Some New Sufficient Conditions on p-Valency for Certain Analytic Functions

Abstract

In the present paper, we develop some implications leading to Carathéodory functions in the open disk and provide some new conditions for functions to be p-valent functions. This work also extends the findings of Nunokawa and others.

1. Introduction and Definitions

The notion of multivalent functions is a natural extension of the injective. A holomorphic function f in an arbitrary domain $\Omega$, a subset of the complex-plane $\mathbb{C}$, is p-valent if it assumes every value a maximum of p-times, which means that the number of roots of the equation similar to $f\left(z\right)=w$ never exceeds in comparison of p. By the geometrical point of discussion, this leads to the fact that all points in the w-plane $\mathbb{C}$ lie, at most, p-times the corresponding Riemann surface, where $w=f\left(z\right)$ maps the domain $\Omega$. If $p=1$, then f is univalent in $\Omega$. The p-valent mappings plays a vital role in the literature of the complex multivalent functions.

Suppose that m is the number of roots $f\left(z\right)=w$ in the set $\Omega$ and let p be a positive number. The function f is said to be p-valent in the mean of circles in the domain $\Omega ,$ if for the number $\rho >0,$ we can write

$$\underset{0}{\overset{2\pi }{\int }}m\rho e^{i\varphi }d\varphi <2\pi p.$$

(1)

From the geometric point of view, the inequality shows that the measure of the circle on the Riemann surface where f maps $\Omega$, along with projecting $\left|w\right|=\rho$, never exceeds p-times the measure of this circle. A function f is termed p-valent in the mean over areas in the domain $\Omega ,$ if we have

$$\underset{0}{\overset{\rho }{\int }}\left(\underset{0}{\overset{2\pi }{\int }}m\varrho e^{i\varphi }d\varphi \right)\varrho d\varrho <\pi p{\rho }^2.$$

(2)

This integral inequality implies that the area of a small segment on the Riemann surface where f takes points from $\Omega$ as well as projecting them on the region defined by $\left|w\right|<R$ and this never exceeds p-times the area of the region $\left|w\right|<R$. Multivalent functions have been under investigation in view of their distortion, as bounds for the coefficient estimates along with various other aspects; see, for example, [1,2,3,4,5].

Any convergent power series is used to represent a holomorphic mapping. If f is holomorphic at a point $z_0$, it is analytic everywhere else in some neighbourhood of $z_0$. Furthermore, if f is entire, then this domain is the finite complex plane. It is a difficult task to deal with the complicated domains in the entire complex plane. As a result, the open unit disc is often used for simplification due to the Riemann mapping theorem. Let $\mathcal{H}$ denote the family of holomorphic functions in $\mathbb{D}:=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$. Let $\mathcal{A}\subseteq \mathcal{H}$ consist of holomorphic functions f satisfying $f^{\prime }\left(0\right)=1$ and $f\left(0\right)=0.$ Further, assume that $\mathcal{S}\subseteq \mathcal{A}$ consist of univalent functions. The analytic description of holomorphic mappings is coupled with the functions that map $\mathbb{D}$ to the right half-plane. Let $\mathcal{P}$ represent the family of functions q that is holomorphic in $\mathbb{D}$ with $q\left(0\right)=1$ and $\Re q\left(\mathbb{D}\right)>0.$ The function $q\in \mathcal{P}$ is called Carathéodory function. It is known that the class $\mathcal{P}$ is compact and normal. In geometric function theory, the Carathéodory function is well-studied and has a lot of applications (see, for example, [6,7,8,9]).

Some known subfamilies of $\mathcal{S}$ are the families ${\mathcal{S}}^{*}$ and $\mathcal{K}$ of starlike and convex mappings, respectively; for detail and further investigations, see [10,11,12,13,14,15]. These families are related to the change in argument of the radius vector and tangent vector of the image of $r{e}^{i\phi }$ as non-decreasing functions of the angle $\phi$, respectively.

Let ${\mathcal{A}}_p\left(n\right)$ denote the class of analytic functions f in the form

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

(3)

In particular, ${\mathcal{A}}_p\left(1\right)={\mathcal{A}}_p$, ${\mathcal{A}}_1\left(n\right)=A\left(n\right)$ and ${\mathcal{A}}_1\left(1\right)=\mathcal{A}.$ A function $f\in {\mathcal{A}}_p$ is called p-valent in $\mathbb{D}$ if f for $\omega \in \mathbb{C}$, the equation $f\left(z\right)=\omega$ has, at most, p roots in $\mathbb{D}$ and there exists a ${\omega }_0\in \mathbb{C}$ such that $f\left(z\right)={\omega }_0$ has exactly p roots in $\mathbb{D}$.

A function $f\in {\mathcal{A}}_p$ is said to be p-valent starlike if

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

(4)

It is known that the p-valent starlike function in $\mathcal{A}\left(p\right)$ is p-valent. For some investigations on properties of p-valent functions, we refer to [16,17,18,19].

It was proved that [20,21] if $f\in \mathcal{A}$ with $f^{\prime }\in \mathcal{P}$; then, the function f is univalent in $\mathbb{D}$. Ozaki [22] further extended the above assertion. They conclude that if f is holomorphic in a convex domain $\Delta \subset \mathbb{C}$ and

$$\frac{e^{i\gamma }{f}^{\left(p\right)}\left(z\right)}{p!}\in \mathcal{P},z\in \Delta ,$$

(5)

for some real $\gamma$, then f is, at most, p-valent in $\Delta$. Thus, if $f\in {\mathcal{A}}_p$ with the condition

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

(6)

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

Recently, Nunokawa et al. [23,24,25,26] found some interesting sufficient conditions for f to be a p-valent function, which improved Ozaki’s condition. Motivated from these works, we aim to develop some new sufficient criteria for Carathéodory functions and obtain certain new conditions for functions to be p-valent.

The following lemmas will be required for our results.

Lemma 1.

(See [27]). Let $q\left(z\right)$ be holomorphic in $\mathbb{D}$ with $q\left(z\right)\ne 0$ and $q\left(0\right)=1$. Suppose also that there is a point $z_0\in \mathbb{D}$ such that $\left|argq\left(z\right)\right|<\frac{\pi }{2}\alpha$ for $\left|z\right|<\left|z_0\right|$ and $\left|argq\left(z_0\right)\right|=\frac{\pi }{2}\alpha$ for some $\alpha >0.$ Then,

$$\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}=ik\alpha ,$$

(7)

where $k\ge \frac{a+a^{-1}}{2}$ when $argq\left(z_0\right)=\frac{\pi }{2}\alpha$ and $k\le -\frac{a+a^{-1}}{2}$ when $argq\left(z_0\right)=-\frac{\pi }{2}\alpha ,$ with

$${\left[q\left(z_0\right)\right]}^{\frac{1}{\alpha }}=\pm ia,a>0.$$

(8)

Lemma 2.

(See [28]). Let $f\in \mathcal{A}\left(p\right)$. If there exists a $(p-s+1)$-valent starlike function g in the form of

$$g\left(z\right)=z^{p-s+1}+\sum _{m=p-s+2}^{\infty }{b}_m{z}^m$$

(9)

such that

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

(10)

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

2. Main Results

Theorem 1.

Let q be a holomorphic function in $\mathbb{D}$ with $q\left(z\right)\ne 0$ and $q\left(0\right)=1$. Suppose also that

$$\left|\frac{1}{n}arg\left\{{\left[q\left(z\right)\right]}^n+n{\left[q\left(z\right)\right]}^{n-1}z{q}^{\prime }\left(z\right)-\beta {\left[q\left(z\right)\right]}^{n-1}\right\}\right|<\frac{\pi }{2}+\frac{1}{n}arctan\left(n\sqrt{\frac{n+2\beta }{n}}\right),$$

(11)

where $0\le \beta <1$. Then, we have

$$\left|arg\left\{q\left(z\right)\right\}\right|<\frac{\pi }{2},z\in \mathbb{D},$$

(12)

or

$$\Re \left(q\left(z\right)\right)>0,z\in \mathbb{D}.$$

(13)

Proof. 

Suppose that we have a point $z_0$ with $\left|z_0\right|<1$ in such a way that

$$\left|arg\left\{q\left(z\right)\right\}\right|<\frac{\pi }{2},\left|z\right|<\left|z_0\right|,$$

(14)

and

$$\left|arg\left\{q\left(z_0\right)\right\}\right|=\frac{\pi }{2}.$$

(15)

Then, by Lemma 1 with $\alpha =1$, we have

$$\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}=ik.$$

(16)

For the case $argq\left(z_0\right)=\frac{\pi }{2}$, $q\left(z_0\right)=ia$ and $a>0,$ we have

$$\begin{array}{cc}\hfill & \frac{1}{n}arg\left\{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)-\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\hfill \\ \hfill =& \frac{1}{n}arg{\left[q\left(z_0\right)\right]}^n+\frac{1}{n}arg\left\{1+\frac{n{z}_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}-\frac{\beta }{\left[q\left(z_0\right)\right]}\right\}\hfill \\ \hfill =& \frac{\pi }{2}+\frac{1}{n}arg\left\{1+nik-\frac{\beta }{ia}\right\}\hfill \\ \hfill =& \frac{\pi }{2}+\frac{1}{n}arg\left\{1+i\left(nk+\frac{\beta }{a}\right)\right\}\hfill \\ \hfill =& \frac{\pi }{2}+\frac{1}{n}arg\left\{1+i\frac{n}{2}\left(a+\frac{n+2\beta }{na}\right)\right\}.\hfill \end{array}$$

Define

$$\vartheta \left(x\right)=\frac{n}{2}\left(x+\frac{n+2\beta }{nx}\right).$$

(17)

Then, this function $\vartheta$ assumes its minimum value for $x=\sqrt{\frac{n+2\beta }{n}}.$ Therefore, in view of the above equality, we see that

$$\frac{1}{n}arg\left\{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)-\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\ge \frac{\pi }{2}+\frac{1}{n}arctan\left(n\sqrt{\frac{n+2\beta }{n}}\right),$$

(18)

which contradicts the hypothesis (11). When $argq\left(z_0\right)=-\frac{\pi }{2}$, using the similar technique, we get that:

$$\frac{1}{n}arg\left\{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)-\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\ge -\frac{\pi }{2}-\frac{1}{n}arctan\left(n\sqrt{\frac{n+2\beta }{n}}\right).$$

(19)

This above inequality also contradicts the hypothesis (11). The proof is thus completed. □

Corollary 1.

Let $p\ge 2.$ If $f\in$${\mathcal{A}}_p\left(n\right)$ satisfying that $f^{(p-1)}\ne 0$ in $\mathbb{D}$ and

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

(20)

where $0\le \beta <1,$ then f is p-valent in $\mathbb{D}$.

Proof. 

Assume that

$${\left[q\left(z\right)\right]}^n=\frac{f^{(p-1)}\left(z\right)}{p!z},q\left(0\right)=1.$$

(21)

Then on simplification, it follows that

$$\begin{array}{cc}\hfill & \left|\frac{1}{n}arg\left\{{\left[q\left(z\right)\right]}^n+n{\left[q\left(z\right)\right]}^{n-1}z{q}^{\prime }\left(z\right)-\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\right|\hfill \\ \hfill =& \left|\frac{1}{n}arg\left\{\frac{f^{\left(p\right)}\left(z\right)}{p!}-\beta {\left(\frac{f^{(p-1)}\left(z\right)}{p!.z}\right)}^{\frac{n-1}{n}}\right\}\right|\hfill \\ \hfill =& \frac{\pi }{2}+\frac{1}{n}arctan\left(n\sqrt{\frac{n+2\beta }{n}}\right).\hfill \end{array}$$

From Theorem 1, we have

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

(22)

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

Taking $n=1$ and $\beta =0$, we easily get the following result obtained by Nunokawa [29].

Corollary 2.

Let $p\ge 2.$ If $f\in$${\mathcal{A}}_p$ and

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

(23)

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

Theorem 2.

Let $q\left(z\right)$ be a holomorphic mapping in $\mathbb{D}$ with $q\left(0\right)=1$ and $q\left(z\right)\ne 0$. Further, suppose that

$$\left|\frac{1}{n}arg\left\{{\left[q\left(z\right)\right]}^n+n{\left[q\left(z\right)\right]}^{n-1}\frac{z{q}^{\prime }\left(z\right)}{q\left(z\right)}+\beta {\left[q\left(z\right)\right]}^{n-1}\right\}\right|<\frac{\pi }{2}-\frac{1}{n}arctan\frac{\beta }{\sqrt{n(n+2)}},$$

(24)

where $0\le$$\beta <\infty$. Then

$$\left|argq\left(z\right)\right|<\frac{\pi }{2},z\in \mathbb{D}.$$

(25)

Proof. 

We suppose that there is a point $z_0$$(\left|z_0\right|<1)$ such that

$$\left|arg\left\{q\left(z\right)\right\}\right|<\frac{\pi }{2},\left|z\right|<\left|z_0\right|,$$

(26)

and

$$\left|arg\left\{q\left(z_0\right)\right\}\right|=\frac{\pi }{2}.$$

(27)

Then, by using Lemma 1 with $\alpha =1$, we have

$$\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}=ik,$$

(28)

For the case $arg\left\{q\left(z_0\right)\right\}=\frac{\pi }{2}$ with $q\left(z_0\right)=ia$ and $a>0,$ we observe that

$$\begin{array}{cc}\hfill & \frac{1}{n}arg\left\{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}+\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\hfill \\ \hfill =& \frac{1}{n}arg{\left[q\left(z_0\right)\right]}^n+\frac{1}{n}arg\left\{1+\frac{n{z}_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}.\frac{1}{q\left(z_0\right)}+\frac{\beta }{q\left(z_0\right)}\right\}\hfill \\ \hfill =& \frac{\pi }{2}+\frac{1}{n}arg\left\{1+nk\frac{1}{a}-i\frac{\beta }{a}\right\}\hfill \\ \hfill =& \frac{\pi }{2}+\frac{1}{n}arctan\left(\frac{-\beta }{a+nk}\right)\hfill \\ \hfill \ge & \frac{\pi }{2}-\frac{1}{n}arctan\left(\frac{\beta }{a+\frac{n}{2}\left(a+\frac{1}{a}\right)}\right).\hfill \end{array}$$

Let

$$\zeta \left(x\right)=x+\frac{n}{2}\left(x+\frac{1}{x}\right).$$

(29)

It is easy to note that $\zeta$ takes the minimum value for $x=\sqrt{\frac{n}{n+2}}$. Therefore, on some simple manipulation, the above equality leads to

$$\frac{1}{n}arg\left\{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}+\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\ge \frac{\pi }{2}-\frac{1}{n}arctan\frac{\beta }{\sqrt{n(n+2)}},$$

which contradicts the hypothesis in (24). For the case $argq\left(z_0\right)=-\frac{\pi }{2}$, applying the same method as the above, we have

$$\begin{array}{cc}\hfill & \frac{1}{n}arg\left\{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}+\beta {\left[q\left(z_0\right)\right]}^{n-1}\right\}\hfill \\ \hfill \ge & -\left(\frac{\pi }{2}-\frac{1}{n}arctan\frac{\beta }{\sqrt{n(n+2)}}\right),\hfill \end{array}$$

which also contradicts the hypothesis as in (24). This completes the proof of Theorem 2. □

Theorem 3.

Let q be a holomorphic function in $\mathbb{D}$ with $q\left(0\right)=1$ and $q\left(z\right)\ne 0$. Suppose that

$$\Re \left\{\frac{1}{n}\sqrt{{\left[q\left(z\right)\right]}^n+n{\left[q\left(z\right)\right]}^{n-1}z{q}^{\prime }\left(z\right)}\right\}>0,z\in \mathbb{D}.$$

(30)

Then we have

$$\left|arg\left\{q\left(z\right)\right\}\right|<\frac{\pi }{2}{\alpha }_1,z\in \mathbb{D},$$

(31)

where ${\alpha }_1$ is the positive zero or root of the equation

$${\alpha }_1+\frac{2}{n\pi }arctan\left(n{\alpha }_1\right)=2.$$

(32)

Proof. 

Assume that there is a point $z_0$$(\left|z_0\right|<1)$ such that

$$\left|argq\left(z\right)\right|<\frac{\pi }{2}{\alpha }_1,\left(\left|z\right|<\left|z_0\right|\right)$$

(33)

and

$$\left|argq\left(z_0\right)\right|=\frac{\pi }{2}{\alpha }_1.$$

(34)

Then, by using Lemma 1 with $\alpha ={\alpha }_1$, we have

$$\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}=i{\alpha }_{1}k,$$

(35)

For the case $argq\left(z_0\right)=\frac{\pi }{2}{\alpha }_1,$ we have

$$\begin{array}{cc}\hfill & \frac{1}{n}arg\sqrt{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)}\hfill \\ \hfill =& \frac{1}{2n}\left(arg{\left[q\left(z_0\right)\right]}^n+arg\left\{1+\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}\right\}\right)\hfill \\ \hfill =& \frac{1}{2n}\left(\frac{n\pi }{2}{\alpha }_1+arg\left\{1+nik{\alpha }_1\right\}\right)\hfill \\ \hfill =& \frac{1}{2}.\frac{\pi }{2}\left({\alpha }_1+\frac{2}{n\pi }arctan\left(n{\alpha }_1\right)\right)\hfill \\ \hfill =& \frac{\pi }{2},\hfill \end{array}$$

which implies that

$$\Re \frac{1}{n}\left\{\sqrt{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)}\right\}\le 0,$$

(36)

and this contradicts the hypothesis as in (30). For $argq\left(z_0\right)=-\frac{\pi }{2}{\alpha }_1,$ using the similar technique yields to

$$\frac{1}{n}arg\sqrt{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)}\le -\frac{\pi }{2},$$

(37)

or

$$\Re \left\{\frac{1}{n}\sqrt{{\left[q\left(z_0\right)\right]}^n+n{\left[q\left(z_0\right)\right]}^{n-1}{z}_0{q}^{\prime }\left(z_0\right)}\right\}\le 0.$$

(38)

This also contradicts the hypothesis in (30) and, therefore, the assertion is concluded. □

Corollary 3.

Suppose that $p\ge 4$ and $f\in$${\mathcal{A}}_p\left(n\right)$ satisfying that $f^{\left(k\right)}\left(z\right)\ne 0(k=p-1,p-2,$$p-3)$ in $\mathbb{D}$. If

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

(39)

then the mapping f is p-valent in $\mathbb{D}$.

Proof. 

Assume that

$${\left[q_1\left(z\right)\right]}^n=\frac{f^{(p-1)}\left(z\right)}{p!z},q_1\left(0\right)=1.$$

(40)

Then a simple simplification leads to

$${\left[q_1\left(z\right)\right]}^n+n{\left[q_1\left(z\right)\right]}^{n-1}z{q}_1^{\prime }\left(z\right)=\frac{f^{\left(p\right)}\left(z\right)}{p!}.$$

(41)

In view of Theorem 3, we obtain that

$$\left|\frac{1}{n}arg\left\{\frac{f^{(p-1)}\left(z\right)}{z}\right\}\right|=\left|\frac{1}{n}arg{\left[q_1\left(z\right)\right]}^n\right|<\frac{\pi }{2}{\alpha }_1,z\in \mathbb{D},$$

(42)

where ${\alpha }_1$ is the positive zero or root of the above equation given by (32). Next, let us put

$${\left[q_2\left(z\right)\right]}^n=\frac{2f^{(p-2)}\left(z\right)}{p!z^2},q_2\left(0\right)=1.$$

(43)

Then a simple calculation leads to

$$2{\left[q_2\left(z\right)\right]}^n+n{\left[q_2\left(z\right)\right]}^{n-1}z{q}_2^{\prime }\left(z\right)=\frac{2f^{(p-1)}\left(z\right)}{p!z}.$$

(44)

Let ${\alpha }_2$ be a positive zero or root of the equation

$$\alpha +\frac{2}{n\pi }arctan\left(\frac{n\alpha }{2}\right)={\alpha }_1.$$

(45)

Suppose that there exists a point $z_1$ with $\left|z_1\right|<1$ such that

$$\left|arg{q}_2\left(z\right)\right|<\frac{\pi }{2}{\alpha }_2,\left|z\right|<\left|z_1\right|$$

(46)

and $\left|arg{q}_2\left(z_1\right)\right|=\frac{\pi }{2}{\alpha }_2,$ then we write

$$\frac{z_1{q}_2^{\prime }\left(z_1\right)}{q_2\left(z_1\right)}=i{\alpha }_{2}k.$$

(47)

For the choice of $arg{q}_2\left(z_1\right)=\frac{\pi }{2}{\alpha }_2,$ we have

$$\begin{array}{cc}\hfill & \frac{1}{n}arg\left\{2{\left[q_2\left(z_1\right)\right]}^n+n{\left[q_2\left(z_1\right)\right]}^{n-1}{z}_1{q}_2^{\prime }\left(z_1\right)\right\}\hfill \\ \hfill =& \frac{1}{n}arg\left\{\frac{f^{(p-1)}\left(z_1\right)}{z_1}\right\}\hfill \\ \hfill =& \frac{1}{n}arg{\left[q_2\left(z_1\right)\right]}^n+\frac{1}{n}arg\left\{2+n\frac{z_1{q}_2^{\prime }\left(z_1\right)}{q_2\left(z_1\right)}\right\}\hfill \\ \hfill =& \frac{\pi }{2}{\alpha }_2+\frac{1}{n}arg\left\{2+nik{\alpha }_2\right\}\hfill \\ \hfill =& \frac{\pi }{2}{\alpha }_2+\frac{1}{n}arctan\left(\frac{n{\alpha }_2}{2}\right)=\frac{\pi }{2}{\alpha }_1,\hfill \end{array}$$

which contradicts the result in (42). For the assumption $arg{q}_2\left(z_1\right)=-\frac{\pi }{2}{\alpha }_2,$ we note that

$$\begin{array}{cc}\hfill & \frac{1}{n}arg\left\{2{\left[q_2\left(z_1\right)\right]}^n+n{\left[q_2\left(z_1\right)\right]}^{n-1}{z}_1{q}_2^{\prime }\left(z_1\right)\right\}\hfill \\ \hfill =& \frac{1}{n}arg\left\{\frac{2f^{(p-1)}\left(z_1\right)}{p!z_1}\right\}=\frac{1}{n}arg\left\{\frac{f^{(p-1)}\left(z_1\right)}{z_1}\right\}\le -\frac{\pi }{2}{\alpha }_1.\hfill \end{array}$$

This also contradicts (42). Hence, we have

$$\left|\frac{1}{n}arg{\left[q_2\left(z_1\right)\right]}^n\right|=\left|\frac{1}{n}arg\left\{\frac{f^{(p-2)}\left(z\right)}{z^2}\right\}\right|<\frac{\pi }{2}{\alpha }_2,z\in \mathbb{D},$$

(48)

where ${\alpha }_2+\frac{2}{n\pi }arctan\left(\frac{n{\alpha }_2}{2}\right)={\alpha }_1.$ Let

$${\left[q_3\left(z\right)\right]}^n=\frac{6f^{(p-3)}\left(z\right)}{p!z^3},q_3\left(0\right)=1.$$

(49)

Then we see that

$$3{\left[q_3\left(z\right)\right]}^n+n{\left[q_3\left(z\right)\right]}^{n-1}z{q}_3^{\prime }\left(z\right)=\frac{6f^{(q-2)}\left(z\right)}{q!.z^2}.$$

(50)

Using the similar approach as adopted above, we note that

$$\begin{array}{cc}\hfill & \left|\frac{1}{n}arg\left\{3{\left[q_3\left(z\right)\right]}^n+n{\left[q_3\left(z\right)\right]}^{n-1}z{q}_3^{\prime }\left(z\right)\right\}\right|\hfill \\ \hfill =& \left|\frac{1}{n}arg{\left[q_3\left(z\right)\right]}^n+\frac{1}{n}arg\left\{3+n\frac{z{q}_3^{\prime }\left(z\right)}{q_3\left(z\right)}\right\}\right|\hfill \\ \hfill =& \left|\frac{1}{n}arg\left\{\frac{6f^{(p-2)}\left(z\right)}{p!z^2}\right\}\right|=\left|\frac{1}{n}arg\left\{\frac{f^{(p-2)}\left(z\right)}{z^2}\right\}\right|<\frac{\pi }{2}{\alpha }_2.\hfill \end{array}$$

This shows that

$$\left|\frac{1}{n}arg\left\{\frac{z{f}^{(p-3)}\left(z\right)}{z^4}\right\}\right|=\left|\frac{1}{n}arg\left\{\frac{z{f}^{(p-3)}\left(z\right)}{z^3}\right\}\right|<\frac{\pi }{2}{\alpha }_3<\frac{\pi }{2},z\in \mathbb{D},$$

(51)

or

$$\Re \left\{\frac{z{f}^{(p-3)}\left(z\right)}{z^4}\right\}>0,z\in \mathbb{D}.$$

(52)

Thus, we note that $g\left(z\right)=z^4$ is a four-valent starlike function in $\mathbb{D}$. Therefore, using the result in (52) and Lemma 2, we observe that f is p-valent in $\mathbb{D}$. This leads to the desired result in Corollary 3. □

3. Conclusions

Analytic p-valent functions were intensively studied recently, as in [30,31,32]. In the present paper, we introduced several sufficient conditions for functions to be p-valent. Some simple criteria on p-valents are obtained. This generalizes some know results and may inspire more effective and concise univalent conditions in geometric function theory.