Second Hankel determinant obtained with New Integral Operator defined by Polylogarithm Function

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1 Global Journal of Pure and Applied Mathematics. ISSN Volume 3, Number 7 (07), pp Research India Publications Second Hankel determinant obtained with New Integral Operator defined by Polylogarithm Function S. M. Patil Department of Applied Sciences, SSVPS B.S. Deore College of Engineering, Deopur, Dhule, Maharashtra, India. S. M. Khairnar Professor & Head, Department of Engineering Sciences, MIT Academy of Engineering, Alandi, Pune-405, Maharashtra, INDIA. Abstract By using polylogarithm function, a new integral operator is introduced. By using this operator a new subclass of analytic functions are introduced for these classes we obtained sharp upper bounds for functional a a 4 a 3. AMS subject classification: Keywords: Integral Operator, Analytic Function, Hankel Determinant, Polylogarithm Function.. Introduction In 966, Pommerenke stated the q th Hankel determinant for q,&n 0as a n a n+ a n+q+ H q (n) = a n+.. a n+q. a n+q (.) where a n s are the coefficients of various power of z in f(z).

2 S. M. Patil and S. M. Khairnar This determinant has also been considered by several authors. For example Noor determined the rate of growth of H q (n) an n for function f, with bounded boundary. One can easily observe that Fekete and Szegö functional H (). Fekete and Szegö then further generalized and estimate a 3 µa where µ is real & f S. We consider the Hankel determinant for the case q=andn=, H () = a a 3 a 3 a 4 = a a 4 a3 (.) We recall here the definition of well known generalization of the polylogarithm function φ(a,z) given by φ(a,z) = k= z k k a (a N,z E) (.3) Let φ δ (a, z) denote the well known generalization of the Riemann Zeta & polylogarithm function or simply the δ th order polylogarithm function given by, φ δ (a, z) = k= z k (k + a) δ (.4) where any term with k+a=0isexcluded. Using the definition of the Gamma function, a simply transformation produces the integral formula, φ δ (a, z) = Ɣ(δ) 0 ( z log ) δ t a δt Ra >, Rδ > (.5) t tz z Note that φ (0,z)= is koebe function for more details about polylogarithms ( z) in theory of univalent functions see Punnusamy & Sabapathy. Recently Khalifa Alshaqsi introduced a certain Integral Operator Ia δ defined by, Ia δ ( + a)δ f(z)= Ɣ(δ) 0 t a ( log t ) δ f (tz)δt a > 0,δ >.z E (.6) We also note that the operator Ia δ f(z) defined by [] can be expressed by the series expansion as follows, Ia δ f(z)= z + ( ) δ + a a k z k (.7) k + a obviously, we have for (δ, λ 0) k= Ia δ (I a λ δ+λ f(z))= Ia f(z) (.8)

3 Second Hankel Determinant with Polylogarithm Function 3 and Moreover from (.7) if follows that, We note that, I δ a (zf (z)) = z(i δ a f(z)) (.9) z(ia δ+ f(z) ) = ()Ia δ f(z) aiδ+ a f(z) (.0) Fora=0andδ = n (n is any integer) the multiplier transformation I n 0 f(z)= I n f(z) was studied by Flett and Salagean. Fora=0andδ =-n(n N 0 {0,,, 3,...}), the differential operator I n 0 f(z) = D n f(z)was studied by Salagen. Fora=andδ = n (n is any integer) the operator I n f(z) = I n f(z)was studied by Uralegaddi and Somanatha. Fora=,themultiplier transformation I f f(z) = I δ f(z) was studied by Jung et al. Fora=k-(k>), the integral operator Ik δ f(z), I k δ f(z) was studied by Komatu using the operator Ia δ, we now introduced the following cases. Definition.. We say that a function f A is in the class S aδ (b) if ) { R + (z (I δa f(z) )} b Ia δf(z) > 0 (a > 0,δ 0,b c\{0}; z E) (.) Definition.. We say that a function f A is in class C aδ (b) if ) { R + z (I δa f(z) } ) p > 0 (a > 0,δ 0,b c\{0}; z E) (.) (I δa f(z) Note that f C aδ (b) zf S aδ (b) (.3) In particular, we have starlike & convex function classes S a0 ()=S and C a0 ()=C respective. Lemma.3. Let p P then c k, k =,,...and the inequality is sharp.

4 4 S. M. Patil and S. M. Khairnar Lemma.4. Let p P then c = c + x(4 c ) 4c 3 = c 3 + xc (4 c ) x c (4 c ) + y( x )(4 c ) (.4) for some x and y such that x, y. Theorem.5. If f S aδ (b) then a a 4 a 3 b ( a + 3 Proof. By the definition of the class S aδ (b), there exist p P such that, + ( z(i δ a f(z)) ) p Ia δf(z) = p(z) z(i δ a f(z)) I δ a f(z) Let I δ a = z + A z + a 3 z +..., where, ( ) δ A = a a + ( ) δ A 3 = a 3 a + ( ) δ A 4 = a 4 a + so that, z{ + A z + 3A 3 z + 4A 3 z } z + A z + A 3 z Simplify and equating the coefficient of z on both side Equating the coefficient of z 3 on both side ) δ = b + bp(z) (.5) = b + b[ + c z + c z + c 3 z 3 ] (.6) A = A A b + A b + bc A = bc ( ) δ (.7) a + A = bc A 3 = b [c + bc ] [ ( ) b a + δ ] (.8) a 3 = (c + bc )

5 Second Hankel Determinant with Polylogarithm Function 5 Equating the coefficients of z 4 It is establish that, 3A 4 = b6c c + b c 3 b c c + bc 3 A 4 = 3b c c + b c 3 + bc 3 6 ( ) a + 4 δ (3b c c + b c 3 a 4 = + bc 3) 6 (.9) ( a + a a 4 a3 = (( a + 3 By using Lemma, ) δ ( ) a + 4 δ (3b c c + b c 4 + bc c 3 ) 6 δ ) b (c + bc ) ) (.0) c = c + x(4 c ) for some x 4c 3 = c 3 + c (4 c )x c (4 c )x + (4 c )( x )z for some real value of zwith z (.) ( ) a + δ ( ) a + 4 δ [ ( a a 4 a3 = c 3b c + x(4 c ) ) ] + b c bc [ c 3 + c (4 c )x c (4 c )x + (4 c )( ] x z) 4 {( ) a + 3 δ [ b c + x(4 c ) + bc ]} (.) Since c,c = c assume without restriction, c [0, ], we obtain by using triangle inequality x = ρ. c = c. [{( a + a a 4 a3 ) δ ( a + 4 ) δ 6 [ c 3 + c(4 c )ρ ρ (4 c )ρ F(ρ) [ 3b c 4 + 3b c (4 c ] )ρ + b c 4 + b c ] ( ) a + 3 δ b [ ρ(4 c ] ) }] + 4 (.3)

6 6 S. M. Patil and S. M. Khairnar {( ) a + δ ( ) a + 4 δ ( F 3b c (4 c ) ) (ρ) = 6 ( ) a + 3 δ b [ ρ(4 c ] ) )} bc [ ] c(4 c ) ρ(4 c )ρ (.4) F (ρ) > 0 for ρ>0, implies that F is an increasing function. The upper bound for (5) are c = 0. ( ) a + 3 δ a a 4 a3 b Corollary.6. If we put δ = 0, b = we get, S a0 () = S a a 4 a 3 this result is coincide the result with Janteng. Corollary.7. If f c aδ (b) then, + ( z(i δ a f(z) ) ) b (Ia δ = p(z) f(z) ) z(i a δf(z) ) (Ia δf(z) ) = bp(z) b (.5) Simplify & equating the coefficients we get, a = bc ( ) a + δ ( ) a + 3 δ [ b c a 3 = + bc ] 6 ( ) a + 4 δ [ 3b c c + b c 3 a 4 = + bc ] 3 4 (.6) and calculate the same a a 4 a3 then we get the result of convex function. After simplification we put δ = 0, b = then we get a a 4 a3 results is coincide the 8 result with Janteng.

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