STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPEDDOMAIN

S. Sivaprasad Kumar1 and Kush Arora2

1,2Department of Applied MathematicsDelhi Technological University

Delhi-110042, Indiaspkumar@dce.ac.in

kush.arora1214@gmail.com

Abstract. This paper deals with some radius results and inclusion relations that are establishedfor functions in a newly defined subclass of starlike functions associated with a petal shapeddomain.

1. Introduction

Let the open unit disk {z ∈ C : |z| < 1} be represented by D and denote the class of allanalytic functions in D by H. Consider An as the class of analytic functions f in D representedby

f(z) = z + an+1zn+1 + an+2z

n+2 + ... (1.1)

In particular, denote A1 := A and let S be the subclass of A such that it involves all univalentfunctions f(z) in D. Let g, h be two analytic functions and ω be a Schwarz function satisfyingω(0) = 0 and |ω(z)| ≤ |z| such that g(z) = h(ω(z)) then g is said to be subordinate to h, org ≺ h. If h is univalent, then g ≺ h iff g(0) = h(0) and g(D) ⊂ h(D). Ma and Minda [11]introduced the univalent function ψ satisfying Reψ(D) > 0, ψ(D) starlike with respect toψ(0) = 1 and ψ′(0) > 0 and the domain ψ(D) being symmetric about the real axis. Further,they gave the definitions for the general subclasses of starlike and convex functions, respectivelyas follows:

S∗(ψ) := {f ∈ S : zf ′(z)/f(z) ≺ ψ(z)} (1.2)and

K(ψ) := {f ∈ S : 1 + zf ′′(z)/f ′(z) ≺ ψ(z)} . (1.3)For different choices of ψ, many subclasses of S∗ and K can be obtained. For example, thenotable classes of Janowski starlike and convex functions [8] are represented by S∗[C,D] :=S∗((1+Cz)/(1+Dz)) and K[C,D] := K((1+Cz)/(1+Dz)) for −1 ≤ D < C ≤ 1, respectively.Further, S∗α := S∗[1 − 2α,−1] and Kα := K∗[1 − 2α,−1] represents the classes of starlikeand convex functions of order α ∈ [0, 1), respectively. Note that S∗ := S∗0 and K := K0represent the well-known classes of starlike and convex functions, respectively. We denoteSS∗(γ) := S∗(((1 + z)/(1− z))γ) and SK(γ) := K(((1 + z)/(1− z))γ) representing the class ofstrongly starlike and strongly convex functions of order γ ∈ (0, 1] respectively.

Recall that for two subfamilies G1 and G2 of A, we say that r0 is the G1 – radius for theclass G2, if r0, (0 < r ≤ r0), is the greatest number which satisfies r−1g(rz) ∈ G1 where g ∈ G2.Moreover, starlike class S∗(ψ) for different ψ(z) were considered by many authors, whose works

2010 Mathematics Subject Classification. Primary 30C80, Secondary 30C45.Key words and phrases. Starlike function, Convex function, Petal shaped domain, Radius problems.

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2 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

examined the geometrical properties, radius results and coefficient estimates of the functions oftheir respective classes. Sokó l and Stankiewicz [20, 21] considered the class S∗L := S∗(

√1 + z)

and Mendiratta et al. [13] worked on the class S∗RL := S∗(√

2− (√

2− 1)((1− z)/((1 + 2(√

2−1)z)))1/2). Sharma et al. [19] studied the class S∗C := S∗(1 + 4z/3 + 2z2/3) while the classS∗s := S∗(1 + sin z) was examined by Cho et al. [6]. The classes S∗e := S∗(ez) and ∆∗ :=S∗(z +

√1 + z2) were considered by Mendiratta et al. [14] and Raina et al. [15], respectively.

Kargar et al. [10] introduced and studied the class BS∗(α) := S∗(1 + z/(1 − αz2)), α ∈ [0, 1],associated with the Booth lemniscate which was also investigated by Cho et al. [4]. Some morerecent work on radius problems can be found in [1, 3, 5, 7, 23].

Motivated by the classes defined in [6, 10, 13–15,19,21], we consider the petal shaped regionΩρ := {w ∈ C : | sinh(w − 1)| < 1}, which is characterised functionally as ρ(z) = 1 + sinh−1(z)to define our class. Clearly, ρ(z) is a Ma-Minda function. See Figure 2 for its boundary curveγ0 which is petal shaped. Note that sinh

−1(z) is a multivalued function and has the branchcuts along the line segments (−i∞,−i)∪ (i, i∞), on the imaginary axis and hence it is analyticin D. Now we introduce a new class of starlike functions

S∗ρ :={f ∈ A : zf

′(z)

f(z)≺ 1 + sinh−1(z)

}(z ∈ D), (1.4)

which is associated with the petal-shaped domain ρ(D). From the above definition, we deducethat f ∈ S∗ρ iff there exists an analytic function q(z) ≺ ρ(z) such that

f(z) = z exp

(∫ z0

q(t)− 1t

dt

). (1.5)

Table 1 presents some functions in the class S∗ρ where qj ≺ ρ.

j qj(z) fj(z)1 1 + z/5 z exp(z/5)2 (5 + 2z)/(5 + z) z + z2/53 (7 + 4z)/(7 + z) z(1 + z/7)3

Table 1. Some functions in the class S∗ρ

Since ρ is univalent in D, qj(D) ⊂ ρ(D) and qj(0) = ρ(0) (j = 1, 2, 3), it follows that eachqj ≺ ρ. Thus the functions fj(z) obtained from (1.5) are in the class S∗ρ . In particular, if wechoose

q(z) = 1 + sinh−1(z) = 1 + z − z3

6+

3z5

40− 5z

7

112...,

then (1.5) gives

f0(z) = z exp

(∫ z0

sinh−1(t)

tdt

)= z + z2 +

z3

2+z4

9− z

5

72− z

6

225· · · , (1.6)

which often acts as the extremal function for the class S∗ρ yielding sharp results.

Remark 1.1. Note that sinh−1(z) = ln(z+√

1 + z2). Let w = zf ′(z)/f(z), where f ∈ S∗ρ . Thenthe class S∗ρ can be alternatively represented by exp(w − 1) ≺ z +

√1 + z2, where z +

√1 + z2

represents the Crescent shaped domain [15]. Thus, there exists an exponential relation amongthe functions in the classes S∗ρ and ∆∗.

In the present investigation, the geometrical properties of the function 1 + sinh−1(z) arestudied and certain inclusion properties as well as radius problems are established for the classS∗ρ .

STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN 3

2. Properties of the function 1 + sinh−1(z)

The current section deals with the study of some geometric properties of the function 1 +sinh−1(z).

Theorem 2.1. The function ρ(z) = 1 + sinh−1(z) is a convex univalent function.

Proof. Let h(z) = sinh−1(z). Clearly, h(0) = 0. Since h′(z) = 1/√

1 + z2 and√

1 + z2 ≺√1 + z ∈ P , where P is the Carathéodory class. Therefore, 1/

√1 + z2 ∈ P which implies that

Reh′(z) > 0. Hence ρ is univalent. Now a calculation yields

1 +zh′′(z)

h′(z)=

1

1 + z2.

Since1

1 + z2≺ 1

1 + z∈ P .

Therefore, Re(1 + zh′′(z)/h′(z)) > 0 which implies that h (and thus ρ) is a convex univalentfunction. � �

Remark 2.1. Note that ρ′(0) > 0 and the function ϕ(z) = z +√

1 + z2 satisfies ϕ(z̄) = ϕ(z).

Therefore, ρ(z̄) = ρ(z) and hence, the domain Ωρ = ρ(D) is symmetric about the real axis.

Theorem 2.2. The domain Ωρ is symmetric about the line Re(w) = 1.

Proof. Since Ωρ is symmetric about the real axis, the condition 0 ≤ θ ≤ π/2 is sufficient toprove our result. As we know that symmetry along imaginary axis for f ∈ A holds if Re(f(θ)) =−Re(f(π − θ)) and Im(f(θ)) = Im(f(π − θ)). Now let h(z) = sinh−1(z) = ln(z +

√1 + z2).

Then Im(h(z)) = arg(z +√

1 + z2). For z = reit, t ∈ [0, π] and fixed r ∈ (0, 1), we have thefollowing expressions for t→ θ

I1 = arg(r(cos θ + i sin θ) +

√1 + r2(cos(2θ) + i sin(2θ))

)= arg

(z +√

1 + z2),

and for t→ π − θ

I2 = arg(r(cos(π − θ) + i sin(π − θ)) +

√1 + r2(cos(2(π − θ)) + i sin(2(π − θ)))

)= arg

(r(− cos θ + i sin θ) +

√1 + r2(cos(2θ)− i sin(2θ))

)= arg

(−z +

√1 + z2

).

Now let us consider (z +√

1 + z2)/(−z +√

1 + z2). On rationalising the denominator, we get

z +√

1 + z2

−z +√

1 + z2=

(z +√

1 + z2)(−z +√

1 + z2)

(−z +√

1 + z2)(−z +√

1 + z2)=

1

| − z +√

1 + z2|2= k > 0,

where k is some real positive constant. Thus,

arg

(z +√

1 + z2

−z +√

1 + z2

)= arg(k) = 0

⇒ arg(z +√

1 + z2)

= arg(−z +

√1 + z2

)⇒ I1 = I2.

Similarly, Re(h(θ)) = −Re(h(π − θ)) for 0 ≤ θ ≤ π/2. Hence, h(z) is symmetric about theimaginary axis and thus, by translation property, ρ(z) is symmetric about the line Re(w) =1. �

4 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

BA

C1

C2

ρ()

-2 -1 0 1 2 3-2

-1

0

1

2

Figure 1. ρ(D) lies in the annular region bounded between the circles C1 and C2.

Now using Theorem 2.2, we obtain the next result:

Corollary 2.1. The disk {w : |w − 1| ≤ sinh−1(r)} is contained in ρ(|z| ≤ r) and is maximal.

Proof. Since min|z|=r | sinh−1(z)| = | sinh−1(−r)| = sinh−1(r) and hence the conclusion canbe drawn at once. �

Theorem 2.3. We find that the following properties hold for ρ(z) = 1 + sinh−1(z):

(i) ρ(−r) ≤ Re ρ(z) ≤ ρ(r) (|z| ≤ r < 1);(ii) | Im ρ(z)| ≤ π/2 (|z| ≤ 1);

(iii) ρ(−r) ≤ |ρ(z)| ≤ ρ(r) (|z| ≤ r < 1);(iv) | arg ρ(z)| ≤ tan−1(1/t) where t = 4

π

√sinh−1(1)(1− sinh−1(1)).

Proof. (i) Since ρ(z) is convex and typically real, the value of Re ρ(z) falls betweenlimθ→0 ρ(re

θ) and limθ→π ρ(reθ), thus the result follows. (ii) Using Theorem 2.2, it suffices

to take θ ∈ [0, π/2]. Then the inequality follows by letting r tending to 1− and observing thatthe function

Im ρ(z) = arg(r cos(θ) +

√1 + r2(cos(2θ) + i sin(2θ)) + ir sin(θ)

)is strictly increasing in the interval [0, π/2] and hence the result follows at once. (iii) Theradially farthest and nearest points in ρ(D) from origin are respectively B and A (see Figure1) and therefore the result obviously holds. Moreover we observe that these points A and Blie on the real line and hence the bounds of |ρ(z)| and Re ρ(z) coincide. The proof of (iv) isevident from Theorem 3.1(iii) so skipped here. �

Next we have the following important result:

Lemma 2.1. For 1− sinh−1(1) < a < 1 + sinh−1(1), let ra be given by

ra =

{a− (1− sinh−1(1)), 1− sinh−1(1) < a ≤ 1;1 + sinh−1(1)− a, 1 ≤ a < 1 + sinh−1(1).

Then {w : |w − a| < ra} ⊂ Ωρ.

We omit the proof of Lemma 2.1 as it directly follows from Theorem 2.2 and Corollary 2.1.

Remark 2.2. Evidently the domain Ωρ is contained inside the disk {w : |w − 1| < π/2}.

STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN 5

3. Inclusion Relations

This section establishes some inclusion results involving the class S∗ρ with some well-knownclasses.

We consider the class M(β), first studied by Uralegaddi et al. [22], given by

M(β) :=

{f ∈ A : Re

(zf ′(z)

f(z)

)< β, z ∈ D, β > 1

},

and another interesting class introduced by Kanas and Wisńiowska [9] of k-starlike functions,denoted by k − ST and defined by

k − ST :={f ∈ A : Re

(zf ′(z)

f(z)

)> k

∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ , z ∈ D, k ≥ 0} .

Note that S∗ = 0−ST and S∗p = 1−ST , where S∗p is the class of parabolic starlike functions [17].We establish the following inclusion relations for the class S∗ρ .

Theorem 3.1. The class S∗ρ satisfies the following relationships:

(i) S∗ρ ⊂ S∗α ⊂ S∗ for 0 ≤ α ≤ 1− sinh−1(1);(ii) S∗ρ ⊂M(β) for β ≥ 1 + sinh−1(1);

(iii) S∗ρ ⊂ SS∗(γ) for (2/π) tan−1(1/t) ≤ γ ≤ 1 where t =4

π

√sinh−1(1)(1− sinh−1(1));

(iv) k − ST ⊂ S∗ρ for k ≥ 1 + 1/ sinh−1(1).

Proof. Consider f ∈ S∗ρ which implies zf ′(z)/f(z) ≺ 1 + sinh−1(z). By Theorem 2.3, it isevident that for z ∈ D,

1− sinh−1(1) = min|z|=1

Re(1 + sinh−1(z)) ≤ Re zf′(z)

f(z)

and

Rezf ′(z)

f(z)≤ max|z|=1

Re(1 + sinh−1(z)) = 1 + sinh−1(1).

This proves (i) and (ii). For (iii), let w ∈ C, X = Re(w), Y = Im(w), and b = 1−sinh−1(1). Nowconsider the parabolic domain ΓP with the boundary curve ∂ΓP = γp : Y

2 = 4a(X − b). Thenthe focus a of the smallest parabola γp which contains Ωρ will touch the peak points 1± iπ/2of S∗ρ is π2/(16 sinh−1(1)). Let P be any point on the parabola γP with parametric coordinates(b+at2, 2at) such that the tangent OE at P passes through origin for some parameter t. Let theequation of the tangent OE be y = mx, where m = dy/dx = (dy/dt)/(dx/dt) = 1/t. Thereforeat P, we have

m =y

x⇒ 1

t=

2at

b+ at2,

which yields

t =

√b

a=

4

π

√sinh−1(1)(1− sinh−1(1)) (3.1)

and the argument of the tangent at P of γp is tan−1(1/t). Since Ωρ ⊂ Γp, it gives∣∣∣∣arg zf ′(z)f(z)

∣∣∣∣ ≤ max|z|=1 arg(ρ(z)) = max arg(γp) = tan−1(1/t),which demonstrates f ∈ SS∗((2/π) tan−1(1/t)), where t is given by (3.1).

6 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

B

γ6 γ5

γ0 γ4

γ7 γ1 γ3

P

A O

F

E C

D

γ2

-1 1 2 3 Re

-2

-1

1

2Im γ0 : w = 1 + sinh

−1(z)γ1 : Re(w) = 1− sinh−1(1)γ2 : Re(w) = 1 + sinh

−1(1)γ3 : | arg(w)| ≤ tan−1(1/t)where t is given by (3.1)γ4 = γp : Y

2 = 4a(X − 1 + sinh−1(1))where a = π2/(16 sinh−1(1))γ5 = γk : Re(w) > k|w − 1|where k = 1 + 1/ sinh−1(1)γ6 = DL : |w − 1| < sinh−1(1)γ7 = DS : |w − 1| < π/2Re(A) = 1− sinh−1(1)Re(B) = 1 + sinh−1(1)Im(C) = − Im(D) = π/2arg(E) = − arg(F) = tan−1(1/t)where t is given by (3.1)O: originP: point of tangency of γ4 w.r.t. OE

Figure 2. Boundary curves, depicting some inclusion relations for w = 1 + sinh−1(z).

To show (iv), consider f ∈ k − ST along with the conic domain Γk = {w ∈ C : Rew >k|w− 1|}. For k > 1, let ∂Γk represent the horizontal ellipse γk : x2 = k2(x− 1)2 + k2y2 whichmay be rewritten as

(x− x0)2

a2+

(y − y0)2

b2= 1,

where x0 = k2/(k2− 1), y0 = 0, a = k/(k2− 1) and b = 1/

√k2 − 1. For γk ⊂ Ωρ, the condition

x0 + a ≤ 1 + sinh−1(1) must hold, or equivalently k ≥ 1 + 1/ sinh−1(1). Since Γk1 ⊆ Γk2 fork1 ≥ k2, it follows that for k ≥ 1 + 1/ sinh−1(1), k − ST ⊂ S∗ρ . Figure 2 clearly depicts theserelations. �

For our next result, we consider Pn[C,D], the class of functions p(z) of the form 1+∑∞

k=n ckzk,

satisfying p(z) ≺ (1 + Cz)/(1 + Dz), where −1 ≤ D < C ≤ 1. Denote by Pn(α) := Pn[1 −2α,−1] and Pn := Pn(0). For n=1, P = P1 is the Carathéodory class. We need the followinglemmas:

Lemma 3.1. [18] For p ∈ Pn(α), we have∣∣∣∣zp′(z)p(z)∣∣∣∣ ≤ 2(1− α)nrn(1− rn)(1 + (1− 2α)rn) , (|z| = r).

Lemma 3.2. [16] For p ∈ Pn[C,D], we have∣∣∣∣p(z)− 1− CDr2n1−D2r2n∣∣∣∣ ≤ (C −D)rn1−D2r2n , (|z| = r).

Especially, for p ∈ Pn(α), we have∣∣∣∣p(z)− 1 + (1− 2α)r2n1− r2n∣∣∣∣ ≤ 2(1− α)rn1− r2n , (|z| = r).

Theorem 3.2. Let −1 < D < C ≤ 1. If either of the following two conditions holds:(i) (1− sinh−1(1))(1−D2) < 1− CD ≤ 1−D2 and C −D ≤ (1−D) sinh−1(1);

(ii) 1−D2 ≤ 1− CD < (1 + sinh−1(1))(1−D2) and C −D ≤ (1 +D) sinh−1(1).

STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN 7

Then S∗[C,D] ⊂ S∗ρ .

Proof. Let f ∈ S∗[C,D] which implies zf ′(z)/f(z) ∈ P [C,D]. Using Lemma 3.2 we have∣∣∣∣zf ′(z)f(z) − 1− CD1−D2∣∣∣∣ ≤ (C −D)1−D2 . (3.2)

Let a = (1−CD)/(1−D2) and assume that (i) holds. Now multiplying 1 +D and dividing by(1−D2) on either sides of the inequality (C−D) ≤ (1−D) sinh−1(1) gives (C−D)/(1−D2) ≤a− (1− sinh−1(1)) on simplification. Also, the inequality (1− sinh−1(1))(1−D2) < 1−CD ≤1−D2 is equivalent to 1− sinh−1(1) < (1− CD)/(1−D2) ≤ 1. Therefore, from (3.2) we findw = zf ′(z)/f(z) is contained inside the disk |w − a| < ra, where ra = a− (1− sinh−1(1)) and1 − sinh−1(1) < a ≤ 1. Hence f ∈ S∗ρ by Lemma 2.1. A similar proof can be shown when (ii)holds. �

4. Radius Problems

In this section, radius results for various subclasses ofA are established. We begin by determiningsharp S∗α (0 ≤ α < 1), M(β) (β > 1) and k − ST -radii (k ≥ 0) for the class S∗ρ . UsingTheorem 3.1, we can establish that RS∗α(S∗ρ) = RM(β)(S∗ρ) = 1 for 0 ≤ α ≤ 1 − sinh

−1(1) and

β > 1 + sinh−1(1).

Theorem 4.1. If f ∈ S∗ρ , then the following results hold:

(i) For 1− sinh−1(1) ≤ α < 1, we have f ∈ S∗α in |z| ≤ sinh(1− α).(ii) For 1 < β ≤ 1 + sinh−1(1), we have f ∈M(β) in |z| ≤ sinh(β − 1).

(iii) For k > 0, we have f ∈ k − ST in |z| ≤ sinh(1/(k + 1)).

The results are sharp.

Proof. Since f ∈ S∗ρ , zf ′(z)/f(z) ≺ 1 + sinh−1(z) and hence for |z| = r < 1 Theorem 2.3gives

1− sinh−1(r) ≤ Re zf′(z)

f(z)≤ 1 + sinh−1(r),

thereby validating the first two parts. Also, the constants sinh(1 − α) and sinh(β − 1) areoptimal for the function f0 given by (1.6). Now to prove (iii), note that f ∈ k−ST in |z| < r,if

Re(1 + sinh−1(w(z))) ≥ k|1 + sinh−1(w(z))− 1| = k| sinh−1(w(z))|.Here w denotes the Schwarz function. Since Re(1 + sinh−1(w(z))) ≥ 1 − sinh−1(r) and| sinh−1(w(z))| ≤ sinh−1(r), the inequality Re(1+sinh−1(w(z)) ≥ k| sinh−1(w(z))| holds whenever1 − sinh−1(r) ≥ k sinh−1(r), which implies r ≤ sinh(1/(1 + k)). For the function f0 given by(1.6) and for z0 = − sinh(1/(1 + k)), we have

Rez0f

′0(z0)

f0(z0)= Re(1 + sinh−1(z0)) =

k

k + 1= k| sinh−1(z0)| = k

∣∣∣∣z0f ′0(z0)f0(z0) − 1∣∣∣∣ .

This concludes the proof. �

Corollary 4.1. Substituting k = 1 in part (iii) above, we find that f ∈ S∗ρ is parabolic starlike[17] in |z| ≤ sinh(1/2).

In the next result, we find the Kα-radius for the class S∗ρ .

8 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

Theorem 4.2. Let f ∈ S∗ρ . Then f ∈ Kα in |z| < rα, where rα is the least positive root of

(1− r2)√

1 + r2(1− sinh−1(r)

) (1− α− sinh−1(r)

)− r = 0 (0 ≤ α < 1). (4.1)

Proof. Let f ∈ S∗ρ and w be a Schwarz function. Then zf ′(z)/f(z) = 1 + sinh−1(w(z)) suchthat

1 +zf ′′(z)

f ′(z)= 1 + sinh−1(w(z)) +

zw′(z)

(1 + sinh−1(w(z)))√

1 + w2(z)

which yields

Re

(1 +

zf ′′(z)

f ′(z)

)≥ Re

(1 + sinh−1(w(z))

)−

∣∣∣∣∣ zw′(z)(1 + sinh−1(w(z)))√1 + w2(z)∣∣∣∣∣ .

We know for the Schwarz function w, the inequality |w′(z)| ≤ (1 − |w(z)|2)/(1 − |z|2) holds.Thus we observe that

Re

(1 +

zf ′′(z)

f ′(z)

)≥ 1− sinh−1(|z|)− |z|(1− |w(z)|

2)

(1− sinh−1(|z|))(1− |z|2)√

1 + |z|2

≥ 1− sinh−1(|z|)− |z|(1− sinh−1(|z|))(1− |z|2)

√1 + |z|2

.

Now consider the function q(r) := 1−sinh−1(r)−r/((1− sinh−1(r))(1− r2)

√1 + r2

). This is a

decreasing function in [0, 1) with q(0) = 1. Therefore Re(1 + zf ′′(z)/f ′(z)) > α in |z| < rα < 1,where rα is given as the least positive root of the equation q(r) = α, which is same as (4.1) andhence the result. �

Remark 4.1. Note for α = 0, r0 ≈ 0.37198 which is not sharp, so the result can be furtherimproved. The sharp K0-radius for the class S∗ρ is r0 ≈ 0.400435, which we can guess graphicallybut a mathematical proof is yet to derive.

For our next theorems 4.3 - 4.5, the following subclasses are required:Let S∗n[C,D] := {f ∈ An : zf ′(z)/f(z) ∈ Pn[C,D]}. Also, let S∗n(α) := S∗n[1 − 2α,−1] =An ∩ S∗α and S∗ρ,n := An ∩ S∗ρ . Further, Ali et al. [2] studied the three classes Sn := {f ∈ An :f(z)/z ∈ Pn}, S∗n[C,D] and

CSn(α) :={f ∈ An :

f(z)

g(z)∈ Pn, g ∈ S∗n(α)

}.

Now we obtain the S∗ρ,n-radii for the classes defined above.

Theorem 4.3. For the class Sn, the sharp S∗ρ,n-radius is given by:

RS∗ρ,n(Sn) =

sinh−1(1)n+

√n2 +

(sinh−1(1)

)21/n .

Proof. Let f ∈ Sn. Define s : D → C by s(z) = f(z)/z. Then s ∈ Pn and we can obtainzf ′(z)/f(z) − 1 = zs′(z)/s(z) from the above definition of s. Using Lemma 2.1 and Lemma3.1, the following holds ∣∣∣∣zf ′(z)f(z) − 1

∣∣∣∣ = zs′(z)s(z) ≤ 2nrn1− r2n ≤ sinh−1(1),or equivalently (sinh−1(1))r2n + 2nrn − sinh−1(1) ≤ 0. Therefore, the S∗ρ,n-radius of Sn is theleast positive root of (sinh−1(1))r2n + 2nrn − sinh−1(1) = 0 for r ∈ (0, 1). We can verify

STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN 9

Re(f0(z)/z) > 0 holds in D where f0(z) = z(1 + zn)/(1− zn). Thus f0 ∈ Sn and zf ′0(z)/f0(z) =1 + 2nzn/(1− z2n). Moreover, the result is sharp since at z = RS∗ρ,n(Sn), we obtain

zf ′0(z)

f0(z)− 1 = 2nz

n

1− z2n= sinh−1(1).

The proof is complete. �

Let F define the class of functions f ∈ A satisfying f(z)/z ∈ P . The radius of univalenceand starlikeness of the class F is

√2− 1, as shown in [12].

Corollary 4.2. For the class F , the S∗ρ -radius is stated as

RS∗ρ (F) = −e+√

1 + e2 ≈ 0.178105.

Theorem 4.4. For the class CSn(α), the sharp S∗ρ,n-radius is given by

RS∗ρ,n(CSn(α)) =

sinh−1(1)n− α + 1 +

√(n− α + 1)2 + (sinh−1(1) + 2(1− α)) sinh−1(1)

1/n .Proof. Let f ∈ CSn(α) and g ∈ S∗n(α). Considering s(z) = f(z)/g(z), clearly indicates

s ∈ Pn. Also, it giveszf ′(z)

f(z)=zs′(z)

s(z)+zg′(z)

g(z).

The use of Lemmas (3.1 – 3.2) gives us∣∣∣∣zf ′(z)f(z) − 1 + (1− 2α)r2n1− r2n∣∣∣∣ ≤ 2(n− α + 1)rn1− r2n . (4.2)

Considering (1 + (1 − 2α)r2n)/(1 − r2n) ≥ 1, the relation f ∈ S∗ρ,n follows from (4.2) andLemma 2.1 if the subsequent inequality is true:

1 + 2(n− α + 1)rn + (1− 2α)r2n

1− r2n≤ 1 + sinh−1(1)

or equivalently, (2− 2α+ sinh−1(1))r2n + 2(n−α+ 1)rn− sinh−1(1) ≤ 0 holds. Thus, the leastpositive root of

(2− 2α + sinh−1(1))r2n + 2(n− α + 1)rn − sinh−1(1) = 0

gives the S∗ρ,n-radius for the class CSn(α). Next examine the following functions

f0(z) =z(1 + zn)

(1− zn)(n+2−2α)/nand g0(z) =

z

(1− zn)2(1−α)/n, (4.3)

which implies f0(z)/g0(z) = (1 + zn)/(1 − zn) and zg′0(z)/g0(z) = (1 + (1 − 2α)zn)/(1 − zn).

Moreover, it is obvious that Re(f0(z)/g0(z)) > 0 and Re(zg′0(z)/g0(z)) > α in the unit disk D.

Hence f0 ∈ CSn(α). At z = RS∗ρ,n(CSn(α)), the function f0 defined in (4.3) satisfies

zf ′0(z)

f0(z)=

1 + 2(n− α + 1)zn + (1− 2α)z2n

1− z2n= 1 + sinh−1(1),

which accomplish sharpness of the result. �

10 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

Theorem 4.5. For the class S∗n[C,D], the S∗ρ,n-radius is given by

RS∗ρ,n(S∗n[C,D]) =

{min{1;R1}, −1 ≤ D < 0 < C ≤ 1;min{1;R2}, 0 < D < C ≤ 1,

where

R1 :=

2 sinh−1(1)C −D +

√(C −D)2 + 4(D2(1 + sinh−1(1))− CD) sinh−1(1)

1/n

and

R2 :=

2 sinh−1(1)C −D +

√(C −D)2 + 4(D2(sinh−1(1)− 1) + CD) sinh−1(1)

1/n .Proof. Let f ∈ S∗n[C,D]. From Lemma 3.2, we have∣∣∣∣zf ′(z)f(z) − b

∣∣∣∣ ≤ (C −D)rn1−D2r2n , (4.4)where b = (1− CDr2n)/(1−D2r2n), |z| = r, represents the center of the disk. We infer b ≥ 1for −1 ≤ D < 0 < C ≤ 1. From Lemma 2.1, f ∈ S∗ρ,n depends on whether following conditionis true:

1 + (C −D)rn − CDr2n

1−D2r2n≤ 1 + sinh−1(1),

which reduces to

r ≤

2 sinh−1(1)C −D +

√(C −D)2 + 4(D2(1 + sinh−1(1))− CD) sinh−1(1)

1/n = R1.Further, taking D = 0, we get b = 1. Then (4.4) yields∣∣∣∣zf ′(z)f(z) − 1

∣∣∣∣ ≤ Crn, (0 < C ≤ 1).Now applying Lemma 2.1 with a = 1 gives f ∈ S∗ρ,n, if r ≤ ((sinh−1(1))/C)1/n.

For 0 < D < C ≤ 1, we have b < 1. Thus, using Lemma 2.1 and (4.4), we have f ∈ S∗ρ,n ifthe following holds:

CDr2n + (C −D)rn − 11−D2r2n

≤ sinh−1(1)− 1,

or equivalently, if

r ≤

2 sinh−1(1)C −D +

√(C −D)2 + 4(D2(sinh−1(1)− 1) + CD) sinh−1(1)

1/n = R2.This concludes the proof. �

The next theorem establishes radius results for some well-known classes mentioned earlier.

Theorem 4.6. The sharp S∗ρ -radii for the classes S∗L,S∗RL,S∗C ,S∗e ,∆∗ andBS∗(α) are:

(i) RS∗ρ (S∗L) = sinh−1(1)(2− sinh−1(1)) ≈ 0.985928.

(ii) RS∗ρ (S∗RL) =(2 + (1 +

√2) sinh−1(1)

)sinh−1(1)

5− 3√

2 +(4(√

2− 1) + 2 sinh−1(1))

sinh−1(1)≈ 0.964694.

STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN 11

(iii) RS∗ρ (S∗C) =1

2

(√2(2 + 3 sinh−1(1)

)− 2)≈ 0.523831.

(iv) RS∗ρ (S∗e ) = ln(1 + sinh−1(1)) ≈ 0.632002.

(v) RS∗ρ (∆∗) =

sinh−1(1)(2 + sinh−1(1))

2(1 + sinh−1(1))≈ 0.674924.

(vi) RS∗ρ (BS∗(α)) =

−1 +√

1 + α(2 sinh−1(1)

)22α sinh−1(1)

, α ∈ [0, 1].

Proof.

(i) Suppose f ∈ S∗L. We have zf ′(z)/f(z) ≺√

1 + z. When |z| = r, we obtain∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ ≤ 1−√1− r ≤ sinh−1(1),

such that r ≤ (2− sinh−1(1)) sinh−1(1) = RS∗ρ (S∗L) holds. Next examine the function

f0(z) =4z(

1 +√

1 + z)2 e2(√1+z−1).

Since zf ′0(z)/f0(z) =√

1 + z, it follows that f0 ∈ S∗L. As zf ′0(z)/f0(z) − 1 = − sinh−1(1)

is obtained at z = −RS∗ρ (S∗L), the result is sharp.(ii) Suppose f ∈ S∗RL, we obtain

zf ′(z)

f(z)≺√

2− (√

2− 1)√

1− z(1 + 2(

√2− 1)z)

.

For |z| = r, the subsequent inequality holds∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ ≤ 1−√2 + (√2− 1)

√1 + r

(1− 2(√

2− 1)r)≤ sinh−1(1),

provided

r ≤(2 + (1 +

√2) sinh−1(1)

)sinh−1(1)

5− 3√

2 +(4(√

2− 1) + 2 sinh−1(1))

sinh−1(1)= RS∗ρ (S

∗RL).

Next observe the following function defined as

f0(z) = z exp

(∫ z0

q0(t)− 1t

dt

),

where

q0(t) =√

2− (√

2− 1)√

1− t(1 + 2(

√2− 1)t)

.

From the definition of f0, at z = −RS∗ρ (S∗RL), we have

zf ′0(z)

f0(z)=√

2− (√

2− 1)√

1− z(1 + 2(

√2− 1)z)

= 1− sinh−1(1).

which confirms the sharpness.

12 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

(iii) Suppose f ∈ S∗C . So zf ′(z)/f(z) ≺ 1 + 4z/3 + 2z2/3. This gives∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ ≤ 4r3 + 2r23 ≤ sinh−1(1), |z| = r,

for r ≤ 12

(√2(2 + 3 sinh−1(1)

)− 2)

= RS∗ρ (S∗C). The sharpness of the result is establishedusing the subsequent function

f0(z) = z exp

(4z + z2

3

),

where zf ′0(z)/f0(z) = 1 + (4z+ 2z2)/3 yields f0 ∈ S∗ρ , and substituting z = RS∗ρ (S∗C) gives

zf ′0(z)/f0(z) = 1 + sinh−1(1), thereby proving the sharpness.

(iv) Suppose f ∈ S∗e , we have zf ′(z)/f(z) ≺ ez, which yields∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ ≤ er − 1 ≤ sinh−1(1) holds in |z| = r,

provided r ≤ ln(1 + sinh−1(1)) = RS∗ρ (S∗e ). Now Consider

f0(z) = z exp

(∫ z0

et − 1t

dt

).

Since zf ′0(z)/f0(z) = ez, f0 ∈ S∗e , and at z = RS∗ρ (S∗e ), we have zf ′0(z)/f0(z) = 1 +

sinh−1(1), which shows the sharpness of the result.(v) Suppose f ∈ ∆∗ which gives zf ′(z)/f(z) ≺ z +

√1 + z2. Then,∣∣∣∣zf ′(z)f(z) − 1

∣∣∣∣ ≤ r +√1 + r2 − 1 ≤ sinh−1(1), |z| = r,for r ≤ sinh

−1(1)(2 + sinh−1(1))

2(1 + sinh−1(1))= RS∗ρ (∆

∗). For sharpness, define f0 as

f0(z) = z exp

(∫ z0

t+√

1 + t2 − 1t

dt

).

Since zf ′0(z)/f0(z) = z +√

1 + z2, f0 ∈ ∆∗, so at z = RS∗ρ (∆∗), we have zf ′0(z)/f0(z)= 1 + sinh−1(1) which shows the sharpness of the result.

(vi) Suppose f ∈ BS∗(α), α ∈ [0, 1], which gives zf ′(z)/f(z) ≺ 1 + z/(1− αz2). Then,∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ ≤ r1− αr2 ≤ sinh−1(1), |z| = r,

for r ≤−1 +

√1 + α

(2 sinh−1(1)

)22α sinh−1(1)

= RS∗ρ (BS∗(α)), α ∈ (0, 1]. For α = 0, r ≤ sinh−1(1).

Next examine the function f0 defined as

f0(z) = z

(1 +√αz

1−√αz

)1/(2√α).

Since zf ′0(z)/f0(z) = 1 + z/(1− αz2), f0 ∈ (BS∗(α)), so at z = −RS∗ρ ((BS∗(α))), we have

zf ′0(z)/f0(z) = 1− sinh−1(1), which ensures sharpness of the result.

Note thatRS∗ρ (BS∗(1)) =

(−1 +

√1 + (2 sinh−1(1))2/(2 sinh−1(1))

)≈ 0.58241 andRS∗ρ (BS

∗(0))

= sinh−1(1) ≈ 0.881374. �

STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN 13

Next we present some radius problems for certain classes of functions expressed as ratio offunctions:

F1 :={f ∈ An : Re

(f(z)

g(z)

)> 0 and Re

(g(z)

z

)> 0, g ∈ An

},

F2 :={f ∈ An : Re

(f(z)

g(z)

)> 0 and Re

(g(z)

z

)> 1/2, g ∈ An

},

and

F3 :={f ∈ An :

∣∣∣∣f(z)g(z) − 1∣∣∣∣ < 1 and Re(g(z)z

)> 0, g ∈ An

}.

Theorem 4.7. For functions in the classes F1, F2 and F3, the sharp S∗ρ,n-radii respectively,are:

(i) RS∗ρ,n(F1) =

√

4n2 + (sinh−1(1))2 − 2nsinh−1(1)

1/n.(ii) RS∗ρ,n(F2) =

√

9n2 + 4 sinh−1(1)(n+ sinh−1(1))− 3n2(n+ sinh−1(1))

1/n.(iii) RS∗ρ,n(F3) = RS∗ρ,n(F2).

Proof.

(i) Let f ∈ F1 and consider the functions s, d : D → C where s(z) = f(z)/g(z) and d(z) =g(z)/z. Clearly, s, d ∈ Pn. As f(z) = zd(z)s(z), applying Lemma 3.1 here gives∣∣∣∣zf ′(z)f(z) − 1

∣∣∣∣ ≤ 4nrn1− r2n ≤ sinh−1(1)such that

r ≤

√

4n2 + (sinh−1(1))2 − 2nsinh−1(1)

1/n = RS∗ρ,n(F1)holds. Next examine the functions

f0(z) = z

(1 + zn

1− zn

)2and g0(z) = z

(1 + zn

1− zn

).

Evidently, Re(f0(z)/g0(z)) > 0 and Re(g0(z)/z) > 0, which implies f0 ∈ F1. Furthercalculation yields at z = RS∗ρ,n(F1)eiπ/n

zf ′0(z)

f0(z)= 1 +

4nzn

1− z2n= 1− sinh−1(1),

which validates the result is sharp.(ii) Let f ∈ F2 and consider the functions s, d : D → C where s(z) = f(z)/g(z) and d(z) =

g(z)/z. Clearly, s ∈ Pn(1/2) and d ∈ Pn. As f(z) = zd(z)s(z), applying 3.1 here gives∣∣∣∣zf ′(z)f(z) − 1∣∣∣∣ ≤ 2nrn1− r2n + nrn1− rn = 3nrn + nr2n1− r2n ≤ sinh−1(1),

whenever

r ≤

√

9n2 + 4 sinh−1(1)(n+ sinh−1(1))− 3n2(n+ sinh−1(1))

1/n = RS∗ρ,n(F2).

14 STARLIKE FUNCTIONS ASSOCIATED WITH A PETAL SHAPED DOMAIN

Therefore, f ∈ S∗ρ,n holds for r ≤ RS∗ρ,n(F2). Next observe that Re(g0(z)/z) > 1/2 whileRe(f0(z)/g0(z)) > 0 for the functions

f0(z) =z(1 + zn)

(1− zn)2and g0(z) =

z

1− zn.

Therefore f0 ∈ F2 which verifies the sharpness for z = RS∗ρ,n(F2) such thatzf ′0(z)

f0(z)− 1 = 3nz

n + nz2n

1− z2n= sinh−1(1).

(iii) Let f ∈ F3 and consider the functions s, d : D → C where s(z) = g(z)/f(z) and d(z) =g(z)/z. Then d ∈ Pn. We can verify that |1/s(z)− 1| < 1 holds whenever Re(s(z)) > 1/2and therefore s ∈ Pn(1/2). As f(z) = zd(z)/s(z), on applying Lemma 3.1, we obtain∣∣∣∣zf ′(z)f(z) − 1

∣∣∣∣ ≤ 3nrn + nr2n1− r2n ≤ sinh−1(1).The rest of the proof is omitted as it is analogous to proof of Theorem 4.7(ii). Thesharpness can be verified as follows. Examine the functions

f0(z) =z(1 + zn)2

1− znand g0(z) =

z(1 + zn)

1− zn.

Using above definitions of f0 and g0, we see that

Re

(g0(z)

f0(z)

)= Re

(1

1 + zn

)>

1

2and Re

(g0(z)

z

)= Re

(1 + zn

1− zn

)> 0,

and therefore, f0 ∈ F3. Now at z = RS∗ρ,n(F3)eiπ/n, we obtainzf ′0(z)

f0(z)− 1 = 3nz

n − nz2n

1− z2n= − sinh−1(1),

which serves as validation for the sharp result.

This concludes the proof. �

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1. Introduction2. Properties of the function 1+sinh-1(z)3. Inclusion Relations4. Radius ProblemsReferences