Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 1, Article 16, 2005
ON CERTAIN SUBCLASS OF p-VALENTLY BAZILEVIC FUNCTIONS
J. PATEL
D EPARTMENT OF M ATHEMATICS
U TKAL U NIVERSITY, VANI V IHAR
B HUBANESWAR -751004, I NDIA
jpatelmath@sify.com
Received 13 December, 2004; accepted 03 February, 2005
Communicated by H.M. Srivastava
A BSTRACT. We introduce a subclass Mp (λ, µ, A, B) of p-valent analytic functions and derive
certain properties of functions belonging to this class by using the techniques of Briot-Bouquet
differential subordination. Further, the integral preserving properties of Bazilevic functions in a
sector are also considered.
Key words and phrases: p-valent; Bazilevic function; Differential subordination.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let Ap be the class of functions of the form
(1.1)
p
f (z) = z +
∞
X
an z n
(p ∈ N = {1, 2, 3, . . . })
n=p+1
which are analytic in the open unit disk E = {z ∈ C : |z| < 1}. We denote A1 = A.
A function f ∈ Ap is said to be in the class Sp∗ (α) of p-valently starlike of order α, if it
satisfies
0 zf (z)
(1.2)
<
> α (0 ≤ α < p; z ∈ E).
f (z)
We write Sp∗ (0) = Sp∗ , the class of p-valently starlike functions in E.
A function f ∈ Ap is said to be in the class Kp (α) of p-valently convex of order α, if it
satisfies
00
zf (z)
(1.3)
< 1+ 0
> α (0 ≤ α < p; z ∈ E).
f (z)
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
The work has been supported by the financial assistance received under DRS programme from UGC, New Delhi.
236-04
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J. PATEL
The class of p-valently convex functions in E is denoted by Kp . It follows from (1.2) and (1.3)
that
f ∈ Kp (α) ⇐⇒ f ∈ Sp∗ (α)
(0 ≤ α < p).
Furthermore, a function f ∈ Ap is said to be p-valently Bazilevic of type µ and order α, if
there exists a function g ∈ Sp∗ such that
0
zf (z)
(1.4)
<
> α (z ∈ E)
f (z)1−µ g(z)µ
for some µ(µ ≥ 0) and α(0 ≤ α < p). We denote by Bp (µ, α), the subclass of Ap consisting
of all such functions. In particular, a function in Bp (1, α) = Bp (α) is said to be p-valently
close-to-convex of order α in E.
For given arbitrary real numbers A and B (−1 ≤ B < A ≤ 1), let
0
zf (z)
1 + Az
∗
(1.5)
Sp (A, B) = f ∈ Ap :
≺p
, z∈E ,
f (z)
1 + Bz
where the symbol ≺ stands for subordination. In particular, we note that Sp∗ 1 − 2α
,
−1
=
p
∗
Sp (α) is the class of p-valently starlike functions of order α(0 ≤ α < p). From (1.5), we
observe that f ∈ Sp∗ (A, B), if and only if
0
zf (z) p(1 − AB) p(A − B)
(1.6)
(−1 < B < A ≤ 1; z ∈ E)
f (z) − 1 − B 2 < 1 − B 2
and
(1.7)
<
0
zf (z)
f (z)
>
p(1 − A)
2
(B = −1; z ∈ E).
Let Mp (λ, µ, A, B) denote the class of functions in Ap satisfying the condition
00
0
0
0
zf (z)
zf (z)
zf (z)
zg (z)
1 + Az
(1.8)
+λ 1+ 0
− (1 − µ)
−µ
≺p
1−µ
µ
f (z) g(z)
f (z)
f (z)
g(z)
1 + Bz
(−1 ≤ B < A ≤ 1; z ∈ E)
for some real µ(µ ≥ 0), λ(λ > 0), and g ∈ Sp∗ . For convenience, we write
2α
Mp λ, µ, 1 −
, −1
p
= Mp (λ, µ, α)
00
0
0
0
zf (z)
zf (z)
zf (z)
zg (z)
= f ∈ Ap : <
+λ 1+ 0
− (1 − µ)
−µ
>α
f (z)1−µ g(z)µ
f (z)
f (z)
g(z)
for some α(0 ≤ α < p) and z ∈ E.
In the present paper, we derive various useful properties and characteristics of the class
Mp (λ, µ, A, B) by employing techniques involving Briot-Bouquet differential subordination.
The integral preserving properties of Bazilevic functions in a sector are also considered. Relevant connections of the results presented here with those obtained in earlier works are pointed
out.
J. Inequal. Pure and Appl. Math., 6(1) Art. 16, 2005
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O N C ERTAIN S UBCLASS OF p-VALENTLY BAZILEVIC F UNCTIONS
3
2. P RELIMINARIES
To establish our main results, we shall require the following lemmas.
Lemma 2.1 ([6]). Let h be a convex function in E and let ω be analytic in E with <{ω(z)} ≥ 0.
If q is analytic in E and q(0) = h(0), then
0
q(z) + ω(z) zq (z) ≺ h(z)
(z ∈ E)
implies
q(z) ≺ h(z)
(z ∈ E).
Lemma 2.2. If −1 ≤ B < A ≤ 1, β > 0 and the complex number γ satisfies <(γ) ≥
−β(1 − A)/(1 − B), then the differential equation
0
1 + Az
zq (z)
q(z) +
=
β q(z) + γ
1 + Bz
(z ∈ E)
has a univalent solution in E given by


z β+γ (1 + Bz)β(A−B)/B
γ


R

 β z tβ+γ−1 (1 + Bt)β(A−B)/B dt − β ,
0
(2.1)
q(z) =

γ
z β+γ exp(β Az)


R

 β z tβ+γ−1 exp(β At)dt − β ,
0
B 6= 0
B = 0.
If φ(z) = 1 + c1 z + c2 z 2 + · · · is analytic in E and satisfies
0
(2.2)
1 + Az
z φ (z)
≺
φ(z) +
βφ(z) + γ
1 + Bz
(z ∈ E),
then
φ(z) ≺ q(z) ≺
1 + Az
1 + Bz
(z ∈ E)
and q(z) is the best dominant of (2.2).
The above lemma is due to Miller and Mocanu [7].
Lemma 2.3 ([12]). Let ν be a positive measure on [0, 1]. Let h be a complex-valued function
defined on E × [0, 1] such that h(·, t) is analytic in E for each t ∈ [0, 1], and h(z, ·) is νintegrable on [0, 1] for all z ∈ E. In addition, suppose that <{h(z, t)} R> 0, h(−r, t) is real
1
and <{1/h(z, t)} ≥ 1/h(−r, t) for |z| ≤ r < 1 and t ∈ [0, 1]. If h(z) = 0 h(z, t) dν(t), then
<{1/h(z)} ≥ 1/h(−r).
For real or complex numbers a, b, c (c 6= 0, −1, −2, . . . ), the hypergeometric function is
defined by
(2.3)
2 F1 (a, b; c; z)
=1+
a·b z
a(a + 1) · b(b + 1) z 2
· +
·
+ ··· .
c
1!
c(c + 1)
2!
We note that the series in (2.3) converges absolutely for z ∈ E and hence represents an analytic
function in E. Each of the identities (asserted by Lemma 2.3 below) is well-known [13].
J. Inequal. Pure and Appl. Math., 6(1) Art. 16, 2005
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J. PATEL
Lemma 2.4. For real numbers a, b, c (c 6= 0, −1, −2, . . . ), we have
Z 1
Γ(b)Γ(c − b)
(2.4)
tb−1 (1 − t)c−b−1 (1 − tz)−a dt =
2 F1 (a, b; c; z) (c > b > 0)
Γ(c)
0
(2.5)
2 F1 (a, b; c; z) = 2 F1 (b, a; c; z)
z
−a
(2.6)
.
2 F1 (a, b; c; z) = (1 − z)
2 F1 a, c − b; c;
z−1
Lemma 2.5 ([10]). Let p(z) = 1 + c1 z + c2 z 2 + · · · be analytic in E and p(z) 6= 0 in E. If
there exists a point z0 ∈ E such that
π
π
(2.7)
| arg p(z)| < η (|z| < |z0 |)
and
| arg p(z0 )| = η (0 < η ≤ 1),
2
2
then we have
0
z0 p (z0 )
= ikη,
p(z0 )
(2.8)
where
(2.9)

k ≥
1
2
x+
1
x
k ≤ − 1 x +
2
when arg p(z0 ) = π2 η,
,
1
x
when arg p(z0 ) = − π2 η,
,
and
1/η
p(z0 )
= ±ix (x > 0).
(2.10)
3. M AIN R ESULTS
Theorem 3.1. Let −1 ≤ B < A ≤ 1, λ > 0 and µ ≥ 0. If f ∈ Mp (λ, µ, A, B), then
0
(3.1)
zf (z)
λ
≺
= q(z)
1−µ
µ
p f (z) g(z)
p Q(z)
(z ∈ E),
where
(3.2)
Q(z) =

R p

 01 s λ −1
1+Bsz
1+Bz

R 1 s λp −1 exp
p(A−B)
λB
p
(s
λ
0
ds,
− 1)Az ds,
B 6= 0,
B = 0,
1
λB
when A = −
, B 6= 0,
1 + Bz
p
and q(z) is the best dominant of (3.1). Furthermore, if A ≤ −λ B/p with −1 ≤ B < 0, then
q(z) =
Mp (λ, µ, A, B) ⊂ Bp (µ, ρ),
(3.3)
where
ρ = ρ(p, λ, A, B) = p
2 F1
p(B − A) p
B
1,
; + 1;
λB
λ
B−1
−1
.
The result is best possible.
Proof. Defining the function φ(z) by
0
(3.4)
zf (z)
φ(z) =
p f (z)1−µ g(z)µ
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(z ∈ E),
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O N C ERTAIN S UBCLASS OF p-VALENTLY BAZILEVIC F UNCTIONS
5
we note that φ(z) = 1+c1 z +c2 z 2 +· · · is analytic in E. Taking the logarithmic differentiations
in both sides of (3.4), we have
0
00
0
0
zf (z)
zf (z)
zf (z)
zg (z)
(3.5)
+λ 1+ 0
− (1 − µ)
−µ
f (z)1−µ g(z)µ
f (z)
f (z)
g(z)
0
zφ (z)
p(1 + Az)
= p φ(z) + λ
≺
φ(z)
1 + Bz
(z ∈ E).
Thus, φ(z) satisfies the differential subordination (2.2) and hence by using Lemma 2.2, we get
1 + Az
(z ∈ E),
1 + Bz
where q(z) is given by (2.1) for β = p/λ and γ = 0, and is the best dominant of (3.5). This
proves the assertion (3.1).
Next, we show that
(3.6)
inf <(q(z)) = q(−1).
φ(z) ≺ q(z) ≺
|z|<1
If we set a = p(B − A)/λB, b = p/λ, c = (p/λ) + 1, then c > b > 0. From (3.2), by using
(2.4), (2.5) and (2.6), we see that for B 6= 0
Z 1
Bz
Γ(b)
a
b−1
−a
.
(3.7)
Q(z) = (1 + Bz)
s (1 + Bsz) ds =
2 F1 1, a; c;
Γ(c)
Bz + 1
0
To prove (3.6), we need to show that <{1/Q(z)} ≥ 1/Q(−1), z ∈ E. Since A < −λ B/p
implies c > a > 0, by using (2.4), (3.7) yields
Z 1
Q(z) =
h(z, s) dν(s),
0
where
h(z, s) =
Γ(b)
1 + Bz
(0 ≤ s ≤ 1) and dν(s) =
sa−1 (1 − s)c−a−1 ds
1 + (1 − s)Bz
Γ(a)Γ(c − a)
which is a positive measure on [0, 1]. For −1 ≤ B < 0, it may be noted that <{h(z, s)} >
0, h(−r, s) is real for 0 ≤ r < 1, 0 ∈ [0, 1] and
1
1 + (1 − s)Bz
1 − (1 − s)Br
1
<
=<
≥
=
h(z, s)
1 + Bz
1 − Br
h(−r, s)
for |z| ≤ r < 1 and s ∈ [0, 1]. Therefore, by using Lemma 2.3, we have
1
1
<
≥
, |z| ≤ r < 1
Q(z)
Q(−r)
and by letting r → 1− , we obtain < 1/Q(z) ≥ 1/Q(−1). Further, by taking A → (−λ B/p)+
for the case A = (−λ B/p), and using (3.1), we get (3.3).
The result is best possible as the function q(z) is the best dominant of (3.1). This completes
the proof of Theorem 3.1.
Setting µ = 1, A = 1 − (2α/p) (p − λ)/2 ≤ α < p and B = −1 in Theorem 3.1, we have
Corollary 3.2. If f ∈ Ap satisfies
0
00
0
zf (z)
zf (z) zg (z)
<
+λ 1+ 0
−
>α
g(z)
f (z)
g(z)
J. Inequal. Pure and Appl. Math., 6(1) Art. 16, 2005
(λ > 0, z ∈ E)
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J. PATEL
for some g ∈ Sp∗ , then f ∈ Bp (κ(p, λ, α)), where
κ(p, λ, α) = p
(3.8)
2 F1
2(p − α) p
1
1,
; + 1;
λ
λ
2
−1
.
The result is best possible.
Taking µ = 0, A = 1 − (2α/p) (p − λ)/2 ≤ α < p and B = −1 in Theorem 3.1, we get
Corollary 3.3. If f ∈ Ap satisfies
0
00
zf (z)
zf (z)
< (1 − λ)
>α
+λ 1+ 0
f (z)
f (z)
(λ > 0, z ∈ E)
then f ∈ Sp∗ (κ(p, λ, α)), where κ(p, λ, α) is given by (3.8). The result is best possible.
Putting λ = 1 in Corollary 3.3, we get
Corollary 3.4. For (p − 1)/2 ≤ α < p, we have
Kp (α) ⊂ Sp∗ (κ(p, α)),
where κ(p, α) = p {2 F1 (1, 2(p − α); p + 1; 1/2)}−1 . The result is best possible.
Remark 3.5.
(1) Noting that
2 F1
1
1, 2(1 − α); 2;
2
−1
=

 22(1−α)1−2α
,
(1−22α−1 )

1
,
2 ln 2
α 6=
1
2
α = 12 ,
Corollary 3.4 yields the corresponding result due to MacGregor [5] (see also [12]) for
p = 1.
(2) It is proved [9] that if p ≥ 2 and f ∈ Kp , then f is p-valently starlike in E but is not
necessarily p-valently starlike of order larger than zero in E. However, our Corollary
3.4 shows that if f is p-valently convex of order at least (p − 1)/2, then f is p-valently
starlike of order larger than zero in E.
Theorem 3.6. If f ∈ Bp (µ, α) for some µ (µ > 0), α (0 ≤ α < p), then f ∈ Mp (λ, µ, α) for
|z| < R(p, λ, α), where λ > 0 and

√
(p+λ−α)− (p+λ−α)2 −p(p−2α)


, α=
6 p2 ;
p−2α
(3.9)
R(p, λ, α) =

 p ,
α = p2 .
p+2λ
The bound R(p, λ, α) is best possible.
Proof. From (1.4), we get
0
(3.10)
zf (z)
= α + (p − α)u(z)
f (z)1−µ g(z)µ
J. Inequal. Pure and Appl. Math., 6(1) Art. 16, 2005
(z ∈ E),
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O N C ERTAIN S UBCLASS OF p-VALENTLY BAZILEVIC F UNCTIONS
7
where u(z) = 1 + u1 z + u2 z 2 + · · · is analytic and has a positive real part in E. Differentiating
(3.10) logarithmically, we deduce that
00
0
0
0
zf (z)
zf (z)
zg (z)
zf (z)
<
+λ 1+ 0
− (1 − µ)
−µ
−α
f (z)1−µ g(z)µ
f (z)
f (z)
g(z)
0
λ z u (z)
= (p − α)< u(z) +
α + (p − α)u(z)
0
λ |z u (z)|
(3.11)
≥ (p − α)< u(z) −
.
|α + (p − α)u(z)|
Using the well-known estimates [5]
0
|z u (z)| ≤
2r
<{u(z)}
1 − r2
and <{u(z)} ≥
1−r
1+r
(|z| = r < 1)
in (3.11), we get
00
0
0
0
zf (z)
zf (z)
zg (z)
zf (z)
<
+λ 1+ 0
− (1 − µ)
−µ
−α
f (z)1−µ g(z)µ
f (z)
f (z)
g(z)
2λ r
≥ (p − α) <{u(z)} 1 −
,
α(1 − r2 ) + (p − α)(1 − r)2
which is certainly positive if r < R(p, λ, α), where R(p, λ, α) is given by (3.9).
To show that the bound R(p, λ, α) is best possible, we consider the function f ∈ Ap defined
by
0
zf (z)
1−z
= α + (p − α)
(0 ≤ α < p, z ∈ E)
1−µ
µ
f (z) g(z)
1+z
for some g ∈ Sp∗ . Noting that
0
00
0
0
zf (z)
zf (z)
zf (z)
zg (z)
<
+λ 1+ 0
− (1 − µ)
−µ
−α
f (z)1−µ g(z)µ
f (z)
f (z)
g(z)
1−z
2λ z
= (p − α)
+
1 + z α(1 − z 2 ) + (p − α)(1 + z)2
=0
for z = −R(p, λ, α), we conclude that the bound is best possible. This proves Theorem 3.6.
For µ = 0 and λ = 1, Theorem 3.6 yields:
Corollary 3.7. If f ∈ Sp∗ (α) (0 ≤ α < p), then f ∈ Kp (α) in |z| < ξ(p, α), where

√ 2

 (p+1−α)− α +2(p−α)+1 , α 6= p ;
p−2α
2
ξ(p, α) =

 p ,
α = p2 .
p+2
The bound ξ(p, α) is best possible.
Theorem 3.8. If f ∈ Ap satisfies
f (z)
<
> 0 and
zp
J. Inequal. Pure and Appl. Math., 6(1) Art. 16, 2005
0
zf (z)
<p
−
p
f (z)1−µ g(z)µ
(0 ≤ µ, z ∈ E)
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J. PATEL
e µ), where
for g ∈ Sp∗ , then f is p-valently convex(univalent) in |z| < R(p,
p
3
+
2µ(p
−
1)
−
(3 + 2µ(p − 1))2 − 4p(2µp − p − 1)
e µ) =
R(p,
.
2(2µp − p − 1)
e µ) is best possible.
The bound R(p,
Proof. Letting
0
h(z) =
zf (z)
−1
p f (z)1−µ g(z)µ
(z ∈ E),
we note that h(z) is analytic in E, h(0) = 0 and |h(z)| < 1 for z ∈ E. Thus, by applying
Schwarz’s Lemma we get
h(z) = z ψ(z),
where ψ(z) is analytic in E and |ψ(z)| ≤ 1 for z ∈ E. Therefore,
0
zf (z) = pf (z)1−µ g(z)µ (1 + zψ(z)).
(3.12)
Making use of logarithmic differentiation in (3.12), we obtain
00
(3.13)
0
0
0
zf (z)
zf (z)
zg (z) z(ψ(z) + zψ (z))
1+ 0
= (1 − µ)
+µ
+
.
f (z)
f (z)
g(z)
1 + zψ(z)
Setting φ(z) = f (z)/z p = 1 + c1 z + c2 z 2 + · · · , <{φ(z)} > 0 for z ∈ E, we get
0
0
zf (z)
zφ (z)
=p+
f (z)
φ(z)
so that by (3.13),
00
(3.14)
0
0
0
zf (z)
zφ (z)
zg (z) z(ψ(z) + zψ (z))
1+ 0
= (1 − µ)p + (1 − µ)
+µ
+
.
f (z)
φ(z)
g(z)
1 + zψ(z)
Now, by using the well-known estimates [1]
0 0 zφ (z)
2r
zg (z)
p(1 − r)
<
≥−
,<
≥−
and
2
φ(z)
1−r
g(z)
1+r
0
ψ(z) + zψ (z)
1
<
≥−
1 + zψ(z)
1−r
for |z| = r < 1 in (3.14), we deduce that
00
zf (z)
(2µp − p − 1)r2 − {3 + 2µ(p − 1)}r + p
< 1+ 0
≥
f (z)
1 − r2
e µ).
which is certainly positive if r < R(p,
e µ) is sharp for the functions f, g ∈ Ap defined in E by
It is easily seen that the bound R(p,
0
zf (z)
1
zp
=
,
g(z)
=
p f (z)1−µ g(z)µ
1+z
(1 + z)2
(0 ≤ µ, z ∈ E).
Choosing µ = 0 in Theorem 3.8, we have
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O N C ERTAIN S UBCLASS OF p-VALENTLY BAZILEVIC F UNCTIONS
9
Corollary 3.9. If f ∈ Ap satisfies
0
zf (z)
f (z)
< p (z ∈ E)
<
>
0
and
−
p
f (z)
zp
np
o.
then f is p-valently convex in |z| <
9 + 4p(p + 1) − 3
2(p + 1). The result is best
possible.
For a function f ∈ Ap , we define the integral operator Fµ,δ as follows:
µ1
Z
δ + pµ z δ−1
µ
(z ∈ E),
(3.15)
Fµ,δ (f ) = Fµ,δ (f )(z) =
t f (t) dt
zδ
0
where µ and δ are real numbers with µ > 0, δ > −pµ.
The following lemma will be required for the proof of Theorem 3.13 below.
n
o
Lemma 3.10. Let g ∈ Sp∗ (A, B), µ and δ are real numbers with µ > 0, δ > max −pµ, − pµ(1−A)
.
(1−B)
∗
Then Fµ,δ (g) ∈ Sp (A, B).
The proof of the above lemma follows by using Lemma 2.2 followed by a simple calculation.
n
o
Theorem 3.11. Let µ and δ be real numbers with µ > 0, δ > max −pµ, − pµ(1−A)
(−1 ≤
(1−B)
B < A ≤ 1) and let f ∈ f ∈ Ap . If
0
π
zf
(z)
< β
arg
−
α
(0 ≤ α < p; 0 < β ≤ 1)
2
f (z)1−µ g(z)µ
for some g ∈ Sp∗ (A, B), then
arg
0
π
z(Fµ,δ ) (f )
< η,
−
α
2
Fµ,δ (f )1−µ Fµ,δ (g)µ
where Fµ,δ (f ) is the operator given by (3.15) and η (0 < η ≤ 1) is the solution of the equation

(1+B)η sin π(1−t(A,B,δ,µ,p))/2

η + 2 tan−1
, B 6= −1;
π
(1+B)δ+µp(1+A)+(1+B)η cos π(1−t(A,B,δ,µ,p))/2
(3.16)
β=


η,
B = −1,
and
2
t(A, B, δ, µ, p) = sin−1
π
(3.17)
µp(A − B)
δ(1 − B 2 ) + µp(1 − AB)
.
Proof. Let us put
1
q(z) =
p−α
0
z(Fµ,δ ) (f )
−β
Fµ,δ (f )1−µ Fµ,δ (g)µ
where
1
Φ(z) =
p−α
δ
µ
z
Z
δ−1
z f (z) − δ
t
µ
=
Φ(z)
,
Ψ(z)
Z
z
f (t) dt − µ α
0
δ−1
t
g(t) dt
µ
0
and
Z
Ψ(z) = µ
z
tδ−1 g(t)µ dt.
0
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10
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Then q(z) is analytic in E and q(0) = 1. By a simple calculation, we get
0
0
Φ (z)
S(z) zq (z)
= q(z) 1 + 0
Ψ0 (z)
zS (z) q(z)
0
zf (z)
1
=
−α .
p − β f (z)1−µ g(z)µ
Since Fµ,δ (g) ∈ Sp∗ (A, B), by (1.6) and (1.7), we have
0
0
zS (z)
z(Fµ,δ ) (g)
=δ+µ
= ρeiπθ/2 ,
S(z)
Fµ,δ (g)
(3.18)
where

δ +
µp(1−A)
1−B
<ρ<δ+
µp(1+A)
1+B

−t(A, B, δ, µ, p) < θ < t(A, B, δ, µ, p) for B 6= −1
when t(A, B, δ, µ, p) is given by (3.17), and

<ρ<∞
δ + µp(1−A)
2

−1 < θ < 1 for B = −1.
0
Further, taking ω(z) = S(z)/zS (z) in Lemma 2.1, we note that q(z) 6= 0 in E. If there exists
a point z0 ∈ E such that the condition (2.7) is satisfied, then (by Lemma 2.5) we obtain (2.8)
under the restrictions (2.9) and (2.10).
1
At first, suppose that q(z0 ) η = ix (x > 0). For the case B 6= −1, by (3.18), we obtain
0
z0 f (z0 )
−α
arg
f (z0 )1−µ g(z0 )µ


0
z0 q (z0 ) 
1
·
= arg q(z0 ) + arg 1 +
z0 (Fµ,δ (g))0 (z0 )
q(z0 )
δ + µ Fµ,δ (g)(z0 )
−1 π
= η + arg 1 + ρeiπθ/2 iηk
2
π
ηk sin(π(1 − θ)/2)
−1
= η + tan
2
ρ + cos(π(1 − θ)/2)
!
η
sin
π(1
−
t(A,
B,
δ,
µ,
p))/2
π
≥ η + tan−1
2
+ η cos π(1 − t(A, B, δ, µ, p))/2
δ + µp(1+A)
1+B
π
= β,
2
where β and t(A, B, δ, µ, p) are given by (3.16) and (3.17), respectively. Similarly, for the case
B = −1, we have
0
zf (z)
π
arg
− α ≥ η.
1−µ
µ
f (z) g(z)
2
This is a contradiction to the assumption of our theorem.
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O N C ERTAIN S UBCLASS OF p-VALENTLY BAZILEVIC F UNCTIONS
11
1
Next, suppose that q(z0 ) η = −ix (x > 0). For the case B 6= −1, applying the same method
as above,we have
0
z0 f (z0 )
arg
−α
f (z0 )1−µ g(z0 )µ




η sin π(1 − t(A, B, δ, µ, p))/2
π

≤ − η − tan−1 


µp(1 + A)
2
δ+
+ η cos π(1 − t(A, B, δ, µ, p))/2
1+B
π
= − β,
2
where β and t(A, B, δ, µ, p) are given by (3.16) and (3.17), respectively and for the case B =
−1, we have
0
zf (z)
π
−
α
≤
−
η,
arg
f (z)1−µ g(z)µ
2
which contradicts the assumption. Thus, we complete the proof of the theorem.
Letting µ = 1, B → A and g(z) = z p in Theorem 3.11, we have
Corollary 3.12. Let δ > −p and f ∈ Ap .
0
arg f (z) − α <
p−1
z
If
π
β
2
(0 ≤ α < p; 0 < β ≤ 1),
then
arg
0
F1,δ (f )
−α
z p−1
!
π
< η,
2
where F1,δ (f ) is the integral operator given by (3.15) for µ = 1 and η (0 < η ≤ 1) is the
solution of the equation
2
η
−1
β = η + tan
.
π
δ+p
Theorem 3.13. Let λ > 0. If f ∈ A satisfies the condition
0
0
0
00
zf (z)
zf (z)
zf (z)
zg (z)
(3.19) γ
+λ 1+ 0
− (1 − µ)
−µ
6= it (z ∈ E)
f 1−µ (z)g µ (z)
f (z)
f (z)
g(z)
p
for some µ(µ ≥ 0), γ(γ > 0) and g ∈ Sp∗ , where t is a real number with |t| ≥ λ(λ + 2pγ),
then
0
zf (z)
<
> 0 (z ∈ E).
f 1−µ (z)g µ (z)
Proof. Let
0
zf (z)
φ(z) =
1−µ
p f (z)g µ (z)
(z ∈ E),
where φ(0) = 1. From (3.19), we easily have φ(z) 6= 0 in E. In fact, if φ has a zero of order m
at z = z1 ∈ E, then φ can be written as
φ(z) = (z − z1 )m q(z)
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12
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where q(z) is analytic in E and q(z1 ) 6= 0. Hence, we have
0
00
0
0
zf (z)
zf (z)
zg (z)
zf (z)
γ
+λ 1+ 0
− (1 − µ)
−µ
f 1−µ (z)g µ (z)
f (z)
f (z)
g(z)
0
zφ (z)
= p γφ(z) + λ
φ(z)
0
zq (z)
mz
+λ
.
= p γ(z − z1 ) q(z) + λ
z − z1
q(z)
m
(3.20)
But the imaginary part of (3.20) can take any infinite values when z → z1 in a suitable direction.
This contradicts (3.19). Thus, if there exists a point z0 ∈ E such that
<{p(z)} > 0 for |z| < |z0 |,
<{p(z0 )} > 0 and p(z0 ) = i` (` 6= 0),
then we have p(z0 ) 6= 0. From Lemma 2.5 and (3.20), we get
0
z0 φ (z0 )
p γφ(z0 ) + λ
= i(p γ` + λ k),
φ(z0 )
p
1 λ
p γ` + λ k ≥
+ (λ + 2p γ)` ≥ λ(λ + 2p γ) when ` > 0,
2 `
and
p
1 λ
p γ` + λ k ≤ −
+ (λ + 2p γ)|`| ≤ − λ(λ + 2p γ) when ` < 0,
2 |`|
which contradicts (3.19). Therefore, we have <{φ(z)} > 0 in E. This completes the proof of
the theorem.
Taking g(z) = z p and µ = 1 in Theorem 3.13, we have
Corollary 3.14. Let λ > 0. If f ∈ Ap satisfies the condition
0
00
f (z)
zf (z)
γ p−1 + λ 1 + 0
− p 6= it
(z ∈ E)
z
f (z)
p
for some γ (γ > 0), where t is a real number with |t| ≥ λ(λ + 2pγ), then
0 f (z)
<
>0
(z ∈ E).
z p−1
Corollary 3.15. Let λ > 0. If f ∈ Ap satisfies the condition
0
00
f
(z)
zf
(z)
γ
−p +λ 1+ 0
− p < λ + γp
p−1
z
f (z)
(z ∈ E)
for some γ (γ > 0), then
<
0
f (z)
z p−1
>0
(z ∈ E).
Remark 3.16. From a result of Nunokawa [9] and Saitoh and Nunokawa [11], it follows that,
if f ∈ Ap satisfies the hypothesis of Corollary
3.14 or Corollary 3.15, then f is p-valent in E
√
and p-valently convex in the disc |z| < ( p + 1 − 1)/p.
Letting γ = 1, µ = 0 in Theorem 3.13, we get the following result due to Dinggong [4] which
in turn yields the work of Cho and Kim [3] for p = 1.
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O N C ERTAIN S UBCLASS OF p-VALENTLY BAZILEVIC F UNCTIONS
13
Corollary 3.17. Let λ > 0. If f ∈ Ap satisfies the condition
0
00
zf (z)
zf (z)
(1 − λ)
+λ 1+ 0
6= it
(z ∈ E),
f (z)
f (z)
p
where t is a real number with |t| ≥ λ(λ + 2p), then f ∈ Sp∗ .
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