). Then O' has exactly two zeros ondU which are of order one. PROOF. By the reflection principle, we have O(-) = O(z) and - -O'(-) = zO'(z). z z z Since <£>(dU) is a bounded real interval, there exists an el(3 and an e11, elf3 ^ en, such that <&(elP) = O'O'7) = 0. Applying the argument principle, we get for p G (1,/?) — / dargzO'- — / dargzO' = 2 - — / dargzO' = 2 Z7T J\z\

LEMMA 2.4. Let f(z) = h(z) - h(-) + 2Aln \z\, A G C z be harmonic on {z : ^ < \z\ < R} and suppose that f is orientation-preserving on VR. Put ^ = ;Â/\A\ i f A ^ o 1 if A = 0, and define 0(z) = eiah(z) +

LEMMA 2.5. Let F be analytic on A = {z : r\ < \z\ < ri\, 0 < r\ < rç, such that (2.18) 0 < R e ^ - ^ < oo on A. Then the following statements are equivalent: (i) F is univalent on A; (ii) $z\=p daigF= 1, for some p G(n,r2); (Hi) p(z) = ^ = 1 + E;€Z\{o} PjZJ e H(A). PROOF, (a) The fact (ii)^(iii) follows from the relation zfiz) dz 1 / p(z) . -$

LEMMA 2.6. Let F be in H(\z\ = 1) such that (i) \F(eil)\ = 1 andF'ie11) f 0 for all te [0,2TT]; (ii) ^ifz|=i dargF= 1. 77ief2 f/zere ejnsta a« 7?i > 1 swc/i r7i£tf F /s univalent on {z:^<\z\<}. PROOF. The fact that ^ G R \ { 0} on 3f/ and that 1 /• 1 /• — <t dargzF — —- i d arg F = 1, https://doi.org/10.4153/CJM-1992-021-x Published online by Cambridge University Press

THEOREM 3.1. Let /(z) = A(z) + ife) + 2Aln|z|, A e C , be a harmonic mapping defined on the unit circle dU — {z : |z| = 1}. Put rl>(z) = zh\z)+A = &(z). Then there exists an exterior neighbourhood VR of Û such thatf is univalent and posi- tively oriented on VR and lim^^j f(z) = 0, if and only if the following conditions are z-Vf 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; (

THEOREM 3.2. Let /(z) = A(z) + £(5 + 2Aln|z|, A G C, Z?e a harmonic mapping defined on the unit circle dU = {z : |z| = 1}. Put ip(z) — zfz(z) = zh!{z) + A and p(z) = ^jfrj. TTz^n r/i^re ex/ste a/i exterior neighbourhood VR of D such thatf is univalent, positively oriented on VR and lim |z|> i f(z) = 0, if and only if the following conditions are satisfied: (a) h and g admit an analytic continuation across dU such that h(j) — —g(z), z G {z:ji < \z\ < R}; (bf) V V ' ) ^ 0 for all t e [0,2TT]; (d)

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 _L_ 2TT J\z\ = \ j> _ darga = -2 andQ g ip(dU) or — i darga = —1 2ir J\Z\ = I fe 2TT ijz| = i 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 Re

Theorem 3.1, i.e. there is an R > 1 such that h G H(l < \z\ < R), oo > Re z^- > k > 0 on V^, 27§[z\=p dargil) = 1 for p G (1,$) and ijj has at most one zero on dU which is of order 1. Put eia=lA/\A\ ifA^O 11 if A = 0 and define: O(z) = eiah(z) + e-iah(-), z G V*, z f(z) = h(z)-h(-) + 2A\n\z\,$

THEOREM 5.1. Let h G H(dU). If Eh is nonempty, then we have the following prop- erties: (a) Eh is a convex set. (b) If in addition, zhf(U) is a convex set, then the bounded component G of C\[zh'(dU)] belongs to -Eh. PROOF, (a) Let A\ and A2 be in Eh. Since Eh is non-empty, we conclude from The- orem 3.1 and Remark 3.5 that (a), (b;) and (c) are satisfied. Observe that (a) and (b7) are https://doi.org/10.4153/CJM-1992-021-x Published online by Cambridge University Press