A subclass of univalent functions

Theorem 2; 8(1944), Theorem 210]. Robertson [14] introduced the concept of quasi-subordination. Let g(z) and G(z) be analytic in E. Let <f>(z) be analytic and |<|>(z)|< 1 in E, such that g(z)/<j>(z) is regular and subordinate to G(z), for z e E. Then g(z) is said to be quasi-subordinate to G(z), written as g(z) <q G(z), z BE. An equivalent condition for this is g(z)=*(z)G(w(z)), \<t>(z)\< l,wGU,zEE. If <j>(z) = 1, then g(z) = G(w(z)) so that g(z) < G(z) in E. If w(z) - z, then g(z) = <f>(z)G(z),

LEMMA 2. Ifg(z) = l^=odkzk <qG{z) = l?=0Dkzk, then (2.2) 2 K I 2 < 2 I'M2 (« = 0,l,2,...). k=0 k=0 This lemma is due to Robertson [14]. https://doi.org/10.1017/S1446788700024733 Published online by Cambridge University Press

LEMMA 3. Forf e R(A, B), iff'(z) = P(z) = 1 + lf=lpkzk, then (2.3) \pn\*(A-B), n>\. The bounds are sharp. PROOF. From (1.5) we have ,k k=\ L k=\ By the application of Lemma 2(2.2), we get n n 1 k=\

LEMMA 4. For w G Uand \z\= r,we have 2 - ' w ( z ) | 2 (2.4) \zw'(z)-w(z)\<^— This result is due to Singh and Goel proved in [15].

LEMMA 5. Let _ , _ , _ ! +Bw(z) https://doi.org/10.1017/S1446788700024733 Published online by Cambridge University Press

THEOREM 3.1. Letf £ R(A, B) then (3.1) | f l J <(l_L!), n > 2 . The bounds are sharp for the functions f(n_ X){z) defined by (3-2) f(H-Jz) ~- PROOF. (3.1) follows on equating the coefficients of z" in (1.4) and then using (2.3).

THEOREM 3.2. / / / £ R(A, B) and if \i is a complex number, then (3.3) \a, The estimate is sharp. PROOF. On equating the coefficients of z2 and z3 in (1.5), we get (3.4) c, = 2a2 ( 3' 5 ) (A-B)'- 3 f , AB -2

THEOREM 4.1. LetfG R(A, B), then for \ z |= r < 1, (4-1) I /'( Z)I <TT^ ; 1 - ^ r (4.2) (4.3) (4-4) Re/'(z) 1 - Br' r — —r 2 ' All the estimates are sharp. = 0, = 0.

THEOREM 5.1. If f £ R(A, 5) then (5.1) |arg/'(z)|<sin- The result is sharp. 1 - ABr2 ' z = r. https://doi.org/10.1017/S1446788700024733 Published online by Cambridge University Press

THEOREM 6.1. Iffandh e R(A, B), then Xf+(\-X)h<=R(A,B), (O PROOF. By definition, (6.1) f'(z (6-2) *'<*-,+&• Since (1 + Az)/{\ + Bz) is convex univalent in E, it follows by a result due to Bernardi [2, page 57, Example 2] that 1 + Az 1 + Bz' 1 +Az Hence Xf'(z) + (I - X)h'(z) < \

THEOREM 7.1. Letf e R(A, B), then (i) for Ao < A < l,/(z) is convex in \z\< r0, where r0 is the smallest positive root of (7.1) ABr2 - 2Ar + 1 = 0; https://doi.org/10.1017/S1446788700024733 Published online by Cambridge University Press

THEOREM 8.1. Let / £ R(A, B) and Lr(f) denotes the length of the image of | z | = r under f(z), 0 < r < 1, then (8-1) Lr(f) •nr A + B B (A-B) 151 log 1 + Br +Arei9\d0, = Q. The results are sharp. https://doi.org/10.1017/S1446788700024733 Published online by Cambridge University Press

THEOREM 8.2. / / / £ R(A, B), and if Ar(f) denotes the area of image of\z\=r under f(z), 0 < r < 1, then (8.4) A,(f) irr mr' 1 + A2r2 = 0. The inequalities are sharp. (8.4) are direct consequences of Lemma 1 (X = 2) and interior area theorem. Equality sign is attained for the function fo(z) defined by (8.3). COROLLARY. For the class R(a), we have from (8.4), This is a result due to Capling and Causey [4].

THEOREM 9.1. If f(z) = z + 2"=2 anz" and h(z) = z + 2~=2 bnz" belong to the class R(A, B), then so does PROOF. Since/ e R(A, B), it follows by (1.6) that \f'(z) - 1 |<| A - Bf'(z) | . It is equivalent to (9-1) \f'{z)-b\<C where b = (1 - AB)/{\ - B2), C = (A - B)/(\ - B2). It is easy to see that 1 - b< C < b. We know that if H(z) = l^=Qhnzn is regular for | z |< 1 and | H(z)\*z M, then, by [11, page 101], (9.2) 2 \hn\2<M2. Applying (9.2) to (9.1), we get (1 - b)2 + 2 ~ = 2 n2 \an\2 < C2 or (9.3)