ABSTRACT
EXAMPLE 2.4a. Let \jl(r, s, t; z) = r + s + t, a = 0, and Q = h(U), where h(z) = (n2 + l)Mz. To use part (i) of Theorem 2.3h we need to show that 'II e 'I'0 [Q,M,O], that is, that admissibility condition (2.3-9) is
satisfied. This follows since
We can use Theorem 2.3f to present a different proof of these results, and to
also show that these results are sharp. The differential equation
2.4 EXAMPLES 37
p(z) + zp'(z) + z2 p"(z) = (n2 + 1)M z
has univalent solution q(z) = Mz . To use Theorem 2.3f we need to show that 'If E 'I'n[h,q], where h(z) = (n2 + 1)Mz and q(z) = Mz . On checking the admissisibi1ity condition (2.3-1) for this h, q and 'I' we have
M(l + m + m2 - m) = (m2 + 1)M ~ (n2 + l)M, when r = Me;, s = mMc;, Re t/s + 1 ~ m, I c; I = 1, and m ~ n . Therefore 'lf(r, s, t; z) ~ Q = h(V), 'If E 'I'n[h , q] , arid from Theorem 2.3f we have:
EXAMPLE 2.4b. Let 'lf(r, s, t; z) = r 2 + r + s, a = 0, and Q = h(U), where h(z ) = nM z. To use part (i) of Theorem 2.3h we need to show that 'I' E 'Pn[.Q,M, 0], that is, that admissibility condition (2.3-9) is satisfied.
