Starlikeness Associated With The Exponential Function

Theorem 1.2. [11, p. 28] Let ψ ∈Ψn(Ω, q) with q(0) = a. If p ∈H[a, n] satisfies ψ(p(z), zp′(z), z2p′′(z); z) ∈Ω then p(z) ≺q(z). Miller and Mocanu [11] in their monograph discussed the class of admissible functions Ψ(Ω, q) when the function q maps D onto a disk or a half-plane. These two special classes together with Theorem 1.2 lead to several important and interesting results in the theory of differential subordinations. However the aim of this paper is to consider differential implications with

Theorem 2.1. Let p ∈H1. (i) If ψ ∈Ψ(Ω, ez), then ψ(p(z), zp′(z), z2p′′(z); z) ∈Ω ⇒ p(z) ≺ez.

Theorem 2.1, it is easy to deduce that for any p ∈H1, we have |p2(z) −p(z) + (1 + e)zp′(z)| < e−1 ⇒ p(z) ≺ez. In the similar fashion, by taking ψ(r, s, t; z) = 1 + s/r2 and Ωas above, it is easily seen that |ψ(r, s, t; z) −1| = |s/r2| = |meiθe−eiθ| = me−cos θ ≥me−1 ≥e−1 whenever r = eeiθ, s = meiθr and Re (1 + t/s) ≥m(1 + cosθ), where θ ∈[0, 2π) and m ≥1. This implies that ψ(r, s, t; z) /∈Ωand hence ψ ∈Ψ(Ω, ez). Thus for any p ∈H1, we have

Theorem 3.1. If n is a non-negative integer, 0 ≤α < 1 and p ∈H1 satisfies the subordination 1 + β zp′(z) pn(z) ≺1 + (1 −α)z, where β ≥ ( e(1 −α) when n = 0 en−1(1 −α) when n ̸= 0 then p(z) ≺ez.

Theorem 2.1 is applicable if we show that ψ ∈Ψ(Ω, ez), that is, ψ(r, s, t; z) ̸∈Ωwhenever r = eeiθ, s = meiθr and Re (1 + t/s) ≥m(1 + cos θ), where z ∈D, θ ∈[0, 2π) and m ≥1. A simple calculation yields |ψ(r, s, t; z) −1| = βmecos θ ≥βe−1 ≥1 −α. Hence ψ(r, s, t; z) /∈Ωwhich gives ψ ∈Ψ(Ω, ez). Using Theorem 2.1, we get p(z) ≺ez. Case (ii). When n ̸= 0, the function ψ(r, s, t; z) = 1 + βs/rn satisfies |ψ(r, s, t; z) −1| = βme−(n−1) cos θ ≥βe−(n−1) ≥1 −α whenever r = eeiθ, s = meiθr and Re (1 + t/s)

Theorem 3.3. If n is any non-negative integer and p ∈H1 satisfies the subordination 1 + β zp′(z) pn+1(z) ≺2 + z 2 −z , where β ≥2en then p(z) ≺ez.

Theorem 3.5. Let α, β be positive real numbers satisfying α(e−1)+βe ≥e and p ∈H1. If the following subordination (1 −α)p(z) + αp2(z) + βzp′(z) ≺1 + z holds, then p(z) ≺ez.

Theorem 3.6. Let p ∈H1. Then both the following conditions are sufficient for p(z) ≺ez: (a) p(z) + βzp′(z) ≺(2 + 2z)/(2 −z) for β ≥(e + 2 − √ 2(e −1))/(e( √ 2 −1)) ≈2.0323. (b) p(z) + βzp′(z)/p(z) ≺(2 + 2z)/(2 −z) for β ≥(e + 2 − √ 2(e −1))/( √ 2 −1) ≈5.52436.

Theorem 4.1. If n is any positive integer and p ∈H1 satisfies the subordination 1 + β(zp′(z))n ≺ez, where β ≥ ( en+1 + en when n is odd en+1 −en when n is even then p(z) ≺ez.

Theorem 4.2. If p ∈H1 satisfies the subordination 1 + β zp′(z) pn+1(z) ≺ez, where β ≥en+1 −en and n is any non-negative integer, then p(z) ≺ez.

Theorem 4.4. Let p ∈H1, then each of the following subordinations are sufficient for p(z) ≺ez: (a) p(z) + βzp′(z) ≺ez for β ≥e2 + 1 ≈8.38906. (b) p(z) + βzp′(z)/p(z) ≺ez for β ≥e + e−1 ≈3.08616.

Theorem 4.5. Let n be any positive integer and βn be a positive root of the equation (4.1) e1+n −x(−1 + n)
2 − e2+2nn −x2n + xe1+n(1 −n + n2)
ln(e + e−nx) = 0. If p ∈H1 satisfies the subordination p(z) + β zp′(z) pn+1(z) ≺ez, for β > βn then p(z) ≺ez.

Theorem 4.6. Let p ∈H1. Then each of the following subordinations are sufficient for p(z) ≺ez: (a) p(z) + β(zp′(z))2 ≺ez for β ≥e3 −e ≈17.3673. (b) p(z) + β(zp′(z))2/p(z) ≺ez for β ≥e2 −1 ≈6.38906. (c) p(z) + β(zp′(z))2/p2(z) ≺ez for β ≥e −e−1 ≈2.3504.

Theorem 4.7. Let p ∈H1. Then each of the following subordinations are sufficient for p(z) ≺ez: (a) p2(z) + βzp′(z) ≺ez for β ≥e2 + e−1 ≈7.75694. (b) p2(z) + βzp′(z)/p(z) ≺ez for β ≥e + e−2 ≈2.85362. (c) p2(z) + βzp′(z)/p2(z) ≺ez for β > β∗≈104.122, where β∗is a positive root of the equation 6e6 + 5xe3 −x2 + (−2e6 −5xe3 + x2) ln(e3 + x) = 0.

Theorem 4.8. Let p ∈H1, then each of the following subordinations are sufficient for p(z) ≺ez: (a) p(z) + βzp(z)p′(z) ≺ez for β ≥e3 + e ≈22.8038. (b) p2(z) + βzp(z)p′(z) ≺ez for β ≥e3 + 1 ≈21.0855. (c) p3(z) + βzp(z)p′(z) ≺ez for β ≥e3 + e−1 ≈20.4534.

Theorem 4.9. Let p ∈H1 and satisfies the subordination p2(z) + p(z) −1 + βzp′(z) ≺ez for β ≥e2 + e−1 −e + 1 ≈6.03865. Then p(z) ≺ez.