Convex meromorphic mappings

Theorem 1. f is a member of C(p) if and only if for z ∈∆, (2.1) Re  1 + p2 −2pz + (z −p)(1 −pz)f ′′(z) f ′(z)  < 0 . P r o o f. If f is a member of S(p) let h(z) = f((z + p)/(1 + pz)); then h has a simple pole at z = 0 and C \ h[∆] = C \ f[∆]. Thus, f is a member of

Lemma 1. Let P(z) satisfy Re P(z) > 0, z ∈∆, and P(0) = 1. If 0 < p < 1, then for z ∈∆, Re (z −p)(1 −pz)P(z) + p z −pz  > 0 . P r o o f. Let 0 < r < 1 and Pr(z) = P(rz). Then Qr(z) = (z −p)(1 −pz)Pr(z) + p z −pz is analytic for |z| ≤1. If |z| = 1, then Qr(z) = (z −p)(1 −pz)Pr(z)

Lemma 2. If Re P(z) > 0 for z ∈∆and P(p) = 1 −p2, then for z ∈∆, Re zP(z) −p + pz2 (z −p)(1 −pz)  > 0 . P r o o f. Let p < r < 1 and α = (r −1)p/(r −p2) and Lr(z) = r(z − α)/(1 −αz). It is easily verified that Lr[∆] = {z : |z| < r} and Lr(p) = p. Let Qr(z) = zP(Lr(z)) −p + pz2 (z −p)(1 −pz) . Qr(z) is analytic for |z| ≤1 and Re P(Lr(z)) > 0 for |z| ≤1. If |z| = 1 then Re Qr(z) = Re 

Theorem 2. C(p) = Σ(p) for 0 < p < 1. P r o o f. Let f be a member of Σ(p) and P(z) = −1 −zf ′′(z) f ′(z) + 1 + pz 1 −pz −z + p z −p . Then Re P(z)>0, z ∈∆, and P(0) = 1. Straightforward computations give 2pz −1 −p2 −(z −p)(1 −pz)f ′′(z) f ′(z) = (z −p)(1 −pz)P(z) + p z −pz . Therefore, by Lemma 1, Re 

Lemma 3. Let P(z) be analytic in ∆and satisfy Re P(z) > 0, z ∈∆, P(p) = 1 −p2 and P ′(p) = 0, 0 < p < 1. If P(z) = (1 −p2) + d2(z −p)2 + d3(z −p)3 + . . . for |z −p| < 1 −p, then |d2| ≤ 2 1 −p2 , (3.1)

Theorem 3. Let f be a member of C(p) and have the expansion (1.2). Then |a1| ≤ p2 (1 −p2)3 , (3.9) |a2| ≤(4 + 9p2)|a−1| 12(1 −p2)3 , 0 < p ≤2/3 , (3.10) |a2| ≤ p (1 −p2)3 |a−1| ≤ p3

Theorem 4. If f is a member of C(p) with expansion (1.2), then p + a0(1 −p2) a−1 ≤1 + p2 p , and the inequality is sharp. P r o o f. Let h(z) = −a−1 (1 −p2)f  p −z 1 −pz  , then h is a member of S(p) and for |z −p| < 1 −p,

Theorem 5. Let f be a member of C(p) with expansion (1.1). Then (4.1) Re b2 ≥ 1 + p4 p(1 + p2) > 1 and (4.2) Re b3 ≥1 −p2 + p4 p2 = 1 + p6 p2(1 + p2) > 1 . Both inequalities are sharp, each being attained by the function f(z) = p(1 + p2)z −2p2z2

Theorem 5 is extremal for all n. That is, we expect that if f is a member of C(p), then Re(bn) ≥(1 + p2n)/(pn−1(1 + p2)) for all n. 5. Starlike functions. Miller [7]–[9] considered functions f of S(p) for which there exists ϱ, 0 < ϱ < 1, so that Re[zf ′(z)/(f(z) −w0)] < 0 for ϱ < |z| < 1 and a fixed w0 ∈C, w0 ̸= 0. These functions map ∆onto the complement of a set which is starlike with respect to w0. This class of functions is a subclass of the class Σ∗(p, w0) defined as the class of functions f

Theorem 6. f is a member of Σs(p, w0) if and only if , for z ∈∆, Re (z −p)(1 −pz)f ′(z) f(z) −w0  < 0 . P r o o f. Suppose f is a member of S(p) and let g(z) = f  z + p 1 + pz  .

Theorem 7. Σs(p, w0) = Σ∗(p, w0) for all p, 0 < p < 1, and all w0 ̸= 0. P r o o f. Let f be a member of Σ∗(p, w0) and P(z) = pz 1 −pz − p z −p − zf ′(z) f(z) −w0 , then Re P(z) > 0 for z ∈∆and P(0) = 1. From this we obtain (5.1) (z −p)(1 −pz)f ′(z) f(z) −w0 = −(z −p)(1 −pz)P(z) + p(1 −z2)

Theorem 8. If f(z) is a member of Σ∗(p, w0) and has expansion (1.2) for |z −p| < 1 −p then |a0 −w0| ≤2 + p 1 −p2 |a−1| (5.4) and |a1| ≤ |a−1| (1 −p2)2 (5.5) Both inequalities are sharp. P r o o f. We first prove inequality (5.5). Let P(z) = −(z −p)(1 −pz)f ′(z) f(z) −w0 ,

Theorem 9. With the notation of Theorem 8, |a−1| ≤p(1 −p) 1 + p |w0| and the inequality is sharp. P r o o f. With P(z) as in the proof of Theorem 8, d dz log(z −p)(f(z) −w0) = (1 −pz) −P(z) (z −p)(1 −pz) . Integrating, we obtain f(z) −w0 = pw0 z −p exp zR 0 (1 −pξ) −P(ξ) (ξ −p)(1 −pξ) dξ .