Volume 9 (2008), Issue 2, Article 32, 6 pp.
SUFFICIENT CONDITIONS FOR STARLIKENESS AND CONVEXITY IN |z| <
1
2
MAMORU NUNOKAWA, SHIGEYOSHI OWA, YAYOI NAKAMURA, AND TOSHIO HAYAMI
U NIVERSITY OF G UNMA
798-8 H OSHIKUKI - MACHI , C HUO - KU , C HIBA - SHI
C HIBA 260-0808, JAPAN
mamoru_nuno@doctor.nifty.jp
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502, JAPAN
owa@math.kindai.ac.jp
yayoi@math.kindai.ac.jp
ha_ya_to112@hotmail.com
Received 18 February, 2008; accepted 04 June, 2008
Communicated by A. Sofo
A BSTRACT. For analytic functions f (z) with f (0) = f 0 (0) − 1 = 0 in the open unit disc E,
T. H. MacGregor has considered some conditions for f (z) to be starlike or convex. The object
of the present paper is to discuss some interesting problems for f (z) to be starlike or convex for
|z| < 12 .
Key words and phrases: Analytic, Starlike, Convex.
2000 Mathematics Subject Classification. Primary 30C45.
1. I NTRODUCTION
Let A denote the class of functions f (z) of the form
f (z) = z +
∞
X
an z n
n=2
which are analytic in the open unit disc E = {z ∈ C : |z| < 1}. A function f ∈ A is said to be
starlike with respect to the origin in E if it satisfies
0 zf (z)
Re
>0
(z ∈ E).
f (z)
We would like to thank the referee for his very useful suggestions which essentially improved this paper.
050-08
2
M AMORU N UNOKAWA , S HIGEYOSHI OWA , YAYOI NAKAMURA , AND T OSHIO H AYAMI
Also, a function f ∈ A is called as convex in E if it satisfies
zf 00 (z)
Re 1 + 0
>0
(z ∈ E).
f (z)
MacGregor [2] has shown the following.
Theorem A. If f ∈ A satisfies
f (z)
<1
−
1
z
(z ∈ E),
then
0
zf (z)
<1
−
1
f (z)
1
|z| <
2
1
|z| <
2
so that
Re
zf 0 (z)
f (z)
>0
.
Therefore, f (z) is univalent and starlike for |z| < 12 .
Also, MacGregor [3] had given the following results.
Theorem B. If f ∈ A satisfies
|f 0 (z) − 1| < 1
(z ∈ E),
then
zf 00 (z)
Re 1 + 0
f (z)
> 0
1
for |z| < .
2
Therefore, f (z) is convex for |z| < 21 .
Theorem C. If f ∈ A satisfies
|f 0 (z) − 1| < 1
then f (z) maps |z| <
origin,
√
2 5
5
(z ∈ E),
= 0.8944 . . . onto a domain which is starlike with respect to the
0
arg zf (z) < π
f (z) 2
or
zf 0 (z)
Re
>0
f (z)
√
2 5
for |z| <
5
√
2 5
for |z| <
.
5
The condition domains of Theorem A, Theorem B and Theorem C are some circular domains
whose center is the point z = 1.
It is the purpose of the present paper to obtain some sufficient conditions for starlikeness or
convexity under the hypotheses whose condition domains are annular domains centered at the
origin.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 32, 6 pp.
http://jipam.vu.edu.au/
C ONDITIONS FOR S TARLIKENESS AND C ONVEXITY
3
2. S TARLIKENESS AND C ONVEXITY
We start with the following result for starlikeness of functions f (z).
Theorem 2.1. Let f ∈ A and suppose that
π2
(2.1)
0.10583 · · · = exp −
4 log 3
0 zf (z) < f (z) 2 π
< exp
= 9.44915 . . .
4 log 3
(z ∈ E).
Then f (z) is starlike for |z| < 12 .
Proof. From the assumption (2.1), we get
f (z) 6= 0
(0 < |z| < 1).
From the harmonic function theory (cf. Duren [1]), we have
0 0 0 Z
ζf (ζ) ζ + z
zf (z)
1
zf (z)
log
=
log dϕ + i arg
f (z)
2π |ζ|=R
f (ζ)
ζ −z
f (z) z=0
0 Z
zf (ζ) ζ + z
1
log dϕ
=
2π |ζ|=R
f (ζ) ζ − z
where |z| = r < |ζ| = R < 1, z = reiθ and ζ = Reiϕ .
It follows that
0 0 Z
1
ζf (ζ) ζ
+
z
zf
(z)
=
arg
log Im
dϕ
f (z)
2π |ζ|=R
f (ζ)
ζ −z
Z 2π 0 ζf (ζ) 2Rr sin(ϕ − θ)
1
dϕ
log ≤
2
2
2π 0
f (ζ)
R − 2Rr cos(ϕ − θ) + r Z 2π
π2
1
2Rr |sin(ϕ − θ)|
<
dϕ
2
4 log 3 2π 0 R − 2Rr cos(ϕ − θ) + r2
π2 2
R+r
=
log
.
4 log 3 π
R−r
Letting R → 1, we have
0
zf
(z)
arg
< π log 1 + r
f (z) 2 log 3
1−r
π
<
log 3
2 log 3
π
1
=
|z| = r <
.
2
2
This completes the proof of the theorem.
Next we derive the following
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 32, 6 pp.
http://jipam.vu.edu.au/
4
M AMORU N UNOKAWA , S HIGEYOSHI OWA , YAYOI NAKAMURA , AND T OSHIO H AYAMI
Theorem 2.2. Let f ∈ A and suppose that
3
(2.2)
0.472367 . . . = exp −
4
f (z) < z 3
< exp
= 2.177 . . .
4
Then we have
0
zf (z)
<1
−
1
f (z)
1
|z| <
2
(z ∈ E).
,
or f (z) is starlike for |z| < 12 .
Proof. From the assumption (2.2), we have
f (z) 6= 0
(0 < |z| < 1).
Applying the harmonic function theory (cf. Duren [1]), we have
Z
f (ζ) ζ + z
1
f (z)
log
=
log dϕ,
z
2π |ζ|=R
ζ ζ −z
where |z| = r < |ζ| = R < 1, z = reiθ and ζ = Reiϕ .
Then, it follows that
Z
f (ζ) zf 0 (z)
1
2ζz
−1=
log dϕ.
f (z)
2π |ζ|=R
ζ
(ζ − z)2
This gives us
0
Z
f (ζ) zf (z)
1
2Rr
dϕ
log f (z) − 1 ≤ 2π
2
ζ R − 2Rr cos(ϕ − θ) + r2
|ζ|=R Z
3 1
2Rr
<
dϕ
2
4 2π |ζ|=R R − 2Rr cos(ϕ − θ) + r2
3 2Rr
=
.
4 R2 − r 2
Making R → 1, we have
0
zf (z)
3 2r
1
|z| = r <
,
f (z) − 1 < 4 1 − r2 < 1
2
which completes the proof of the theorem.
For convexity of functions f (z), we show the following corollary without the proof.
Corollary 2.3. Let f ∈ A and suppose that
3
3
0
(2.3)
0.472367 · · · = exp −
< |f (z)| < exp
= 2.117 . . .
4
4
(z ∈ E).
Then f (z) is convex for |z| < 12 .
Next our result for the convexity of functions f (z) is contained in
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 32, 6 pp.
http://jipam.vu.edu.au/
C ONDITIONS FOR S TARLIKENESS AND C ONVEXITY
5
Theorem 2.4. Let f ∈ A and suppose that
0 zf (z) 1
< exp 1 = 1.28403 . . .
(2.4)
0.778801 · · · = exp −
< 4
f (z) 4
(z ∈ E).
Then f (z) is convex for |z| < 21 .
Proof. From the condition (2.4) of the theorem, we have
zf 0 (z)
6= 0
in E.
f (z)
Then, it follows that
Z
zf 0 (z)
1
ζf 0 (ζ) ζ + z
(2.5)
log
=
log
dϕ,
f (z)
2π |ζ|=R
f (ζ) ζ − z
where |z| = r < |ζ| = R < 1, z = reiθ and ζ = Reiϕ .
Differentiating (2.5) and multiplying by z, we obtain that
0 Z
ζf (ζ) zf 0 (z)
1
2ζz
zf 00 (z)
1+ 0
=
+
log dϕ.
f (z)
f (z)
2π |ζ|=R
f (ζ)
(ζ − z)2
In view of Theorem 2.1, f (z) is starlike for |z| <
Re
zf 0 (z)
1−r
≥
f (z)
1+r
1
2
and therefore, we have
1
|z| = r <
.
2
Then, we have
0 Z
ζf (ζ) zf 00 (z)
zf 0 (z)
1
2ζz
1 + Re 0
= Re
+
log Re
dϕ
f (z)
f (z)
2π |ζ|=R
f (ζ) (ζ − z)2
Z
1−r
1
1 2Rr
>
−
dϕ
1 + r 2π |ζ|=R 4 |ζ − z|2
1 − r 1 2Rr
=
−
.
1 + r 4 R2 − r 2
Letting R → 1, we see that
zf 00 (z)
1 − r 1 2r
1 + Re 0
>
−
f (z)
1 + r 4 1 − r2
1 1 4
= − ·
3 4 3
1
=0
|z| = r <
,
2
which completes the proof of our theorem.
Finally, we prove
Theorem 2.5. Let f ∈ A and suppose that
π2
0.10583 . . . = exp −
4 log 3
0 2 zf (z) π
< exp
< = 9.44915 . . .
f (z)
4 log 3
(z ∈ E).
Then f (z) is convex in |z| < r0 where r0 is the root of the equation
(4 log 3)r2 − 2(4 log 3 + π 2 )r + 4 log 3 = 0,
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 32, 6 pp.
http://jipam.vu.edu.au/
6
M AMORU N UNOKAWA , S HIGEYOSHI OWA , YAYOI NAKAMURA , AND T OSHIO H AYAMI
p
π 2 − 4 log 3 − π π 2 + 8 log 3
r0 =
= 0.15787 . . . .
4 log 3
Proof. Applying the same method as the proof of Theorem 2.5, we have
0 Z
ζf (ζ) zf 00 (z)
zf 0 (z)
1
2ζz
= Re
+
log Re
dϕ
1 + Re 0
f (z)
f (z)
2π |ζ|=R
f (ζ) (ζ − z)2
>
1−r
π2
2Rr
−
2
1 + r 4 log 3 R − r2
where |z| = r < |ζ| = R < 1, z = reiθ and ζ = Reiϕ .
Putting R → 1, we have
1 + Re
zf 00 (z)
1−r
π2
2r
>
−
0
f (z)
1 + r 4 log 3 1 − r2
n
o
1
2
2
=
(4 log 3)r − 2(4 log 3 + π )r + 4 log 3
(1 − r2 )4 log 3
>0
(|z| < r0 ).
Remark 1. The condition in Theorem A by MacGregor [2] implies that
f (z)
0 < Re
<2
(z ∈ E).
z
However, the condition in Theorem 2.2 implies that
f (z)
−2.117 · · · < Re
< 2.117 . . .
(z ∈ E).
z
Furthermore, the condition in Theorem B by MacGregor [3] implies that
0 < Re f 0 (z) < 2
(z ∈ E).
However, the condition in Corollary 2.3 implies that
−2.117 · · · < Re f 0 (z) < 2.117 . . .
(z ∈ E).
R EFERENCES
[1] P. DUREN, Harmonic mappings in the plane, Cambridge Tracts in Mathematics 156, Cambridge
Univ. Press, 2004.
[2] T.H. MacGREGOR, The radius of univalence of certain analytic functions. II, Proc. Amer. Math.
Soc., 14(3) (1963), 521–524.
[3] T.H. MacGREGOR, A class of univalent functions, Proc. Amer. Math. Soc., 15 (1964), 311–317.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 32, 6 pp.
http://jipam.vu.edu.au/