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On certain univalent functions with missing coefficients

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Abstract

The main object of the present paper is to show certain sufficient conditions for univalency of analytic functions with missing coefficients.

MSC:30C45, 30C55.

1 Introduction

Let A(n) be the class of functions of the form

f(z)=z+ a n z n + a n + 1 z n + 1 +(n=2,3,),
(1.1)

which are analytic in the unit disk U={z:|z|<1}. We write A(2)=A.

A function f(z)A is said to be starlike in |z|<r (r1) if and only if it satisfies

Re z f ( z ) f ( z ) >0 ( | z | < r ) .
(1.2)

A function f(z)A is said to be close-to-convex in |z|<r (r1) if and only if there is a starlike function g(z) such that

Re z f ( z ) g ( z ) >0 ( | z | < r ) .
(1.3)

Let f(z) and g(z) be analytic in U. Then we say that f(z) is subordinate to g(z) in U, written f(z)g(z), if there exists an analytic function w(z) in U, such that |w(z)||z| and f(z)=g(w(z)) (zU). If g(z) is univalent in U, then the subordination f(z)g(z) is equivalent to f(0)=g(0) and f(U)g(U).

Recently, several authors showed some new criteria for univalency of analytic functions (see, e.g., [17]). In this note, we shall derive certain sufficient conditions for univalency of analytic functions with missing coefficients.

For our purpose, we shall need the following lemma.

Lemma (see [8, 9])

Let f(z) and g(z) be analytic in U with f(0)=g(0). If h(z)=z g (z) is starlike in U and z f (z)h(z), then

f(z)f(0)+ 0 z h ( t ) t dt.
(1.4)

2 Main results

Our first theorem is given by the following.

Theorem 1 Let f(z)=z+ a n z n +A(n) with f(z)0 for 0<|z|<1. If

| ( z f ( z ) ) ( n ) |β(zU),
(2.1)

where 0<β2[1(n2)| a n |], then f(z) is univalent in U.

Proof

Let

p(z)= ( z f ( z ) ) ( n ) (zU),
(2.2)

then p(z) is analytic in U. By integration from 0 to z n-times, we obtain

z f ( z ) =1 a n z n 1 + 0 z d w n 0 w n d w n 1 0 w 2 p( w 1 )d w 1 (zU).
(2.3)

Thus, we have

f(z)= z 1 a n z n 1 + φ ( z ) (zU),
(2.4)

where

φ(z)= 0 z d w n 0 w n d w n 1 0 w 2 p( w 1 )d w 1 (zU).
(2.5)

It is easily seen from (2.1), (2.2) and (2.5) that

| φ ( n ) ( z ) | β(zU)
(2.6)

and, in consequence,

| φ ( z ) | β(zU).

Since

( φ ( z ) z ) = 1 z 2 0 z w φ (w)dw(zU),

we get

| ( φ ( z ) z ) |=| 1 z 2 0 z w φ (w)dw| β 2 (zU)

and so

| φ ( z 2 ) z 2 φ ( z 1 ) z 1 |=| z 1 z 2 ( φ ( w ) w ) dw| β 2 | z 2 z 1 |
(2.7)

for z 1 , z 2 U and z 1 z 2 .

Now it follows from (2.4) and (2.7) that

Hence, f(z) is univalent in U. The proof of the theorem is complete. □

Let S n (β) denote the class of functions f(z)=z+ a n z n +A(n) with f(z)0 for 0<|z|<1, which satisfy the condition (2.1) given by Theorem 1.

Next we derive the following.

Theorem 2 Let f(z)=z+ a n z n + S n (β). Then, for zU,

(2.8)
(2.9)
(2.10)

Proof In view of (2.1), we have

z ( z f ( z ) ) ( n ) βz(zU).
(2.11)

Applying Lemma to (2.11), we get

( z f ( z ) ) ( n 1 ) +(n1)! a n βz(zU).
(2.12)

By using the lemma repeatedly, we finally have

( z f ( z ) ) +(n1) a n z n 2 βz(zU).
(2.13)

According to a result of Hallenbeck and Ruscheweyh [[1], Theorem 1], (2.13) gives

1 z 0 z [ ( t f ( t ) ) + ( n 1 ) a n t n 2 ] dt β 2 z(zU),
(2.14)

i.e.,

z f ( z ) =1 a n z n 1 + β 2 zw(z)(zU),
(2.15)

where w(z) is analytic in U and |w(z)| | z | n 1 (zU).

Now, from (2.15), we can easily derive the inequalities (2.8), (2.9) and (2.10). □

Finally, we discuss the following theorem.

Theorem 3 Let f(z) S n (β) and have the form

f(z)=z+ a n + 1 z n + 1 + a n + 2 z n + 2 +(zU).
(2.16)
  1. (i)

    If 2 5 β2, then f(z) is starlike in |z|< 2 β n 1 5 2 n ;

  2. (ii)

    If 3 1β2, then f(z) is close-to-convex in |z|< 3 1 β n .

Proof

If we put

p(z)= z 2 f ( z ) f 2 ( z ) =1+ p n z n +(zU),
(2.17)

then by (2.1) and the proof of Theorem 2 with a n =0, we have

z p (z)= z 2 ( z f ( z ) ) βz.
(2.18)

It follows from the lemma that

p(z)1+βz,
(2.19)

which implies that

| z 2 f ( z ) f 2 ( z ) 1|β | z | n (zU).
(2.20)
  1. (i)

    Let 2 5 β2 and

    |z|< r 1 = 2 β n 1 5 2 n .
    (2.21)

Then by (2.20), we have

|arg z 2 f ( z ) f 2 ( z ) |<arcsin 2 5 .
(2.22)

Also, from (2.8) in Theorem 2 with a n =0, we obtain

| z f ( z ) 1|< β 2 r 1 n
(2.23)

and so

|arg z f ( z ) |< 1 5 .
(2.24)

Therefore, it follows from (2.22) and (2.24) that

| arg z f ( z ) f ( z ) | | arg z 2 f ( z ) f 2 ( z ) | + | arg z f ( z ) | < arcsin 2 5 + arcsin 1 5 = π 2

for |z|< r 1 . This proves that f(z) is starlike in |z|< r 1 .

  1. (ii)

    Let 3 1β2 and

    |z|< r 2 = 3 1 β n .
    (2.25)

Then we have

| arg f ( z ) | | arg z 2 f ( z ) f 2 ( z ) | + 2 | arg z f ( z ) | < arcsin ( β r 2 n ) + 2 arcsin ( β 2 r 2 n ) = arcsin ( 3 1 ) + 2 arcsin ( 3 1 2 ) = π 2 .

Thus, Re f (z)>0 for |z|< r 2 . This shows that f(z) is close-to-convex in |z|< r 2 . □

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

We would like to express sincere thanks to the referees for careful reading and suggestions, which helped us to improve the paper.

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Correspondence to Jin-Lin Liu.

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The authors declare that they have no competing interests.

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The authors have made the same contribution. All authors read and approved the final manuscript.

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Keywords

  • analytic
  • univalent
  • subordination