On Oblique Domains of Janowski Functions

Theorem 2.1

Let h ∈ ℋ₁ and A, B ∈ ℂ with A ≠ B and |B| ≤ 1. Further if

for some 0 < α ≤ 1, γ ∈ ℂ \ {1}, then for |z| = r < 1, we have

1. |arg⁡(h(z)−γ1−γ)−αtan−1⁡Im(AB¯)r2Re(AB¯)r2−1|<αsin−1⁡|A−B|r|1−AB¯r2|,whenever |A|≤1;

2. (|1−AB¯r2|−|A−B|r1−|B|2r2)α≤|h(z)−γ1−γ|≤(|1−AB¯r2|+|A−B|r1−|B|2r2)α;
3. min{M(t1)cos⁡(N(t1)),M(t1+π)cos⁡(N(t1+π))}≤Re(h(z)−γ1−γ)≤max{M(t1)cos⁡(N(t1)),M(t1+π)cos⁡(N(t1+π))};
4. min{M(t2)sin⁡(N(t2)),M(τ−t2)sin⁡(N(τ−t2))}≤Im(h(z)−γ1−γ)≤max{M(t2)sin⁡(N(t2)),M(τ−t2)sin⁡(N(τ−t2))},

with

Further t₁ and t₂ are the roots of

and

respectively. Further, all the above bounds are sharp.

Proof

Let H(z):=((h(z)−γ)/(1−γ))1α. Since h(z) ≠ 0 and h(0) = 1, we have H ∈ ℋ₁ and

As A, B ∈ ℂ, clearly H(z) is contained in the oblique circle shown in the Figure 1, whose radius and center are given by R := |A − B|r/(1 − |B|²r² and C:=1−AB¯r2/(1−|B|2r2), respectively, with angles τ(r):=tan−1⁡(Im(AB¯)r2/ (Re(AB¯)r2−1)) and ζ(r):=sin−1⁡(|A−B|r2/(|1−AB¯|r2)). By taking argument estimate of (2.4) and using the Figure 1 with the fact that circle is symmetric about the line passing through origin and center, we obtain (i). Let |z| = r < 1 and t ∈ [0, 2π), we have (h(reit)−γ)/(1−γ)∈Ω:=((1+Az)/(1+Bz))α, which implies ∂Ω:(C+Reit)α:=M(t)eiN(t). To find modulus, real and imaginary parts estimate, we need the critical points. By a simple computation, we obtain M′(t) = 0 at t = τ(1) = τ and τ + π, (M(t) cos N(t))′ = 0 at the roots t₁ and t₁ + π of the equation (2.2) and (M(t) sin N(t))′ = 0 at the roots t₂ and τ − t₂ of the equation (2.3), all these values eventually yield (ii), (iii) and (iv), respectively. □

Figure 1

The image of ∂𝔻 under (1 + Az)/(1 + Bz),.

Remark 1

1. When γ = 0, we can obtain from Theorem 2.1 various bound estimates for functions in 𝒫(A, B, α).

2. By taking α = 1/2, γ = 0, A = 1 and B = 0 in Theorem 2.1, we have h(z)≺1+z. Therefore the estimates are |arg h(z)| ≤ sin⁻¹ r/2 and 1−r≤|h(z)|≤1+r. If r = 1, we have t₁ = 0 and t₂ = 2π/3, which implies 0<Reh(z)<2 and −0.5 < Im h(z) < 0.5.

3. Note that if w(z) = (1 + Az)/(1 + Bz), then w(0) = 1 ∈ w(𝔻). Thus the image domain w(𝔻) will always intersect real axis even if it is an oblique domain, non-symmetric with respect to real axis.

Corollary 2.1.1

Let h ∈ ℋ₁ and A, B ∈ ℂ with A ≠ B, |A| ≤ 1 and |B| ≤ 1. Further if

then for |z| = r < 1, we have

1. |arg⁡h(z)−tan−1⁡Im(AB¯)r2Re(AB¯)r2−1|<sin−1⁡|A−B|r|1−AB¯r2|;

2. |1−AB¯r2|−|A−B|r1−|B|2r2≤|h(z)|≤|1−AB¯r2|+|A−B|r1−|B|2r2;

3. 1−Re(AB¯)r2−|A−B|r1−|B|2r2≤Reh(z)≤1−Re(AB¯)r2+|A−B|r1−|B|2r2;

4. Im(AB¯)r2+|A−B|r|B|2r2−1≤Imh(z)≤Im(AB¯)r2−|A−B|r|B|2r2−1.

Theorem 2.2

For 0 < α ≤ 1 and −1 < m < 1, then the function

is analytic, univalent and convex in 𝔻 with

Proof

By using Theorem 2.1, the function h(z) given in (2.7) satisfy

which gives the desired result. □

Remark 2

1. From the domain (2.8) we have:

   2.9 (h(D))1/α={w∈C:Ree−imπ/2w>0}.

2. For 0 < α₁ ≤ 1 and 0 < α₂ ≤ 1, if m = (α₁ − α₂)/(α₁ + α₂) and α = (α₁ + α₂)/2 in (2.7), then Theorem 2.2 reduces to [9: Lemma 3].

Theorem 2.3

Let h ∈ ℋ₁ and 0 ≤ b ≤ 1 with b + e^imπ ≠ 0, where −1 ≤ m ≤ 1. Also, if

then

where λ=tan−1⁡(bsin⁡(mπ)bcos⁡(mπ)+1).

Proof

To obtain (2.11), it suffices to show that |arg(e^−iλw)| < π/2 or |arg w − λ| < π/2, where w = h(z). By using Theorem 2.1 with α = 1, the function h(z) given in (2.10) satisfies

which leads to the desired result. □

Remark 3

1. When b = 1, then Theorem 2.3 provides sufficient condition for functions to be in the class 𝒫_−λ.

2. Let |A| ≤ 1 and |B| ≤ 1 with A ≠ B. Assume a(α):=arg⁡((1+Az)/(1+Bz))α. Then from Theorem 2.1, we observe that max a(α₁) ≤ max a(α₂) and min a(α₁) ≥ min a(α₂), whenever 0 < α₁ ≤ α₂ ≤ 1. Therefore we have

   (1+Az1+Bz)α1≺(1+Az1+Bz)α2(0<α1≤α2≤1).

Theorem 2.4

Let |A_j| ≤ 1 and |B_j| ≤ 1 (j = 1, 2) with A₁ ≠ B₁ and A₂ ≠ B₂. Let C_j and R_j be the centres and radii of (1 + A_jz/(1 + B_jz) = ϕ_j(z) (j = 1, 2), respectively. Then

1. 𝒫(A₁, B₁) ⊆ 𝒫(A₂, B₂) if and only if the line segment C₁C₂ lies entirely in the domain ϕ₁(𝔻), whenever |C₁ − C₂| ≤ R₁.

2. 𝒫(A₁, B₁) ⊆ 𝒫(A₂, B₂) if and only if the line segment C₁C₂ does not lie entirely in the domain ϕ₁(𝔻), whenever |C₁ − C₂| ≥ R₁.

Corollary 2.4.1

If 𝒫(A₁, B₁) ⊆ 𝒫(A₂, B₂), then for 0 < α ≤ 1, we also have

Lemma 3.1. ([11])

Let h ∈ ℋ[1, n]. If there exists a point z₀ ∈ 𝔻 such that

for some β > 0, then we have

for some k ≥ n(a + a⁻¹)/2 > n, where p(z₀)^1/β = ±ia, and a > 0.

Lemma 3.2. ([13])

Let h(z) be analytic in 𝔻 with h(0) = 1 and h(z) ≠ 0. If there exist two points z₁, z₂ ∈ 𝔻 such that

for α₁, α₂ ∈ (0, 2] and |z| < |z₁| = |z₂|, then we have

where

Theorem 3.1

Let α ∈ (0, 1], l, m ∈ [−1, 1] and a, b, c, d ∈ [0, 1] such that ae^ilπ + b ≠ 0 and ce^imπ + d ≠ 0. Let Q ∈ ℋ[1, n] satisfy

and

for p ∈ ℋ₁. If

then

where γ = (tan⁻¹ A − tan⁻¹ B)/μ with A=cdsin⁡(mπ)cdcos⁡(mπ)+1 and B=absin⁡(lπ)abcos⁡(lπ)+1.

Proof

From Theorem 2.1, (3.1) yields

Similarly, from (3.2), we obtain

After some computations using (3.4) and (3.5), we obtain

which eventually yields

that completes the proof. □

Theorem 3.2

Let f, g ∈ 𝒜 and 0 < α ≤ 1. For some m ∈ [−1, 1) and β ∈ (0, 1), let a=itan⁡mπ4 and |g(z)/(zg′(z))| < β. If

then

where

Proof

Let p(z) := f(z)/g(z). Then in view of Theorem 2.2, to prove (3.8), it is sufficient to show −α(1 − m)π/2 ≤ arg p(z) ≤ α(1 + m)π/2. On the contrary, if there exist two points z₁, z₂ ∈ 𝔻 such that

for |z| < |z₁| = |z₂|, then by Lemma 3.2, we have

Since p(z) = f(z)/g(z), thus we have

and

For z = z₁, we have

which contradicts (3.7). Similarly, for z = z₂, we have

which again contradicts (3.7), that completes the proof. □

Corollary 3.2.1

Let f ∈ 𝒜 and 0 < α ≤ 1. For some m ∈ [−1, 1), let a=itan⁡mπ4. If

then

where μj=1+(−1)jm+2απtan−1⁡α(1−|a|)1+|a| (j = 1, 2) and μ=μ2−μ1μ1+μ2.

Further, f ∈ 𝒮𝒮^*(β), where β=απ+tan−1⁡α(1−|a|)1+|a|.

Proof

The subordination (3.9) holds, by using Theorem 3.2. Now just to prove f ∈ 𝒮𝒮^*(β). Since

using Theorem 2.2, we obtain

which implies f ∈ 𝒮𝒮^*(β). □

Remark 4

If m = 0, then Corollary 3.2.1 reduces to [14: Theorem 1.7].

Corollary 3.2.2

Let f ∈ 𝒜, g ∈ 𝒞 and g ∈ ℛ𝒮^*(β), where 0 ≤ β < 1. Suppose 0 < α ≤ 1 and for some m ∈ [−1, 1), let a=itan⁡mπ4. If

then

where μj=1−m+2απtan−1⁡αβ(1−|a|)1+|a|+α(1−|a|) (j = 1, 2) and μ=μ2−μ1μ1+μ2.

Proof

Since g ∈ 𝒞, from Marx-Strohhäcker’s theorem, we have Re(zg′(z)/g(z)) < 1/2, which implies that |g(z)/(zg′(z)) − 1| < 1. Thus we obtain

Now using (3.10) and the methodology of Theorem 3.2, the result follows at once. □

Remark 5

When we take m = 0 in Corollary 3.2.2, then the result reduces to [12: Theorem 1].

Theorem 3.3

Let p ∈ ℋ₁ and m ∈ [−1, 1). Also, for a fixed γ ∈ [0, 1] and α > 0, let β > β₀(≥ 0), where β₀ is the solution of the equation

and for a suitable fixed η ≥ 0, let λ(z) : 𝔻 → ℂ be a function satisfying

If

then

where μj=αβ(1+(−1)jm)+2γπtan−1⁡η (j = 1, 2), δ=μ1+μ22 and μ=μ2−μ1μ1+μ2.

Proof

According to Theorem 2.2, to prove (3.14), it is sufficient to show that −β(1−m)π2<arg⁡p(z)<β(1+m)π2. On the contrary if there exists two points z₁, z₂ ∈ 𝔻 such that

for |z| < |z₁| = |z₂|, then from Lemma 3.2, we have

where k≥1−|a|1+|a| and a=itan⁡mπ4. For z = z₁, using (3.12) we have

which contradicts (3.13). Similarly, for z = z₂, we have

which also contradicts (3.13) and this completes the proof of the theorem. □

Corollary 3.3.1

Let f ∈ 𝒜, g ∈ 𝒜 and λ(z) = g(z)/(zg′(z)), which satisfies (3.12). Also, for a fixed γ ∈ [0, 1], m ∈ [−1, 1), α > 0 and β be as defined in (3.11). If

then

where μ and δ are same as defined in Theorem 3.3.

Proof

Let p(z) = f(z)/g(z). By simple computation we have

Hence p ∈ ℋ₁, thus result follows from Theorem 3.3. □

Corollary 3.3.2

Let f ∈ 𝒜, g ∈ 𝒞 and g ∈ ℛ𝒮^*(ζ), where 0 ≤ ζ < 1. Also, for a fixed γ ∈ [0, 1], m ∈ [−1, 1) and α > 0. If

then

where μ and δ are same as defined in Theorem 3.3 and β > β₀(≥ 0) is the solution of the equation

Proof

Since g ∈ 𝒞, from Marx-Strohhäcker’s theorem, we have Re(zg′(z)/g(z)) > 1/2, which implies that |g(z)/(zg′(z)) − 1| < 1. Thus we obtain

which satisfies (3.12) with η = βζ/(1 + β). Now letting p(z) = f(z)/g(z), the result follows at once, by following the proof of Corollary 3.3.1. □

Lemma 4.1. ([10: Theorem 3.1d, p. 76])

Let h be analytic and starlike univalent in 𝔻 with h(0) = 0. If g is analytic in 𝔻 and zg′(z) ≺ h(z), then

Lemma 4.2. ([10: Theorem 3.4h, p. 132])

Let g(z) be univalent in 𝔻, Φ and Θ be analytic in a domain Ω containing g(𝔻) such that Φ(w) ≠ 0, when w ∈ g(𝔻). Now letting G(z) = zg′(z)·Φ(g(z)), h(z) = Θ(g(z)) + G(z) and either h or g is convex. Further, if

as well as p is analytic in 𝔻, with p(0) = g(0), p(𝔻) ⊂ Ω and

then p ≺ g, and g is the best dominant.

Theorem 4.1

Let p(z) ∈ ℋ₁, A ∈ ℂ and 0 ≤ b ≤ 1 with |A| ≤ 1, A + b ≠ 0 and Re(1 + Ab) ≥ |A + b|. Further let α, γ are two real parameters lying in [0, 1] and μ, δ, ρ and η are complex parameters such that Re(μ/η) > 0, Re δ > 0 and Re ρ ≥ 0. If