New Sufficient Conditions for p-Valent Functions

Abstract

A function that is holomorphic in a set E, in the complex plane, is called p-valent in E if, for every complex number, w, the equation f (z) = w has, at most, p roots in E. In this paper, we established some sufficient conditions for holomorphic functions in the unit disk |z| < 1 to be at most p-valent in the unit disk or p-valent or p-valent starlike in the unit disk.

1. Introduction

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

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], S. 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

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

and then, f is at most p-valent in D. Applying Ozaki’s theorem, we find that, if $f\in {\mathcal{A}}_p$ and

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

(3)

then f is at most p-valent in $\mathbb{D}$. Let $\mathcal{A}={\mathcal{A}}_1$. If $f\in \mathcal{A}$ satisfies $f\left(0\right)=0$, $f^{\prime }\left(0\right)=1$, and

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

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

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

${\mathcal{S}}_p^{*}$ is called the class of p-valent starlike functions, and ${\mathcal{C}}_p$ is called the class of p-valent convex functions. Note that ${\mathcal{S}}_1^{*}={\mathcal{S}}^{*}$ and ${\mathcal{C}}_1=\mathcal{C}$, where ${\mathcal{S}}^{*}$ 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

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

(4)

In [4] (Theorem 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 ${\mathcal{C}}_p\subset {\mathcal{S}}_p^{*}\subset {\mathcal{K}}_p$, we have from Theorem 1 the condition that p-valent starlike functions and p-valent convex functions are at most p-valent in $\mathbb{D}$, too.

In [5], it was proven that, if $F,G\in {\mathcal{A}}_p$ and there exists a positive integer, k, $2\le k\le p$, and a real $\beta$, $0<\beta <1$, for which

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

where G satisfies

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

and then

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

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

2. Materials and Methods

In this paper, we need the following lemmas. In [6], Mocanu proved the following result.

Lemma 1.

Let $f\left(z\right)=z+{\sum }_{k=n+1}^{\infty }{a}_k{z}^k$ be analytic in $\mathbb{D}$, and suppose that

$$|arg{f}^{\prime }\left(z\right)|<{\alpha }_n{\displaystyle \frac{\pi }{2}},(z\in \mathbb{D}),$$

(5)

where ${\alpha }_n$ is the unique root of the equation

$$2{tan}^{-1}\left[n(1-\alpha )\right]+\pi (1-2\alpha )=0.$$

(6)

Then, $f\in {\mathcal{S}}^{*}$ or

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

Lemma 2

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

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

Lemma 3

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

$$|arg\left\{p\left(z\right)\right\}|<\pi \beta \mathrm{for}|z|<|z_0|$$

and

$$|arg\left\{p\left(z_0\right)\right\}|=\pi \beta$$

for some $0<\beta$, with

$${\left\{p\left(z_0\right)\right\}}^{1/\left(2\beta \right)}=\pm ia,anda>0,$$

and then we have

$${\displaystyle \frac{z_0{p}^{\prime }\left(z_0\right)}{p\left(z_0\right)}}=2ik\beta ,$$

where

$$k\ge {\displaystyle \frac{1}{2}}\left(a+{\displaystyle \frac{1}{a}}\right)whenarg\left\{p\left(z_0\right)\right\}=\pi \beta$$

and

$$k\le -{\displaystyle \frac{1}{2}}\left(a+{\displaystyle \frac{1}{a}}\right)whenarg\left\{p\left(z_0\right)\right\}=-\pi \beta .$$

In [6], Mocanu proved the following result.

Lemma 4.

Let $f\left(z\right)=z+{\sum }_{k=n}^{\infty }{a}_k{z}^k$ be analytic in $\mathbb{D}$, and suppose that

$$|f^{\prime }\left(z\right)-1|<{\displaystyle \frac{n+1}{\sqrt{{(n+1)}^2+1}}},(z\in \mathbb{D}),$$

(7)

or suppose that

$$|f^{\prime \prime }\left(z\right)|<{\displaystyle \frac{n(n+1)}{\sqrt{{(n+1)}^2+1}}},(z\in \mathbb{D}),$$

(8)

and then $f\in {\mathcal{S}}^{*}$ or

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

In this paper, we prove several sufficient conditions for a function, g, to be p-valent in $\mathbb{D}$. These symmetric results are under several different assumptions, like the bounds on $|arg\{g^{\left(p\right)}\left(z\right)\left\}\right|$ or $\left|{\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}-1\right|$. The results obtained in the paper are new, and they do not constitute a direct improvement of some existing results. Also, in our opinion, these sufficient conditions for g to be p-valent are not sharp. In this paper, we note some related results in this topic that are symmetric in some sense.

3. Results

Theorem 2.

Let $g\left(z\right)=z^p+{\sum }_{k=p+n}^{\infty }{a}_k{z}^k$ be analytic in $\mathbb{D}$, where $p,n$ are some positive integers, and suppose that

$$\left|arg\left\{{\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}\right\}\right|<{\alpha }_n{\displaystyle \frac{\pi }{2}},(z\in \mathbb{D}),$$

(9)

where ${\alpha }_n$ is the unique root of Equation (6). Then, we have the case in which g is p-valent starlike and

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

for $k=1,2,\dots ,p$, and so, for the case $k=1$, we have $g\in {\mathcal{S}}^{*}$.

Proof. 

If we assume

$$f\left(z\right)={\displaystyle \frac{g^{(p-1)}\left(z\right)}{p!}}=z+{\displaystyle \frac{(p+n)!}{p!(n+1)!}}{a}_{p+n}{z}^{n+1}+\dots ,$$

(10)

then

$$f^{\prime }\left(z\right)={\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}=1+{\displaystyle \frac{(p+n)!}{p!n!}}{a}_{p+n}{z}^n+\dots .$$

Therefore, from (9), the function f satisfies (5). Then, from Lemma 1, we have

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

We have also

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right\}=\mathfrak{Re}\left\{{\displaystyle \frac{z{g}^{\left(p\right)}\left(z\right)}{g^{(p-1)}\left(z\right)}}\right\},(z\in \mathbb{D})$$

and so, we have

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

(11)

thus, g is p-valent starlike. Inequality (11) and Nunokawa’s Lemma 2 imply that

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

for $k=1,2,\dots ,p-1$, too, so, for the case $k=1$, we have $g\in {\mathcal{S}}^{*}$. □

In (9), we assumed some bounds for $|arg\{g^{\left(p\right)}\left(z\right)\left\}\right|$. In the next theorem, we will obtain a symmetric result but under some assumptions on $\left|{\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}-1\right|$.

Theorem 3.

Let

$$g\left(z\right)=z^p+\sum _{k=n+p}^{\infty }{a}_k{z}^k$$

be analytic in $\mathbb{D}$, where p and n are some positive integers, and suppose that

$$\left|{\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}-1\right|<{\displaystyle \frac{n+1}{\sqrt{{(n+1)}^2+1}}},(z\in \mathbb{D}),$$

(12)

or suppose that

$$\left|{\displaystyle \frac{g^{(p+1)}\left(z\right)}{p!}}\right|<{\displaystyle \frac{n(n+1)}{\sqrt{{(n+1)}^2+1}}},(z\in \mathbb{D}),$$

(13)

Then, g is p-valent starlike and

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

(14)

for $k=1,2,\dots ,p$, and so, for the case $k=1$, we have $g\in {\mathcal{S}}^{*}$.

Proof. 

Assume again that

$$f\left(z\right)={\displaystyle \frac{g^{(p-1)}\left(z\right)}{p!}}=z+{\displaystyle \frac{(p+n)!}{p!(n+1)!}}{a}_{p+n}{z}^{n+1}+\dots .$$

If condition (12) holds, then we have

$$|f^{\prime }\left(z\right)-1|=\left|{\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}-1\right|<{\displaystyle \frac{n+1}{\sqrt{{(n+1)}^2+1}}},(z\in \mathbb{D}).$$

Therefore, through Lemma 4, we have

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

(15)

If condition (13) holds, then we have

$$|f^{\prime \prime }\left(z\right)|=\left|{\displaystyle \frac{g^{(p+1)}\left(z\right)}{p!}}\right|<{\displaystyle \frac{n(n+1)}{\sqrt{{(n+1)}^2+1}}},(z\in \mathbb{D}),$$

and so, again, through Lemma 4, we have (15). Now, inequality (15), together with Nunokawa’s Lemma 2, imply that, for all $k=1,2,\dots ,p-1$

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

holds. In the special case $k=1$, we have $g\in {\mathcal{S}}^{*}$. □

For example, if $p=3$, $n=2$, and $g_3\left(z\right)=z^3+a_5{z}^5$ with $|a_5|\le 1/\left[20\sqrt{10}\right]$, then $g_3\left(z\right)$ satisfies (12), so, according to Theorem 3, the function $g_3$ satisfies (14) with $p=3$ and $n=2$.

Theorem 4.

Let $g\left(z\right)=z^p+{\sum }_{n=p+1}^{\infty }{a}_n{z}^n$ be analytic in $\mathbb{D}$, where p is a positive integer, and suppose that

$$\left|{\displaystyle \frac{1}{p!}}\left(g^{\left(p\right)}\left(z\right)+\delta z{g}^{(p-1)}\left(z\right)\right)-1\right|<1,(z\in \mathbb{D}),$$

(16)

for some $\delta >1/2$. Then, we have

$$\left|{\displaystyle \frac{z{g}^{\left(p\right)}\left(z\right)}{g^{(p-1)}\left(z\right)}}-1\right|<1,(z\in \mathbb{D}).$$

(17)

Proof. 

Recall that [9] Mocanu proved that, if f is of the form $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n$ in $\mathbb{D}$ and

$$\left|\left(f^{\prime \prime }\left(z\right)+\delta z{f}^{\prime }\left(z\right)\right)-1\right|<1,(z\in \mathbb{D}),$$

(18)

for some $\delta >1/2$, then we have

$$\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right|<1,(z\in \mathbb{D}).$$

(19)

Assume that

$$f\left(z\right)={\displaystyle \frac{g^{(p-1)}\left(z\right)}{p!}}=z+\dots .$$

With this notation, condition (16) becomes (18), and through (19), we obtain (17). □

Theorem 5.

Let $g\left(z\right)=z^p+{\sum }_{n=p+1}^{\infty }{a}_n{z}^n$ be analytic in $\mathbb{D}$, where p is a positive integer, and suppose that

$$\left|\left((1-\alpha )g^{(p-1)}\left(z\right)/z+\alpha g^{\left(p\right)}\left(z\right)\right)-p!\right|<M\left(\alpha \right)p!,(z\in \mathbb{D}),$$

(20)

for some $\alpha >1/2$, and

$$M\left(\alpha \right)=(\alpha +1)\sqrt{(2\alpha -1)/({\alpha }^2+4\alpha )}.$$

Then, we have the case in which g is p-valent starlike and

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

(21)

for $k=1,2,\dots ,p$, and so, for the case $p=1$, we have $g\in {\mathcal{S}}^{*}$.

Proof. 

In [9], Mocanu proved that, if f has the form $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n$ in $\mathbb{D}$ and

$$\left|\left((1-\alpha )f\left(z\right)/z+\alpha f^{\prime }\left(z\right)\right)-1\right|<M\left(\alpha \right),(z\in \mathbb{D}),$$

(22)

for some $\alpha >1/2$, then we have

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

(23)

Assume that

$$f\left(z\right)={\displaystyle \frac{g^{(p-1)}\left(z\right)}{p!}}=z+\dots .$$

With this notation, condition (20) becomes (22), and through (23), we obtain

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

Applying Nunokawa’s Lemma 2 gives (21). □

If we consider, for example, the case $\alpha =3/2$, $p=3$, and the function $g_4\left(z\right)=z^3+a_4{z}^4$ with $|a_4|\le \sqrt{2/33}$, then $g_4$ satisfies (21) because, in this case, condition (20) becomes

$$|a_{4}z|<\sqrt{2/33},M(3/2)=5\sqrt{2/33}$$

in $\mathbb{D}$. This gives $|a_4|\le \sqrt{2/33}$. For $\alpha =1$, Theorem 5 becomes the following corollary.

Corollary 1.

Let $g\left(z\right)=z^p+{\sum }_{n=p+1}^{\infty }{a}_n{z}^n$ be analytic in $\mathbb{D}$, where p is a positive integer, and suppose that

$$\left|g^{\left(p\right)}\left(z\right)-p!\right|<{\displaystyle \frac{2p!}{\sqrt{5}}},(z\in \mathbb{D}),$$

Then, we have the case in which g is p-valent starlike and

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

for $k=1,2,\dots ,p$, and so, for the case $p=1$, we have $g\in {\mathcal{S}}^{*}$.

Theorem 6.

If $f\in {\mathcal{A}}_p$ and

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

(24)

for some $\alpha \in (0,1]$, and

$${\displaystyle \frac{f^{(p-1)}\left(z\right)}{z}}\ne 0,(z\in \mathbb{D}),$$

then

$$\left|arg\left\{{\displaystyle \frac{f^{(p-1)}\left(z\right)}{z}}\right\}\right|<\beta \pi <\alpha \pi ,(z\in \mathbb{D}),$$

where β is the unique root of the equation

$$\beta \pi +{tan}^{-1}\left(2\beta \right)=\alpha \pi .$$

(25)

Proof. 

Consider the function

$$s\left(z\right)={\displaystyle \frac{f^{(p-1)}\left(z\right)}{z}},(z\in \mathbb{D}),$$

which satisfies

$$f^{\left(p\right)}\left(z\right)=s\left(z\right)+z{s}^{\prime }\left(z\right),(z\in \mathbb{D}).$$

(26)

We want to show that $|arg\left\{s\left(z\right)\right\}|<\beta \pi$ in the unit disk>>> $\mathbb{D}$. If not, then there exists a point $z_0\in \mathbb{D}$, such that

$$|arg\left\{s\left(z\right)\right\}|<\beta \pi ,(\left|z\right|<|z_0\left|\right)$$

and

$$|arg\left\{s\left(z_0\right)\right\}|=\beta \pi .$$

From Lemma 3, we have for this case

$${\displaystyle \frac{z_0{s}^{\prime }\left(z_0\right)}{s\left(z_0\right)}}=2ik\beta ,$$

where

$$k\ge {\displaystyle \frac{a^2+1}{2a}}\ge 1\mathrm{when}arg\left\{s\left(z_0\right)\right\}=\pi \beta$$

and

$$k\le -{\displaystyle \frac{a^2+1}{2a}}\le -1\mathrm{when}arg\left\{s\left(z_0\right)\right\}=-\pi \beta ,$$

for some $a>0$ such that

$${\left\{s\left(z_0\right)\right\}}^{1/\left(2\beta \right)}=\pm ia,\mathrm{and}a>0.$$

Therefore, from (26), for the case when $arg\left\{s\left(z_0\right)\right\}=\pm \pi \beta$, we have

$$\begin{array}{ccc}\hfill \left|arg\left\{f^{\left(p\right)}\left(z_0\right)\right\}\right|& =& \left|arg\left\{s\left(z_0\right)\left(1+{\displaystyle \frac{z_0{s}^{\prime }\left(z_0\right)}{s\left(z_0\right)}}\right)\right\}\right|\hfill \\ & =& \left|arg\left\{s\left(z_0\right)\right\}+arg\left\{1+{\displaystyle \frac{z_0{s}^{\prime }\left(z_0\right)}{s\left(z_0\right)}}\right\}\right|\hfill \\ & =& \left|\beta \pi +arg\left\{1+{\displaystyle \frac{z_0{s}^{\prime }\left(z_0\right)}{s\left(z_0\right)}}\right\}\right|\hfill \\ & =& \left|\beta \pi +arg\left\{1+2i\beta k\right\}\right|\hfill \\ & \ge & \beta \pi +{tan}^{-1}\left(2\beta \right)\hfill \\ & =& \alpha \pi \hfill \end{array}$$

because of (25), but this contradicts the hypothesis (24). This shows that

$$\left|arg\left\{s\left(z\right)\right\}\right|=\left|{\displaystyle \frac{f^{(p-1)}\left(z\right)}{z}}\right|<\beta \pi ,(z\in \mathbb{D}).$$

□

If we take $\alpha =1/2$ then Theorem 6 becomes the following corollary.

Corollary 2.

If $f\in {\mathcal{A}}_p$ and

$$\left|arg\left\{f^{\left(p\right)}\left(z\right)\right\}\right|<{\displaystyle \frac{\pi }{2}},(z\in \mathbb{D})$$

and

$${\displaystyle \frac{f^{(p-1)}\left(z\right)}{z}}\ne 0,(z\in \mathbb{D}),$$

then

$$\left|arg\left\{{\displaystyle \frac{f^{(p-1)}\left(z\right)}{z}}\right\}\right|<\beta \pi <{\displaystyle \frac{\pi }{2}},(z\in \mathbb{D}),$$

where $\beta =0.319161\dots$, is the unique root of the equation

$$\beta \pi +{tan}^{-1}\left(2\beta \right)={\displaystyle \frac{\pi }{2}}.$$

Theorem 7.

Let $f\in {\mathcal{A}}_p$, $p\ge 2$, $0\ge \alpha <1$, and let it satisfy the following condition

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\left(p\right)}\left(z\right)}{f^{(p-1)}\left(z\right)}}\right\}>{\displaystyle \frac{2{\alpha }^2+3\alpha -1}{2(1+\alpha )}},(z\in \mathbb{D}),$$

(27)

Then we have

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

(28)

Proof. 

Let $f\in {\mathcal{A}}_p$, $p\ge 2$. Assume for a moment that $0\ge \alpha <1$ and

$$p\left(z\right)={\displaystyle \frac{z{f}^{(p-1)}\left(z\right)}{2f^{(p-2)}\left(z\right)}}\ne \alpha (z\in \mathbb{D}).$$

(29)

Then the function

$$w\left(z\right)={\displaystyle \frac{1-p\left(z\right)}{\alpha -p\left(z\right)}}(z\in \mathbb{D})$$

is analytic in $\mathbb{D}$ with $w\left(0\right)=0$. It is known from [10], that if $w\left(z\right)$ is a non-constant and analytic function in the unit disk $\mathbb{D}$ with $w\left(0\right)=0$ and if $\left|w\right(z\left)\right|$ attains its maximum value 1 on the circle $\left|z\right|=r$ at the point $z_0$, then $z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right)$ and $k\ge 1$. Therefore, we have also

$$p\left(z\right)={\displaystyle \frac{z{f}^{(p-1)}\left(z\right)}{2f^{(p-2)}\left(z\right)}}={\displaystyle \frac{1-\alpha w\left(z\right)}{1-w\left(z\right)}},\left|w\left(z_0\right)\right|=1$$

(30)

so $\mathfrak{Re}\left\{2p\left(z_0\right)-1\right\}=\alpha$, which gives the following equation

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

Then $z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right)$ and $k\ge 1$ follow that

$$\begin{array}{ccc}\hfill & & \mathfrak{Re}\left\{{\displaystyle \frac{z_0{f}^{\left(p\right)}\left(z_0\right)}{f^{(p-1)}\left(z_0\right)}}\right\}\hfill \\ & =& \mathfrak{Re}\left\{-{\displaystyle \frac{\alpha z_0{w}^{\prime }\left(z_0\right)}{1-\alpha w\left(z_0\right)}}+{\displaystyle \frac{z_0{w}^{\prime }\left(z_0\right)}{1-w\left(z_0\right)}}+2p\left(z_0\right)-1\right\}\hfill \\ & =& \mathfrak{Re}\left\{k{\displaystyle \frac{-\alpha w\left(z_0\right)}{1-\alpha w\left(z_0\right)}}\right\}+\mathfrak{Re}\left\{{\displaystyle \frac{kw\left(z_0\right)}{1-w\left(z_0\right)}}\right\}+\mathfrak{Re}\left\{2p\left(z_0\right)-1\right\}\hfill \\ & =& \mathfrak{Re}\left\{-{\displaystyle \frac{k}{2}}{\displaystyle \frac{1+\alpha w\left(z_0\right)}{1-\alpha w\left(z_0\right)}}\right\}+{\displaystyle \frac{k}{2}}+\mathfrak{Re}\left\{{\displaystyle \frac{kw\left(z_0\right)}{1-w\left(z_0\right)}}\right\}+\mathfrak{Re}\left\{2p\left(z_0\right)-1\right\}\hfill \\ & =& \mathfrak{Re}\left\{-{\displaystyle \frac{k}{2}}{\displaystyle \frac{1+\alpha w\left(z_0\right)}{1-\alpha w\left(z_0\right)}}\right\}+{\displaystyle \frac{k}{2}}-{\displaystyle \frac{k}{2}}+\alpha \hfill \\ & \le & -{\displaystyle \frac{1-\alpha }{2(1+\alpha )}}+\alpha \hfill \\ & =& {\displaystyle \frac{2{\alpha }^2+3\alpha -1}{2(1+\alpha )}}.\hfill \end{array}$$

(31)

Hence, if we would assume that

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\left(p\right)}\left(z\right)}{f^{(p-1)}\left(z\right)}}\right\}>{\displaystyle \frac{2{\alpha }^2+3\alpha -1}{2(1+\alpha )}},(z\in \mathbb{D}),$$

(32)

then (31) would imply that $\left|w\right(z\left)\right|<1$ in $\mathbb{D}$ and by (30)

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{(p-1)}\left(z\right)}{2f^{(p-2)}\left(z\right)}}\right\}>{\displaystyle \frac{1+\alpha }{2}}(z\in \mathbb{D}).$$

Furthermore, condition (29) is not necessary. Inequality (32) implies (29). To show this, we apply a result from [11] (p. 26) which says that if $p\left(z\right)=1+c_{1}z+\dots$, $p\left(z\right)\not\equiv 1$, $z_1\in \mathbb{D}$ and

$$\mathfrak{Re}\left\{p\left(z_1\right)\right\}=min\left\{\mathfrak{Re}\left\{p\left(z\right)\right\}:\left|z\right|<|z_1|\right\},$$

(33)

then $z_1{p}^{\prime }\left(z_1\right)$ is a negative real and

$$z_1{p}^{\prime }\left(z_1\right)\le -{\displaystyle \frac{1}{2}}{\displaystyle \frac{|1-p\left(z_1\right){|}^2}{\mathfrak{Re}\left\{1-p\left(z_1\right)\right\}}}.$$

From that, if we suppose that there exists $z_1\in \mathbb{D}$ such that (33) holds with $\mathfrak{Re}\left\{p\left(z_1\right)\right\}=\alpha$, then we have

$$\begin{array}{ccc}\hfill {\left.\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\left(p\right)}\left(z\right)}{f^{(p-1)}\left(z\right)}}\right\}\right|}_{z=z_1}& =& {\left.\mathfrak{Re}\left\{{\displaystyle \frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}}+2p\left(z\right)-1\right\}\right|}_{z=z_1}\hfill \\ & \le & -{\displaystyle \frac{1}{2\alpha }}{\displaystyle \frac{|1-p\left(z_1\right){|}^2}{1-\alpha }}+2\alpha -1\hfill \\ & \le & -{\displaystyle \frac{1}{2\alpha }}{\displaystyle \frac{{(1-\alpha )}^2}{1-\alpha }}+2\alpha -1\hfill \\ & \le & -{\displaystyle \frac{1-\alpha }{2\alpha }}+2\alpha -1\hfill \\ & \le & {\displaystyle \frac{2{\alpha }^2+3\alpha -1}{2(1+\alpha )}}\hfill \end{array}$$

which negates (32). □

For $p=2$, $\alpha =0$, we have

Corollary 3.

Let $f\in {\mathcal{A}}_2$, and let it satisfy the following condition

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}}\right\}>-{\displaystyle \frac{1}{2}},(z\in \mathbb{D}),$$

Then we have

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

Putting

$$\beta ={\displaystyle \frac{2{\alpha }^2+3\alpha -1}{2(1+\alpha )}},0\le \alpha <1,$$

(34)

then $-1/2\le \beta \le 1$, and Theorem 7 is equivalent to the following result.

Theorem 8.

Assume that $f\in {\mathcal{A}}_p$, $p\ge 2$, $-1/2\le \beta <1$. If the function

$${\displaystyle \frac{f^{(p-1)}\left(z\right)}{p!}}=z+\dots$$

is starlike of order β or

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

then

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{(p-1)}\left(z\right)}{f^{(p-2)}\left(z\right)}}\right\}>{\displaystyle \frac{1}{4}}\left\{2\beta +1+\sqrt{4{\beta }^2+4\beta +17}\right\},(z\in \mathbb{D}),$$

or

$$f\in \mathcal{S}\left(p,p-1,\beta \right)\Rightarrow f\in \mathcal{S}\left(p-1,p-2,\left\{2\beta +1+\sqrt{4{\beta }^2+4\beta +17}\right\}/4\right),$$

where we use for simplification the notation

$$\mathcal{S}\left(k,k-1,\beta \right)=\left\{f\in {\mathcal{A}}_p:\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\left(k\right)}\left(z\right)}{f^{(k-1)}\left(z\right)}}\right\}>\beta \right\},$$

for an integer k, $0<k\le p$.

4. Discussion

In this paper, we prove several sufficient conditions for a function g to be p-valent in $\mathbb{D}$. These symmetric results are under several different assumptions like the bounds on $|arg\{g^{\left(p\right)}\left(z\right)\left\}\right|$ or on $\left|{\displaystyle \frac{g^{\left(p\right)}\left(z\right)}{p!}}-1\right|$. In this paper, we note some related results in this topic, symmetric in some sense. Recall here the well-known Noshiro-Warschawski theorem and some related results. The Noshiro-Warschawski theorem [1,2], says that if $f\in \mathcal{H}$ satisfies $\mathfrak{Re}\left\{e^{i\alpha }{f}^{\prime }\left(z\right)\right\}0$ for all $z\in \mathbb{D}$ and for some real $\alpha$, then f is univalent in $\mathbb{D}$. Ozaki [3] generalized the above theorem for $f\in {\mathcal{A}}_p$: if $\mathfrak{Re}\left\{e^{i\alpha }{f}^{\left(p\right)}\left(z\right)\right\}>0$ for all $z\in \mathbb{D}$ and for some real $\alpha$, then f is at most p-valent in $\mathbb{D}$. The results obtained in the paper are new and do not constitute a direct improvement of some existing results. Also, in our opinion, these sufficient conditions for g to be p-valent are not sharp. The future direction of the work is to establish whether the results that were obtained are sharp.