On Some Sufficient Conditions for a Function to Be p-Valent Starlike

Abstract

A function f analytic in a domain $D\in \mathbb{C}$ is called p-valent in D, if for every complex number w, the equation $f\left(z\right)=w$ has at most p roots in D, so that there exists a complex number $w_0$ such that the equation $f\left(z\right)=w_0$ has exactly p roots in D. The aim of this paper is to establish some sufficient conditions for a function analytic in the unit disc $\mathbb{D}$ to be p-valent starlike in $\mathbb{D}$ or to be at most p-valent in $\mathbb{D}$. Our results are proved mainly by applying Nunokawa’s lemmas.

1. Introduction

A function f analytic in a domain $D\in \mathbb{C}$ is called p-valent in D, if for every complex number w, the equation $f\left(z\right)=w$ has at most p roots in D, so that there exists a complex number $w_0$ such that the equation $f\left(z\right)=w_0$ has exactly p roots in D. We denote by $\mathcal{H}$ the class of functions f which are holomorphic in the open unit unit $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$. Denote by ${\mathcal{A}}_p$, $p\in \mathbb{N}=\{1,2,\dots \}$, the class of functions $f\in \mathcal{H}$ given by

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

(1)

Let $\mathcal{A}={\mathcal{A}}_1$. The well known Noshiro-Warschawski univalence condition, (see [1,2]) indicates that if f is analytic in a convex domain $D\subset \mathbb{C}$ and

$$\mathfrak{Re}\left\{e^{i\theta }{f}^{\prime }\left(z\right)\right\}>0,z\in D,$$

(2)

for some real $\theta$, then f is univalent in D. In [3] Ozaki extended the above result by showing that if f of the form (1) is analytic in a convex domain D and for some real $\theta$ we have

$$\mathfrak{Re}\left\{e^{i\theta }{f}^{\left(p\right)}\left(z\right)\right\}>0,z\in D,$$

then f is at most p-valent in D. In [4] it was proved that if $f\in {\mathcal{A}}_p$, $p\ge 2$, and

$$\left|arg\left\{f^{\left(p\right)}\left(z\right)\right\}\right|<\frac{3\pi }{4},z\in \mathbb{D},$$

then f is at most p-valent in $\mathbb{D}$.

If $f\in \mathcal{H}$ satisfies $f\left(0\right)=0$, $f^{\prime }\left(0\right)=1$ and

$$\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>0,z\in \mathbb{D},$$

then f is said to be starlike with respect to the origin in $\mathbb{D}$ and it is denoted by $f\in {\mathcal{S}}^{\ast }$. It is known that ${\mathcal{S}}^{\ast }\subset \mathcal{S}$, where $\mathcal{S}$ denotes the class of all functions in $\mathcal{A}$ which are univalent in $\mathbb{D}$. Moreover, let ${\mathcal{S}}_p^{\ast }$ and ${\mathcal{C}}_p$ be the subclasses of ${\mathcal{A}}_p$ defined as follows

$$\begin{array}{ccc}\hfill {\mathcal{S}}_p^{\ast }& =& \left\{f\in {\mathcal{A}}_p:\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>0,z\in \mathbb{D}\right\},\hfill \\ \hfill {\mathcal{C}}_p& =& \left\{f\in {\mathcal{A}}_p:z{f}^{\prime }\left(z\right)/p\in {\mathcal{S}}_p^{\ast }\right\}.\hfill \end{array}$$

${\mathcal{S}}_p^{\ast }$ is called the class of p-valent starlike functions and ${\mathcal{C}}_p$ is called the class of p-valent convex functions. Note that ${\mathcal{S}}_1^{\ast }={\mathcal{S}}^{\ast }$ and ${\mathcal{C}}_1=\mathcal{C}$, where ${\mathcal{S}}^{\ast }$ and $\mathcal{C}$ are usual classes of starlike and convex functions, respectively. A function $f\in {\mathcal{A}}_p$ is said to be an element of the class ${\mathcal{K}}_p$ of p-valent close-to-convex functions if there exists a function $g\in {\mathcal{C}}_p$ for which

$$\mathfrak{Re}\left\{\frac{f^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}\right\}>0,z\in \mathbb{D}.$$

(3)

In [5] (Th.1) Umezawa proved the following theorem.

Theorem 1.

If $f\in {\mathcal{K}}_p$, then f is at most p-valent in $\mathbb{D}$.

Because ${\mathcal{C}}_p\subset {\mathcal{S}}_p^{\ast }\subset {\mathcal{K}}_p$, we have from Theorem 1 that p-valent starlike functions and p-valent convex functions are at most p-valent in $\mathbb{D}$ too.

2. Preliminaries

In this paper we need the following lemmas.

Lemma 1

([6] (Th.5)). If $f\in {\mathcal{A}}_p$, then for all $z\in \mathbb{D}$, we have

$$\mathfrak{Re}\left\{\frac{z{f}^{\left(p\right)}\left(z\right)}{f^{(p-1)}\left(z\right)}\right\}>0\Rightarrow \forall k\in \{1,\dots ,p\}:\mathfrak{Re}\left\{\frac{z{f}^{\left(k\right)}\left(z\right)}{f^{(k-1)}\left(z\right)}\right\}>0.$$

Lemma 2

([7]). Let p be an analytic function in $\left|z\right|<1$, with $p\left(0\right)=1$. If there exists a point $z_0$, $|z_0|<1$, such that

$$\mathfrak{Re}\left\{p\left(z\right)\right\}>0for\left|z\right|<|z_0|$$

and

$$p\left(z_0\right)=\pm ia$$

for some $a>0$, then we have

$$\frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}=\frac{2ikarg\left\{p\left(z_0\right)\right\}}{\pi },$$

(4)

for some $k\ge (a+a^{-1})/2\ge 1$.

Corollary 1.

Under the assumptions of Lemma 2, we have from (4)

$$z_0{p}^{\prime }\left(z_0\right)=-ka\le -\frac{1}{2}(a+a^{-1})a=-\frac{1}{2}\left(1+|p\left(z_0\right){|}^2\right).$$

(5)

Lemma 3

([8] (p. 200)). Assume that $q\left(z\right)$ is univalent in $\mathbb{D}$, $q\left(\mathbb{D}\right)$ is a convex set and F, G are analytic in $\mathbb{D}$. If

$$\frac{F^{\prime }\left(z\right)}{G^{\prime }\left(z\right)}\prec q\left(z\right),z\in \mathbb{D},$$

(6)

where G satisfies $G\left(0\right)=F\left(0\right)$ and

$$\mathfrak{Re}\left\{\frac{z{G}^{\prime }\left(z\right)}{G\left(z\right)}\right\}>0,z\in \mathbb{D},$$

then we have

$$\frac{F\left(z\right)}{G\left(z\right)}\prec q\left(z\right),z\in \mathbb{D}.$$

Here ≺ means the subordination.

Corollary 2.

Let $\alpha <1$ be real number. If $f^{(p-1)}\left(z\right),g^{(p-1)}\left(z\right)$ are analytic in $\mathbb{D}$, $f^{(p-1)}\left(0\right)=g^{(p-1)}\left(0\right)$ and

$$\mathfrak{Re}\left\{\frac{f^{\left(p\right)}\left(z\right)}{g^{\left(p\right)}\left(z\right)}\right\}>\alpha ,z\in \mathbb{D},$$

where g satisfies

$$\mathfrak{Re}\left\{\frac{z{g}^{\left(p\right)}\left(z\right)}{g^{(p-1)}\left(z\right)}\right\}>0,z\in \mathbb{D},$$

then we have

$$\mathfrak{Re}\left\{\frac{f^{(p-1)}\left(z\right)}{g^{(p-1)}\left(z\right)}\right\}>\alpha ,z\in \mathbb{D}.$$

3. Main Results

Theorem 2.

Let $f,g\in {\mathcal{A}}_p$. Assume that

$$\mathfrak{Re}\left\{\frac{g\left(z\right)}{z{g}^{\prime }\left(z\right)}\right\}>\beta ,z\in \mathbb{D}$$

(7)

for some β, $0<\beta <1$. If

$$\left|arg\left\{\frac{f^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}\right\}\right|\le \pi -{tan}^{-1}\left\{\frac{2(1-\beta )\left|z\right|+1-{\left|z\right|}^2}{\beta (1-|z{|}^2)}\right\},z\in \mathbb{D},$$

(8)

then we have

$$\mathfrak{Re}\left\{\frac{f\left(z\right)}{g\left(z\right)}\right\}>0,z\in \mathbb{D}.$$

(9)

Proof. 

If we put

$$q\left(z\right)=\frac{f\left(z\right)}{g\left(z\right)},q\left(0\right)=1,$$

then it follows that

$$f\left(z\right)=q\left(z\right)g\left(z\right),f^{\prime }\left(z\right)=g^{\prime }\left(z\right)q\left(z\right)+q^{\prime }\left(z\right)g\left(z\right),$$

and

$$\frac{f^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}=q\left(z\right)+q^{\prime }\left(z\right)\frac{g\left(z\right)}{g^{\prime }\left(z\right)}=q\left(z\right)+z{q}^{\prime }\left(z\right)\frac{g\left(z\right)}{z{g}^{\prime }\left(z\right)}.$$

If there exists a point $z_0\in \mathbb{D}$, such that

$$\mathfrak{Re}\left\{q\left(z\right)\right\}>0,\left(\right|z|<|z_0|<1)$$

and

$$\mathfrak{Re}\left\{q\left(z_0\right)\right\}=0,$$

then by (5), we have

$$z_0{q}^{\prime }\left(z_0\right)=\mathfrak{Re}\left\{z_0{q}^{\prime }\left(z_0\right)\right\}\le -\frac{1}{2}(1+|q\left(z_0\right){|}^2)<0.$$

(10)

This shows that $z_0{q}^{\prime }\left(z_0\right)$ is a negative real number. Furthermore, by (4), we have

$$\left|\frac{q\left(z_0\right)}{z_0{q}^{\prime }\left(z_0\right)}\right|\le 1.$$

(11)

Then it follows that

$$\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}=q\left(z_0\right)+z_0{q}^{\prime }\left(z_0\right)\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}=\pm ia+z_0{q}^{\prime }\left(z_0\right)\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)},$$

where $q\left(z_0\right)=\pm ai$, $a>0$ and (7), (10) give

$$\begin{array}{ccc}\mathfrak{Re}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}\hfill & =\hfill & \mathfrak{Re}\left\{z_0{q}^{\prime }\left(z_0\right)\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}=z_0{q}^{\prime }\left(z_0\right)\mathfrak{Re}\left\{\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}\hfill \\ \hfill & <\hfill & -\frac{\beta }{2}(1+a^2)<0.\hfill \end{array}$$

Next, we have

$$\mathfrak{Im}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}=\mathfrak{Im}\left\{\pm ia+z_0{q}^{\prime }\left(z_0\right)\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}=\pm a+z_0{q}^{\prime }\left(z_0\right)\mathfrak{Im}\left\{\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}.$$

We will consider the four cases:

(i) 

$arg\left\{q\left(z_0\right)\right\}=\pi /2$ (i.e., $q\left(z_0\right)=ia$, $a>0$) and $\mathfrak{Im}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}\ge 0$,

(ii) 

$arg\left\{q\left(z_0\right)\right\}=\pi /2$ (i.e., $q\left(z_0\right)=ia$, $a>0$) and $\mathfrak{Im}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}<0$,

(iii) 

$arg\left\{q\left(z_0\right)\right\}=-\pi /2$ (i.e., $q\left(z_0\right)=-ia$, $a>0$) and $\mathfrak{Im}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}\ge 0$,

(iv) 

$arg\left\{q\left(z_0\right)\right\}=-\pi /2$ (i.e., $q\left(z_0\right)=-ia$, $a>0$) and $\mathfrak{Im}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}<0$.

Let us put

$$G\left(z\right)=\frac{pg\left(z\right)}{z{g}^{\prime }\left(z\right)},G\left(0\right)=1.$$

Then from the hypothesis, we have

$$\frac{G\left(z\right)-\beta }{1-\beta }\prec \frac{1+z}{1-z},z\in \mathbb{D},$$

and so we have

$$G\left(z\right)\prec \beta +(1-\beta )\frac{1+z}{1-z},z\in \mathbb{D},$$

and so

$$\left|\mathfrak{Im}\left\{G\right(z\left)\right\}\right|=\left|\mathfrak{Im}\frac{pg\left(z\right)}{z{g}^{\prime }\left(z\right)}\right|\le (1-\beta )\frac{2\left|z\right|}{1-{\left|z\right|}^2},z\in \mathbb{D}.$$

(12)

In the case (i) we have $arg\left\{q\left(z_0\right)\right\}=\pi /2$, $q\left(z_0\right)=ia$, $a>0$ and

$$\mathfrak{Im}\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}=\left|q\left(z_0\right)\right|+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\left\{\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}\right)\ge 0.$$

Therefore, we have

$$\begin{array}{ccc}arg\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}\hfill & =\hfill & arg\left[z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Re}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right)+i\left\{|q\left(z_0\right)|+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right)\right\}\right]\hfill \\ \hfill & =\hfill & \pi -{tan}^{-1}\left\{\frac{|q\left(z_0\right)|+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right)}{-z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Re}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right)}\right\}\hfill \\ \hfill & >\hfill & \pi -{tan}^{-1}\left\{\frac{|q\left(z_0\right)|+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right)}{-\beta z_0{q}^{\prime }\left(z_0\right)}\right\}\hfill \\ \hfill & =\hfill & \pi -{tan}^{-1}\left\{-\frac{\left|q\right(z_0\left)\right|}{\beta z_0{q}^{\prime }\left(z_0\right)}-\frac{\mathfrak{Im}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}}{\beta }\right\}\hfill \\ \hfill & \ge \hfill & \pi -{tan}^{-1}\left\{\left|\frac{q\left(z_0\right)}{\beta z_0{q}^{\prime }\left(z_0\right)}\right|+\left|\frac{\mathfrak{Im}\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}}{\beta }\right|\right\}.\hfill \end{array}$$

Then, by (11) and (12), we have

$$\begin{array}{ccc}arg\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}\hfill & >\hfill & \pi -{tan}^{-1}\left\{\frac{1}{\beta }+\frac{2(1-\beta )|z_0|}{(1-|z_0{|}^2)\beta }\right\}\hfill \\ \hfill & =\hfill & \pi -{tan}^{-1}\left[\frac{1}{\beta (1-|z_0{|}^2)}\left\{2(1-\beta )|z_0|+1-|z_0{|}^2\right\}\right].\hfill \end{array}$$

This contradicts hypothesis (8). In the case (ii) when $arg\left\{q\left(z_0\right)\right\}=\pi /2$, $q\left(z_0\right)=ia$, $a>0$, and

$$arg\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}=q\left(z_0\right)+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\left\{\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}\right)<0$$

applying the same method as the above, we have

$$\begin{array}{ccc}\hfill arg\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}& <& -\pi +{tan}^{-1}\left[\frac{1}{\beta (1-|z_0{|}^2)}\left\{2(1-\beta )|z_0|+1-|z_0{|}^2\right\}\right].\hfill \end{array}$$

This is also a contradiction. In the case (iii) when $arg\left\{q\left(z_0\right)\right\}=-\pi /2$, $q\left(z_0\right)=-ia$, $a>0$, and

$$q\left(z_0\right)+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\left\{\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}\right)>0$$

and in the case (iv) when $arg\left\{q\left(z_0\right)\right\}=-\pi /2$, $q\left(z_0\right)=-ia$, $a>0$, and

$$q\left(z_0\right)+z_0{q}^{\prime }\left(z_0\right)\left(\mathfrak{Im}\left\{\frac{g\left(z_0\right)}{z_0{g}^{\prime }\left(z_0\right)}\right\}\right)<0,$$

applying the same method as in the proof of case (i) gives

$$\begin{array}{ccc}\hfill \left|arg\left\{\frac{f^{\prime }\left(z_0\right)}{g^{\prime }\left(z_0\right)}\right\}\right|& >& \pi -{tan}^{-1}\left[\frac{1}{\beta (1-|z_0{|}^2)}\left\{2(1-\beta )|z_0|+1-|z_0{|}^2\right\}\right].\hfill \end{array}$$

This is a contradiction. This completes the proof. □

Inequalities (9) and (2) show that the assumptions of Theorem 2 are sufficient for

$${\int }_0^z\frac{f\left(\zeta \right)}{g\left(\zeta \right)}\mathrm{d}\zeta$$

to be univalent in $\mathbb{D}$.

Theorem 3.

Let $F,G\in {\mathcal{A}}_p$. Assume that there exist a positive integer k, $2\le k\le p$ and a real β, $0<\beta <1$, for which

$$\left|arg\left\{\frac{F^{\left(k\right)}\left(z\right)}{G^{\left(k\right)}\left(z\right)}\right\}\right|<\pi -{tan}^{-1}\left\{\frac{2(1-\beta )\left|z\right|+1-{\left|z\right|}^2}{\beta (1-|z{|}^2)}\right\},z\in \mathbb{D},$$

where G satisfies

$$\mathfrak{Re}\left\{\frac{G^{(k-1)}\left(z\right)}{z{G}^{\left(k\right)}\left(z\right)}\right\}>\beta ,z\in \mathbb{D}.$$

(13)

Then

$$\forall n\in \{1,\dots ,k-1\}:\mathfrak{Re}\left\{\frac{F^{\left(n\right)}\left(z\right)}{G^{\left(n\right)}\left(z\right)}\right\}>0,z\in \mathbb{D}.$$

(14)

and $F\in {\mathcal{K}}_p$, F is at most p-valent in $\mathbb{D}$.

Proof. 

If we put $f=F^{(k-1)}$ and $g=G^{(k-1)}$ in Theorem 2 we immediately obtain

$$\mathfrak{Re}\left\{\frac{F^{(k-1)}\left(z\right)}{G^{(k-1)}\left(z\right)}\right\}>0,z\in \mathbb{D}.$$

(15)

Then, by Lemma 1, we obtain (14). For $n=1$ the condition (14) is of the form

$$\mathfrak{Re}\left\{\frac{F^{\prime }\left(z\right)}{G^{\prime }\left(z\right)}\right\}>0,z\in \mathbb{D},$$

where G satisfies (13). Therefore, by Lemma 1 we have also

$$\mathfrak{Re}\left\{\frac{z{G}^{\prime }\left(z\right)}{G\left(z\right)}\right\}>0,z\in \mathbb{D},$$

which by (3) implies $F\in {\mathcal{K}}_p$. By Theorem 1, F is at most p-valent in $\mathbb{D}$. □

Theorem 4.

Assume that $f\in {\mathcal{A}}_p$, $2\le p$, and that there exists a positive integer k, $2\le k\le p$ for which

$$\mathfrak{Re}\left\{\frac{z{f}^{\left(k\right)}\left(z\right)}{f^{(k-1)}\left(z\right)}\right\}>-1,z\in \mathbb{D},$$

then we have

$$\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>0,z\in \mathbb{D},$$

or f is p-valent starlike in $\mathbb{D}$.

Proof. 

Let us put

$$q_1\left(z\right)=\frac{1}{p-k+2}\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)},q_1\left(0\right)=1.$$

(16)

By (16) we have

$$\frac{z{q}_1^{\prime }\left(z\right)}{q_1\left(z\right)}=1+\frac{z{f}^{\left(k\right)}\left(z\right)}{f^{(k-1)}\left(z\right)}-\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)}$$

and so

$$1+\frac{z{f}^{\left(k\right)}\left(z\right)}{f^{(k-1)}\left(z\right)}=\frac{z{q}_1^{\prime }\left(z\right)}{q_1\left(z\right)}+(p-k+2)q_1\left(z\right).$$

By the hypothesis, we have

$$1+\mathfrak{Re}\left\{\frac{z{f}^{\left(k\right)}\left(z\right)}{f^{(k-1)}\left(z\right)}\right\}=\mathfrak{Re}\left\{\frac{z{q}_1^{\prime }\left(z\right)}{q_1\left(z\right)}+(p-k+2)q_1\left(z\right)\right\}>0,z\in \mathbb{D}.$$

(17)

If there exists a point $z_1\in \mathbb{D}$, such that

$$\mathfrak{Re}\left\{q_1\left(z\right)\right\}>0,\left(\right|z|<|z_1|<1)$$

and

$$\mathfrak{Re}\left\{q_1\left(z_1\right)\right\}=0,$$

then by Lemma 2, we have

$$\mathfrak{Re}\left\{\frac{z_1{q}_1^{\prime }\left(z_1\right)}{q_1\left(z_1\right)}\right\}=0,\frac{z_1{q}_1^{\prime }\left(z_1\right)}{q_1\left(z_1\right)}=i{k}_1$$

for some real $k_1$, $|k_1|\ge 1$. This gives

$$1+\mathfrak{Re}\left\{\frac{z{f}^{\left(k\right)}\left(z_1\right)}{f^{(k-1)}\left(z_1\right)}\right\}=\mathfrak{Re}\left\{\frac{z_1{q}_1^{\prime }\left(z_1\right)}{q_1\left(z_1\right)}+(p-k+2)q_1\left(z_1\right)\right\}=0.$$

It is contrary to inequality (17) and therefore, we have

$$\mathfrak{Re}\left\{\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)}\right\}>\mathfrak{Re}\left\{\frac{1}{p-k+2}\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)}\right\}=\mathfrak{Re}\left\{q_1\left(z\right)\right\}>0,z\in \mathbb{D}.$$

(18)

Next, let us put

$$q_2\left(z\right)=\frac{1}{p-k+3}\frac{z{f}^{(k-2)}\left(z\right)}{f^{(k-3)}\left(z\right)},q_2\left(0\right)=1,$$

then it follows that

$$\frac{z{q}_2^{\prime }\left(z\right)}{q_2\left(z\right)}=1+\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)}-\frac{z{f}^{(k-2)}\left(z\right)}{f^{(k-3)}\left(z\right)}$$

and so

$$1+\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)}=\frac{z{q}_2^{\prime }\left(z\right)}{q_2\left(z\right)}+(p-k+3)q_2\left(z\right).$$

(19)

By (18) and (19), we have

$$1+\mathfrak{Re}\left\{\frac{z{f}^{(k-1)}\left(z\right)}{f^{(k-2)}\left(z\right)}\right\}=\mathfrak{Re}\left\{\frac{z{q}_2^{\prime }\left(z\right)}{q_2\left(z\right)}+(p-k+3)q_2\left(z\right)\right\}>0,z\in \mathbb{D}.$$

(20)

If there exists a point $z_2\in \mathbb{D}$, such that

$$\mathfrak{Re}\left\{q_2\left(z\right)\right\}>0,\left(\right|z|<|z_2|<1)$$

and

$$\mathfrak{Re}\left\{q_2\left(z_2\right)\right\}=0,$$

then by Lemma 2, we have

$$\mathfrak{Re}\left\{\frac{z_2{q}_2^{\prime }\left(z_2\right)}{q_2\left(z_2\right)}\right\}=0,\frac{z_2{q}_2^{\prime }\left(z_2\right)}{q_2\left(z_2\right)}=i{k}_2$$

for some real $k_2$, $|k_2|\ge 1$. Then, we have

$$1+\mathfrak{Re}\left\{\frac{z_2{f}^{(k-1)}\left(z_2\right)}{f^{(k-2)}\left(z_2\right)}\right\}=\mathfrak{Re}\left\{\frac{z_2{q}_2^{\prime }\left(z_2\right)}{q_2\left(z_2\right)}+(p-k+3)q_2\left(z_2\right)\right\}=0.$$

It is contrary to (20) and therefore, we have

$$\mathfrak{Re}\left\{\frac{z{f}^{(k-2)}\left(z\right)}{f^{(k-3)}\left(z\right)}\right\}=\mathfrak{Re}\left\{(p-k+3)q_2\left(z\right)\right\}>0,z\in \mathbb{D}.$$

Applying the same method many times in succession we are able to obtain

$$\mathfrak{Re}\left\{\frac{z{f}^{(k-3)}\left(z\right)}{f^{(k-4)}\left(z\right)}\right\}>0,\mathfrak{Re}\left\{\frac{z{f}^{(k-4)}\left(z\right)}{f^{(k-5)}\left(z\right)}\right\}>0\dots ,\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>0,z\in \mathbb{D}$$

This shows that f is p-valent starlike in $\mathbb{D}$. □

For some related conditions for starlikeness we refer to our papers [9,10].