TAMKANG JOURNAL OF MATHEMATICS

Lemma 1.1. If p ∈P , then |pn| É 2 for n Ê 1, and ¯¯¯p2 −µ 2 p2 1 ¯¯¯ É max{2, 2|µ−1|} = ( 2, 0 É µ É 2, 2|µ−1|, elsewhere.

Lemma 1.2. Let p ∈P . If 0 É B É 1 and B(2B −1) É D É B, then ¯¯p3 −2Bp1p2 +Dp3 1 ¯¯ É 2.

Lemma 1.3. If p ∈P , and 0 É B É 1, then ¯¯p3 −2Bp1p2 +Bp3 1 ¯¯ É 2.

Lemma 1.4. If p ∈P , then ¯¯p3 −(1+µ)p1p2 +µp3 1 ¯¯ É max{2, 2|2µ−1|} = ( 2, 0 É µ É 1, 2|2µ−1|, elsewhere. In the following, the methods of proof develop those employed in [1, 2], and in the inter- ests of brevity, we omit much of the elementary algebra. Main Results 2. Coefficients of functions in M(α,β) and M ∗(γ,β)

Theorem 2.1. Let f ∈M(α,β) and be given by (1.1), then |a2| É 2β 1+α, |a3| É          

Lemma 1.1. The first inequality for |a3| is sharp when p1 = 0 and p2 = 2, and the second inequality is sharp when p1 = p2 = 2. For a4, first write Λ1(α) = 2(1+α)(1+2α) (1+4α)(5+α) , and note that the coefficients of p1, p2 and p3 1 are positive when Λ1(α) É β É 1. So using |pn| É 2 for n = 1,2 and 3, gives the second inequality for |a4| when Λ1(α) É β É 1.

Theorem 2.2. Let f ∈M ∗(γ,β) and be given by (1.1), then |a2| É 2β 1+γ, |a3| É           

Theorem 3.1. Let f ∈M(α,β) and the coefficients of the inverse function f −1 be given by (1.4), then |A2| É 2β 1+α, |A3| É         

Theorem 2.1 and use Lemma 1.2 so that A4 = β 3(1+3α) ³ p3 −2B2p1p2 +D2p3 1 ´ , with B2 = 1+2α2(1−β)+5β+3α(1+4β) 2(1+α)(1+2α) , and D2 =

Lemma 1.4 and using the fact that |p1| É 2, we obtain the second inequality for |A4| on the interval Λ2(α) É β É 1, after substituting for D2 and µ. The first inequality for |A4| is sharp when p1 = p2 = 0 and p3 = 2, and the second inequal- ity is sharp when p1 = p2 = p3 = 2. □ We note again that when β = 1, the results in Theorem 3.1 correspond to the estimates found in [6], when α = 0 to those in [1], and when α = 1 to those in [12].

Theorem 3.2. Let f ∈M ∗(γ,β) and the coefficients of the inverse function f −1 be given by (1.4), then |A2| É 2β 1+γ, |A3| É          