An application of differential subordinations and some criteria for univalency

LEMMA 1. ([2]). Let q{z) be univalent in the unit disc U, and let 0(w) and <j)(w) be analytic in a domain D containing q(U), with <f>(w) ^ 0 when w G q(U). Set Q(z) = zq'(z)(j>(q{z)), h{z) = 0{q{z)) + Q(z), and suppose that (i) Q{z) is starlike in the unit disc U, and (ii) IU{(zh'(z))/(Q(z))} = Re{(fl'(g(z)))/(^(g(z))) + (zQ'(^))/(Q(z))} >o (zeu). If p(z) is analytic in U, with p(0) = g(0), p(U) C D and (1.4) 6(p(z)) + zp'(z)<f>(p(z)) * 6(q(z)) + zq'{z)4>(q(z)) = h(z), then p(z) -< q(z), and q(z

LEMMA 2. ([3]). If f(z) eA and (1.5) | a r g ( / ' ( z ) ) | < ^ =0.968... (z e U), where 70 = 0.6165 ... is the unique root of the equation 2 arctan (1 - 7) + TT(1 - 27) = 0, then f(z) 6 5*. 2. DIFFERENTIAL SUBORDINATION AND SOME CRITERIA FOR UNIVALENCY We first prove: https://doi.org/10.1017/S0004972700018360 Published online by Cambridge University Press

THEOREM 1. Let p(z) be analytic in U, with p(0) = 1, and let 0 < A ^ 1. If (2.1) (1 - A)p(z) + XzP'(z) -< (l± 0 < 7 ^ 1, then and this is the best dominant of (2.1). PROOF: We choose q(z) = ((1 + z)/(l - z))7, 0 < 7 ^ 1, <f>(w) = A and 6(w) = (1 — X)w in Lemma 1. Then the function q(z) is convex in U and g(0) = 1. Further Q(z) = zq'{z)<f>(q(z)) = \zq\z) is starlike, and for the function h(z) = 0(q(z)) + Q(z) = (1 - X)q(z) + Xzq'(z), Therefore the conditions (i) and (ii) in Lemma 1 are satisfied.

THEOREM 2. Let p{z) be analytic in U with p(0) = 1, and 0 < A ^ 1. If (2.2) Re{(l - A)P(z) + Xzp'(z)} > - | (26 U), then Re{p(z)} > 0 (z G U). PROOF: Taking 7 = 1 in Theorem 1, we have q(z) = (1 + z)/(l - z) and (2.3) fc(,) = (i_A)l±i+A^-jy. https://doi.org/10.1017/S0004972700018360 Published online by Cambridge University Press

COROLLARY 1. Let f(z) e A and o < A < l. If Re{(l - X)f'(z) + Xzf'(z)} > -± (z G U), then Re{/'(2)} > 0 (z £ U). If we take A = 1 and zf'(z)/f(z) or 1 + zf"(z)/f'(z) instead of p(z) in Theorem 2, we obtain:

COROLLARY 2. I / / ( z ) e 4 a n d then f(z) e S*.

COROLLARY 3 . If f(z) e A and > o where {/(z), 2} denotes the Schwarzian derivative defined by then /(zJeJfif. By a similar method to that used in Theorem 2, we may obtain the answer for the case 0 < 7 < 1, but this is more complicated than the case 7 = 1. Namely, we have: https://doi.org/10.1017/S0004972700018360 Published online by Cambridge University Press

THEOREM 3 . Let p(z) be analytic in U with p(0) - 1, and let 0 < A < 1 and 0 < 7 < 1. If (2.4) where (2.5) Re{(l - X)p(z) + \zp'(z)} > C{\, 7) U), , 7) = j 7 1-A - 7 ) «n (7*72), o = to = cot (TT/2), a + \/a2 + 1 - 72

COROLLARY 4 . Let f(z) £ A, and let 0 < A < 1 and 0 < 7 < 1. If (2.13) Re{(l - A)/'(z) + Az/"(z)} > <?(A, 7o) (z G U), where y0 is as in Lemma 2 and C(A, 7) is defined by (2.5), then f(z) £ 5*. EXAMPLE 2. For A = 1 in Theorem 3, we have that if p(z) is analytic in U with p(0) = 1 and 0 < 7 < 1, then the following implication Re{zp'(z)} > C(l, 7) ^ ' — ' - ' - « ' - 7 7 r where is true. Therefore, from Corollary 4, we obtain Re{zf"(z)} > C(l, 70) - -0.414076... = * f(z) where 70 is as in Lemma 2.

Lemma 2, this type of problem was treated by Mocanu [3]. By using the result of

Theorem 3, we may find other subsets which imply starlikeness, whenever f'{z) belong to them for all z g U.

THEOREM 4 . Let f(z) e A, and let f'(z) satisfy (2.14) |arg (f'(z))\ < ^ (70 < 7 < 1; * € U), https://doi.org/10.1017/S0004972700018360 Published online by Cambridge University Press