Mapping properties of the zero-balanced hypergeometric functions
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MAPPING PROPERTIES OF THE ZERO-BALANCEDHYPERGEOMETRIC FUNCTIONS
LI-MEI WANG
Abstract.
In the present paper, the order of convexity of z F ( a, b ; c ; z ) is first givenunder some conditions on the positive real parameters a, b and c . Then we show thatthe image domains of the unit disc D under some shifted zero-balanced hypergeometricfunctions z F ( a, b ; a + b ; z ) are convex and bounded by two horizontal lines which solvesthe problem raised by Ponnusamy and Vuorinen in [9]. Introduction and main results
The
Gaussian hypergeometric function is defined by the power series F ( a, b ; c ; z ) = ∞ X n =0 ( a ) n ( b ) n ( c ) n n ! z n for z ∈ D := { z ∈ C : | z | < } , where ( a ) n is the Pochhammer symbol ; namely, ( a ) = 1and ( a ) n +1 = ( a ) n ( a + n ) = a ( a + 1) · · · ( a + n ) for all n ∈ N := { , , , . . . } . Here a, b and c are complex constants with − c N . Hypergeometric functions can be analyticallycontinued along any path in the complex plane that avoids the branch points 1 and ∞ .If c = a + b , it is called the zero-balanced one and z F ( a, b ; c ; z ) is usually called a shifted hypergeometric function. For instance the function z F (1 ,
1; 2; z ) = − log(1 − z ) is ashifted zero-balanced hypergeometric function. In the present paper, we only restrict tothe real parameters a, b and c . For the basic properties of hypergeometric functions werefer to [1], [8] and [17].For a function f analytic in D and normalized by f (0) = f ′ (0) − order ofconvexity and the order of starlikeness (with respect to the origin) of f are defined by κ = κ ( f ) := 1 + inf z ∈ D Re zf ′′ ( z ) f ′ ( z ) ∈ [ −∞ , σ = σ ( f ) := inf z ∈ D Re zf ′ ( z ) f ( z ) ∈ [ −∞ , κ ( f ) = −∞ only if f ′ is zero-free in D and Re [ zf ′′ ( z ) /f ′ ( z )] is not bounded from below in D , whereas κ ( f )is regarded to be not defined if f ′ has zeros in D . The corresponding convention can alsobe made for σ ( f ) = −∞ . It is known that f is convex , i.e. κ ( f ) ≥ f is Mathematics Subject Classification.
Primary 30C45; Secondary 33C05.
Key words and phrases.
Gaussian hypergeometric function, order of convexity, zero-balanced one,Ramanujan’s formula.This research is supported by National Natural Science Foundation of China (No. 11901086) ” andthe Fundamental Research Funds for the Central Universities” in UIBE (No. 18YB02). univalent in D and f ( D ) is a convex domain; and f is starlike , i.e. σ ( f ) ≥ f is univalent in D and f ( D ) is a starlike domain with respect to the origin. It is alsotrue that if κ ( f ) ≥ − /
2, then f is univalent in D and f ( D ) is convex in (at least) onedirection, see [16] and [10, p.17, Thm.2.24; p.73]. If f is convex in D , then the order ofstarlikeness of f is at least 1 / a, b and c for some shiftedhypergeometric functions. As an application, we solve the following problem on the zero-balanced hypergeometric functions which is raised by Ponnusamy and Vuorinen in [9], seealso [3]. Problem 1.1.
Do there exist positive numbers δ , δ such that for a ∈ (0 , δ ) and b ∈ (0 , δ ) the normalized function z F ( a, b ; a + b ; z ) ( z F ( a, b ; a + b ; z ) resp. ) satisfies theproperty that maps the unit disc D into a strip domain? We restrict our attention only on the mapping z z F ( a, b ; a + b ; z ) in the presentpaper. In view of Ramanujan’s formula:(1.1) F ( a, b ; a + b ; z ) = 1B( a, b ) (cid:0) R ( a, b ) − log(1 − z ) (cid:1) + O (cid:18) | − z | log 1 | − z | (cid:19) , as z → D , where R ( a, b ) = 2 ψ (1) − ψ ( a ) − ψ ( b ) and ψ ( x ) = Γ ′ ( x ) / Γ( x ) and B( a, b )denote the digamma function and the Beta function respectively, the zero-balanced hy-pergeometric function behaves like the function − a,b ) log(1 − z ). Thus it is easy to seethat it maps the unit disc D into a strip domain. It seems that it is trivial to consider onlyProblem 1.1 along this direction. In fact, our aim is to find conditions on the pair ( a, b )such that the image domain of D under the mapping z z F ( a, b ; a + b ; z ) is convex andlies in a parallel strip. Along this line, the author in [15, Corollary 1.2] (cf. [5, Corollary8, Corollary 9]) got the following result: Theorem A. If a and b are real constants satisfying − a N , − b N and ab < , thenthe shifted zero-balanced hypergeometric function z F ( a, b ; a + b ; z ) is not convex in D . If furthermore, we suppose a and b are both positive. According to Theorem A, inorder to consider the convexity of the shifted zero-balanced hypergeometric functions thecondition ab ≥ a ≥ b ≥
1. Furthermore if a + b = 2, it follows from 2 = a + b ≥ √ ab that ab ≤ ab ≥ ab = 1. Thus when investigating the convexity of the shifted zero-balancedhypergeometric functions z F ( a, − a ; 2; z ) with a >
0, it is sufficient to consider onlythe function z F (1 ,
1; 2; z ) = − log(1 − z ).For 0 < a ≤ b <
1, even though z F ( a, b ; a + b ; z ) is not convex, it is natural to ask thequestion: whether the function z F ( a, b ; a + b ; z ) possesses the other geometric properties, APPING PROPERTIES OF THE ZERO-BALANCED HYPERGEOMETRIC FUNCTIONS 3 for example starlikeness and univalence and so on. In fact, K¨ustner in [5, Theorem 1] (cf.[4, Theorem 1.1, Remark 2.3]) got the following result on the order of starlikenss of theshifted hypergeometric functions.
Theorem B. If < a ≤ b ≤ c , then σ ( z F ( a, b ; c ; z )) = 1 − F ′ ( a, b ; c ; − F ( a, b ; c ; − ≥ − abb + c . As a consequence of Theorem B, the function z F ( a, b ; a + b ; z ) is starlike thus univalentin D if 0 < a ≤ b < a ≥
1. First if a = 1, we proved thenext result in [15] on the convexity of z F (1 , b ; c ; z ). Theorem C. ([15, Corollary 4.3])
Assume b and c are real constants satisfying < b ≤ c ,the function z F (1 , b ; c ; z ) is convex in D if one of the following conditions holds: (1) 0 ≤ b ≤ and c ≥ ; (2) 1 ≤ c < min { , b } and c ≥ − b . For general a >
1, by making use of Riemman-Stieltjes type integral of the ratio ofhypergeometric functions, we first obtain the order of convexity of the shifted hypergeo-metric functions as follows.
Theorem 1.2.
Let a, b and c be real constants satisfying < a ≤ b ≤ , c > and b ≤ c ≤ b ≤ a ) . The order of convexity of the function z F ( a, b ; c ; z ) is κ ( z F ( a, b ; c ; z )) = 5 − c − a − b c − − (1 − a )(1 − b )2 (cid:18) − a + a F ( a + 1 , b ; c ; − F ( a, b ; c ; − (cid:19) . In particular, F ′ ( a, b ; c ; z ) = 0 for all z ∈ D . We remark that in [15], the order of convexity of z F ( a, b ; c ; z ) has already been dealtwith, but the condition 0 < a ≤ a >
1. By virtue of Theorems C and 1.2, we obtain the next theorem whichnot only gives the convexity of some shifted zero-balanced hypergeometric functions butalso demonstrates the explicit bounds of the strip domains of the image domains. Thuswe provide a strong version of solution to Problem 1.1.
Theorem 1.3.
Let a and b be real constants. If one of the following conditions holds: (1) a = 1 and ≤ b ≤ ; (2) 1 < a ≤ b ≤ min { , a, a a − } ,then the shifted zero-balanced hypergeometric function z F ( a, b ; a + b ; z ) is convex andmaps the unit disc D into a strip domain which is bounded by two horizontal lines Im w = π a,b ) . The boundary lines are optimal. L.-M. WANG Some lemmas
In this section, we list some auxiliary lemmas which play important roles in the proofsof the main results. The first lemma is due to Ruscheweyh, Salinas and Sugawa [11].Later, Liu and Pego in [6] pointed out that one condition can be deduced from the others.So the lemma can be finally characterized in the following form.
Lemma 2.1.
Let F ( z ) be analytic in the slit domain C \ [1 , + ∞ ) . Then F ( z ) = Z dµ ( t )1 − tz for some probability measure µ on [0 , , if and only if the following conditions are fulfilled: (1) F (0) = 1 ; (2) F ( x ) ∈ R for x ∈ ( −∞ , ; (3) Im F ( z ) ≥ for Im z > ; (4) lim sup x → + ∞ F ( − x ) ≥ .The measure µ and the functions F are in one-to-one correspondence. The forthcoming four lemmas describe the properties of the ratio to two hypergeometricfunctions in different aspects.
Lemma 2.2. ([4, Thm. 1.5] , [17, p.337-339 and Thm.69.2]) If − ≤ a ≤ c and ≤ b ≤ c = 0 , the ratio of two hypergeometric functions can be written in integral as F ( a + 1 , b + 1; c + 1; z ) F ( a + 1 , b ; c ; z ) = Z dµ ( t )1 − tz , z ∈ C \ [1 , + ∞ ) where µ : [0 , → [0 , is nondecreasing with µ (1) − µ (0) = 1 . Lemma 2.3.
Let a, b and c be real constants with < a ≤ b ≤ a , c − a
6∈ − N and c − b
6∈ − N . Then lim x → + ∞ x F ( a + 1 , b + 1; c + 1; − x ) F ( a + 1 , b ; c ; − x ) = ∞ . Proof.
Denote H ( z ) = F ( a +1 , b +1; c +1; z ) and G ( z ) = F ( a +1 , b ; c ; z ) for convenience.In order to show the claimed equation, the next two linear transforms (see [1, p.559, 15.3.3,15.3.5, 15.3.7]) are required: F ( a, b ; c ; z ) = Γ( c )Γ( b − a )Γ( b )Γ( c − a ) ( − z ) − a F ( a, − c + a ; 1 − b + a ; 1 /z )+ Γ( c )Γ( a − b )Γ( a )Γ( c − b ) ( − z ) − b F ( b, − c + b ; 1 − a + b ; 1 /z ) , ( | arg ( − z ) | < π )(2.1)and(2.2) F ( a, b ; c ; z ) = (1 − z ) − b F (cid:18) b, c − a ; c ; zz − (cid:19) , ( | arg (1 − z ) | < π ) . APPING PROPERTIES OF THE ZERO-BALANCED HYPERGEOMETRIC FUNCTIONS 5
Then the proof can be separated into three cases according to the relationship between a and b .Case I: Assume that a < b < a . By virtue of (2.1), we obtain H ( − x ) = Γ( c + 1)Γ( b − a )Γ( b + 1)Γ( c − a ) x − a − + O ( x − b − )and G ( − x ) = Γ( c )Γ( a + 1 − b )Γ( a + 1)Γ( c − b ) x − b + O ( x − a − )as x → + ∞ . It follows from the above asymptotic behaviors that lim x → + ∞ xH ( − x ) /G ( − x ) = ∞ since a < b < a .Case II: Assume b = a . We deduce from the identity (2.2) and Ramanujan’s formula(1.1) that H ( − x ) = F ( a + 1 , a + 1; c + 1; − x ) = (1 + x ) − a − F (cid:18) a + 1 , c − a − c ; xx + 1 (cid:19) = (1 + x ) − a − (cid:18) log(1 + x )B( a + 1 , c − a −
1) + O (1) (cid:19) , as x → + ∞ . Similarly, applying the identity (2.2) and Gauss summation formula:lim x → − F ( a, b ; c ; x ) = F ( a, b ; c ; 1) = Γ( c )Γ( c − a − b )Γ( c − a )Γ( c − b ) , if c − a − b > x → + ∞ (1 + x ) a G ( − x ) = lim x → + ∞ F (cid:18) a, c − a − c ; xx + 1 (cid:19) = Γ( c )Γ( c − a )Γ( a + 1) . Thus by substituting the proceeding descriptions of H ( − x ) and G ( − x ) as x → + ∞ , wededuce that lim x → + ∞ xH ( − x ) G ( − x ) = lim x → + ∞ x (1 + x ) − a − (cid:16) log(1+ x )B( a +1 ,c − a − + O (1) (cid:17) G ( − x )= lim x → + ∞ x (1 + x ) − (cid:16) log(1+ x )B( a +1 ,c − a − + O (1) (cid:17) (1 + x ) a G ( − x ) = ∞ in this case.Case III: Assume that b = 1 + a . Applying the same techniques in Case II, we find thatlim x → + ∞ (1 + x ) a +1 H ( − x ) = Γ( c + 1)Γ( c − a )Γ( a + 2) , and G ( − x ) = (1 + x ) − a − (cid:18) log(1 + x )B( a + 1 , c − a −
2) + O (1) (cid:19) , x → + ∞ , which yields the required claim again.We complete all these three cases, thus the proof is done. (cid:3) L.-M. WANG
Lemma 2.4.
Let a, b and c be real constants with a, b, c
6∈ − N , c − a
6∈ − N and c − b
6∈ − N . (1) If c < a + b , then F ( a + 1 , b ; c ; z ) F ( a, b ; c ; z ) = a + b − ca (1 − z ) + O (cid:0) | − z | a + b − c (cid:1) . (2) If c = a + b , then F ( a + 1 , b ; c ; z ) F ( a, b ; c ; z ) = 1 − a (1 − z ) log(1 − z ) + O (cid:18) log 1 | − z | (cid:19) . (3) If a + b < c < a + b + 1 , then F ( a + 1 , b ; c ; z ) F ( a, b ; c ; z ) = A (1 − z ) − α + O ( | − z | ε − ) where A = Γ( a + b + 1 − c )Γ( c − a )Γ( c − b )Γ( a + 1)Γ( b )Γ( c − a − b ) ,α = c − a − b ∈ (0 , and ε = min { α, } . (4) If c = 1 + a + b , then (2.3) F ( a + 1 , b ; c ; z ) F ( a, b ; c ; z ) = − b log(1 − z ) + O (1) . Proof.
The first three assertions can be found in [15, Lemma 2.3]. We need only to provethe last one. First recall the linear transform F ( a, b ; c ; z ) = Γ( c )Γ( c − a − b )Γ( c − a )Γ( c − b ) F ( a, b ; a + b − c + 1; 1 − z )(2.4) + (1 − z ) c − a − b Γ( c )Γ( a + b − c )Γ( a )Γ( b ) F ( c − a, c − b ; c − a − b + 1; 1 − z ) . Applying the identity above and Ramanujan’s formula (1.1), we have F ( a + 1 , b ; c ; z ) F ( a, b ; c ; z )= a,b ) (cid:0) R (1 + a, b ) − log(1 − z ) (cid:1) + O (cid:16) | − z | log | − z | (cid:17) Γ( c )Γ( c − a − b )Γ( c − a )Γ( c − b ) + O ( | − z | )= − b log(1 − z ) + O (1)as z → D , since c = 1 + a + b . (cid:3) Lemma 2.5. ([15] , see also [4, Ramark 2.3]) If − ≤ a ≤ c and ≤ b ≤ c = 0 , then cb + c ≤ F ( a + 1 , b ; c ; − F ( a, b ; c ; − ≤ c − b c . APPING PROPERTIES OF THE ZERO-BALANCED HYPERGEOMETRIC FUNCTIONS 7
Lemma 2.6. ( see [14, Lemma 3.2]) Let Ω be an unbounded convex domain in C whoseboundary is parametrized positively by a Jordan curve w ( t ) = u ( t ) + iv ( t ) , < t < , with w (0 + ) = w (1 − ) = ∞ . Suppose that u (0 + ) = + ∞ and that v ( t ) has a finite limit as t → + . Then v ( t ) ≤ v (0 + ) for < t < . In the end of this section, the behavior of a special zero-balanced hypergeometric func-tion F (1 , , z ) around z = 1 is given. Lemma 2.7.
For θ ∈ (0 , π ) , we have lim θ → + Re [ − e iθ log(1 − e iθ )] = + ∞ and lim θ → + Im [ − e iθ log(1 − e iθ )] = π . Proof.
For z = e iθ with θ ∈ (0 , π ), we do the following computations: − e iθ log(1 − e iθ ) = − e iθ (cid:20) (cid:18) sin θ (cid:19) + i θ − π (cid:21) = − θ log (cid:18) sin θ (cid:19) + θ − π θ − i (cid:20) θ log (cid:18) sin θ (cid:19) + θ − π θ (cid:21) . We deduce the required identities by letting θ → + in the above form. (cid:3) Proofs of the main results
Proof of Theorem 1.2.
Let F ( z ) = F ( a, b ; c ; z ), G ( z ) = F ( a + 1 , b ; c ; z ) and H ( z ) = F ( a + 1 , b + 1; c + 1; z ) for simplicity. In order to obtain the order of convexity of zF ( z ),we should first compute its second logarithmic derivative and then evaluate the real partof this derivative in the unit disc D . According to the equation (3.2) in [15], we get(3.1) W ( z ) := 1 + z ( zF ) ′′ ( zF ) ′ = 3 − c + ( a + b − z − z + c − − a )(1 − b ) z (1 − z )(1 − a + aG/F ) . We will show that the infimum real part of W ( z ) in D is attained at z = − F ( a + 1 , b ; c ; z ) F ( a, b ; c ; z ) = 11 − bc z F ( a + 1 , b + 1; c + 1; z ) F ( a, b ; c ; z ) , z ∈ C \ [1 , + ∞ )that the denominator of the second term in W ( z ) can be transformed into11 − a + aG/F = 11 − a + a/ [1 − b/czH ( z ) /G ( z )]= 11 − a + aa − ( a − bc zHG . L.-M. WANG
Substituting the equation above into equation (3.1), we have W ( z ) = 3 − c + ( a + b − z − z + c − − a )(1 − b ) z (1 − a )(1 − z )+ a ( a − − z ) c − − a )(1 − b ) z ( a − bc zHG = 1 − a + a ( a + 1 − c ) a − − z + a ( a − − z ) c − − a )(1 − b ) z ( a − bc zHG . Denote M ( z ) = c − − a )(1 − b ) z − z , M ( z ) = 1 + τ zHG and M ( z ) = M ( z ) M ( z ) for simplicity where τ = ( a − bc .Next we show that under the conditions of this theorem, there exists a probabilitymeasure ν on [0 ,
1] such that(3.2) M ( z ) = ( c − Z dν ( t )1 − tz . By virtue of Lemma 2.1, it is sufficient to verify the four conditions there.The first and second ones are easy to check since a, b and c are real constants with c >
2. We proceed to verify the third one. It follows from Lemma 2.2 that there exists aprobability measures µ on [0 ,
1] such that H ( z ) G ( z ) = Z dµ ( t )1 − tz which implies that M ( z ) = Z τ − t ) z − tz dµ ( t ) . Thus for z = x + iy with y >
0, an elementary computation generates thatIm M ( z ) | M ( z ) | = − Re M Im M + Im M Re M = − Z τ y | − tz | c − − (1 − a )(1 − b ) | z | − [ c − − (1 − a )(1 − b )] x | − z | dµ ( t )+ Z t − τ t ) | z | + ( τ − t ) x | − tz | [ c − − a )(1 − b )] y | − z | dµ ( t ):= y Z I ( t, x, y ) | − z | | − tz | dµ ( t ) . APPING PROPERTIES OF THE ZERO-BALANCED HYPERGEOMETRIC FUNCTIONS 9
Letting p = c − − a )(1 − b ) and q = c − − (1 − a )(1 − b ) for simplicity, thenumerator of the integrand can be simplified into I ( t, x, y )= p ( t − τ t ) | z | + τ (1 − a )(1 − b ) | z | + p ( τ − t ) x + τ qx + p − τ ( c − τ (1 − a )(1 − b ) + p ( t − τ t )] | z | + [ p ( τ − t ) + τ q ] x + p − τ ( c − Q ( t ) y + Q ( t ) (cid:18) x + τ ( c − − ptQ ( t ) (cid:19) + [ p − τ ( c − Q ( t ) − [ τ ( c − − pt ] Q ( t ):= Q ( t ) y + Q ( t ) (cid:18) x + τ ( c − − ptQ ( t ) (cid:19) + S ( t ) Q ( t ) , since Q ( t ) < a ≤ b . Thus for y = Im z >
0, Im M ( z ) ≥ ( Q ( t ) = τ (1 − a )(1 − b ) + p ( t − τ t ) ≥ S ( t ) = [ p − τ ( c − Q ( t ) − [ τ ( c − − pt ] ≥ , hold for any t ∈ [0 , ( Q (0) = τ (1 − a )(1 − b ) > Q (1) = c − ( a − b − c ) c ≥ . Note also that the conditions 1 < a ≤ b ≤ b ≤ c imply 0 < τ ≤
3. Thus if2 ≤ τ ≤
3, the equations in (3.3) yield that the quadratic function Q ( t ) is nonnegative in[0 , < τ <
2, and in this case the inequality Q ( t ) ≥ Q (cid:16) τ (cid:17) = τ b [4( b − c − pb ] , always holds for t ∈ [0 , Q ( t ) on [0 , b − c − pb ≥
0. As 1 < a ≤ b implies4( b − c − pb = (3 b − c + 2 b − ( a − b − b ≥ (3 b − c + 2 b − ( b − b := h ( b, c ) , we divide the proof into two cases according to the sign of 3 b − < b ≤ /
3. Then h ( b, c ) ≥ b (3 b −
4) + 2 b − ( b − b = − b ( b − b − > c ≤ b .Case II: Assume that b > /
3. Then h ( b, c ) ≥ (3 b − b +2 b − ( b − b = − b ( b − b +3) ≥ b ≤ c and 4 / < b ≤ Q ( t ) ≥ t ∈ [0 ,
1] under the assumptions of this theorem.We next verify the property of S ( t ). After computing, we get S ( t ) = − pτ [( c − t + p − ( c − τ + 2)] t + [ p − τ ( c − τ (1 − a )(1 − b ) − ( c − τ and S (1) = 0. Furthermore an elementary computation yields that S (0) = pcτ (2 b − c ),which is nonnegative as c ≤ b . Therefore the quadratic function S ( t ) ≥ t ∈ [0 , x → + ∞ M ( − x ) ≥ x → + ∞ M ( − x ) = ∞ which in conjunction with lim x → + ∞ M ( − x ) = − ( a − b − = 0 generates that lim x → + ∞ M ( − x ) = 0under the condition a ≤ b ≤ a .We thus obtain the integral form (3.2) which immediately yields that W ( z ) = 1 − a + a ( a + 1 − c ) a − − z + a ( c − a − Z dν ( t )1 − tz , where ν is a probability measure on [0 , W ( z ) is analytic on D and since a > c >
2, then(3.4) Re W ( z ) ≥ W ( − , for | z | = 1 . Next we will prove the above inequality for all z ∈ D which is exactly the assertion of thistheorem. If c ≤ a + 1 in addition, it is obvious that inequality (3.4) holds for z ∈ D sinceRe 1 / (1 − z ) > / z ∈ D . We need only to deal with the case c > a + 1. Firstwe observe that c ≤ a + b as c ≤ b and b ≤ a . As a direct consequence of Lemma2.4, we conclude that lim z → − z ∈ R G ( z ) F ( z ) = + ∞ , for 1 + a < c ≤ a + b . Thus it follows from the above equation that W ( z ) = 3 − c + ( a + b − z − z + c − − a )(1 − b ) z (1 − z )(1 − a + aG/F )= 3 − c + ( a + b − z − z + o (1)1 − z = 1 + a + b − c + o (1)1 − z as z → D . Since W ( z ) is analytic in D , we obtain that (3.4) holdsfor all z ∈ D if c < a + b . As for the case c = 1 + a + b , by virtue of equation (2.3) inLemma 2.4, we conclude that W ( z ) = 2 − a − b + c − − a )(1 − b ) z − z − a − ab log(1 − z ) + O (1) → + ∞ , as z → D . Thus the inequality (3.4) holds for all z ∈ D if c = 1 + a + b , since W ( z ) is analytic in D .We complete the proof. (cid:3) Proof of Theorem 1.3.
We first prove that z F ( a, b ; a + b ; z ) is convex in D underthe assumptions. If a = 1 and 1 ≤ b ≤
4, the convexity of z F ( a, b ; a + b ; z ) is aconsequence of the case (2) in Theorem C. While if 1 < a ≤ b ≤
3, the order of convexity
APPING PROPERTIES OF THE ZERO-BALANCED HYPERGEOMETRIC FUNCTIONS 11 of z F ( a, b ; a + b ; z ) is κ = 5 − a − b a + b ) − − ab (cid:18) − a + a F ( a + 1 , b ; a + b ; − F ( a, b ; a + b ; − (cid:19) according to Theorem 1.2. It is easy to see that 2( a + b ) − − ab ≥ < a ≤ b ≤ κ ≥ − a − b a + b ) − − ab (cid:16) − a + a a + b a + b ) (cid:17) = (4 − a ) b + 4 a a + 2 b − ab ) . By observing the numerator of the above lower bound, we conclude that if 1 < a ≤ b ≤ a/ (5 a − κ is non-negative, which means z F ( a, b ; a + b ; z )is convex in D .On the other hand, we consider the image domain Ω of D under the mapping z z F ( a, b ; a + b ; z ). For 0 < t <
1, let w ( t ) = u ( t ) + iv ( t ) stand for the boundary e πit F ( a, b ; a + b ; e πit ) of the domain G , which is a Jordan convex curve. In viewof Lemma 2.7 and Ramanujan’s formula (2.2), we have lim t → + w ( t ) = lim t → − w ( t ) = ∞ ,lim t → + u ( t ) = + ∞ and lim t → + v ( t ) = π a, b ) . Therefore it follows from Lemma 2.6 that v ( t ) ≤ v (0 + ) holds for 0 < t <
1. Since theimage domain G is symmetric with respect to the real axis, we finally have | v ( t ) | ≤ π a, b ) , for 0 < t < , which means that the image domain of D under the function z F ( a, b ; a + b ; z ) is boundedby two horizontal lines Im w = ± π a,b ) .The proof is finished. (cid:3) Acknowledgements.
The author would like to appreciate Professor Toshiyuki Sugawafor discussions and guidance, without whom I could’t finish this work. The author is alsoindebted to Professor Matti Vuorinen for drawing my attention to this problem.
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