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On Class of Analytic Function Defined by Generalized Ruscheweyh 

Derivative 

 

Rashmi B T.*and Dileep L.** 

*Adichunchanagiri Institute Of Technology, Chikkamagaluru, India-577101 

**Vidyavardhaka College Of Engineering, Mysuru, India-570 002 

Visvesvaraya Technological University, Belagavi 

 

Article History: 

Received: 12-01-2025 

Revised: 15-02-2025 

Accepted: 01-03-2025 

Abstract: The aim of this paper is to introduce a class of analytic functions defined by using 

generalized Ruscheweyh derivative. The coefficient bound, inclusion result and a radius 

problem has been discussed in this paper. 

 Keywords: Univalent functions, Analytic function, Generalized Ruscheweyh Operator, 

coefficient inequalities, convex domain. 

AMS Classification: Primary 30C45; Secondary 30C50;30C80 

Conclusion: Here, in our present investigation, we have successfully introduced a new 

subclass of analytic functions 𝒱𝑘,𝜆
𝑚 [𝐴, 𝐵, 𝛼, 𝑏] using the Generalized Ruscheweyh derivative 

operator. Many properties and characteristics of this newly defined function class such as 

coefficient estimates, inclusion bounds and radius problem have been studied. 

 

1. Introduction 

Let 𝒜 be the class of functions of the form 

(1.1)                                                             f(z)= z+∑ 𝑎𝑛𝑧𝑛∞
𝑛=2  

which are analytic in the open unit disk 𝒰 = {z : |z| < 1}. If f and g are analytic in 𝒰, we say that f is 

subordinate to g, written f ≺ g or f(z) ≺ g(z), if there exists a Schwarz function 𝜔(z) in 𝒰 such that 

f(z) = g(𝜔 (z)). 

Let P[A, B] be the class of functions h, analytic in 𝒰 with h(0) = 1 and 

ℎ(𝑧) ≺
1+𝐴𝑧

1+𝐵𝑧
,     −1 ≤ B < A ≤ 1. 

This class was introduced by Janowski [18]. The class P [A, B] is connected with the class P of 

functions with positive real parts by the relation 

(1.2)                                                       ℎ ∈ P[A, B] ⟺
(𝐵−1)ℎ−(𝐴−1)

(𝐵+1)ℎ−(𝐴+1)
∈ 𝑃. 

 

Later Polato�̌�lu [19] defined the class P[A,B,α] as: 

Let P[A,B,α] be the class of functions 𝑝1, analytic in 𝒰 with 𝑝1 (0) = 1 and 



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(1.3)                                   𝑝1(𝑧) ≺
1+{(1−𝛼)𝐴+𝛼𝐵}𝑧

1+𝐵𝑧
,    where   −1 ≤ B < A ≤ 1,   0 ≤ α < 1. 

From (1.3), it can easily be seen that, 𝑝1 ∈ P[A,B,α], if and only if, there exists h ∈ P[A,B] such that 

(1.4)                                                  𝑝1(z) = (1 − α)h(z) + α, 0 ≤ α < 1, z ∈ 𝒰. 

It is also noted that P [1,−1,0] ≡ P, the well-known class of analytic functions in 𝒰 with positive real 

part. Noor [5] considered the generalized class 𝑃𝑘[𝐴, 𝐵, α] of Janowski functions which is defined as 

follows. 

A function 𝑝 is said to be in the class 𝑃𝑘[𝐴, 𝐵, α], if and only if, 

(1.5)                                                               𝑝(𝑧) = (
𝑘

4
+

1

2
) 𝑝1(z) - (

𝑘

4
−

1

2
) 𝑝2(z)                    

where 

𝑝1,   𝑝2 ∈ P [A, B, α], −1 ≤ B < A ≤ 1, k ≥ 2 and 0 ≤ α < 1. It is clear that 𝑃2[𝐴, 𝐵, α] ≡ P[A,B,α] and 

𝑃𝑘[1, −1, 0] ≡ 𝑃𝑘,  the well-known class given and studied by Pinchuk[3]. 

For any two analytic functions 

𝑓1(z)= ∑ 𝑎𝑛𝑧𝑛∞
𝑛=0    and   𝑓2(z)= ∑ 𝑏𝑛𝑧𝑛∞

𝑛=0    (𝑧 ∈  𝒰) 

the convolution of 𝑓1 and 𝑓2 is defined by 

 

(1.6)                                                 ((𝑓1 ∗ 𝑓2)(𝑧) = ∑ 𝑎𝑛𝑏𝑛𝑧𝑛∞
𝑛=1         

 

The Generalized Ruscheweyh Derivative 𝒟𝜆
𝑚 [4] is defined as follows, 

For f ∈ 𝒜, λ ≥ 0 and m ∈ℝ, m > −1, we have 

 

(1.7)                                              𝐷𝜆
𝑚𝑓(𝑧) =

𝑧

(1−𝑧)𝑚+1*𝐷𝜆𝑓(𝑧),   𝑧 ∈ 𝒰. 

 

(1.8)                                      (𝑚 + 1)𝒟𝜆
𝑚+1𝑓(𝑧) = 𝑚𝒟𝜆

𝑚𝑓(𝑧) + 𝑧(𝒟𝜆
𝑚𝑓(𝑧))′ 

For function f ∈ 𝒜 of the form (1.1), we obtain the power series expansion of the 

form, 

(1.9)                     𝒟𝜆
𝑚𝑓(𝑧) = 𝑧 + ∑ [1 + (𝑛 − 1)𝜆]

(𝑚+1)(𝑛−1)

(1)(𝑛−1)
𝑎𝑛𝑧𝑛∞

𝑛=2 ,      𝑧𝜖 𝒰         

where  

(a)n =
Γ(a + n)

Γ(a)
= {

 1,                                                      for n = 0              
a(a + 1)(a + 2) … (a + n − 1),        for n ∈ ℕ

 

 



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Definition 1.1.  A function f ∈ 𝒜  is in the class 𝒱𝑘,𝜆
𝑚 [𝐴, 𝐵, 𝛼, 𝑏] if and only if , 

(1 −
2

𝑏
+

2

𝑏

𝐷𝜆
𝑚+1𝑓(𝑧)

𝐷𝜆
𝑚𝑓(𝑧)

) ∈ 𝑃𝑘[𝐴, 𝐵, 𝛼],     z 𝜖 𝒰 , 

where  k≥ 2,  m ≥0,  -1≤ 𝐵 < 𝐴 ≤1,  0 ≤ 𝛼<1 and b ∈  ℂ − {0}. 

Assigning certain values to different parameters, we have different well-known classes of analytic 

functions as can be seen below. 

Special cases 

   (i)  For a parametric value λ = 1; we get the class studied by S.N. Malik, M. Arif, K.I. Noor and  

         M. Raza.[15] 

  (ii)  𝒱𝑘
λ[1, −1, 𝛼, 𝑏] ≡  𝑉𝑘(𝑎, 𝑏, λ), the well-known class defined by Latha and Nanjunda Rao in 

[14]. 

  (iii)  𝒱2
1[𝐴, 𝐵, 𝛼, 1] ≡ C[A, B, 𝛼] 

       𝒱2
0[𝐴, 𝐵, 𝛼, 2] ≡ 𝑆∗[𝐴, 𝐵, 𝛼] ,the well-known class defined by Polato�̌�lu [19] 

  (iv)  𝒱𝑘
1[𝐴, 𝐵, 0,1] ≡  𝑉𝑘[𝐴, 𝐵], 

          𝒱2
0[𝐴, 𝐵, 0,2] ≡ 𝑅𝑘[𝐴, 𝐵] , where 𝑉𝑘[𝐴, 𝐵] and 𝑅𝑘[𝐴, 𝐵]  denote the class of janowski functions 

with bounded boundary and bounded radius rotations respectively, given by Noor [9]. 

  

2. PRELIMINARY RESULTS 

We need the following results to obtain our main results. 

Lemma 2.1. Let  𝑝(𝑧) = 1 + ∑ 𝑞𝑛𝑧𝑛∞
𝑛=1 ∈ 𝑃𝑘[𝐴, 𝐵, 𝛼].  Then, for all n ≥ 1, 

(2.1)                                                   |𝑞𝑛| ≤
𝑘(𝐴−𝐵)(1−𝛼)

2
 

This inequality is sharp. 

The proof follows from (1.4), (1.5) and the coefficient bound of h ∈ P[A, B] given by Aouf [13]. 

Lemma 2.2.  [17] Let  𝑢 = 𝑢1 + 𝑖𝑢2, 𝑣 = 𝑣1 + 𝑖𝑣2 and 𝜓(𝑢, 𝑣) be a complex valued 

function satisfying the conditions: 

(i) 𝜓(𝑢, 𝑣) is continuous in a domain , 𝐷 ⊂ ℂ2 

(ii) (1,0) ∈𝐷 and Re 𝜓(1,0)> 0, 

(iii) Re 𝜓(𝑖𝑢2, 𝑣1)≤ 0, whenever (𝑖𝑢2, 𝑣1) ∈𝐷 and 𝑣1 ≤ −
1

2
(1 + 𝑢2).  . 

If h(z) = 1 + c1z + ... is a function analytic in 𝒰 such that (h(z),zh′(z)) ∈ D and Re 𝜓 (h(z),zh′(z)) > 0 

for z ∈ 𝒰, then Reh(z) > 0 in 𝒰. 

Lemma 2.3.  Let 𝑝 ∈  𝑃𝑘[𝐴, 𝐵, 0] with k ≥ 2.  Then, for |z| = r < 1, 



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(2.2)                                
2−(𝐴−𝐵)𝑘𝑟−2𝐴𝐵𝑟2

2(1−𝐵2𝑟2)
≤ 𝑅𝑒 𝑝(𝑧) ≤ |𝑝(𝑧)| ≤

2+(𝐴−𝐵)𝑘𝑟−2𝐴𝐵𝑟2

2(1−𝐵2𝑟2)
 

The proof is immediate by using (1.5) and the growth result of h ∈ P [A, B], see [13]. 

Lemma 2.4. Let ∈  𝑃𝑘[𝐴, 𝐵, 0] with k ≥ 2. Then, for |z| = r < 1.  

(2.3)                                                        |𝑧𝑝′(𝑧)| ≤
𝑟{(𝐴−𝐵)𝑘−4𝐵(𝐴−𝐵)𝑟+𝐵2(𝐴−𝐵)𝑘𝑟2} 𝑅𝑒 𝑝(𝑧)

(1−𝐵2𝑟2)(2+(𝐴−𝐵)𝑘𝑟−2𝐴𝐵𝑟2)
 

The result follows directly by using Lemma 2.3 

 

3. MAIN RESULTS  

Theorem 3.1. Let  f ∈ 𝒱𝑘,𝜆
𝑚 [𝐴, 𝐵, 𝛼, 𝑏] with -1≤ 𝐵 < 𝐴 ≤1, m ≥0,  0 ≤ 𝛼<1 and    

    b ∈  ℂ − {0}. Then  

    (3.1)                                        |𝑎𝑛| ≤
(𝜎)𝑛−1

(𝑛−1)! 𝜙𝑛(𝑚)
,     ∀ 𝑛 ≥ 2, 

    Where  

𝜎 =
𝑘|𝑏|(𝐴−𝐵)(1−𝛼)(𝑚+1)

4
     and    𝜙𝑛(𝑚)=[1+(n-1)𝜆]

(𝑚+1)𝑛−1

(1)𝑛−1
. 

    This result is sharp 

Proof:  Let 

(3.2)                                                  1 −
2

𝑏
+

2

𝑏

𝐷𝜆
𝑚+1𝑓(𝑧)

𝐷𝜆
𝑚𝑓(𝑧)

= 𝑝(𝑧) 

so that 𝑝 ∈  𝑃𝑘[𝐴, 𝐵, 𝛼]. Let 𝑝(𝑧) = 1 + ∑ 𝑞𝑛𝑧𝑛∞
𝑛=1 .      Then (3.2) can be written as 

2(𝐷𝜆
𝑚+1𝑓(𝑧) − 𝐷𝜆

𝑚𝑓(𝑧))=𝑏𝐷𝜆
𝑚𝑓(𝑧) ∑ 𝑞𝑛𝑧𝑛∞

𝑛=1  

which implies that 

2𝜙𝑛(𝑚)(𝑛−1)𝑎𝑛

(𝑚+1)
= 𝑏(𝑞𝑛−1 + 𝜙2(𝑚)𝑎2𝑞𝑛−2 + ⋯ + 𝜙𝑛−1(𝑚)𝑎𝑛−1𝑞1). 

Using Lemma 2.1, we obtain 

|𝑎𝑛| ≤
𝑘|𝑏|(𝐴 − 𝐵)(1 − 𝛼)(𝑚 + 1)

4(𝑛 − 1)𝜙𝑛(𝑚)
(1 + 𝜙2(𝑚)|𝑎2| + ⋯ + 𝜙𝑛−1(𝑚)|𝑎𝑛−1|) 

 = 
𝜎

(𝑛−1)𝜙𝑛(𝑚)
(1 + ∑ 𝜙𝑖(𝑚)|𝑎𝑖|

𝑛−1
𝑖=2 ). 

 

For n=2,   |𝑎2| ≤
𝜎

𝜙2(𝑚)
=

(𝜎)2−1

(2−1)! 𝜙2(𝑚)
 

   

Therefore (3.1) holds for n=2.   Assume that (3.1) is true for n = l and consider 



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                           |𝑎𝑙+1| ≤
𝜎

𝑙 𝜙𝑙+1(𝑚)
(1 + ∑ 𝜙𝑖(𝑚)|𝑎𝑖|

𝑙
𝑖=2 ) 

                                         ≤
𝜎

𝑙 𝜙𝑙+1(𝑚)
(1 + ∑

(𝜎)𝑖−1

(𝑖−1)!

𝑙
𝑖=2 ) 

                                         =  
𝜎

𝑙 𝜙𝑙+1(𝑚)
(1 + ∑ 𝜎 ∏ (1 +

𝜎

𝑗
)𝑖−1

𝑗=1
𝑙
𝑖=2 ) 

                                        = 
𝜎

𝑙 𝜙𝑙+1(𝑚)
∏ (1 +

𝜎

𝑗
)𝑙−1

𝑗=1  

                                        =
(𝜎)𝑙

𝑙 ! 𝜙𝑙+1(𝑚)
.                                

Therefore, the result is true for n = l + 1. Using mathematical induction, (3.1) holds true for all n ≥ 2. 

This result is sharp for m ≥ 0, 0 ≤ α < 1, b ∈ ℂ− {0} and k ≥ 2 as can be seen 

from the functions 𝑓0(𝑧) which are given as 

           1 −
2

𝑏
+

2

𝑏

𝐷𝜆
𝑚+1𝑓0(𝑧)

𝐷𝜆
𝑚𝑓0(𝑧)

= (1 − 𝛼) [(
𝑘

4
+

1

2
)

1+𝐴𝑧

1+𝐵𝑧
− (

𝑘

4
−

1

2
)

1−𝐴𝑧

1−𝐵𝑧
] + 𝛼. 

For different values of A, B, α, b and λ, we obtain the following results [16]. 

Corollary 3.2. If  f ∈ 𝒱𝑘,𝜆
0 [1, −1, 𝛼, 2] = 𝑅𝑘(𝛼) , then 

|𝑎𝑛| ≤
(𝑘(1 − 𝛼))

𝑛−1

(𝑛 − 1)!
,     ∀ 𝑛 ≥ 2 

this result is sharp. 

Corollary 3.3. If  f ∈ 𝒱𝑘,𝜆
1 [1, −1, 𝛼, 1] = 𝑉𝑘(𝛼), then 

|𝑎𝑛| ≤
(𝑘(1 − 𝛼))

𝑛−1

(𝑛)!
,     ∀ 𝑛 ≥ 2. 

this result is sharp. 

Theorem 3.4. For real b > 0, 

𝒱𝑘,𝜆
𝑚+1[𝐴, 𝐵, 𝛼, 𝑏] ⊆ 𝒱𝑘,𝜆

𝑚 [1, −1, 𝛽, 𝑏 + 1],   z 𝜖 𝒰, 

where β (0 ≤ β < 1) is one of the roots of 

(3.3)       𝜂1𝜂2𝑏2(𝑚 + 2)2(1 − 𝛼)2 − 𝑏(𝑚 + 2)(1 − 𝛼)[𝜂1(𝐵 + 1) + 𝜂2(𝐵 − 1)] + (𝐵2 − 1) = 0,  

where 

 (3.4)                                         𝜂1 =
(1−𝑏)+𝛽(1+𝑏)

(1+𝑏)(1−𝛽)
[Δ(B − 1) − (A − 1)]          

 

(3.5)                                          𝜂2 =
(1−𝑏)+𝛽(1+𝑏)

(1+𝑏)(1−𝛽)
[Δ(B + 1) − (A + 1)]        

and 



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Δ = 1 −
(1 + 𝑏)(𝑚 + 1)(1 − 𝛽)

𝑏(𝑚 + 2)(1 − 𝛼)
. 

Proof: Suppose f ∈ 𝒱𝑘,𝜆
𝑚+1[𝐴, 𝐵, 𝛼, 𝑏]  and set  

  (3.6)                                           𝑝(𝑧) =   1 −
2

𝑏+1
+

2

𝑏+1

𝐷𝜆
𝑚+1𝑓(𝑧)

𝐷𝜆
𝑚𝑓(𝑧)

 

where 𝑝 is analytic in 𝒰 with 𝑝(0) = 1. Then, by simple computations together with 

(3.6) and (1.9) yield 

(3.7)                                       1 −
2

𝑏
+

2

𝑏

𝐷𝜆
𝑚+2𝑓(𝑧)

𝐷𝜆
𝑚+1𝑓(𝑧)

= (1 − 𝜇1) + 𝜇1 [𝑝(𝑧) +
𝜇2𝑧𝑝′(𝑧)

𝑝(𝑧)+𝜇3
],    

where 𝜇1 =
𝑚+1

𝑚+2

𝑏+1

𝑏
,  𝜇2 =

2

(𝑚+1)(𝑏+1)
,  𝜇3 =

2

𝑏+1
− 1. 

Since f ∈ 𝒱𝑘,𝜆
𝑚+1[𝐴, 𝐵, 𝛼, 𝑏], it follows that (1 − 𝜇1) + 𝜇1 [𝑝(𝑧) +

𝜇2𝑧𝑝′(𝑧)

𝑝(𝑧)+𝜇3
] ∈  𝑃𝑘[𝐴, 𝐵, 𝛼], 

Or, equivalently 

(3.8)                                               
(1−𝛼−𝜇1)

(1−𝛼)
+

𝜇1

1−𝛼
[𝑝(𝑧) +

𝜇2𝑧𝑝′(𝑧)

𝑝(𝑧)+𝜇3
]  ∈  𝑃𝑘[𝐴, 𝐵]. 

Define  

𝜙(𝑧) =
1

(1 + 𝜇3)

𝑧

(1 − 𝑧)𝜇2
+

𝜇3

(1 + 𝜇3)

𝑧

(1 − 𝑧)𝜇2+1
, 

and by using convolution techniques given by Noor[3], we have 

𝑝(𝑧) +
𝜇2𝑧𝑝′(𝑧)

𝑝(𝑧) + 𝜇3
= (

𝑘

4
+

1

2
) (𝑝1(𝑧) +

𝜇2𝑧𝑝1
′(𝑧)

𝑝1(𝑧) + 𝜇3
) − (

𝑘

4
−

1

2
) (𝑝2(𝑧) +

𝜇2𝑧𝑝2
′(𝑧)

𝑝2(𝑧) + 𝜇3
) 

By using (3.8), we see that 

(1 − 𝛼 − 𝜇1)

(1 − 𝛼)
+

𝜇1

1 − 𝛼
[𝑝𝑖(𝑧) +

𝜇2𝑧𝑝𝑖
′(𝑧)

𝑝𝑖(𝑧) + 𝜇3
]  ∈  𝑃[𝐴, 𝐵], 

where   z 𝜖 𝒰, i=1,2.  

Now, want to show that 𝑝𝑖 ∈ 𝑃[𝐴, 𝐵, 𝛽], where 𝛽(0 ≤ 𝛽 < 1) is one of the root of (3.3). 

Let  𝑝𝑖(𝑧) = (1 − 𝛽)ℎ𝑖(𝑧) + 𝛽,  i=1,2. 

Then, 

(1 − 𝛼 − 𝜇1)(1 − 𝛽)

(1 − 𝛼)
+

𝜇1(1 − 𝛽)

1 − 𝛼
[ℎ𝑖(𝑧) +

𝜇2

(1 − 𝛽)
𝑧ℎ𝑖

′(𝑧)

ℎ𝑖(𝑧) +
𝜇3 + 𝛽
(1 − 𝛽)

] ∈ 𝑃[𝐴, 𝐵] 

Using the fact illustrated in (1.2), we have 



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{
(𝐵 − 1)[(𝜂 + 𝜇ℎ𝑖(𝑧))(ℎ𝑖(𝑧) + 𝜔2) + 𝜔1𝜇𝑧ℎ𝑖

′(𝑧)] − (𝐴 − 1)(ℎ𝑖(𝑧) + 𝜔2)

(𝐵 + 1)[(𝜂 + 𝜇ℎ𝑖(𝑧))(ℎ𝑖(𝑧) + 𝜔2) + 𝜔1𝜇𝑧ℎ𝑖
′(𝑧)] − (𝐴 + 1)(ℎ𝑖(𝑧) + 𝜔2)

} ∈ 𝑃, 

where 𝜔1 =
𝜇2

1−𝛽
,    𝜔2 =

𝜇3+𝛽

1−𝛽
,  𝜂 =

1−𝛼−𝜇1(1−𝛽)

1−𝛼
 and 𝜇 =

𝜇1(1−𝛽)

1−𝛼
 . We now form the functional 

𝜓(𝑢, 𝑣) by choosing 𝑢 = ℎ𝑖(𝑧), 𝑣 = 𝑧ℎ𝑖
′(𝑧) and note that the first two conditions of Lemma 2.2 are 

clearly satisfied. We check condition (iii) as follows. 

𝜓(𝑢, 𝑣) =
{(𝐵 − 1)[(𝜂 + 𝜇𝑢)(𝑢 + 𝜔2) + 𝜔1𝜇𝑣] − (𝐴 − 1)(𝑢 + 𝜔2)}

{(𝐵 + 1)[(𝜂 + 𝜇𝑢)(𝑢 + 𝜔2) + 𝜔1𝜇𝑣] − (𝐴 + 1)(𝑢 + 𝜔2)}
 

            

=  
{𝜂1+𝜔1𝜇(𝐵−1)𝑣+[𝜂+𝜇(𝑢+𝜔2))(𝐵−1)−(𝐴−1)]𝑢}

{𝜂2+𝜔1𝜇(𝐵+1)𝑣+[(𝜂+𝜇(𝑢+𝜔2))(𝐵+1)−(𝐴+1)]𝑢}
 

where  𝜂1 =   𝜔2[𝜂(𝐵 − 1) − (𝐴 − 1)] and  𝜂2 =   𝜔2[𝜂(𝐵 + 1) − (𝐴 + 1)]. 

Now, 

𝜓(𝑖𝑢2, 𝑣1) =
{𝜂1 + 𝜇(𝜔1𝑣1 − 𝑢2

2)(𝐵 − 1) + [(𝜂 + 𝜇𝜔2)(𝐵 − 1) − (𝐴 − 1)]𝑖𝑢2}

{𝜂2 + 𝜇(𝜔1𝑣1 − 𝑢2
2)(𝐵 + 1) + [(𝜂 + 𝜇𝑢2)(𝐵 + 1) − (𝐴 + 1)]𝑖𝑢2}

 

Taking real part of   𝜓(𝑖𝑢2, 𝑣1), we have 

Re ψ(iu2, v1)

=
[−η1 + μ(ω1v1 − u2

2)(1 − B)][η2 + μ(ω1v1 − u2
2)(B + 1)] − [(η + μω2)(B − 1) − (A − 1)][(η + μω2)(B + 1) − (A + 1)]𝑢2

2

−[η2 + μ(ω1v1 + u2)(B + 1)]2 − [(η + μω2)(B + 1) − (A + 1)]2𝑢2
2  

As 𝜔1 > 0, 𝜇 > 0, so applying 𝑣1 ≤ −
1

2
(1 + 𝑢2

2) and after a little simplification, we have  

(3.9)                                    Re ψ(iu2, v1) ≤
A1+B1u2

2+C1u2
4

D1
   

where  

𝐴1 =
1

4
[2𝜂1 − 𝜔1𝜇(𝐵 − 1)][2𝜂2 − 𝜔1𝜇(𝐵 + 1],  

𝐵1 = −
1

2
𝜇(𝜔1 + 2)[𝜂1(𝐵 + 1) − 𝜔1𝜇(𝐵2 − 1) + 𝜂2(𝐵 − 1)] + (𝜂 + 𝜇𝜔2)2(𝐵2 − 1)

− 2(𝜂 + 𝜇𝜔2)(𝐴𝐵 − 1) + (𝐴2 − 1), 

𝐶1 = −
1

4
𝜇2(1 − 𝐵2)(𝜔1 + 2)2, 

and 

𝐷1 = [𝜂2 + 𝜇(𝜔1𝑣1 + 𝑢2)(𝐵 + 1)]2 + [(𝜂 + 𝜇𝜔2)(𝐵 + 1) − (𝐴 + 1)]2𝑢2
2. 

The right hand side of (3.9) is negative if 𝐴1 ≤ 0 and 𝐵1 ≤ 0. From 𝐴1 ≤ 0, we have 𝛽 to be one of 

the roots of  

𝜂1𝜂2𝑏2(𝑚 + 2)2(1 − 𝛼)2 − 𝑏(𝑚 + 2)(1 − 𝛼)[𝜂1(𝐵 + 1) + 𝜂2(𝐵 − 1)] + (𝐵2 − 1) = 0 



Communications on Applied Nonlinear Analysis 

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1345 
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With 0 ≤ β < 1 and also for 0 ≤  β < 1, we have 𝐵1≤ 0. 

Since all the conditions of Lemma 2.2 are satisfied, it follows that ℎ𝑖 ∈ 𝑃,  i=1,2 and consequently 

𝑝 ∈  𝑃𝑘[1, −1, 𝛽]. Hence from (3.6), 

𝑓 ∈ 𝒱𝑘,𝜆
𝑚 [1, −1, 𝛽, 𝑏 + 1]. 

By choosing the parameters A = 1, B = −1, b = 1, m = 0, we obtain the following known result, 

proved in [11]. 

Corollary 3.5. Let 𝑓 ∈ 𝑉𝑘,𝜆(𝛼). Then 𝑓 ∈ 𝑅𝑘,𝜆(𝛽), where 𝛽 is a root of  

2𝛽2 − (2𝛼 − 1)𝛽 − 1 = 0 

With 0 ≤ β < 1, which is  

β =
1

4
[(2𝛼 − 1) + √4𝛼2 − 4𝛼 + 9]. 

For 𝛼 = 0, 𝑘 = 2 in Corollary 3.5, we have the following well known results [2]. 

𝑉2,𝜆(0) = 𝐶 ⊆ 𝑅2,𝜆 (
1

2
) = 𝑆∗ (

1

2
),  for z ∈ 𝒰. 

Theorem 3.6. Let f ∈  𝒱𝑘,𝜆
𝑚 [𝐴, 𝐵, 0, 𝑏],   𝑚 ≥ 0, b>0(real), k≥ 2 and  0 < 𝑎 =

𝑏(𝑚+1)

2
≤ 1. Then 

𝐷𝜆
𝑚𝑓(𝑧) maps |𝑧| < 𝑟0 onto a convex domain, where 𝑟0 is the least  positive root of the equation  

  (3.10)      𝑎1𝑟4 + 𝑎2𝑟3 + 𝑎3𝑟2 + 𝑎4𝑟 + 4(2𝑎 − 1) = 0 with 0≤ 𝑟 < 1, 

  where  

        𝑎1 = 4𝑎2𝐴2𝐵2 − 4(𝑎 − 1)2𝐵4  

         𝑎2 = 2a(2a − 1)(B − A)𝐵2𝑘 

         𝑎3 = 8𝑎2(𝑎 − 2) + 8𝑎(1 − 𝑎)𝐴𝐵 − 𝑎2(𝐴 − 𝐵)2𝑘2 

         𝑎4 = 2𝑎(2𝑎 − 3)(𝐴 − 𝐵)𝑘. 

 This result is sharp. 

Proof:  Since f ∈  𝒱𝑘,𝜆
𝑚 [𝐴, 𝐵, 0, 𝑏] then 

(3.11)                                                        
Dλ

m+1f(z)

Dλ
mf(z)

=
b(p(z)−1)+2

2
  

where p ∈ 𝑃𝑘[𝐴, 𝐵, 0]. Using the identity (1.9), we have from (3.11), 

(3.12)                                                    
𝑧(𝐷𝜆

𝑚𝑓(𝑧))′

𝐷𝜆
𝑚𝑓(𝑧)

=
𝑏(𝑝(𝑧)−1)(𝑚+1)+2

2
 

Logarithmic differentiation of (3.12) yields 

(𝒛(𝐷𝜆
𝑚𝑓(𝑧))′)′

(𝐷𝜆
𝑚𝑓(𝑧))′

= 𝑎𝑝(𝑧) − 𝑎 + 1 +
𝑧𝑝′(𝑧)

𝑝(𝑧) − 1 +
1
𝑎

 



Communications on Applied Nonlinear Analysis 

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1346 
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where  𝑎 =
𝑏(𝑚+1)

2
. Then we have  

Re (1 +
z(𝐷𝜆

𝑚𝑓(𝑧))"

(𝐷𝜆
𝑚𝑓(𝑧))′

) ≥ a Re p(z) + (1 − a) −
|𝑧𝑝′(𝑧)|

|𝑝(𝑧) − 1 +
1
𝑎|

, 

and hence, by using Lemma 2.3 and Lemma 2.4, 

Re (1 +
z(𝐷𝜆

𝑚𝑓(𝑧))"

(𝐷𝜆
𝑚𝑓(𝑧))′

) 

≥ 𝑅𝑒 𝑝(𝑧) {𝑎 +
2(1 − 𝑎)(1 − 𝐵2𝑟2)

2 + (𝐴 − 𝐵)𝑘𝑟 − 2𝐴𝐵𝑟2
−

2𝑎𝑟{(𝐴 − 𝐵)𝑘 − 4𝐵(𝐴 − 𝐵)𝑟 + 𝐵2(𝐴 − 𝐵)𝑘𝑟2}

(2 + (𝐴 − 𝐵)𝑘𝑟 − 2𝐴𝐵𝑟2)𝜉
} 

= 𝑅𝑒 𝑝(𝑧) {
𝑎1𝑟4 + 𝑎2𝑟3 + 𝑎3𝑟2 + 𝑎4𝑟 + 4(2𝑎 − 1)

(2 + (𝐴 − 𝐵)𝑘𝑟 − 2𝐴𝐵𝑟2)𝜉
} > 0, 

provided 

𝑇(𝑟) = 𝑎1𝑟4 + 𝑎2𝑟3 + 𝑎3𝑟2 + 𝑎4𝑟 + 4(2𝑎 − 1) > 0, 

where 

         𝑎1 = 4𝑎2𝐴2𝐵2 − 4(𝑎 − 1)2𝐵4  

         𝑎2 = 2a(2a − 1)(B − A)𝐵2𝑘 

         𝑎3 = 8𝑎2(𝑎 − 2) + 8𝑎(1 − 𝑎)𝐴𝐵 − 𝑎2(𝐴 − 𝐵)2𝑘2 

         𝑎4 = 2𝑎(2𝑎 − 3)(𝐴 − 𝐵)𝑘 

and 

𝜉 = 2(2𝑎 − 1) − 𝑎(𝐴 − 𝐵)𝑘𝑟 + 2(𝐵2 − 𝑎(𝐴 + 𝐵)𝐵)𝑟2. 

We have 𝑇(0) > 0 𝑎𝑛𝑑 𝑇(1) < 0. Therefore, 𝐷𝜆
𝑚𝑓(𝑧) maps |𝑧| < 𝑟0 onto a convex domain, where  

𝑟0 is the least positive root of the equation 𝑇(𝑟) = 0, lying in (0,1). 

For 𝐷𝜆
𝑚𝑓1(𝑧) such that 

𝐷𝜆
𝑚+1𝑓1(𝑧)

𝐷𝜆
𝑚𝑓1(𝑧)

=
𝑏(𝑝𝑘(𝑧) − 1) + 2

2
 

where  𝑝𝑘(𝑧) =
2+(𝐴−𝐵)𝑘𝑧−2𝐴𝐵𝑧2

2(1−𝐵2𝑧2)
, we have 

(𝒛(𝐷𝜆
𝑚𝑓1(𝑧))′)′

(𝐷𝜆
𝑚𝑓1(𝑧))′

=  
𝑎1𝑟4 + 𝑎2𝑟3 + 𝑎3𝑟2 + 𝑎4𝑟 + 4(2𝑎 − 1)

(2 + (𝐴 − 𝐵)𝑘𝑟 − 2𝐴𝐵𝑟2)𝜉
= 0 

for 𝑧 = 𝑟0.  Hence this radius  𝑟0 is sharp.  

By choosing the parameters 𝐴 = 1, 𝐵 = −1, 𝑘 = 2, 𝑏 = 2 𝑎𝑛𝑑 𝑚 = 0, we obtain the following 

known result, see[2]. 



Communications on Applied Nonlinear Analysis 

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1347 
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Corollary 3.7: Let 𝑓 ∈ 𝑆′. Then 𝑓 maps |𝑧| < 𝑟0 onto a convex domain, where 𝑟0 is the least 

positive root of the equation 

𝑟4 − 2𝑟3 − 6𝑟2 − 2𝑟 + 1 = 0 

with 0 ≤ 𝑟 < 1, which is  𝑟0 = 2 − √3. This is also sharp.        

 

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