Vol. 7, No. 3, September 2019, pp.1114-1125 Available online at: http://pen.ius.edu.ba
Bounds of the initial coefficient for sakaguchi function in the conical domain
S. Balaji1, B. Srutha Keerthi2
1,2Department of Mathematics, School of Advanced Sciences, VIT Chennai Campus
Article Info ABSTRACT
Received May 27, 2019 In this paper, we consider a new class of sakaguchi type functions which is defined by Ruscheweyh q-differential operator. We investigated of co- efficient inequalities and other interesting properties of this class.
Keyword:
Conical domain Analytic functions Sakaguchi type functions k-uniformly starlike functions q-differential Operator.
Corresponding Author:
B. SruthaKeerthi
Department of Mathematics, School of Advanced Sciences VIT Chennai Campus, Chennai - 600 048, India
E-mail: [email protected]
1. Introduction
Let A denotes the class of functions of the form
2 n
n nz a z
f(z) (1)
which are analytic in the open unit disk D = {z C : |z| <1}. If f and g are analytic function in D, then thefunction f is said to be subordinate to g, and write f(z) g(z), if there exists a function analytic in D with(0) = 0 and |(z)| < 1 for all z D, such thatf(z) = g((z)), z D. Moreover, if the function g isunivalent in D, then f(z) g(z) if and only if f(0) =g(0) and f(D) g(D).
Kanas and Wisniowka [5, 6], established the conic kind of domain k, k 0 as
k
{u iv : u
k(u 1)
2 v
2}
We note that k is a region in the right half-plane, symmetric with respect to real axis, and contains the point (1,0). More precisely for k=0, 0 is the right half-plane, for 0<k<1, kis an unbounded region having boundary k, a rectangular hyperbola for k=1, 1 is still an unbounded region where 1 is a parabola, and for k>1, k is a bounded region enclosed by an ellipse. The extremalfunction for these conic regions are
k 1 , k 1
t k 1 t 1
dk 2k(κk
sin π 1 k
1
1 k 0 k ,
z 1 1
z log 1 arccosk π
cosh 2 k 1
1
1 k ,
z 1
z log 1 π 1 2
0 k z ,
1 z 1
(z) p
2 κ 2
u(z)
0 2 2 2
2
2 2 2
2
2
k
k k (2)
where ,
kz 1
k u(z) 2
zD and k (0,1) is chosen such that . κ) 4K(
κ) ( πK cosh
k
Here K() is Legendres’s
complete elliptic integral of first kind and K(κ)K( 1k2)and K(t) is the complementary integral of K(t) for details see [1, 5, 6] and more recently [9,12,14]. If Pk(z)=1+M1(k)z+M2(k)z2+, zD, then it was shown in [6] that for (2) one can have,
1 k t t) (1 (t) 4K
π
1 π k
8
1 k k 0
1 2A
(k) M
2 2
2 2
2 2
1 (3)
M2(k)= E(k) M1(k) where
1 k t
t) (1 24K(t)
π 1) 6t (t (4K(t))
1 π k
8
1 k 3 0
2 A E(k)
2
2 2
2 2
2
(4)
with arccosk π
A 2
Further more a function p is said to be in the class kP[A,B ] if and only if 0
k (z), q
p(z) k
1) (B (z) 1)p (B
1) (A (z) 1)p (z) (A
q where
k k
k
(5)
where pk is defined in (2) and 1 B < A 1. Geometrically the function p kP[A,B] takes all the values from the domain k[A,B], 1 B < A 1, k 0, which is defined as:
k[A, B] = { : ((c()) >k|c()|}
1) (A 1)ω (B
1) (A 1)ω c(ω( (B
where
or equivalently k[A,B] is a set of numbers = u+iv such that [(B2 1)(u2 + v2) 2(AB 1)u + (A2 1)]2
>k[ 2(B + 1)(u2 + v2) + 2(A + B + 2)u 2(A + 1)2 + 4(A B)2v2]
This domain represents the conic type of regions for detail see [11]. For any n positive integer n, the q-integer number n, [n,q] is defined by
(0,1) q
0, q]
[0, q q
q 1 1
q q] 1
[n,
n 1n
(6)q-differential operator be defined by
D) (z 1)z , (q
f(z) f(qz)
qf(z)
It is easy to observe that for n N := {1,2,3...} and z D
qzn = [n,q]zn1 Let the q-generated pochhammer symbol be defined as
[r,q]n =[r,q][r+1,q][r+2,q]...[r+n1,q]
and for r>0 let the q-gamma function be defined as
q(r+1)=[r] q(r) and q(1)=1
These kind of operators see [2,3,13], play great in GFT. Kanas et al, defined Rucheweyh q-differential operator as follows:
Definition 1. [7]
For the function fA is in the form (1) , the Rucheweyh q-differential operator is:
λ 1) , (z (z), F f(z) f(z)
Rqλ q,λ1 D (7)
where
n 2 n
1 n
n 2
n
1 n
n 2
n q
q λ 1
q,
z z
q]! z 1, [n
q]
1, z [λ
λ)z Γ (1 q]!
1, [n
(n λ) z Γ
(z) F
(8)
where
q]!
1, [n
q]
1, [λ λ) Γ (1 q]!
1, [n
(n λ)
Γ n 1
q q 1
n
from (7) we get that
f(z) z f(z) R f(z), f(z)
R0q q q and
) q]! (m
[m, f(z)) (z f(z) z
R
1 m m m q
q N
Using (7) and (8), the power series Rqλf(z) is given by
n n 2
n
1 n
n n 2
n q
q q
z q]! a 1, [n
q]
1, z [λ
z λ)a Γ (1 q]!
1, [n
λ) Γ (n z
f(z) R
(9)
Note that
λ 1 λ 1
1 q,
q (1 z)
(z) z
limΓ
and
λ 1 λ
1 q
q (1 z)
f(z) z f(z) R
lim
When q1 see[16], we observe that
(z) q F
λ,q]
(z) [ q F
λ,q]
1 [ (z)) (F
z q,λ1 λ q,λ2 λ q,λ1
(10)
making use of (7), (10) and the properties of hadamard product we obtain the following equality f(z)
q R λ,q]
f(z) [ q R
λ,q]
1 [ f(z)) (R
z λq λ λq 1 λ λq
(11)
If q1, the equality (11) implies
z(Rf(z))=(1+) R+1f(z) Rf(z)
which is the familiar recurrent formula for the above operator. we now defined the following classes of functions.
Definition 2.
A function f(z) A is said to be in the class kUSq(,A,B,t), k 0, 0, t C with |t| 1
1 B <A 1, if and only if
(c(X( z )))>k |c(X(z))|
f(tz)) R f(z) (R
f(z)) R t)(z
X(z) (1 λ
q λ
q
λ q q
orequvalently,
B]
P[A, f(tz)) k
R f(z) (R
f(z)) R t)(z
X(z) (1 λ
q λ
q
λ q
q
Definition 3.
A function f(z) A is said to be in the class kUCq(,A,B,t), k 0, 0, t C with |t| 1,1 B <A 1, if and only if
(c(Y(z)))>k |c(Y(z))|
f(tz)) R t f(z) R z(
f(z)) R z f(z) R t)(z
Y(z) (1
λq q λ
q q
λ q 2 q λ 2
q q
orequvalently,
λ q q q λ
q q
λ q 2 q λ 2
q
q
k UC
f(tz)) R t f(z) R z(
f(z)) R z f(z) R t)(z
Y(z) (1
Definition 4.
A function f(z) A is said to be in the class kUSq(,A,B,,t), k 0, 0, t C with |t| 1
1 B <A 1, if and only if
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z
G(z) (1 λ
q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
orequvalently,
B]
P[A, f(tz)) k
R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z
G(z) (1 λ
q q λ
q q λ
q λ
q
λ q 2 q λ 2
q
q
(12)
Remark 5.
It is easily see that limq 1 kUSq(0, A, B, 0, 0) = kST(A, B) where kST(A, B) is a functions class, investigated by Noor and sarfraz [11]
c(
G(
z) k
c(
G(
z))
Lemma 6.[15]
Let
1 n
n nz c 1
h(z) be subordinate to H(z) 1 C z .
1 n
n
n
If H(z) is univalent in D and H(E) is convex, then
|cn| |C1|, n 1 Lemma 7. [8,10]
If q(z)=1+c1z+c2z2+ is an analytic function with positive real part in D then,
|}
1 2v
| 2max{1, vc
c2 12
The result is sharp for the function
z 1
z q(z) 1 z (or)
1 z q(z) 1
2 2
Lemma 8. [8]
If the function D is in the form
(z)=c1z+c2z2+ zD Then,
2 1 2
1
2 vc 1 (v 1)c
c
where v is the complex number Lemma 9. [11]
Let k [0,) be fixed and qk(z) in the form (5)then
qk(z)=1+H1(k)z+H2(k)z2+, zD and
(k) 2 M
B (k) A
H :
H1 1 1
(k) }M 1)H (B {2E(k) 4
B (k) A
H :
H2 2 1 1
where M1(k) and E(k) are defined in (3) and (4) 2. Main Results
Theorem 10:
A function fA and of the form (1) is in the class kUSq(,A,B,, t), if it satisfies the condition
A B a
|}
} γ[n,q]u 1){(1 γ)u
(A
q]}
1, γ[n,q][n
q]
1){[n, (B
| q]}
1, γ[n,q][n
q]
1)[n, u
γ)u (γ 1){(1 {2(k
n 1 n n n
2 n
n n
(13)
roof.
Assume (13) is hold, then it suffices to show that
(A 1) 1 1
1) (B
1) (A
1) (B
1 1) (A
1) (B
1) (A
1) (B k
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
we have
1 1) (A
1) (B
1) (A
1) (B
1 1) (A
1) (B
1) (A
1) (B k
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
(1
f(z)) R γz f(z) R t)(z (1
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
f(tz))] 1 R t f(z) R γz(
f(tz)) R γ)(R f(z)
1)[(1 (A f(z)) R γz f(z) R t)(z 1)(1 (B
f(tz))]
R t f(z) R γz(
f(tz)) R γ)(R f(z)
1)[(1 (A f(z)) R γz f(z) R t)(z 1)(1 1)(B
(k λ
q q λ
q q λ
q λ
q λ
q 2 q λ 2
q q
λ q q λ
q q λ
q λ
q λ
q 2 q λ 2
q
q
f(tz))]
R t f(z) R γz(
f(tz)) R γ)(R f(z)
1)[(1 (A f(z)) R γz f(z) R t)(z 1)(1 (B
f(z)) R γz f(z) R t)(z (1 f(tz)) R t f(z) R γz(
f(tz)) R γ)(R f(z)
1) (1
2(k λ
q q λ
q q λ
q λ
q λ
q 2 q λ 2
q q
λ q 2 q λ 2
q q λ
q q λ
q q λ
q λ
q
2 n
n n 1 n n n
2 n
n n 1 n n
n
z a }}
γ[n,q]u γ)u
1){(1 (A
q]}
1, γ[n,q][n
q]
1){[n, {(B
A)z (B
z a q]}
1, γ[n,q][n
q]
1)[n, u
γ)u (γ 1) {(1
2(k
2
n n n n 1 n
2
n n n n 1 n
a }}
γ[n,q]u γ)u
1){(1 (A
q]}
1, γ[n,q][n
q]
1){[n, {(B
A B
a q]}
1, γ[n,q][n
q]
1)[n, u
γ)u (γ {(1 1)
2(k
< 1 by (13) □
When q1 and =0 we have, Corollary 11.
A function f A and of the form (1) is in the class kUSq(,A,B,t), if it satisfies the condition, A
B a } 1) (A u 1) n(B n) 1)(u {2(k
2 n
n 1 n n
n
When q 1 and =1 we have, Corollary 12.
A function f A and of the form (1) is in the class kUCq(,A,B,t), if it satisfies the condition,
A B a } 1) (A nu 1)n (B 1)}
n(n 1) 1){n(u {2(k
2 n
n 1 n n
2
n
Theorem 13.
If f(z) kUSq(,A,B,,t), and is of the form (1). then
2 n ] ,
q]}u γ[j 2,
γ) {(1 q]
2, q]}[j γ[j 1,
2[{1
]B q]}u γ[j 1,
γ) {(1 q]
1, q]}[j γ[j, 2[{1 q]}
γ[j 1, γ) {(1 B)u (A M a
2 n
0
j j 2 n 1
1 j 1
j 1
n
(14) where M1(k) is defined by (3)
Proof.
Let
f(tz)) p(z) R t f(z) R γz(
f(tz)) R γ)(Rf(z)
(1
f(z)) R γz f(z) R t)(z (1
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q
q
(15)
then
2 2 k
3 2 2 k
1 k
k k
(z)) 1) (p
(B
1) 1)(B (A 1)
(B
1) 1)(B (z)) (A
1) (p (B
1) (A 1)
(B
1) 1)(B (A 1) (B
1) (A
1)]
(B (z) 1)p 1)][(B (A
(z) 1)p [(A
(z) q p(z)
By taking pk(z)=1+M1(k)z+M2(k)z2+
after some simplification, we obtain,
1 n
1 1
n
1 n
1 n n
1 n
(k) 1) M
(B
1) B)(B 2n(A
1) (B
1)
p(z) 2(B
Now we see that the series
1
n n
1 n
1) (B
1)
2(B and
1
n n 1
1 n
1) (B
1) B)(B 2n(A
are convergent and converge to 1 and 2
B A
respectively. Therefore,
M (k)z 2
B 1 A
p(z) 1
Now if p(z) 1 c z ,
1 n
n
n
then by lemma [6], we have 1 n (k), 2 M
B
cn A 1 (16)
Now from (15) we have
) f(tz))]p(z R
t f(z) R γz(
f(tz)) R γ)(R f(z)
[(1 f(z)]
R γz f(z) R t)[z
(1 q qλ 2q2 λq λq λq q λq q λq by simple calculation provides us,
1 a ] ,
γ[n,q]}u γ)
{(1 q]
q]}[n, γ[n 1,
[{1
c a γ[j,q]]u
γ) [(1
a 1
1 n n 1
n 1
j j j 1 j n j
n
1 a ] ,
γ[n,q]}u γ)
{(1 q]
q]}[n, γ[n 1,
2[{1
a γ[j,q]]u
[(1 γ) (k)
M B) (A
a 1
1 n n 1
n 1
j j j 1 j
1
n
(17) Now we prove that
2 n
0
j j 2 n1
1 j 1
j 1
1 n n 1
n 1
j j j1 j
1
] q]}u γ[j 2,
{(1 γ) q]
2, q]}[j γ[j 1,
2[{1
]B q]}u γ[j 1,
{(1 γ) q]
1, q]}[j 2[{1 γ[j,
q]}
γ[j 1, {(1 γ)
B)u (k)(A M
] γ[n,q]}u {(1 γ)
q]
q]}[n, γ[n 1,
2[{1
a γ[j,q]]u
γ) [(1 (k)
M B) (A
(18)
For this we use the induction method For n=2 From (17), we have
1 2 1
2 2[(1 γ)[2,q] {(1 γ) γ[2,q]}u ] (k)
M B) a (A
From (14) we have
1 2 1
2 2[(1 γ)[2,q] {(1 γ) γ[2,q]}u ] (k)
M B) a (A
For n=3 From (17) we have
] γ[2,q]}u
γ) {(1 γ)[2,q]
2[(1
] γ[2,q]}u
γ) {(1 γ)[2,q]
2[(1 γ[2,q]]
γ) [(1 B)u (A
(k) M
] q])}u γ[3, γ
{(1 q]
q]}[3, γ[2,
2[{1
M B) a (A
2
2 2
1
2 3 1
3
from (14) we have
] γ[3,q]}u
γ) {(1 q]
q]}[3, γ[2,
2[{1
] γ[2,q]}u
γ) {(1 γ)[2,q]
2[(1 γ[2,q]]
γ) [(1 B)u (A
(k) M
] γ[2,q]}u
γ) {(1 γ)[2,q]
2[(1
M B) a (A
3
2 2
1
2 2 1
3
Let the hypothesis be true for n=m from (16) we have,
1 a ] ,
γ[m,q]}u {(1 γ)
q]
q]}[m, γ[m 1,
2[{1
a γ[j,q]]u
γ) [(1 (k)
M B) (A
a 1
1 m m 1
m 1
j j j 1 j
1
m
From(14) we have
m 2
0
j j 2 m 1
1 j 1
j 1
m 2[{1 γ[j 1,q]}[j 2,q] {(1 γ) γ[j 2,q]}u ]
] q]}u γ[j 1,
γ) {(1 q]
1, q]}[j γ[j, 2[{1 q]}
γ[j 1, γ) {(1 B)u (A a M
By induction hypothesis, we have
2 m
0
j j 2 m 1
1 j 1
j 1
1 m m 1
m 1
j j j1 j
1
] q]}u γ[j 2,
γ) {(1 q]
2, q]}[j γ[j 1,
2[{1
] q]}u γ[j 1,
{(1 γ) q]
1, q]}[j 2[{1 γ[j,
q]}
γ[j 1, {(1 γ)
B)u (A M
] γ[m,q]}u
{(1 γ) q]
q]}[m, γ[m 1,
2[{1
a γ[j,q]]u
γ) [(1 (k)
M B) (A
(19) Multiplying both sides (19) by
] q]}u γ[m 1,
{(1 γ) q]
1, q]}[m 2[{1 γ[m,
] γ[m,q]}u
{(1 γ) q]
q]}[m, γ[m 1,
2[{1 γ[m,q]}
{(1 γ) B)u (A M
1 m
m m
1
we have
2 m
0
j j 2 m 1
1 j 1
j 1
] q]}u γ[j 2,
{(1 γ) q]
2, q]}[j γ[j 1,
2[{1
] q]}u γ[j 1,
{(1 γ) q]
1, q]}[j 2[{1 γ[j,
q]}
γ[j 1, {(1 γ)
B)u (A M
1 m m 1
m 1
j j j 1 j
1 m
1 m
m m
1
] γ[m,q]}u
γ) {(1 q]
q]}[m, γ[m 1,
2[{1
a γ[j,q]]u
[(1 γ) (k)
M B) (A a
] q]}u γ[m 1,
{(1 γ) q]
1, q]}[m 2[{1 γ[m,
] γ[m,q]}u
{(1 γ) q]
q]}[m, γ[m 1,
2[{1 γ[m,q]}
{(1 γ) B)u (A
M
1
m
1 j
m 1 m m
j 1 j
1 m 1 m 1
a γ[m,q]}
γ) {(1 u a γ[j,q]]
γ) [(1
] q]}u γ[m 1,
γ) {(1 q]
1, q]}[m γ[m, 2[{1
M B) (A
m
1 j
j 1 j
1 m 1 m 1
a γ[j,q]]
γ) [(1
] q]}u γ[m 1,
γ) {(1 q]
1, q]}[m γ[m, 2[{1
M B) (A
That is
2 m
0
j j 2 m 1
1 j 1
j 1
1 m 1 m 1
m 1
j j j 1 j
1
] q]}u γ[j 2,
γ) {(1 q]
2, q]}[j γ[j 1,
2[{1
] q]}u γ[j 1,
γ) {(1 q]
1, q]}[j γ[j, 2[{1 q]}
γ[j 1, γ) {(1 B)u (A M
] q]}u γ[m 1,
γ) {(1 q]
1, q]}[m γ[m, 2[{1
a γ[j, q]]u
γ) [(1 (k)
M B) (A
Which gives (19). □
When q 1and =0 we have, Corollary 14.
A function f A and of the form (1) is in kUSq(,A,B,t), if,
n 2
0
j j 2 n 1
1 j 1
j 1
n 2[j 2 u ]
]B u 1 2[j B)u
(k)(A a M
is satisfied.
When q 1 and =1 we have, Corollary 15.
A function f A and of the form (1) is said to be in the class kUCq(,A,B,t), if,
n 2
0
j j 2 n 1
1 j 1
n 2(j 2)[2 j u ]
]B u j 1)[1 2(j 1) B)(j (k)(A a M
is true.
Theorem 16.
Let 1 B < A1and t=0, 0 k < be fixed and let f(z) kUSq(,A,B,, t) and is of the form (1) then for a complex number
δ ) (μ Ru
δ ) μ (δ
δ ) (μ , Ru
2 μa
a
) 2 }(
Ru {P
} μ{Q Su 2 } Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(k) } Su 2{Q
(k) B)M (A
2 1
} Su 2{Q
(k) B)M (A
) 1 }(
Ru {P
} μ{Q Su } 2
Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(k) }
Su 2{Q
(k) B)M (A
2 2 3
2 1 2
2 3 2
1 1
2 3
1 2 3
1
2 1 2
2 3 2
1 1
2 3
1
(20)
where
} (k)Ru B)M (A } Ru (k)}{P B)M (1 2E(k) }{{2
Su (k){Q B)M (A
} Ru {P )
δ ( 1 2 1 2
3 1
2
2 2
1
1
(21)
} (k)Ru B)M (A } Ru 2}{P (k) B)M (1 {{2E(k) }
Su (k){Q B)M (A
} Ru {P )
δ ( 1 2 1 2
3 1
2
2 2
1
2
(22)
and
P=(1+)[2,q] Q=(1+[2,q])[3,q]
R= (1)+[2,q] S=(1)+[3,q]
and M1(k), E(k) are defined in (3) and (4) Proof.
If f(z) kUSq(,A,B,,t) then it follows that
2 1 1
1
λ k q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
(k)z 4 M
B) (k)}(A B)M
(1 {2E(k) (k)z
2 M B) 1 (A
(z) f(t z)) q
R t f(z) R γz(
f(t z)) R
γ)(R f(z) (1
f(z)) R
γz f(z) R t )(z (1
(23)
Now by the definition of subordination there exists a function analytic in D with (0)=0 and
|(z)|<1 such that
(z) (k)ω 4 M
B) (k)}(A B)M
(1 {2E(k) (z)
(k)ω 2 M
B) 1 (A
f(t z)) R
t f(z) R γz(
f(t z)) R
γ)(R f(z) (1
f(z)) R
γz f(z) R t )(z (1
2 1
1 1
λ q q λ
q q λ
q λ
q
λ q 2 q λ 2
q q
(24)
Now from lemma 8, equation (23) and equation (24), We have
1 2
1 1
2 2[P Ru ]
(k)c B)M a (A
and
12
2 1 2
1 2
2 3
1
3 c
] Ru 2[P
(k) B)M (A Ru 2
(k)]
B)M (1 [2E(k) ] c
Su 2[Q
(k) B)M a (A
, ) c }(
Ru {P
} μ{Q Su } Ru
Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(K) } c
Su 2{Q
(k) B)M a (A
a μ 2 12
1 2
2 3 2
2 1 1
2 2 3 2 1
2
3
(25)
which gives
(26) Suppose that >1 then using the estimates c2c12 1 from lemma 8 and the well known estimate |c1| 1 of the Schewarz lemma, we obtain
) , }(
Ru {P
} μ{Q Su
} Ru Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(k) } 2
Su 2{Q
(k) B)M a (A
μ
a 2
1 2
2 3 2
2 1 1
2 3 2 1
2
3
(27)The inequality (27) is our required assertion (20) for >1 on other hand if <2 then (25) gives,
,
| c ) | }(
Ru {P
} μ{Q Su
} Ru Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(k) } c
Su 2{Q
(k) B)M a (A
μ
a
2 121 2
2 3 2
2 1 1
2 2 3 2 1
2
3
, ) c }(
Ru {P
} μ{Q Su } Ru
Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(K) 1
c } c
Su 2{Q
(k) B)M a (A
a μ 2 12
1 2
2 3 2
2 1 2 1
1 2 2 3 2 1
2
3
(28) Applying the estimates |c2| 1|c1|2 of lemma 8 and |c1| 1, We have
2
1 2
2 3 2
2 1 1
2 3 2 1
2
3 {P Ru }( )
} μ{Q Su } Ru
Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(K) }
Su 2{Q
(k) B)M a (A
μ
a
This is last inequality in (20). Finally if 1<<2 then
) 1 }(
Ru {P
} μ{Q Su
} Ru Ru 2{P
(k) B)M (A 2
(k) B)M (1 2E(k)
2 1 2
2 3 2
2 1
1
Therefore (25) yields
} c c } { Su 2{Q
(k) B)M a (A
μ
a 2 12
2 3 2 1
2
3
} c c 1 } { Su 2{Q
(k) B)M a (A
a μ 2 12
2 3 2 1
2
3
2 3 2 1
2
3 2{Q Su }
(k) B)M a (A
μ
a
We get the middle inequality in(20). This completes the proof. □
Theorem 17.
Let k <, 1 B < A 1 and t=0 be fixed and let f(z) kUSq(,A,B, , t) and is of the form (1). Then for a complex number
} 1 2 , 1 } max{
Su {Q
(k) M B) a (A
μ a
2 3
1 2
2
3
v
where v is given by (31)
Proof.
From (25) we have
, ) c }(
Ru {P
} μ{Q Su
} Ru Ru 2{P
(k) B)M (A 2
2E(k) (k)
B)M c (1
} Su 2{Q
(k) B)M a (A
μ
a
2 121 2
2 3 2
2 1 1
2 2 3 2 1
2
3
(29)
2 1 2 2 3
1 c vc
} Su 2{Q
(k) B)M
(A
(30)
where
21 2
2 3 2
2 1 1
) }(
Ru {P
} μ{Q Su
} Ru Ru 2{P
(k) B)M (A 2
2E(k) (k)
B)M (1
v
(31) Applying the lemma 7 on the equation (30), we obtain the required result.
□ References
[1] N.I. Ahiezer, “Elements of theory of elliptic functions”, Moscow, 1970.
[2] S. Hussain, S. Khan, M.A. Zaighum and M.Darus, “Certain subclass of analytic functions related with conic domains and associated with salagean q-differential operator”, AIMS Math.,vol.2, no. 4, pp. 622- 634, 2017.
[3] S.Hussain, S. Khan, M.A. Zaighum, M.Darus and Z.Shareef, “Coefficient Bounds for Certain Subclass of Biunivalent Functions Associated with Rucheweyh q-differential Operator”, J. Complex Anal., 2017 (2017).
