Internat. J. Math. & Math. Sci.
Vol. 23, No. 12 (2000) 855–857
S0161171200001824
© Hindawi Publishing Corp.
ON A CLASS OF UNIVALENT FUNCTIONS
VIKRAMADITYA SINGH
(Received 3 March 1998)
Abstract. We consider the class of univalent functions f (z) = z + a3 z3 + a4 z4 + · · · analytic in the unit disc and satisfying |(z2 f (z)/f 2 (z))−1| < 1, and show that such functions
√
are starlike if they satisfy |(z2 f (z)/f 2 (z)) − 1| < (1/ 2).
Keywords and phrases. Analytic, univalent and starlike functions.
2000 Mathematics Subject Classification. Primary 30C45.
Let A denote the class of functions which are analytic in the unit disc U = {z : |z| < 1}
and have Taylor series expansion
f (z) = z + a2 z2 + a3 z3 + · · · ,
and let T be the univalent [3] subclass of A which satisfy
z2 f (z)
2
− 1 < 1, z ∈ U.
f (z)
(1)
(2)
By T2 we denote the subclass of T for which f (0) = 0. In this paper, we prove the
following theorem.
Theorem 1. If f ∈ T2 , then
(i) Re(f (z)/z) > 1/2, z ∈ U ,
√
4
(ii) f is starlike in |z| < 1/ 2 = 0.840896 . . . ,
√
(iii) Re f (z) > 0 for |z| < 1/ 2.
Items (i) and (iii) are improvements of results in [2], and (ii) is the same as in [2] but
has a different proof. Furthermore, (i) and (iii) are sharp as shown by the function
f (z) =
z
,
1 − z2
(3)
but the sharpness of (ii) is difficult to establish by a direct example. We also prove the
following theorem which partially answers a question raised in [1].
Theorem 2. If T2,µ is the subclass of T2 which satisfies
f (z)
2
< µ < 1,
z
−
1
f 2 (z)
√
then T2,µ is a subclass of starlike functions if 0 ≤ µ ≤ 1/ 2.
(4)
856
VIKRAMADITYA SINGH
We define by B the class of functions ω analytic in U and satisfying
|ω(z)| < 1,
ω(0) = ω (0) = 0.
z ∈ U,
(5)
From Schwarz’s lemma it then follows that
|ω(z)| ≤ |z|2 .
(6)
Proof of Theorem 1. If f ∈ T2 and satisfies (2), then
z2
f (z)
− 1 = ω(z),
f 2 (z)
z ∈ U, ω ∈ B,
(7)
and by direct integration
z
= 1−
f (z)
From (8), we obtain
and this gives
1
0
ω(tz)
dt,
t2
z ∈ U, ω ∈ B.
(8)
z
2
f (z) − 1 ≤ |z| < 1,
(9)
1 − f (z) ≤ f (z) ,
z
z (10)
which is equivalent to (Re f (z)/z) > 1/2, This proves (i).
Furthermore, from (9), we obtain
arg f (z) ≤ sin−1 |z|2 .
z (11)
From (7), we obtain
z
f (z) f (z)
=
1 + ω(z)
f (z)
z
(12)
and, therefore,
f (z)
zf (z) ≤ 2 sin−1 |z|2 .
=
arg
arg
+
arg
1
+
ω(z)
f (z) z
This gives (ii).
In order to prove (iii), we notice that (7) yields
2
f (z) f (z) =
1 + ω(z)
z
(13)
(14)
and, therefore,
arg f (z) = 2 arg f (z) + arg 1 + ω(z) ≤ 3 sin−1 |z|2 .
z
(15)
But this is equivalent to (iii).
Proof of Theorem 2. If f ∈ T2,µ , we obtain from (4)
z
f (z)
− 1 = µω(z),
f 2 (z)
ω ∈ B, z ∈ U
and
z
= 1−µ
f (z)
1
0
ω(tz)
dt.
t2
(16)
ON A CLASS OF UNIVALENT FUNCTIONS
857
Hence
z
1 + µω(z)
f (z)
=
.
1
f (z)
1 − µ 0 (ω(tz)/t 2 ) dt
(17)
Now Re z (f (z)/f (z)) > 0 is equivalent to the condition
z
f (z)
1 + µω(z)
=
≠ −iT ,
1
f (z)
1 − µ 0 (ω(tz)/t 2 )dt
T ∈ Re .
(18)
Relation (18) is equivalent to
µ
2
Let
M=
ω(z) +
1
0
1
ω(tz)
ω(tz)
1 − iT
ω(z)
−
dt
+
dt
≠ −1.
t2
1 + iT
t2
0
1
1
ω(tz)
ω(tz)
1 − iT
,
ω(z) +
ω(z)
−
dt
+
dt
2
2
t
1 + iT
t
0
0
z∈U,ω∈B,T ∈Re
sup
(19)
(20)
then, in view of the rotation invariance of B, it follows that
Re z
f (z)
> 0,
f (z)
if µ ≤
2
.
M
However, from (20), we notice that
1
1
ω(tz) ω(tz) M ≤ sup
dt + ω(z) −
dt ω(z) +
t2
t2
0
0
z∈U,ω∈B
2 1 ω(tz) 2
ω(z) + ≤ 2 sup
dt ≤ 2 2.
2
t
0
z∈Ui ,ω∈B
(21)
(22)
Inequality (22) follows from the parallelogram law and the last step from (6). And (21)
√
shows that µ ≤ 1/ 2.
References
[1]
[2]
[3]
M. Obradović, Starlikeness and certain class of rational functions, Math. Nachr. 175 (1995),
263–268. MR 96m:30016. Zbl 845.30005.
M. Obradović, N. N. Pascu, and I. Radomir, A class of univalent functions, Math. Japon. 44
(1996), no. 3, 565–568. MR 97i:30016. Zbl 902.30008.
S. Ozaki and M. Nunokawa, The Schwarzian derivative and univalent functions, Proc. Amer.
Math. Soc. 33 (1972), 392–394. MR 45#8821. Zbl 233.30011.
Singh: 3A/95 Azad Nagar, Kanpur 208002, India