ISSN (e): 2250-3021, ISSN (p): 2278-8719
Vol. 08, Issue 9 (September. 2018), ||V (IV) || PP 48-52
Characteristic Properties Of Subclasses For The
Meromorphically Multivalent Functions
Mr. Vidyadhar Sharma and Dr. Nisha Mathur
Research Scholar and Assistant Professor,Department of Mathematics, M .L.V. Government P G College Bhilwara-311001
Corresponding Author: Mr. Vidyadhar Sharma
Abstract:
In the present paper, we introduce and investigate some new classes of multivalent functions involving linear operator and derive useful characteristic properties for Meromorphically multivalent functions. Several results are presented exhibiting relevant connections to some other results proved here and those obtained in earlier works.Keywords:
Analytic functions, Meromorphic functions, multivalent functions, linear operator and Hyper geometric function.--- --- Date of Submission: 13-09-2018 Date of acceptance: 28-09-2018 --- ---
I.
INTRODUCTION
For any integer𝑚 > −𝑝, let 𝒯𝑝,𝑚 denote the class of all meromorphic functions 𝑓(𝑧) of the form
𝑓 𝑧 = 𝑧−𝑝 + 𝑎 𝑟𝑧𝑟
∞
𝑟=𝑚 𝑝 ∈ ℕ (1)
which are analytic and p-valent in the punctured unit disk 𝔻∗= 𝑧: 𝑧 ∈ ℂ and 0 < 𝑧 < 1 = 𝔻/ 0 .
Now we define the general integral operator𝜒𝑝𝑛 𝛼, 𝛽 𝑓(𝑧) which is as follows
𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 =
𝛽 𝛼𝑧
−𝑝− 𝛽
𝛼 𝑡
𝛽
𝛼+𝑝−1 𝜒𝑝𝑛−1
𝑧
0
𝛼, 𝛽 𝑓 𝑡 𝑑𝑡
= 𝜒𝑝1 𝛼, 𝛽 1
𝑧𝑝(1−𝑧) ∗ 𝜒𝑝1 𝛼, 𝛽 1
𝑧𝑝 1−𝑧 ∗ … ∗ 𝜒𝑝1 𝛼, 𝛽 1
𝑧𝑝 1−𝑧 ∈ 𝑓 𝑧 , 𝑧 ∈ 𝔻
∗and 𝑝 ∈ ℕ(2)For the sake of
convenience, in particular cases
(i) If 𝑛 = 0 then the above integral operator 𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 converts into𝜒𝑝0 𝛼, 𝛽 𝑓 𝑧 = 𝑓(𝑧).
(ii) If 𝑛 = 1 then the above integral operator 𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 converts
into𝜒𝑝1 𝛼, 𝛽 𝑓 𝑧 = 𝛽 𝛼 𝑧
−𝑝− 𝛽𝛼
𝑡 𝛽 𝛼+𝑝−1
𝑧
0 𝑓 𝑡 𝑑𝑡.
(iii) If 𝑛 = 2 then the above integral operator 𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 converts
into𝜒𝑝2 𝛼, 𝛽 𝑓 𝑧 = 𝛽 𝛼 𝑧
−𝑝− 𝛽
𝛼 𝑡
𝛽
𝛼+𝑝−1 𝜒𝑝1 𝛼, 𝛽
𝑧
0 𝑓 𝑡 𝑑𝑡.
Hence, if 𝑓(𝑧) ∈ 𝒯𝑝,𝑚 , we obtained the following results
𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 = 1 𝑧𝑝+
𝛽 𝛽 +𝛼 𝑟+𝑝
𝑛
𝑎𝑟
∞
𝑟=𝑚 𝑧𝑟 (3)
Therefore from (3), it is easy to see that
𝛼𝑧 𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧
′
= 𝛽𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 − 𝛼𝑝 + 𝛽 𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 (𝛼 > 0) (4)
We observe that (i) 𝜒𝑝𝜆 1,1 𝑓 𝑧 =Ρ𝑝𝜆𝑓 𝑧 (see, [4]), (ii)𝜒1𝜆 1, 𝑗 𝑓 𝑧 =Ρ𝑗𝜆𝑓 𝑧
We also see that
(i) 𝜒𝑝𝑛 1, 𝛽 𝑓 𝑧 = 𝜒𝑝 ,𝛽𝑛 𝑓 𝑧 where 𝜒𝑝,𝛽𝑛 𝑓 𝑧 = 1 𝑧𝑝+
𝛽 𝛽 + 𝑟+𝑝
𝑛
𝑎𝑟
∞
𝑟=𝑚 𝑧𝑟
(ii) 𝜒𝑝𝑛 𝛼, 1 𝑓 𝑧 = 𝜒𝑝 ,𝛼𝑛 𝑓 𝑧 where 𝜒𝑝,𝛼𝑛 𝑓 𝑧 = 1 𝑧𝑝+
1 1+𝛼 𝑟+𝑝
𝑛
𝑎𝑟
∞
𝑟=𝑚 𝑧𝑟
(iii)𝜒𝑝𝑛 1,1 𝑓 𝑧 = 𝜒𝑝𝑛𝑓 𝑧 where 𝜒𝑝𝑛𝑓 𝑧 = 1 𝑧𝑝+
1 1+ 𝑟+𝑝
𝑛
𝑎𝑟
∞
𝑟=𝑚 𝑧𝑟
II.
DEFINITIONS
Let 𝓣𝒑,𝒎𝒏+𝟏(η,δ,μ,λ) be the class of functions 𝑓(𝑧) ∈ 𝓣𝑝,𝑚 which is satisfies the condition
ℜ𝑒 1 − 𝜆 𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 𝜇
+ 𝜆𝜒𝑝𝑛 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 𝜇 −1
International organization of Scientific
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49 | P a g e
where𝑔(𝑧) ∈ 𝓣𝑝fulfill the following conditionℜ𝑒𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 > 𝛿, 𝛿 ∈ [0,1) (6)
Here 𝜇 𝑎𝑛𝑑 𝜂 belongs to real numbers such that 0 ≤ 𝜂 < 1, 𝜇 > 0and 𝜆 ∈ 𝑐 with ℜ𝑒(𝜆) > 0.
Currently many scholars and researchers studied similar classes for various types of different operators including Aouf and Mostafa [2], Al-Ashwah [6], [7] et al.
III.
PRELIMINARY LEMMAS
The following lemmas will be required in our present investigation
Lemma i. Let Ω be a set in the complex plane c and let the function Ψ: 𝑐2→ 𝑐 satisfy the condition
Ψ 𝑖𝑟2, 𝑠1 ∉ Ωfor allreal 𝑟2, 𝑠1≤ − 1+𝑟12
2 . If 𝑞 𝑧 is analytic in 𝑤𝑖𝑡𝑞 0 = 1 and
Ψ 𝑞 𝑧 , 𝑧𝑞′(𝑧) ∈Ω, 𝑧 ∈ 𝔻 , then ℜ 𝑞 𝑧 > 0 𝑧 ∈ 𝔻 .
Lemma ii. Let 𝑞(𝑧) is analytic function in disk 𝔻 with 𝑞 0 = 1 and if 𝛼 ∈ 𝑐/{0} with ℜ{𝛼} ≥ 0, then ℜ𝑒 𝑞 𝑧 + 𝛼𝑧𝑞′(𝑧) ≥ 𝜆, 𝜆 ∈ [0,1) ⟹ ℜ𝑒 𝑞 𝑧 > 𝜆 + 1 + 𝜆 2𝛾 − 1 where𝛾is given by 𝛾 = 𝛾 ℜ𝑒 𝛼 = 1 + 𝑡1 ℜ𝑒 𝛼 −1𝑑𝑡
0 which isa function of ℜ𝑒 𝛼 𝑎𝑛𝑑 1
2≤ 𝛾 < 1.The estimate sharp in the sense
that bound cannot be improved.
For the complex or a real number a,b and c (c≠ 0and not negative integer) the Gauss hyper geometric function is defined by
2𝐹1(𝑎, 𝑏; 𝑐; 𝑧) = 1 +𝑎𝑏
𝑐 𝑧 +
𝑎 𝑎+1 𝑏(𝑏+1) 𝑐 𝑐+1 2! 𝑧
2+ ⋯
The above series converges absolutely for 𝑧 ∈ 𝔻 and2𝐹1(𝑎, 𝑏; 𝑐; 𝑧) is an analytic function in the unit disk 𝔻. Each of the identities is well known cf.; e.g., [7].
Lemma iii. For a real or complex parameters a, b and c (c ≠ 0)
𝑡01 𝑏−1 1 − 𝑡 𝑐−𝑏−1 1 − 𝑡𝑧 −𝑎𝑑𝑡 =Γ 𝑏 Γ 𝑐−𝑏
Γ(𝑐) 2
𝐹1 𝑎, 𝑏; 𝑐; 𝑧 , ℜ(𝑐) > 𝑅(𝑏) > 0 (7)
1 − 𝑡𝑧 −𝑎2𝐹1 𝑎, 𝑐 − 𝑏; 𝑐; 𝑧
𝑧−1 = 2
𝐹1(𝑎, 𝑏; 𝑐; 𝑧) (8)
2ln 2 = 2𝐹1 1,1; 2;1
2
(9)
IV.
MAIN RESULTS
If it is not mentioned in other sense, let us assume that 𝛼, 𝛽 > 0, 𝑝 ∈ 𝑁, 𝑛 ∈ ℕ0𝑎𝑛𝑑 − 1 ≤ 𝐵 < 𝐴 ≤ 1.
Theorem i. let 𝑓(𝑧) ∈ 𝓣𝑝,𝑚(η,δ,μ,λ) thenℜ𝑒
𝜒𝑝𝑛 +1 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 𝜇
> 2𝜇𝜂𝛽 +𝛼𝛽𝛿
2𝜇𝛽 +𝛼𝛽𝛿 , 𝑔 𝑧 ∈ 𝓣𝑝,𝑚(η,δ,μ,λ)
(10)
Proof:let =2𝜇𝜂𝛽 +𝛼𝛽𝛿
2𝜇𝛽 +𝛼𝛽𝛿 , and we define the function 𝑞(𝑧) given by
𝑞 𝑧 = 1
(1−𝛾)
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 − 𝛾 (11) Then 𝑞(𝑧) is analytic in 𝔻 and 𝑞 0 = 1.
If we consider the function (𝑧) defined by 𝑧 = 𝜒𝑛 +1 𝛼 ,𝛽 𝑔 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 (12) then by the our assumption (8), ℜ𝑒 (𝑧) > 𝛿.
Differentiating (12) with respect to 𝑧 and by the (6), we obtain 1−𝛾 𝛼𝑧 𝑞 𝑧 ′
𝛽 𝜇 = 𝜒𝑝
𝑛+1 𝛼, 𝛽 𝑔 𝑧 ∗ 𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 −
𝜒𝑝𝑛+1𝛼,𝛽𝑓𝑧∗𝜒𝑝𝑛𝛼,𝛽𝑔𝑧∗𝜒𝑝𝑛+1𝛼,𝛽𝑓𝑧𝜒𝑝𝑛𝛼,𝛽𝑔𝑧 (13)
now using (13), we have
1 − 𝛾 𝑞 𝑧 + 𝛾 +𝛼𝜆 1 − 𝛾 𝑧𝑞 𝑧 ′ 𝑧
𝜇𝛽
= 𝜆 𝜒𝑝
𝑛 𝛼, 𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼, 𝛽 𝑔 𝑧
−𝜒𝑝
𝑛+1 𝛼, 𝛽 𝑓 𝑧
𝜒𝑝𝑛+1 𝛼, 𝛽 𝑔 𝑧
+ 𝜒𝑝
𝑛+1 𝛼, 𝛽 𝑓 𝑧
𝜒𝑝𝑛+1 𝛼, 𝛽 𝑔 𝑧 𝜇
⟹ 1 − 𝛾 𝑞 𝑧 + 𝛾 +𝛼𝜆 1 − 𝛾 𝑧𝑞 𝑧 ′ 𝑧
𝜇𝛽
= 1 − 𝜆 𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 𝜇
+ 𝜆Γ 𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼,𝛽 𝑔 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 𝜇 −1
(14)
Again, we consider the functionΨ 𝑟, 𝑠 = 𝑟 − 𝑟 − 1 𝛾 +𝛼𝜆 1−𝛾
𝛽𝜇 𝒮(𝑧) (15)
⇒ 𝑤 ∈ 𝑐: ℜ𝑒 𝑤 > 𝜂 . Now for real 𝑟2, 𝑠1≤ − 1+𝑟22
2 , we have
ℜ𝑒 Ψ 𝑖𝑟2, 𝑠1
=ℜ𝑒 𝑧 (1−γ)αλs1
𝜇𝛽 + 𝛾 ≤ −
𝛿 1+𝑟22 𝛼𝜆 (1−𝛾) 2𝜇𝛽 ≤ −
𝜆 (1−𝛾)𝛿
2𝜇𝛽 = 𝜂.Therefore ∀𝑧 ∈ 𝔻, Ψ(𝑟2, 𝑠1) ∉Ω .
Thus from the lemma [1], we have ℜ𝑒𝑞 𝑧 > 0 (𝑧 ∈ 𝔻) and hence
ℜ𝑒 𝜒
𝑛+1 𝛼, 𝛽 𝑓(𝑧)
𝜒𝑛+1 𝛼, 𝛽 𝑔(𝑧) 𝜇
> 𝛾 (𝑧 ∈ 𝔻)
Theorem ii. Let the function 𝑓 𝑧 𝑎𝑛𝑑 𝑔(𝑧) be in the class 𝓣𝑝,𝑚 and let the function 𝑔(𝑧) satisfy the
condition (7).If 𝛼 ≥ 1 and ℜ𝑒 1 − 𝜆 𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 + 𝛼
𝜒𝑝𝑛 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 > 𝜂Where 𝜂 ∈ 0,1 and (𝑧 ∈ 𝔻) .
ℜ𝑒 𝜒𝑝
𝑛 𝛼, 𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼, 𝛽 𝑔 𝑧
> 𝛾 = 2𝛽 + 𝛼𝛿 + 𝛼𝜆 𝜆 − 1 2𝛽 + 𝛼𝛽𝜆
Proof: If we set𝜆 𝜒𝑝𝑛 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 = 𝜆 − 1
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 + 1 − 𝜆
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧 𝜒𝑝𝑛 +1 𝛼,𝛽 𝑔 𝑧 + 𝜆
𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧 𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧
By the theorem i and ii we have, 𝜆 𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 ≥ 𝜆 − 1 2𝜂𝛽 +𝛼𝜆𝛿
2𝛽 +𝛼𝛽𝛿 + 𝜂
⇒ 𝜆 𝜒𝑝
𝑛 𝛼, 𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼, 𝛽 𝑔 𝑧
≥ 𝛼𝛿 𝜆 − 1 + 𝜂 𝜆𝛿 + 2𝛽
2𝛽 + 𝛼𝛿𝜆 , 𝜇 = 1 and 𝛼 ≥ 1, (𝑧 ∈ 𝔻) Corollary i :Let ∈ 𝐶/ 0 𝑤𝑖𝑡 ℜ𝑒(𝜆) ≥ 0satisfies the following condition If 𝑓(𝑧) ∈ 𝓣𝑝,𝑚
𝕽𝒆 1 −λ (zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝝁
+λ(zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 ) (zp𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝜇 −1
> 𝜂
Then ℜ𝑒 𝑧𝑝𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝜇
> 𝛼ℜ𝑒 𝜆 +2𝜇𝛽𝜂
𝛼ℜ𝑒 𝜆 +2𝜇𝛽 Further if 𝜆 ≥ 1 𝑎𝑛𝑑 𝑓(𝑧) ∈ 𝓣𝑝,𝑚 ,satisfies
𝕽𝒆 1 −λ zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 +λ(zp𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 > 𝜂.
Then ℜ𝑒 zp𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 >
2𝛽𝜂 +𝛼𝜂 +𝛼 (𝜆−1) 2𝛽 +𝛼𝜆
Proof: If we taking 𝑔 𝑧 = 1
𝑧𝑝 in theorem i and ii with ℜ𝑒 𝛿 = 1, then the result is obvious. Theorem iii. Let 𝜆 ∈ 𝐶 𝑤𝑖𝑡 ℜ𝑒(𝜆) ≥ 0. If 𝑓(𝑧) ∈ 𝓣𝑝,𝑚 satisfies the following condition:
𝕽𝒆 1 −λ (zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝝁
+λ(zp𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 ) (zp𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝜇 −1
> 𝜂
Then𝕽𝒆 (zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝝁
> 𝜂 + 𝟏 − 𝜼 (𝟐𝝆 − 𝟏), 𝜌 =1
2𝐹2
1(1,1; 𝜇𝛽 𝛼ℜ𝑒 𝜆 + 1;
1 2)
Proof: let 𝑞 𝑧 = (zp𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝝁
(16)
Then 𝑞 𝑧 is analytic with the condition 𝑞 0 = 1.Differentiating (16) with respect to z and using the identity
(6), we get𝑞′ 𝑧 𝛼𝑧
𝜇𝛽 = z p 𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝜇 −1
zp 𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 − 𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧
Now, by the definition of 𝑞 𝑧 ,we have𝑞 𝑧 + 𝑧𝑞′ 𝑧 𝛼𝑧
𝜇𝛽
= zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝝁
+ 𝜶 (zp𝜒
𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 𝝁−𝟏
zp𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 − zp𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 Solving above
mention equation, we get𝒒 𝒛 + 𝒛𝒒′ 𝒛 𝜶𝒛
𝝁𝜷 = 𝟏 − 𝛌 𝐳𝐩𝝌 𝒑 𝒏+𝟏 𝜶, 𝜷 𝒇 𝒛 𝝁+ 𝝀 𝐳𝐩𝝌 𝒑 𝒏 𝜶, 𝜷 𝒇 𝒛 (𝐳𝐩𝝌 𝒑 𝒏+𝟏 𝜶, 𝜷 𝒇 𝒛 𝝁−𝟏
Therefore , by the assumption of theorem iii, we haveℜ𝑒 𝒒 𝒛 + 𝒛𝒒′ 𝒛 𝜶𝒛
𝝁𝜷 > 𝜂
By the lemma ii, ℜ𝑒 𝑞 𝑧 > 𝜂 + 1 − 𝜂 (2𝜌 − 1)Where 𝜌 = 𝜌(ℜ𝑒 𝜆 = 1 + 𝑡𝛼 ℜ𝑒 𝜆 𝛽𝜇 −1 𝑑𝑡 1 0 = 𝜇𝛽
𝛼ℜ𝑒 𝜆 (1 + 𝜇) −1 1
0 𝜇
𝜇𝛽 𝛼 ℜ𝑒 𝜆 −1𝑑𝜇
Taking ℜ𝑒 𝜆 = 𝜆1 and𝜆1 > 0. Therefore 𝜌 = 𝜌(ℜ𝑒 𝜆 = 1 + 𝑡𝛼 𝜆𝛽𝜇1 −1
𝑑𝑡
1 0
Again using lemma iii, we have 𝜌 =1
2𝐹2
1(1,1; 𝜇𝛽 𝛼𝜆1 + 1;
1 2
Corollary ii: let 𝜆 ∈ Real number with 𝜆 ≥ 1. If 𝑓(𝑧) ∈ 𝓣𝑝,𝑚 satisfies ℜ𝑒 1 −λ (zp𝜒𝑝𝑛+1 𝛼, 𝛽 𝑓 𝑧 +
λ(zp𝜒𝑝𝑛𝛼,𝛽𝑓𝑧>𝜂Then ℜ𝑒 zp𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 > 𝜂 + 1 − 𝜂 2𝜌1− 1 1 − 1
𝛼 where𝜌1= 1 2𝐹1
2(1,1;𝛽 𝛼𝜆+ 1;
1 2) .
Proof : let us consider 𝜆ℜ𝑒 zp𝜒
𝑝𝑛 𝛼, 𝛽 𝑓 𝑧
= ℜ𝑒 1 −λ (zp𝜒
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By the theorem iii and above corollary ℜ𝑒 zp𝜒𝑝𝑛 𝛼, 𝛽 𝑓 𝑧 = 𝜂 + 1 − 𝜂 (2𝜌1− 1) 1 − 1
𝛼 where 𝜌1= 1
2𝐹1 2(1,1;𝛽
𝛼𝜆+ 1; 1 2) .
Theorem iv. Let 𝑓 𝑧 𝑎𝑛𝑑 𝑔(𝑧) belongs to the class 𝓣𝑝,𝑚 and 𝑔(𝑧) satisfies the condition
ℜ𝑒 1 − 𝜆 𝜒𝑝
𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 𝜇
+ 𝜆𝜒𝑝 𝑛 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧 𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧
𝜇 −1
> 𝜂 .
ifℜ𝑒 𝜒𝑝𝑛 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 −
𝜒𝑝𝑛 +1 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 ≥ −𝛿 (1−𝜂)
2𝛽 (17)
For some 𝜂(0 ≤ 𝜂 < 1 then ℜ𝑒 𝜒𝑝
𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 > 𝜂 (18)
andℜ𝑒 𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼,𝛽 𝑔 𝑧 >
𝜂 2𝛽 +𝛿 −𝛿
2𝛽 (19)
Proof: let 𝑧 = 1
(1−𝜂 )
𝜒𝑝𝑛 +1 𝛼,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 − 𝜂 , then 𝑞(𝑧) is analytic function in𝔻 with 𝑞 0 = 1. Now , if we
setting 𝜑(𝑧) =𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 , then form (7), we have ℜ𝑒 𝜑(𝑧) > 𝛿 and 𝛿 ∈ 𝔻.A easy calculation show that
1 −η q′ z =χpn +1α,β g z χpn +1α,β f z
′
−χpn +1α,β f z χ p
n +1α,β g z ′
χpn +1α,β g z 2
Solving above and using the relation (4), we haveαz 1 −η q′ z =χn +1p α,β g z χpn α,β f z −χpn +1α,β f z χpn α,β g z χpn +1α,β g z 2
Now by the definition of 𝜑(𝑧), we get αz 1−η q′ z β 𝜑 𝑧 =
χpn α,β f z χpn α,β g z −
χpn +1α,β f z
χpn +1α,β g z = 𝜑 𝑞 𝑧 , 𝑧𝑞′ 𝑧 , where Ψ 𝑟, 𝑠 = 1−𝜂 𝑠𝜑 𝑧
𝛽 .
So by the our assumption
Ψ 𝑞 𝑧 , 𝑧𝑞′ 𝑧 ; 𝑧 ∈ 𝔻 ⊂Ω= 𝜔 ∈ 𝐶: ℜ𝑒 𝜔 > − 1 − 𝜂)𝛿 2𝛽
ℜ𝑒 Ψ ir2, s1 =
𝑟1 1 − 𝜂 ℜ𝑒 𝜑 𝑧
𝛽 ≤ −
(1 − 𝜂)𝛿 2𝛽
This is shows that Ψ(ir2, s1) ∉Ω for each 𝑧 ∈ 𝔻. Hence by the lemma [1] we get ℜ𝑒 𝑞(𝑧) > 0 (𝑧 ∈ 𝔻). This
is the proof of (18). Now for the proof of (19), ℜ𝑒 𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 𝛼 ,𝛽 𝑔 𝑧 = ℜ𝑒
𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧 𝜒𝑝𝑛 𝛼 ,𝛽 𝑓 𝑧 −
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧 𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧 +
ℜ𝑒 𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑓 𝑧
𝜒𝑝𝑛 +1 𝛼 ,𝛽 𝑔 𝑧 ≥ − 1−𝜂 𝛿
2𝛽 + 𝜂
≥𝜂 2𝛽 +𝛿 −𝛿
2𝛽 which is the complete proof of (19).
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