Reciprocal classes of p-valently spirallike and p-valently Robertson functions

Abstract

For p-valently spirallike and p-valently Robertson functions in the open unit disk $U$, reciprocal classes ${\mathcal{S}}_p\left(\alpha ,\beta \right)$, and ${\mathcal{C}}_p\left(\alpha ,\beta \right)$ are introduced. The object of the present paper is to discuss some interesting properties for functions f(z) belonging to the classes ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ and ${\mathcal{C}}_p\left(\alpha ,\beta \right)$.

2010 Mathematics Subject Classification

Primary 30C45

1 Introduction

Let ${\mathcal{A}}_p$ be the class of functions f(z) of the form

$$f\left(z\right)=z^p+\sum _{n=p+1}^{\infty }{a}_n{z}^n$$

(1.1)

which are analytic in the open unit disk $U=\left\{z\in ℂ:\mid z\mid <1\right\}$.

For $f\left(z\right)\in {\mathcal{A}}_p$, we say that f(z) belongs to the class ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ if it satisfies

$$\mathsf{\text{Re}}\left({\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\beta \left(z\in U\right)$$

(1.2)

for some real $\alpha \left(\mid \alpha \mid <\frac{\pi }{2}\right)$ and β (β > p cos α).

When α = 0, the class ${\mathcal{S}}_p\left(0,\beta \right)$ was studied by Polatoglu et al. [1], and the classes ${\mathcal{S}}_1\left(0,\beta \right)$ and ${\mathcal{C}}_1\left(0,\beta \right)$ were introduced by Owa and Nishiwaki [2].

Further, let ${\mathcal{C}}_p\left(\alpha ,\beta \right)$ denote the subclass of ${\mathcal{A}}_p$ consisting of functions f(z), which satisfy

$$\mathsf{\text{Re}}\left\{{\mathsf{\text{e}}}^{i\alpha }\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)\right\}<\beta \left(z\in U\right)$$

(1.3)

for some real $\alpha \left(\mid \alpha \mid <\frac{\pi }{2}\right)$ and β (β > p cos α).

We note that $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$ if and only if $\frac{z{f}^{\prime }\left(z\right)}{p}\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$, and that, $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$ if and only if $p{\int }_0^z\frac{f\left(t\right)}{t}\mathsf{\text{d}}t\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$.

Remark 1 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\mathsf{\text{Re}}\left({\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>0\left(z\in U\right),$$

then we say that f(z) is p-valently spirallike in $U$ (cf. [1]). Also, if $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\mathsf{\text{Re}}\left\{{\mathsf{\text{e}}}^{i\alpha }\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)\right\}>0\left(z\in U\right),$$

then f(z) is said to be p-valently Robertson function in $U$ (cf. [3, 4]). Therefore, ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ defined by (1.2) is the reciprocal class of p-valently spirallike functions in $U$, and ${\mathcal{C}}_p\left(\alpha ,\beta \right)$ defined by (1.3) is the reciprocal class of p-valently Robertson functions in $U$.

Let $\mathcal{P}$ be the class of functions p(z) of the form

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n\left(z\in U\right)$$

(1.4)

that are analytic in $U$ and satisfy $\mathsf{\text{Re}}p\left(z\right)>0\left(z\in U\right)$. A function $p\left(z\right)\in \mathcal{P}$ is called the Carathéodory function and satisfies

$$\mid c_n\mid \leqq 2\left(n=1,2,3,\dots \right)$$

(1.5)

with the equality for $p\left(z\right)=\frac{1+z}{1-z}$ (cf. [5]).

For analytic functions g(z) and h(z) in $U$, we say that g(z) is subordinate to h(z) if there exists an analytic function w(z) in $U$ with w(0) = 0 and $\mid w\left(z\right)\mid <1\left(z\in U\right)$, and such that g(z) = h(w(z)). We denote this subordination by

$$g\left(z\right)\prec h\left(z\right)\left(z\in U\right).$$

(1.6)

If h(z) is univalent in $U$, then this subordination (1.6) is equivalent to g(0) = h(0) and $g\left(U\right)\subset h\left(U\right)$ (cf. [5]).

2 Subordinations for classes

We consider subordination properties of function f(z) in the classes ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ and ${\mathcal{C}}_p\left(\alpha ,\beta \right)$.

Theorem 1 A function f(z) belongs to the class ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ if and only if

$$e^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec 2\beta -p{e}^{-i\alpha }+\frac{2\left(pcos\alpha -\beta \right)}{1-z}\left(z\in U\right)$$

(2.1)

for some real $\alpha \left(\mid \alpha \mid <\frac{\pi }{2}\right)$ and β (β > p cos α).

The result is sharp for f(z) given by

$$f\left(z\right)=\frac{z^p}{{\left(1-z\right)}^{2e^{-i\alpha }\left(pcos\alpha -\beta \right)}}.$$

(2.2)

Proof. Let $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$. If we define the function w(z) by

$$\frac{\beta -{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+ipsin\alpha }{\beta -pcos\alpha }=\frac{1+w\left(z\right)}{1-w\left(z\right)}\left(w\left(z\right)\ne 1\right),$$

(2.3)

then we know that w(z) is analytic in $U$, w(0) = 0, and

$$\mathsf{\text{Re}}\left(\frac{1+w\left(z\right)}{1-w\left(z\right)}\right)>0\left(z\in U\right).$$

(2.4)

Therefore, we have that $\mid w\left(z\right)\mid <1\left(z\in U\right)$. If follows from (2.3) that

$${\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=2\beta -p{\mathsf{\text{e}}}^{-i\alpha }+\frac{2\left(pcos\alpha -\beta \right)}{1-w\left(z\right)}\left(z\in U\right),$$

(2.5)

which is equivalent to the subordination (2.1).

Conversely, we suppose that the subordination (2.1) holds true. Then, we have that

$${\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=2\beta -p{\mathsf{\text{e}}}^{-i\alpha }+\frac{2\left(pcos\alpha -\beta \right)}{1-w\left(z\right)}\left(z\in U\right),$$

(2.6)

for some Shwarz function w(z), which is analytic in $U$, w(0) = 0, and $\mid w\left(z\right)\mid <1\left(z\in U\right)$. It is easy to see that the equality (2.6) is equivalent to the equality (2.3). Since

$$\mathsf{\text{Re}}\left(\frac{1+w\left(z\right)}{1-w\left(z\right)}\right)=\mathsf{\text{Re}}\left\{\frac{\beta -{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+ipsin\alpha }{\beta -pcos\alpha }\right\}>0\left(z\in U\right),$$

(2.7)

we conclude that

$$\mathsf{\text{Re}}\left(\beta -{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)>0\left(z\in U\right),$$

(2.8)

which shows that $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$.

Finally, we consider the function f(z) given by (2.2). Then, f(z) satisfies

$${\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=2\beta -p{\mathsf{\text{e}}}^{-i\alpha }+\frac{2\left(pcos\alpha -\beta \right)}{1-z}.$$

(2.9)

This completes the proof of the theorem.   □

Noting that $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$ if and only if $\frac{z{f}^{\prime }\left(z\right)}{p}\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$, we also have

Corollary 1 A function f(z) belongs to the class ${\mathcal{C}}_p\left(\alpha ,\beta \right)$ if and only if

$$e^{i\alpha }\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)\prec 2\beta -p{e}^{-i\alpha }+\frac{2\left(pcos\alpha -\beta \right)}{1-z}\left(z\in U\right)$$

(2.10)

for some real $\alpha \left(\mid \alpha \mid <\frac{\pi }{2}\right)$ and β (β > p cos α).

The result is sharp for f(z) given by

$$f^{\prime }\left(z\right)=\frac{p{z}^{p-1}}{{\left(1-z\right)}^{2e^{-i\alpha }\left(pcos\alpha -\beta \right)}}.$$

(2.11)

3 Coefficient inequalities

Applying the properties for Carathéodory functions, we discuss the coefficient inequalities for f(z) in the classes ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ and ${\mathcal{C}}_p\left(\alpha ,\beta \right)$.

Theorem 2 If f(z) belongs to the class ${\mathcal{S}}_p\left(\alpha ,\beta \right)$, then

$$\mid a_{p+k}\mid \leqq \frac{1}{k!}\prod _{j=0}^{k-1}\left(2\left(\beta -pcos\alpha \right)+j\right)\left(k=1,2,3,\dots \right).$$

(3.1)

The result is sharp for

$$f\left(z\right)=\frac{z^p}{{\left(1-z\right)}^{2\left(p-\beta \right)}}$$

(3.2)

for α = 0.

Proof. In view of Theorem 1, we can consider the function w(z) given by (2.3) for $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$. Since w(z) is the Schwarz function, the function q(z) defined by

$$q\left(z\right)=\frac{\beta -{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+ipsin\alpha }{\beta -pcos\alpha }$$

(3.3)

is the Carathéodory function. If we write that

$$q\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n,$$

(3.4)

then we see that

$$\mid c_n\mid \leqq 2\left(n=1,2,3,\dots \right)$$

and the equality holds true for $q\left(z\right)=\frac{1+z}{1-z}$ and its rotation. It is to be noted that the equation (3.3) is equivalent to

$${\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\beta +ipsin\alpha -\left(\beta -pcos\alpha \right)q\left(z\right).$$

(3.5)

This gives us that

$$\begin{array}{c}{\mathsf{\text{e}}}^{i\alpha }\left(p{z}^p+\sum _{n=p+1}^{\infty }n{a}_n{z}^n\right)\\ =\left\{p{\mathsf{\text{e}}}^{i\alpha }-\left(\beta -pcos\alpha \right)\left(\sum _{n=1}^{\infty }{c}_n{z}^n\right)\right\}\left(z^p+\sum _{n=p+1}^{\infty }{a}_n{z}^n\right),\end{array}$$

(3.6)

which implies that

$${\mathsf{\text{e}}}^{i\alpha }\left(n-p\right)a_n=-\left(\beta -pcos\alpha \right)\left(c_{n-1}+a_2{c}_{n-2}+\cdots +a_{n-1}{c}_1\right).$$

(3.7)

It follows from (3.7) that

$$\mid a_n\mid \leqq \frac{2\left(\beta -pcos\alpha \right)}{n-p}\left(1+\mid a_2\mid +\mid a_3\mid +\cdots +\mid a_{n-1}\mid \right).$$

(3.8)

If n = p + 1, then we have that

$$\mid a_{p+1}\mid \leqq 2\left(\beta -pcos\alpha \right).$$

(3.9)

If n = p + 2, then we also have that

$$\begin{array}{c}\mid a_{p+2}\mid \leqq \frac{2\left(\beta -pcos\alpha \right)}{2}\left(1+\mid a_2\mid \right)\\ \leqq \left(\beta -pcos\alpha \right)\left(1+2\left(\beta -pcos\alpha \right)\right).\end{array}$$

(3.10)

Thus, the coefficient inequality (3.1) is true for n = p + 1 and n = p + 2. Next, we suppose that (3.1) holds true for n = p + 1, p + 2, p + 3, ..., p + k - 1. Then

$$\begin{array}{c}\mid a_{p+k}\mid \leqq \frac{2\left(\beta -pcos\alpha \right)}{k}\left(1+\mid a_2\mid +\mid a_3\mid +\cdots +\mid a_{p+k-1}\mid \right)\\ \leqq \frac{2\left(\beta -pcos\alpha \right)}{k}\left\{1+2\left(\beta -pcos\alpha \right)+\frac{2\left(\beta -pcos\alpha \right)}{2}\left(1+2\left(\beta -pcos\alpha \right)\right)\right.\\ +\frac{2\left(\beta -pcos\alpha \right)}{3}\left(1+2\left(\beta -pcos\alpha \right)\right)\left(1+\frac{2\left(\beta -pcos\alpha \right)}{2}\right)\\ \left.+\cdots +\frac{1}{\left(k-1\right)!}\prod _{j=0}^{k-2}\left(2\left(\beta -pcos\right)+j\right)\right\}\\ =\frac{2\left(\beta -pcos\alpha \right)}{k}\left(1+2\left(\beta -pcos\alpha \right)\right)\left\{1+\frac{2\left(\beta -pcos\alpha \right)}{2}\right.\\ +\left(1+\frac{2\left(\beta -pcos\alpha \right)}{2}\right)\frac{2\left(\beta -pcos\alpha \right)}{3}\\ \left.+\cdots +\frac{2\left(\beta -pcos\alpha \right)}{\left(k-1\right)!}\prod _{j=1}^{k-2}\left(2\left(\beta -pcos\alpha \right)+j\right)\right\}\\ =\frac{1}{k!}\prod _{j=0}^{k-1}\left(2\left(\beta -pcos\alpha \right)+j\right).\end{array}$$

(3.11)

This means that the inequality (3.1) holds true for n = p + k. Therefore, by the mathematical induction, we prove the coefficient inequality (3.1).

Finally, let us consider the function f(z) given by (3.2). Then, f(z) can be written by

$$\begin{array}{c}f\left(z\right)=z^p\left(\sum _{j=0}^{\infty }\left(\begin{array}{c}\hfill 2\left(\beta -p\right)\hfill \\ \hfill j\hfill \end{array}\right){\left(-z\right)}^j\right)\\ =z^p+2\left(\beta -p\right)z^{p+1}+\cdots +\left(\frac{1}{k!}\prod _{j=0}^{k-1}\left(2\left(\beta -p\right)+j\right)z^{p+k}\right)+\dots .\end{array}$$

(3.12)

Thus, this function f(z) satisfies the equality in (3.1).   □

Corollary 2 If f(z) belongs to the class ${\mathcal{C}}_p\left(\alpha ,\beta \right)$, then

$$\mid a_{p+k}\mid \leqq \frac{1}{\left(k-1\right)!}\prod _{j=0}^{k-1}\left(2\left(\beta -pcos\alpha \right)+j\right)\left(k=1,2,3,\dots \right).$$

(3.13)

The result is sharp for f(z) defined by

$$f^{\prime }\left(z\right)=\frac{p{z}^{p-1}}{{\left(1-z\right)}^{2\left(p-\beta \right)}}$$

(3.14)

for α = 0.

Remark 2 We know that the extremal functions for $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$ is f(z) given by (2.2) and for $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$ is f(z) given by (2.11). But, we see that

$$\mid a_{p+k}\mid \leqq \frac{1}{k!}\prod _{j=0}^{k-1}\left|2{\mathsf{\text{e}}}^{-i\alpha }\left(\beta -pcos\alpha \right)+j\right|$$

(3.15)

and

$$\mid a_{p+k}\mid \leqq \frac{1}{\left(k-1\right)!}\prod _{j=0}^{k-1}\left|2{\mathsf{\text{e}}}^{-i\alpha }\left(\beta -pcos\alpha \right)+j\right|$$

(3.16)

for such functions.

Therefore, the extremal functions for $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$ and $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$ do not satisfy the equalities in (3.1) and (3.13), respectively.

Furthermore, if we consider α = 0 in Theorem 2, then we obtain the corresponding result due to Polatoglu et al. [1].

4 Inequalities for the real parts

We discuss some problems of inequalities for the real parts of $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$.

Theorem 3 If $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$ , then we have

$$\frac{pcos\alpha -\left(2\beta -pcos\alpha \right)r}{1-r}\leqq \mathsf{\text{Re}}\left(e^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)\leqq \frac{pcos\alpha +\left(2\beta -pcos\alpha \right)r}{1+r}$$

(4.1)

for | z | = r < 1. The equalities hold true for f(z) given by (2.2).

Proof. By virtue of Theorem 1, we consider the function g(z) defined by

$$g\left(z\right)=2\beta -p{\mathsf{\text{e}}}^{-i\alpha }+\frac{2\left(pcos\alpha -\beta \right)}{1-z}\left(z\in U\right).$$

(4.2)

Letting z = r e^iθ (0 ≦ r < 1), we see that

$$\mathsf{\text{Re}}g\left(z\right)=2\beta -pcos\alpha +\frac{2\left(pcos\alpha -\beta \right)\left(1-rcos\theta \right)}{1+r^2-2rcos\theta }.$$

(4.3)

Let us define

$$h\left(t\right)=\frac{1-rt}{1+r^2-2rt}\left(t=cos\theta \right).$$

(4.4)

Then, we know that h'(t) ≧ 0. This implies that

$$2\beta -pcos\alpha +\frac{2\left(pcos\alpha -\beta \right)}{1-r}\leqq \mathsf{\text{Re}}g\left(z\right)\leqq 2\beta -pcos\alpha +\frac{2\left(pcos\alpha -\beta \right)}{1+r},$$

(4.5)

which is equivalent to

$$\frac{pcos\alpha -\left(2\beta -pcos\alpha \right)r}{1-r}\leqq \mathsf{\text{Re}}g\left(z\right)\leqq \frac{pcos\alpha +\left(2\beta -pcos\alpha \right)r}{1+r}.$$

(4.6)

Noting that ${\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec g\left(z\right)\left(z\in U\right)$ by Theorem 1 and g(z) is univalent in $U$, we prove the inequality (4.1). Since the subordination (2.1) is sharp for f(z) given by (2.2), we say that the equalities in (4.1) are attained by the function f(z) given by (2.2).   □

Taking α = 0 in Theorem 3, we have

Corollary 3 If $f\left(z\right)\in {\mathcal{S}}_p\left(0,\beta \right)$ , then

$$\frac{p-\left(2\beta -p\right)r}{1-r}\leqq \mathsf{\text{Re}}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)\leqq \frac{p+\left(2\beta -p\right)r}{1+r}$$

(4.7)

for | z | = r < 1. The equalities hold true for

$$f\left(z\right)=\frac{z^p}{{\left(1-z\right)}^{2\left(p-\beta \right)}}.$$

(4.8)

Corollary 4 If $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$ , then we have

$$\frac{pcos\alpha -\left(2\beta -pcos\alpha \right)r}{1-r}\leqq \mathsf{\text{Re}}\left\{e^{i\alpha }\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)\right\}\leqq \frac{pcos\alpha +\left(2\beta -pcos\alpha \right)r}{1+r}$$

(4.9)

for | z | = r < 1. The equalities hold true for f(z) defined by (2.11).

Corollary 5 If $f\left(z\right)\in {\mathcal{C}}_p\left(0,\beta \right)$ , then

$$\frac{p-\left(2\beta -p\right)r}{1-r}\leqq \mathsf{\text{Re}}\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)\leqq \frac{p+\left(2\beta -p\right)r}{1+r}$$

(4.10)

for | z | = r < 1. The equalities hold true for f(z) defined by

$$f^{\prime }\left(z\right)=\frac{p{z}^{p-1}}{{\left(1-z\right)}^{2\left(p-\beta \right)}}.$$

(4.11)

5 Sufficient conditions

We consider some sufficient conditions for f(z) to be in the classes ${\mathcal{S}}_p\left(0,\beta \right)$ and ${\mathcal{C}}_p\left(0,\beta \right)$.

To discuss our sufficient conditions, we have to recall here the following lemma by Miller and Mocanu [6] (also due to Jack [7]).

Lemma 1 Let w(z) be analytic in $U$ with w(0) = 0. If there exists a point $z_0\in U$ such that

$$\underset{\mid z\mid \leqq \mid z_0\mid }{max}\mid w\left(z\right)\mid =\mid w\left(z_0\right)\mid ,$$

(5.1)

then we can write

$$z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right),$$

(5.2)

where k is real and k ≧ 1.

Applying Lemma 1, we derive

Theorem 4 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\mathsf{\text{Re}}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)>\frac{p+\beta }{2\beta }\left(z\in U\right)$$

(5.3)

for some real β > p, then $f\left(z\right)\in {\mathcal{S}}_p\left(0,\beta \right)$.

Proof. Let us define the function w(z) by

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\frac{p+\left(p-2\beta \right)w\left(z\right)}{1-w\left(z\right)}\left(w\left(z\right)\ne 1\right).$$

(5.4)

Then we see that w(z) is analytic in $U$ and w(0) = 0.

It follows from (5.4) that

$$\begin{array}{c}\mathsf{\text{Re}}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)=\mathsf{\text{Re}}\left(1-\frac{\left(p-2\beta \right)z{w}^{\prime }\left(z\right)}{p+\left(p-2\beta \right)w\left(z\right)}-\frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}\right)\\ >\frac{p+\beta }{2\beta }\left(z\in U\right).\end{array}$$

(5.5)

We suppose that there exists a point $z_0\in U$ such that

$$\underset{\mid z\mid \leqq \mid z_0\mid }{max}\mid w\left(z\right)\mid =\mid w\left(z_0\right)\mid =1.$$

Then, Lemma 1 gives us that w(z₀) = e^iθ and z₀w'(z₀) = k e^iθ. For such a point z₀, we have that

$$\begin{array}{c}\mathsf{\text{Re}}\left(\frac{z_0{f}^{\prime }\left(z_0\right)}{f\left(z_0\right)}-\frac{z_0{f}^{″}\left(z_0\right)}{f^{\prime }\left(z_0\right)}\right)=\mathsf{\text{Re}}\left(1-\frac{\left(p-2\beta \right)k{\mathsf{\text{e}}}^{i\theta }}{p+\left(p-2\beta \right){\mathsf{\text{e}}}^{i\theta }}-\frac{k{\mathsf{\text{e}}}^{i\theta }}{1-{\mathsf{\text{e}}}^{i\theta }}\right)\\ =1+\frac{\left(2\beta -p\right)k\left(pcos\theta +p-2\beta \right)}{p^2+{\left(p-2\beta \right)}^2+2p\left(p-2\beta \right)cos\theta }+\frac{k}{2}\\ \leqq 1-\frac{\left(2\beta -p\right)k}{2\beta }+\frac{k}{2}\\ =1-\frac{\left(\beta -p\right)k}{2\beta }\leqq \frac{p+\beta }{2\beta }.\end{array}$$

(5.6)

This contradicts our condition (5.3). Therefore, there is no $z_0\in U$ such that |w (z₀) | = 1. This implies that $\mid w\left(z\right)\mid <1\left(z\in U\right)$, that is, that

$$\left|\frac{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-p}{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}+\left(p-2\beta \right)}\right|<1\left(z\in U\right).$$

(5.7)

Thus, we observe that $f\left(z\right)\in {\mathcal{S}}_p\left(0,\beta \right)$.   □

Further, we derive

Theorem 5 If $f\left(z\right)\in {\mathcal{S}}_p\left(0,\beta \right)$ for some real $\beta \geqq p+\frac{1}{2}$ , then

$$\mathsf{\text{Re}}\left(\frac{z^p}{f\left(z\right)}\right)>\frac{1}{2\beta -2p+1}\left(z\in U\right).$$

(5.8)

Proof. We consider the function w(z) such that

$$\frac{z^p}{f\left(z\right)}=\frac{1+\left(1-2\gamma \right)w\left(z\right)}{1-w\left(z\right)}\left(w\left(z\right)\ne 1\right)$$

(5.9)

for $\gamma =\frac{1}{2\beta -2p+1}$ and for $f\left(z\right)\in {\mathcal{S}}_p\left(0,\beta \right)$.

Then, we know that

$$\mathsf{\text{Re}}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)=\mathsf{\text{Re}}\left(p-\frac{\left(1-2\gamma \right)z{w}^{\prime }\left(z\right)}{1+\left(1-2\gamma \right)w\left(z\right)}-\frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}\right)<\beta$$

(5.10)

for $z\in U$.

Since w(z) is analytic in $U$ and w(0) = 0, we suppose that there exists a point $z_0\in U$ such that

$$\underset{\mid z\mid \leqq \mid z_0\mid }{max}\mid w\left(z\right)\mid =\mid w\left(z_0\right)\mid =1.$$

Then, applying Lemma 1, we can write that w(z₀) = e^iθ and z₀w'(z₀) = k e^iθ (k ≧ 1). This gives us that

$$\begin{array}{c}\mathsf{\text{Re}}\left(\frac{z_0{f}^{\prime }\left(z_0\right)}{f\left(z_0\right)}\right)=\mathsf{\text{Re}}\left(p-\frac{\left(1-2\gamma \right)k{\mathsf{\text{e}}}^{i\theta }}{1+\left(1-2\gamma \right){\mathsf{\text{e}}}^{i\theta }}-\frac{k{\mathsf{\text{e}}}^{i\theta }}{1-{\mathsf{\text{e}}}^{i\theta }}\right)\\ =p+\frac{\left(1-2\gamma \right)k}{2\gamma }+\frac{k}{2}\\ \geqq p+\frac{1-\gamma }{2\gamma }=\beta ,\end{array}$$

(5.11)

which contradicts the inequality (5.10). Thus, there is no point $z_0\in U$ such that |w (z₀) | = 1. This means that $\mid w\left(z\right)\mid <1\left(z\in U\right)$, and that,

$$\mathsf{\text{Re}}\left(\frac{z^p}{f\left(z\right)}\right)>\frac{1}{2\beta -2p+1}\left(z\in U\right).$$

This completes the proof of the theorem.   □

Letting $\frac{z{f}^{\prime }\left(z\right)}{p}$ instead of f(z) in Theorem 5, we have

Corollary 6 If $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$ for some $\beta \geqq p+\frac{1}{2}$ , Then

$$\mathsf{\text{Re}}\left(\frac{p{z}^{p-1}}{f^{\prime }\left(z\right)}\right)>\frac{1}{2\beta -2p+1}\left(z\in U\right).$$

(5.12)

Finally, we consider the coefficient estimates for functions f(z) to be in the classes ${\mathcal{S}}_p\left(\alpha ,\beta \right)$ and ${\mathcal{C}}_p\left(\alpha ,\beta \right)$.

Theorem 6 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\sum _{n=p+1}^{\infty }\left(\mid n{e}^{i\alpha }-k\mid +\mid n{e}^{i\alpha }-\left(2\beta -k\right)\mid \right)\mid a_n\mid \leqq \mid p{e}^{i\alpha }-\left(2\beta -k\right)\mid -\mid p{e}^{i\alpha }-k\mid$$

(5.13)

for some real $\alpha \left(\mid \alpha \mid <\frac{\pi }{2}\right)$, β (β > p cos α), and k (0 ≦ k ≦ p cos α), then $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$

Proof. It is to be noted that if $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\left|\frac{{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-k}{{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\left(2\beta -k\right)}\right|<1\left(z\in U\right),$$

(5.14)

Then $f\left(z\right)\in {\mathcal{S}}_p\left(\alpha ,\beta \right)$. It follows that

$$\begin{array}{c}\left|\frac{{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-k}{{\mathsf{\text{e}}}^{i\alpha }\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\left(2\beta -k\right)}\right|=\left|\frac{{\mathsf{\text{e}}}^{i\alpha }z{f}^{\prime }\left(z\right)-kf\left(z\right)}{{\mathsf{\text{e}}}^{i\alpha }-\left(2\beta -k\right)f\left(z\right)}\right|\\ <\frac{\mid p{\mathsf{\text{e}}}^{i\alpha }-k\mid +\sum _{n=p+1}^{\infty }\mid n{\mathsf{\text{e}}}^{i\alpha }-k\parallel \parallel a_n\mid }{\mid p{\mathsf{\text{e}}}^{i\alpha }-\left(2\beta -k\right)\mid -\sum _{n=p+1}^{\infty }\mid n{\mathsf{\text{e}}}^{i\alpha }-\left(2\beta -k\right)\parallel \parallel a_n\mid }.\end{array}$$

Therefore, if f(z) satisfies the coefficient estimate (5.13), then we know that f(z) satisfies the inequality (5.14). This completes the proof of the theorem.   □

Letting α = 0 and k = p in Theorem 6, we have

Corollary 7 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\sum _{n=p+1}^{\infty }\left(n-\beta \right)\mid a_n\mid \leqq \left(\beta -p\right)$$

for some real $\beta \left(p<\beta <p+\frac{1}{2}\right)$, then $f\left(z\right)\in {\mathcal{S}}_p\left(0,\beta \right)$.

Further, we have

Theorem 7 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\sum _{n=p+1}^{\infty }n\left(\mid n{e}^{i\alpha }-k\mid +\mid n{e}^{i\alpha }-\left(2\beta -k\right)\mid \right)\mid a_n\mid \leqq p\left(\mid p{e}^{i\alpha }-\left(2\beta -k\right)\mid -\mid p{e}^{i\alpha }-k\mid \right)$$

for some real $\alpha \left(\mid \alpha \mid <\frac{\pi }{2}\right)$, β (β > p cos α) and k (0 ≦ k ≦ p cos α), then $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$

Corollary 8 If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\sum _{n=p+1}^{\infty }n\left(n-\beta \right)\mid a_n\mid \leqq p\left(\beta -p\right)$$

for some real $\beta \left(p<\beta <p+\frac{1}{2}\right)$, then $f\left(z\right)\in {\mathcal{C}}_p\left(\alpha ,\beta \right)$.