ISSN 2319-8133 (Online)
(An International Research Journal), www.compmath-journal.org
A Certain Class of p-Valent Analytic Function and Partial Sums
Vinod Kumar
1and Prachi Srivastava
21, 2
Faculty of Mathematical & Statistical Sciences,
Shri Ramswaroop Memorial University, Barabanki, 225003, U.P., INDIA.
email:[email protected], [email protected].
(Received on: June 1, 2018) ABSTRACT
In this paper, we define and study some subclasses of p-valent analytic function of higher order in unit disc. These classes generalize some classes which were previously studied. We obtain coefficient conditions; partial sums fn(z) of functions f(z) in these classes are considered and sharp lower bounds for the ratios of real part of f(z) to fn(z) and f′(z) to fn′(z) be determined.
2000 Mathematics Subject Classification: 30A10, 30C45.
Keywords: Analytic Functions, Hadamard Product, Starlike Function, Convex Function, Close-to-Convex Functions, Partial Sums.
1. INTRODUCTION
Let 𝐴
𝑝denotes a class of functions of the form:
1
( )
p p k p k, {1, 2,3,...}
k
f z z a z p
𝑝 ∈ ℂ = {1,2,3, … … } (1.1) which are analytic and p - valent in the open unit disk△= {𝑧 ∈ ℂ, |𝑧| < 1}.
Also a subclass of 𝐴
𝑝denoted by 𝑇
𝑝consisting of functions of the form:
1
( )
p p k p kk
f z z a z
(1.2) Let 𝑔 ∈ 𝐴
𝑝be of the form:
0 1
( )
p p k p k,
p kk
g z z b z b
(1.3)
Where 𝑏
𝑝+𝑘≤ 𝑏
𝑝+𝑘+1, 𝑘 ∈ ℕ
A function 𝑓(𝑧) ∈ 𝐴
𝑝is said to be p-valently starlike of order α in ∆, if it satisfies the inequality 𝑅𝑒 { 𝑧𝑓
′(𝑧)
𝑓(𝑧) } > 𝛼 (𝑧 ∈ ∆; 0 ≤ 𝛼 < 𝑝; 𝑝 ∈ ℕ)
The class of all p-valent starlike functions of order α is denoted by 𝑆
𝑝∗(𝛼) and we write 𝑆
1∗(𝛼) ≡ 𝑆
∗(𝛼).
On the other hand, a function 𝑓(𝑧) ∈ 𝐴
𝑝is said to p-valently convex of order α in ∆, if it satisfies the inequality
𝑅𝑒 {1 + 𝑧𝑓
′′(𝑧)
𝑓
′(𝑧) } > 𝛼 (𝑧 ∈ ∆; 0 ≤ 𝛼 < 𝑝; 𝑝 ∈ ℕ)
The class of all p-valent convex functions of order α is denoted by K
p(α) and write K
1(α)≡K(α).
The classes 𝑆
𝑝∗(𝛼) and 𝐾
𝑝( 𝛼) were studied by Patil and Thakre
3and Owa
2.
Furthermore, a function 𝑓(𝑧) ∈ 𝐴
𝑝is said to be p-valently close-to-convex of order α in ∆, if it satisfies the inequality
𝑅𝑒{𝑧
1−𝑝𝑓
′(𝑧)} > 𝛼 (𝑧 ∈ ∆; 0 ≤ 𝛼 < 𝑝; 𝑝 ∈ ℕ)
The class of all p-valent close-to-convex of order α is denoted by 𝐶𝐾
𝑝(𝛼), 𝐶𝐾
𝑝(0) ≡ 𝐶𝐾
𝑝and denote 𝐶𝐾
1(0) ≡ 𝐶𝐾.
A Hadamard product of two p-valent analytic functions 𝑓 & 𝑔 of the form (1.1) and (1.3) respectively, is defined by
(𝑓 ∗ 𝑔)(𝑧) = 𝑧
𝑝+ ∑
∞𝑘=1𝑎
𝑝+𝑘𝑏
𝑝+𝑘𝑧
𝑝+𝑘= (𝑔 ∗ 𝑓)(𝑧) (1.4) Involving derivatives of the Hadamard product defined in (1.4), the subclass 𝐾(𝑝, 𝑔, 𝛼, 𝑚) of
the class 𝐴
𝑝is defined as follows:
For 0 ≤ 𝛼 < 𝑝, 𝑝 > 𝑚, 𝑚 ∈ ℕ
0𝐾(𝑝, 𝑔
,𝛼, 𝑚): = {𝑓(𝑧) ∈ 𝐴
𝑝: 𝑅𝑒 {1 + 𝑧(𝑓 ∗ 𝑔)
𝑚+2(𝑧)
(𝑓 ∗ 𝑔)
𝑚+1(𝑧) + 𝑚} > 𝛼, 𝑧 ∈ ∆}
where (𝑓 ∗ 𝑔)
𝑟(𝑧) denotes the r
thderivative of (𝑓 ∗ 𝑔)(𝑧) and is given by (𝑓 ∗ 𝑔)
𝑟(𝑧) =
𝑝!(𝑝−𝑟)!
𝑧
𝑝−𝑟+ ∑
(𝑝+𝑘)!(𝑝+𝑘−𝑟)!
𝑎
𝑝+𝑘𝑏
𝑃+𝑘𝑧
𝑝+𝑘−𝑟∞𝑘=1
, 𝑟 ∈ ℕ
0(1.5) We also let 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚) = 𝐾(𝑝, 𝑔, 𝛼, 𝑚) ∩ 𝑇
𝑝and for m=0 and 𝑔(𝑧) =
𝑧𝑝1−𝑧
then the class 𝐾(𝑝, 𝑔, 𝛼, 𝑚) coincides with the classes studied by Srivastava et al. in [5, 6].We note that taking 𝑔(𝑧) =
𝑧𝑝1−𝑧
, 𝐾(𝑝, 𝑔
,𝛼, 0) ≡ 𝐾
𝑝(𝛼).
Following the idea of P. Sharma and P.Srivastava
5we define the class 𝐾(𝑝, 𝑔
,𝛼, 𝑚)
and obtained the coefficient conditions for these classes. Further, results based on partial sums
of functions belonging to these classes are derived.
2. MAIN RESULTS
In this section, we obtain coefficient conditions for functions in the classes defined above.
Theorem 1 Let 𝑔 ∈ 𝐴
𝑝of the form (1.3), then for 0 ≤ 𝛼 < 𝑝, 𝑝 > 𝑚, 𝑚 ∈ ℕ
0and 𝜁 ∈ ℂ/{−1} with |𝜁| = 1, then 𝑓 ∈ 𝐴
𝑝of the form (1.1) belongs to the class 𝐾(𝑝, 𝑔, 𝛼, 𝑚) if and only if
1 + ∑
∞𝑘=1𝐴
𝑝+𝑘𝑧
𝑘≠ 0 (𝑧 ∈ ∆/{0}) Where
𝐴
𝑝+𝑘= (𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! [𝑘(1 + 𝜁) + 2(𝑝 − 𝛼)]
2(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝 ! 𝑎
𝑝+𝑘𝑏
𝑝+𝑘Proof: Set
𝑝(𝑧) =
1 +
𝑧(𝑓∗𝑔)(𝑓∗𝑔)𝑚+1𝑚+2(𝑧)(𝑧)+ 𝑚 − 𝛼 𝑝 − 𝛼
which is analytic in ∆ with 𝑝(0) = 1, then 𝑓 ∈ 𝐾(𝑝, 𝑔, 𝛼, 𝑚) if and only if 𝜁 ∈ ℂ/{−1},
|𝜁| = 1, 𝑧 ∈ ∆
1 +
𝑧(𝑓∗𝑔)𝑚+2(𝑧)(𝑓∗𝑔)𝑚+1(𝑧)
+ 𝑚 − 𝛼
𝑝 − 𝛼 ≠ 𝜁 − 1
𝜁 + 1 Or,
(1 + 𝜁)𝑧(𝑓 ∗ 𝑔)
𝑚+2(𝑧) − [2𝛼 + 𝜁(𝑝 − 𝑚) − (𝑝 + 𝑚 + 𝜁 + 1)](𝑓 ∗ 𝑔)
𝑚+1(𝑧) ≠ 0 On using series expansion of (𝑓 ∗ 𝑔)
𝑚+2and (𝑓 ∗ 𝑔)
𝑚+1from (1.5), it is equivalent that
𝑝! 2(𝑝 − 𝛼)
(𝑝 − 𝑚 − 1)!𝑧𝑝−𝑚−1{1 + ∑(𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! [𝑘(1 + 𝜁) + 2(𝑝 − 𝛼)]
2(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝! 𝑎𝑝+𝑘𝑏𝑝+𝑘𝑧𝑘≠ 0
∞
𝑘=1
}
and for z 0 , it is equivalent that
{1 + ∑ (𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! [𝑘(1 + 𝜁) + 2(𝑝 − 𝛼)]
2(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝! 𝑎
𝑝+𝑘𝑏
𝑝+𝑘𝑧
𝑘≠ 0
∞
𝑘=1
} This completes the proof.
Taking p=1, 𝑔(𝑧) =
𝑧𝑝1−𝑧
and m = 0 in Theorem 1, we get special case as follows.
Corollary 1 A function 𝑓 ∈ 𝐴
𝑝of the form (1.1) is in the class 𝐾(𝛼) if and only if 1 + ∑ 𝐴
𝑘𝑧
𝑘−1≠ 0 (𝑧 ∈ ∆)
∞
𝑘=2
Where A
k=
k2(ζ+1)−2α−k(ζ−1)2−2α
a
k, ζ ∈ ℂ/{−1}, |ζ| = 1
Theorem 2 Let 𝑔 ∈ 𝐴
𝑝of the form (1.3) and 0 ≤ α < p, p > m, m ϵ ℕ
0a necessary and
sufficient condition for a function 𝑓 ∈ 𝑇
𝑝of the form (1.2) belongs to 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚) is that:
∑
(𝑝+𝑘)!(𝑝+𝑘−𝛼)(𝑝−𝛼)𝑝!
𝜙
𝑘𝑚𝑏
𝑝+𝑘|𝑎
𝑝+𝑘| ≤ 1
∞𝑘=1
, (2.1) Where 𝜙
𝑘𝑚=
(𝑝−𝑚−1)!(𝑝+𝑘−𝑚−1)!
Proof. To prove the sufficient condition for 𝑓𝜖𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚), we consider 𝑝(𝑧) =
1 +
𝑧(𝑓∗𝑔)(𝑓∗𝑔)𝑚+1𝑚+2(𝑧)(𝑧)+ 𝑚 − 𝛼 𝑝 − 𝛼
Now it needs to show that
|𝑝(𝑧) − 1| − |𝑝(𝑧) + 1| ≤ 0 We see that
|𝑝(𝑧) − 1| = |𝑧
𝑝−𝑚−1|
(𝑝 − 𝛼)𝛼|(𝑓 ∗ 𝑔)
𝑚+1(𝑧)| |− ∑ (𝑝 + 𝑘)! 𝑘𝑎
𝑝+𝑘𝑏
𝑝+𝑘(𝑝 + 𝑘 − 𝑚 − 1)!
∞
𝑘=1
𝑧
𝑘|
≤ |𝑧
𝑝−𝑚−1|
(𝑝 − 𝛼)𝛼|(𝑓 ∗ 𝑔)
𝑚+1(𝑧)| ∑ (𝑝 + 𝑘)! 𝑘|𝑎
𝑝+𝑘|𝑏
𝑝+𝑘(𝑝 + 𝑘 − 𝑚 − 1)!
∞
𝑘=1
and
|𝑝(𝑧) + 1| =
= |𝑧
𝑝−𝑚−1|
(𝑝 − 𝛼)𝛼|(𝑓 ∗ 𝑔)
𝑚+1(𝑧)| | 2(𝑝 − 𝛼)𝑝!
(𝑝 − 𝑚 − 1)! − ∑ (𝑝 + 𝑘)! [𝑘+2(𝑝 − 𝛼)]𝑎
𝑝+𝑘𝑏
𝑝+𝑘(𝑝 + 𝑘 − 𝑚 − 1)!
∞
𝑘=1
𝑧
𝑘|
≥ |𝑧
𝑝−𝑚−1|
(𝑝 − 𝛼)𝛼|(𝑓 ∗ 𝑔)
𝑚+1(𝑧)| { 2(𝑝 − 𝛼)𝑝!
(𝑝 − 𝑚 − 1)! − ∑ (𝑝 + 𝑘)! [𝑘+2(𝑝 − 𝛼)|𝑎
𝑝+𝑘|𝑏
𝑝+𝑘(𝑝 + 𝑘 − 𝑚 − 1)!
∞
𝑘=1
} Thus
|𝑝(𝑧) − 1| − |𝑝(𝑧) + 1|
≤ |𝑧
𝑝−𝑚−1|
(𝑝 − 𝛼)𝛼|(𝑓 ∗ 𝑔)
𝑚+1(𝑧)| {− 2(𝑝 − 𝛼)𝑝!
(𝑝 − 𝑚 − 1)! + ∑ (𝑝 + 𝑘)! 2[𝑘+𝑝 − 𝛼]|𝑎
𝑝+𝑘|𝑏
𝑝+𝑘(𝑝 + 𝑘 − 𝑚 − 1)!
∞
𝑘=1
} ≤ 0,
If (2.1) holds. Conversely, to show that condition (2.1) is necessary for𝑓(𝑧) ∈ 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚), let
𝑅𝑒 {
1 +
𝑧(𝑓∗𝑔)(𝑓∗𝑔)𝑚+1𝑚+2(𝑧)(𝑧)+ 𝑚 − 𝛼
𝑝 − 𝛼 } > 0(𝑧 ∈ ∆)
Applying series expansion of (𝑓 ∗ 𝑔)
𝑚+2and (𝑓 ∗ 𝑔)
𝑚+1from (1.5), we get {
(𝑝−𝛼)𝑝!
(𝑝−𝑚−1)!
𝑧
𝑝−𝑚−1− ∑
(𝑝+𝑘)!(𝑝+𝑘−𝛼)(𝑝+𝑘−𝑚−1)!
|𝑎
𝑝+𝑘|𝑏
𝑝+𝑘𝑧
𝑝+𝑘−𝑚−1∞𝑘=1
(𝑝 − 𝛼)(𝑓 ∗ 𝑔)
𝑚+1(𝑧) } > 0 (𝑧 ∈ ∆)
Or,
𝑅𝑒 {1 − ∑ (𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! (𝑝 + 𝑘 − 𝛼)
(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝! |𝑎
𝑝+𝑘|𝑏
𝑝+𝑘𝑧
𝑘∞
𝑘=1
} > 0 (𝑧 ∈ ∆), Letting 𝑧 → 1
−along the real axis, we get
{1 − ∑ (𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! (𝑝 + 𝑘 − 𝛼)
(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝! |𝑎
𝑝+𝑘|𝑏
𝑝+𝑘𝑧
𝑘∞
𝑘=1
} ≥ 0 Which is (2.1). This proves Theorem 2.
Corollary 2 Let 𝑔 ∈ 𝐴
𝑝of the form (1.3), 0 ≤ 𝛼 < 𝑝 and 𝑝 > 𝑚, 𝑚 ∈ ℕ
0if 𝑓 ∈ 𝑇
𝑝of the form (1.2) belongs to the class 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚), then
|𝑎
𝑝+𝑘| ≤ (𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝!
(𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! (𝑝 + 𝑘 − 𝛼)𝑏
𝑝+𝑘(𝑘, 𝑝𝜖 ℕ, 𝑏
𝑝+𝑘> 0) The result is sharp for the function 𝑓
𝑘(𝑧) given by
𝑓
𝑘(𝑧) = (𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑚 − 1)! 𝑝!
(𝑝 + 𝑘)! (𝑝 − 𝑚 − 1)! (𝑝 + 𝑘 − 𝛼)𝑏
𝑝+𝑘𝑧
𝑝+𝑘(𝑘, 𝑝 ∈ ℕ, 𝑏
𝑝+𝑘> 0) Taking 𝑔(𝑧) =
𝑧𝑝1−𝑧
, m=0 in Theorem 2, we get special case as follows:
Corollary 3 A necessary and sufficient conditions for a function 𝑓 ∈ 𝑇
𝑝of the form (1.2) to be in the class 𝐾
𝑝(𝛼) (0 ≤ 𝛼 < 1) is that
∑
(𝑝+𝑘−𝛼)(𝑝+𝑘) (𝑝−𝛼)𝑝|𝑎
𝑝+𝑘|
∞𝑘=1
≤ 1.
Remark1. The result obtained in corollary 3 coincides with the result obtained by P.Srivastava
5, if we put 𝑝 = 1 and 𝑘 = 𝑛 − 1 in the result of corollary 3, then the above result coincides with the result obtained by Silverman
8and S.Topkaya
9.
3. SOME CONSEQUENCES OF MAIN RESULTS
A consequence of the theorem 1 follows as a sufficient condition for 𝐾(𝑝, 𝑔, 𝛼, 𝑚) Class:
Theorem 3 Let 𝑔 ∈ 𝐴
𝑝of the form (1.3), then for 0 ≤ 𝛼 < 𝑝, 𝑝 > 𝑚, 𝑚 ∈ ℕ
0and 𝜁 ∈ ℂ/{−1} with |𝜁| = 1, then 𝑓 ∈ 𝐴
𝑝of the form (1.1) satisfies:
1 + ∑
∞𝑘=1𝐵
𝑘𝑧
𝑘≠ 0 , (𝑧 ∈ ∆ / {0}) (3.1) Where
𝐵
𝑘= 𝛽
𝑘𝑘! + ∑ 𝛽
𝑗𝑗!
𝑘−1
𝑗=0
(𝑝 + 𝑘 − 𝑗)! (𝑝 − 𝑚 − 1)! [(𝑘 − 𝑗)(1 + 𝜁) + 2(𝑝 − 𝛼)]
2(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑗 − 𝑚 − 1)! 𝑝! 𝑎
𝑝+𝑘−𝑗𝑏
𝑝+𝑘−𝑗Then 𝑓(𝑧) ∈ 𝐾(𝑝, 𝑔, 𝛼, 𝑚).
Proof. Let
𝐴
𝑝+𝑘−𝑗= (𝑝 + 𝑘 − 𝑗)! (𝑝 − 𝑚 − 1)! [(𝑘 − 𝑗)(1 + 𝜁) + 2(𝑝 − 𝛼)]
2(𝑝 − 𝛼)(𝑝 + 𝑘 − 𝑗 − 𝑚 − 1)! 𝑝! 𝑎
𝑝+𝑘−𝑗𝑏
𝑝+𝑘−𝑗, the condition (3.1) can be written as
1 + ∑ (
𝛽𝑘𝑘!
+ ∑
𝛽𝑗𝑗!
𝐴
𝑝+𝑘−𝑗𝑘−1𝑗=0
) 𝑧
𝑘≠ 0
∞𝑘=1
which is equivalent to
(1 + ∑ 𝐴
𝑝+𝑘𝑧
𝑘∞
𝑘=1
) (1 + ∑ 𝛽
𝑘𝑘! 𝑧
𝑘∞
𝑘=1
) ≠ 0 Since
1 + ∑
𝛽𝑘𝑘!
𝑧
𝑘= 𝑒
𝛽𝑧≠ 0 , 𝛽 ∈ ℝ.
∞𝑘=1
Thus (2.2) implies
(1 + ∑ 𝐴
𝑝+𝑘𝑧
𝑘∞
𝑘=1
) ≠ 0 which by Theorem 1 proves that 𝑓(𝑧) ∈ 𝐾(𝑝, 𝑔, 𝛼, 𝑚).
Theorem 4 Let 𝑔 ∈ 𝐴
𝑝with positive coefficients, m<p, 0 ≤ 𝛼 < 𝑝, 𝑝 > 𝑚, 𝑚 ∈ ℕ
0, then 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚 + 1) ⊂ 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚).
Proof Let 𝑓 ∈ 𝑇
𝑝of the form (1.2) belongs to 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚 + 1) and 𝑔 ∈ 𝐴
𝑝of the form (1.3), then by Theorem 2, we get
∑ (𝑘 + 𝑝)! (𝑝 + 𝑘 − 𝛼)
(𝑝 − 𝛼) 𝑝! 𝜑
𝑘𝑚+1𝑏
𝑝+𝑘|𝑎
𝑝+𝑘|
∞
𝑘=1
≤ 1, Where
𝜑
𝑘𝑚+1=
(𝑝+𝑘−𝑚−2)!(𝑝−𝑚−2)!. By simple calculation, we see that 𝜑
𝑘𝑚≤ 𝜑
𝑘𝑚+1and hence,
∑ (𝑘 + 𝑝)! (𝑝 + 𝑘 − 𝛼)
(𝑝 − 𝛼) 𝑝! 𝜑
𝑘𝑚𝑏
𝑝+𝑘|𝑎
𝑝+𝑘|
∞
𝑘=1
≤ ∑ (𝑘 + 𝑝)! (𝑝 + 𝑘 − 𝛼)
(𝑝 − 𝛼) 𝑝! 𝜑
𝑘𝑚+1𝑏
𝑝+𝑘|𝑎
𝑝+𝑘|
∞
𝑘=1
≤ 1,
Which again by Theorem 2 it is proved that 𝑓(𝑧) ∈ 𝑇𝐾(𝑝, 𝑔, 𝛼, 𝑚). this proves the Theorem 4.
4. PARTIAL SUMS FOR THE CLASS 𝑲(𝒑, 𝒈, 𝜶, 𝒎).
Following the earlier works by Silverman
8, Silvia
7, T. Rosy
4, Zhi-Gang Wang
10and P. Srivastava
5on partial sums of analytic functions. We consider in this section partial sums of functions in the class 𝐾(𝑝, 𝑔, 𝛼, 𝑚) and 𝐾
𝜆(𝑝, 𝑔, 𝑚) and obtain sharp lower bounds for the ratios of real part of f(z) to f
n(z) and f
′(z) to f
n′(z) .
Consider the partial sums𝑓(𝑧) 𝜖 𝐴
𝑝of of the form (1.1) be as follows:
( )
pf z z and
1
1
( ) , 2.
n
p p k
n p k
k
f z z a z n
(4.1) Theorem 5 Let 𝑔(𝑧) 𝜖 𝐴
𝑝be of the form (1.2) and for 0 ≤ 𝛼 < 𝑝, 𝑚 ∈ ℕ
0,the function
( )
pf z A of the form (1.1) satisfies
∑
𝑛−1𝑘=1𝑐
𝑝+𝑘|𝑎
𝑝+𝑘| ≤ 1, 𝑘 ∈ ℕ
0(4.2) Where 𝑐
𝑝+𝑘= (𝑝−𝑚−1)!(𝑝+𝑘)!(𝑘+𝑝−𝛼)(𝑝−𝛼)𝑝!(𝑝+𝑘−𝑚−1)
𝑏
𝑝+𝑘,Then 𝑓𝜖 𝐾(𝑝, 𝑔, 𝛼, 𝑚) and 𝑅𝑒 {
𝑓(𝑧)𝑓𝑛(𝑧)
} > 1 −
𝑐1𝑝+𝑛
, (4.3) 𝑅𝑒 {
𝑓𝑛(𝑧)𝑓(𝑧)
} >
𝑐𝑝+𝑛1+𝑐𝑝+𝑛
(4.4) Proof: Let (4.1) holds, from Theorem 2, we get that𝑓𝜖 𝐾(𝑝, 𝑔, 𝛼, 𝑚). Again it is easy to verify
that c
p k c
p k 1, k , From (4.2), we have
∑
𝑛−1𝑘=1|𝑎
𝑝+𝑘| + 𝑐
𝑝+𝑛∑
∞𝑘=𝑛|𝑎
𝑝+𝑘| ≤ ∑
∞𝑘=1𝑐
𝑝+𝑘|𝑎
𝑝+𝑘| ≤ 1. (4.5) Set
𝑤
1= 𝑐
𝑝+𝑛{ 𝑓(𝑧)
𝑓
𝑛(𝑧)− (1 − 1 𝑐
𝑝+𝑛)}
= 1 + 𝑐
𝑝+𝑛∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘1 + ∑
𝑛−1𝑘=1𝑎
𝑝=𝑘𝑧
𝑘On using (4.5), we get
| 𝑤
1(𝑧) − 1
𝑤
1(𝑧) + 1 | = | 𝑐
𝑝+𝑛∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘2 + 2 ∑
𝑛−1𝑘=1𝑎
𝑝+𝑘𝑧
𝑘+ 𝑐
𝑝+𝑛∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘|
≤ | 𝑐
𝑝+𝑛∑
∞𝑘=𝑛|𝑎
𝑝+𝑘|
2 − 2 ∑
𝑛−1𝑘=1|𝑎
𝑝+𝑘| − 𝑐
𝑝+𝑘∑
∞𝑘=𝑛|𝑎
𝑝+𝑘| | ≤ 1, This readily yields the assertion (4.3) of Theorem 5.
Similarly, if we take
𝑤
2(𝑧) = (1 + 𝑐
𝑝+𝑛) { 𝑓
𝑛(𝑧)
𝑓(𝑧) − 𝑐
𝑝+𝑛1 + 𝑐
𝑝+𝑛}
= 1 − (𝑐
𝑝+𝑘+ 1) ∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘1 + ∑
∞𝑘=1𝑎
𝑝+𝑘𝑧
𝑘We get with the use of (4.5)
| 𝑤
2(𝑧) − 1
𝑤
2(𝑧) + 1 | = | −(𝑐
𝑝+𝑛+ 1) ∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘2 + 2 ∑
∞𝑘=1𝑎
𝑝+𝑘𝑧
𝑘− (𝑐
𝑝+𝑘+ 1) ∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘|
≤ (𝑐
𝑝+𝑛− 1) ∑
∞𝑘=𝑛|𝑎
𝑝+𝑘|
2 − 2 ∑
𝑛−1𝑘=1|𝑎
𝑝+𝑘| − (𝑐
𝑝+𝑛− 1) ∑
∞𝑘=𝑛|𝑎
𝑝+𝑘| ≤ 1, Which proves (4.4) of Theorem 5.
Theorem 6 If 𝑓(𝑧) of the form (1.1) satisfies the condition (4.1) then 𝑅𝑒 {
𝑓′(𝑧)𝑓𝑛′(𝑧)
} > 1 −
𝑛+𝑝𝑐𝑝+𝑛
, (4.6) 𝑅𝑒 {
𝑓𝑛,(𝑧)𝑓′(𝑧)
} >
𝑐𝑝+𝑛𝑛+𝑝+𝑐𝑝+𝑛
, (4.7) Proof: By setting
𝑤
3= 𝑐
𝑝+𝑛𝑝 + 𝑛 { 𝑓
′(𝑧)
𝑓
𝑛,(𝑧) − (1 − 𝑝 + 𝑛 𝑐
𝑝+𝑛)}
= 1 +
𝑐𝑝+𝑛
𝑝+𝑛
∑
(𝑝+𝑘)𝑝
𝑎
𝑝+𝑘𝑧
𝑘∞𝑘=𝑛
1 + ∑
(𝑝+𝑘)𝑝
𝑎
𝑝+𝑘𝑧
𝑘𝑛−1𝑘=1
We get
| 𝑤
3(𝑧) − 1
𝑤
3(𝑧) + 1 | = | −(𝑐
𝑝+𝑛+ 1) ∑
∞𝑘=𝑛𝑎
𝑝+𝑘𝑧
𝑘2 + 2 ∑
𝑝+𝑘𝑝
𝑎
𝑝+𝑘𝑧
𝑘+
𝑐𝑝+𝑘𝑝+𝑛
∑
𝑝+𝑘𝑝
𝑎
𝑝+𝑘𝑧
𝑘∞𝑘=𝑛 𝑛−1𝑘=1
|
≤
𝑐𝑝+𝑘
𝑝+𝑛
∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘|
∞𝑘=𝑛
2 − 2 ∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘| −
𝑐𝑝+𝑛𝑝+𝑘∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘|
∞𝑘=𝑛 𝑛−1𝑘=1
.
| 𝑤
3(𝑧) − 1 𝑤
3(𝑧) + 1 | ≤ 1 If
𝑐
𝑝+𝑛𝑝 + 𝑛 ∑ 𝑝 + 𝑘 𝑝 |𝑎
𝑝+𝑘|
∞
𝑘=𝑛
≤ 2 − 2 ∑ 𝑝 + 𝑘
𝑝 |𝑎
𝑝+𝑘| − 𝑐
𝑝+𝑘𝑝 + 𝑛
𝑛−1
𝑘=1
∑ 𝑝 + 𝑘 𝑝 |𝑎
𝑝+𝑘|
∞
𝑘=𝑛
Or,
∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘|
𝑛−1𝑘=1
+
𝑐𝑝+𝑛𝑝+𝑛
∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘|
∞𝑘=𝑛
≤ 1 (4.8)
Since, the left hand side of (4.6) is bounded above by ∑
∞𝑘=1𝑐
𝑝+𝑘|𝑎
𝑝+𝑘| if
∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘| +
𝑐𝑝+𝑛𝑝+𝑛∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘| ≤ ∑
∞𝑘=1𝑐
𝑝+𝑘|𝑎
𝑝+𝑘|
∞𝑘=𝑛
𝑛−1𝑘=1
or,
∑
𝑛−1𝑘=1(𝑐
𝑝+𝑘−
𝑝+𝑘𝑝) |𝑎
𝑝+𝑘| + ∑
∞𝑘=𝑛(𝑐
𝑝+𝑘−
(𝑝+𝑛)𝑐𝑝+𝑛 𝑝+𝑘𝑝) |𝑎
𝑝+𝑘| ≥ 0 (4.9)
This is true. This proves assertion (4.6) of Theorem 6.
Again by setting,
'
4 '
( ) ( ) ( )
( )
n n p n p
n p
f z c
w z n p c
n p c f z
1
1
1 1
1
n p k
p k k n
n
k p k k
c p k
a z
n p p
p k a z p
Now,
| 𝑤
4(𝑧) − 1 𝑤
4(𝑧) + 1 | ≤
(1 +
𝑐𝑛+𝑝𝑛+𝑝
) ∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘|
∞𝑘=𝑛
2 − 2 ∑
𝑝+𝑘𝑝
𝑛−1𝑘=1
|𝑎
𝑝+𝑘| − (1 +
𝑐𝑛+𝑝𝑛+𝑝
) ∑
𝑝+𝑘𝑝
|𝑎
𝑝+𝑘|
∞𝑘=𝑛