Confluent Hypergeometric Functions It is cl ear that - Project Euclid

Nihonkai Math. J. vol. 3 (1992), 109-114

The Order of Starlikeness and Convexity of Confluent Hypergeometric Functions KHALIDA INAYAT NOOR

ABSTRACT We determi ne the condi ti ons for whi ch confl uent hypergeometri functi ons

are convex

and starlike of order

$\alpha$

$c$

.

functi ons, convex, starl ike of order , convol ution, 1 inear operator, radius of convexi ty. 198O Mathematics Classification (1985 Revision), $33A3O,3OC45$

Key Words and Phrases: Confl uent hypergeometri

$c$

$\alpha$

.

1.

INTRODUCTION Let A denote the ass of analyti functi ons in the unit di sk $E=\{z:|z|{\rm Re}$

$\phi(a;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}$

a

$>0$ ,

then

$\int_{0}1t^{a-1}(1-t)^{c-a-1}e^{tz}dt\Rightarrow$

$\int_{0}^{1}e^{tz}d,1(t)$

,

(1.7)

where $t^{a-1}(1-t)^{c-a-1}$

$||(t)=\frac{r(c)}{r(a)r(c-a)}$

is a probability

measure on

$\int_{0}1d_{W}(t)=\frac{r(c)}{r(a)r(c-a)}$

where

2. $C(\alpha)$

$B$

$[0,1]$

.

$B$

.

In fact

( , c-a) $a$

$=1$

,

(1.8)

is the beta function.

MAIN RESULTS We shall now determi ne conditi ons on a and and

$S^{*}(\alpha)$

Theorem 2. 1 Let $a,$

$c>N(a, \alpha)$

$c$

.

and

$\alpha$

be real numbers with a

$c$

so

$\neq 0,0\alpha C1$

that

$\phi$

bel ongs to

and sati sfy

, where

$N(a, \alpha)\Rightarrow\left\{\begin{array}{l}\frac{1-2_{\alpha}^{2}+2_{\alpha}\star 2|a|}{2(1-\alpha)}|\#a|>\frac{(1-\alpha)^{2}}{(3-2\alpha)}\\\\\frac{4(1-\alpha)^{2}(1+a)-\alpha 2}{2(3-2\alpha)(1-\alpha)}+\frac{(3-2\alpha)(\alpha\succ a)^{2}}{2(1-\alpha)}|\theta a|\frac{(1-\alpha)^{2}}{(3-2_{\alpha})}\end{array}\right.$

Then

$\phi(a;c;z)is$

convex of

order

$\alpha in$

E.

To prove this, we follow the techni que of Miller and Mocanu [2] and we need the following result which is a special case of Theorem 1 in [3].

–110

–

$\frac{LeIN\mathfrak{n}a2.1}{Let}D$

be a set in the compl ex pl ane

$H:C^{2}xE\rightarrow C$

$H(is;t;z)\not\in D$

analyti

in

$c$

sati sfy the condi for

on

and for real

$z\in E$

wi th $p(0)=1$ and in E.

$E$

${\rm Re} p(z)>0$

$ti$

$H$

a functi on

and 1 et

$C$

with

$s,t$

( $p(z)$ ; zp’

$t$

$(z):z$ )

$\in E,$

.

If $pis$ $z\in E$ , then

$\frac{-(1+s^{2})}{2}$

Proof of Theorem 1 Let $(\frac{Z\phi^{I\prime}(Z)}{\phi^{1}(z)}+1)=(1-\alpha)p(z)+\alpha$

(2.2)

,

Cl earl the functi on $p\{z$ ) is satisfies the di fferential Since (2.2) in (1.3) to have

see

where $\phi^{\prime}(z)=\phi^{I}(a;c;z)\neq 0$ , analytic in with $p(0)=1$

.

$E$

equation (1.3), we use

[2].

$y$

$\phi$

{zp’ $(z)+(1-\alpha)p^{2}(z)+(1-\alpha)(c-z+\alpha-2)p(z)$ $[(1-a)c+z(a+\alpha)-(1-\alpha)^{2}]$

-

}

$=0$

(2.3)

Let $H$

(

$w_{1}$

;

$w_{2}$

; z)

$=w_{2}+(1-\alpha)w_{1}^{2}+(1-\alpha)(c-z+\alpha- 2)w_{1}$

$-[(1-\alpha)c+z(a+\alpha)-(1-\alpha)^{2}]$

and

$D=\{0\}$

$H$

We shall

( $p(z)$ ; zp

$|(z)$

; z)

use Lemma 2.1 to prove that

$z=x+iy$

Let

as

, then (2.3) can be written

.

$\in D$

${\rm Re} p(z)>0$

.

Then

${\rm Re} H(is;t;z)=t-(1-\alpha)s^{2}+(1-\alpha)ys-(1-\alpha)(c-1+\alpha)-(\alpha\vdash a)x$

‘

)

$-\frac{(1+s^{2}}{2}-$

)

$-\frac{2(1-\alpha)s^{2}}{2}+\frac{2(1-\alpha)ys-2(\alpha\vdash a)x}{2}-\frac{2(1-\alpha)(c-1+\alpha}{2}$

$=-\frac{1}{2}[(3-2_{\alpha})s^{2}-2(1-\alpha)ys+2(\alpha\vdash a)x+2(1-\alpha)(c-1-\alpha)+1]$

$\equiv Q(s)$

Now

of

$Q(s)$

$Q(s)$

$