# On the multivalency of certain analytic functions

## Abstract

We prove several relations of the type $|arg\left\{z{f}^{″}\left(z\right)/{f}^{\prime }\left(z\right)\right\}|\le |arg\left\{{f}^{\prime }\left(z\right)\right\}|$ for functions satisfying some geometric conditions.

MSC:30C45, 30C80.

## 1 Introduction

Let p be positive integer and let $\mathcal{A}\left(p\right)$ be the class of functions

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

which are analytic in the unit disk $\mathbb{D}=\left\{z\in \mathbb{C}:|z|<1\right\}$ and denote $\mathcal{A}=\mathcal{A}\left(1\right)$.

The subclass of $\mathcal{A}\left(p\right)$ consisting of p-valently starlike functions is denoted by ${\mathcal{S}}^{\ast }\left(p\right)$. An analytic description of ${\mathcal{S}}^{\ast }\left(p\right)$ is given by

${\mathcal{S}}^{\ast }\left(p\right)=\left\{f\in \mathcal{A}\left(p\right):|arg\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}|<\frac{\pi }{2},z\in \mathbb{D}\right\}.$

The subclass of $\mathcal{A}\left(p\right)$ consisting of p-valently and strongly starlike functions of order α, $0<\alpha \le 1$ is denoted by ${\mathcal{S}}_{\alpha }^{\ast }\left(p\right)$. An analytic description of ${\mathcal{S}}_{\alpha }^{\ast }\left(p\right)$ is given by

${\mathcal{S}}_{\alpha }^{\ast }\left(p\right)=\left\{f\in \mathcal{A}\left(p\right):|arg\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}|<\frac{\alpha \pi }{2},z\in \mathbb{D}\right\}.$

The subclass of $\mathcal{A}\left(p\right)$ consisting of p-valently convex functions and p-valently strongly convex functions of order α, $0<\alpha \le 1$, are denoted by $\mathcal{C}\left(p\right)$ and ${\mathcal{C}}_{\alpha }\left(p\right)$, respectively. The analytic descriptions of $\mathcal{C}\left(p\right)$ and ${\mathcal{C}}_{\alpha }\left(p\right)$ are given by

$\mathcal{C}\left(p\right)=\left\{f\in \mathcal{A}\left(p\right):|arg\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}|<\frac{\pi }{2},z\in \mathbb{D}\right\}$

and

${\mathcal{C}}_{\alpha }\left(p\right)=\left\{f\in \mathcal{A}\left(p\right):|arg\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}|<\frac{\alpha \pi }{2},z\in \mathbb{D}\right\}.$

For $p=1$ the classes ${\mathcal{S}}_{\alpha }^{\ast }\left(p\right)$ and ${\mathcal{C}}_{\alpha }\left(p\right)$ become the well known classes ${\mathcal{S}}_{\alpha }^{\ast }$ and ${\mathcal{C}}_{\alpha }^{\ast }$ of strongly starlike and strongly convex functions of order α, respectively. The concept of strongly starlike and strongly convex functions of order α was introduced in  and  with their geometric interpretation. For $\alpha =1$ the classes ${\mathcal{S}}_{\alpha }^{\ast }$ and ${\mathcal{C}}_{\alpha }^{\ast }$ become the classes ${\mathcal{S}}^{\ast }$ and ${\mathcal{C}}^{\ast }$ of starlike and convex functions; see for example . In this paper, we need the following lemma.

Lemma 1.1 Assume that $f\in \mathcal{A}$ with $f\left(z\right)/z\ne 0$ in $\mathbb{D}.$. Assume also that for all θ, $0\le \theta <2\pi$, the function f satisfies the following condition:

$\begin{array}{rcl}\left(\mathfrak{Im}\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)sin\theta & =& \rho \left(\frac{\mathrm{d}arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}}{\mathrm{d}\rho }\right)sin\theta \\ =& \frac{1}{2}\mathfrak{Re}\left\{\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\left(\overline{z}-z\right)\right\}\\ =& \frac{1}{2}\mathfrak{Re}\left\{\frac{{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\left({|z|}^{2}-{z}^{2}\right)\right\}\ge 0,\end{array}$
(1.1)

where $z=\rho {e}^{i\theta }$, $0<\rho \le r<1$. Then we have

$|arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|\le |arg\left\{{f}^{\prime }\left(z\right)\right\}|,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Proof First we note that from

$0\le arg\left\{{z}_{1}\right\}\le arg\left\{{z}_{2}\right\}\le \pi \phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}arg\left\{{z}_{1}\right\}\le arg\left\{{z}_{1}+{z}_{2}\right\}\le arg\left\{{z}_{2}\right\},$

the implication

$0\le arg\left\{{z}_{1}\right\}\le \cdots \le arg\left\{{z}_{n}\right\}\le \pi \phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}arg\left\{{z}_{1}\right\}\le arg\left\{\sum _{k=1}^{n}{z}_{k}\right\}\le arg\left\{{z}_{n}\right\}$
(1.2)

follows by mathematical induction.

For the case $0\le \theta <\pi$, $z=r{e}^{i\theta }\in \mathbb{D}$, we have

$\begin{array}{rcl}arg\left\{\frac{f\left(z\right)}{z}\right\}& =& arg\left\{\frac{1}{r{e}^{i\theta }}{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right){e}^{i\theta }\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}\\ =& arg\left\{{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}.\end{array}$
(1.3)

Let $0={\rho }_{0}<{\rho }_{1}<\cdots <{\rho }_{n-1}<{\rho }_{n}=r$, $\mathrm{\Delta }{\rho }_{k}={\rho }_{k}-{\rho }_{k-1}$, $k=1,\dots ,n$. By (1.1) $arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}$ is an increasing function with respect to ρ, thus

$0=arg\left\{{f}^{\prime }\left({\rho }_{0}{e}^{i\theta }\right)\right\}\le arg\left\{{f}^{\prime }\left({\rho }_{1}{e}^{i\theta }\right)\right\}\le \cdots \le arg\left\{{f}^{\prime }\left({\rho }_{n}{e}^{i\theta }\right)\right\}=arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}.$
(1.4)

Therefore, by (1.2) and by (1.4), we have

$arg\left\{\sum _{k=1}^{n}{f}^{\prime }\left({\rho }_{k}{e}^{i\theta }\right)\right\}\le arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}.$
(1.5)

Using (1.5) in (1.3), we obtain

$\begin{array}{rcl}arg\left\{\frac{f\left(z\right)}{z}\right\}& =& arg\left\{{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}\\ =& arg\left\{\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}{f}^{\prime }\left({\rho }_{k}{e}^{i\theta }\right)\mathrm{\Delta }{\rho }_{k}\right\}\\ =& \underset{n\to \mathrm{\infty }}{lim}arg\left\{\sum _{k=1}^{n}{f}^{\prime }\left({\rho }_{k}{e}^{i\theta }\right)\mathrm{\Delta }{\rho }_{k}\right\}\\ \le & arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}\end{array}$
(1.6)

or we have

$0\le arg\left\{\frac{f\left(z\right)}{z}\right\}\le arg\left\{{f}^{\prime }\left(z\right)\right\}$
(1.7)

for $z=r{e}^{i\theta }$ and $0\le \theta \le \pi$.

For the case $\pi \le \theta <2\pi$, from the hypothesis (1.1), we find that $arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}$ is an decreasing function with respect to ρ, thus

$0=arg\left\{{f}^{\prime }\left({\rho }_{0}{e}^{i\theta }\right)\right\}\ge arg\left\{{f}^{\prime }\left({\rho }_{1}{e}^{i\theta }\right)\right\}\ge \cdots \ge arg\left\{{f}^{\prime }\left({\rho }_{n}{e}^{i\theta }\right)\right\}=arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}$

and

$arg\left\{\sum _{k=1}^{n}{f}^{\prime }\left({\rho }_{k}{e}^{i\theta }\right)\right\}\ge arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}.$

Therefore, in a similar way to above, we obtain

$\begin{array}{rcl}arg\left\{\frac{f\left(z\right)}{z}\right\}& =& arg\left\{{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}\\ \ge & arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}\end{array}$

and we also have

$0\ge arg\left\{\frac{f\left(z\right)}{z}\right\}\ge arg\left\{{f}^{\prime }\left(z\right)\right\}$
(1.8)

for $z=r{e}^{i\theta }$ and $\pi \le \theta \le 2\pi$. From (1.7) and (1.8), we have

$\begin{array}{rcl}|arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|& =& |arg\left\{{f}^{\prime }\left(z\right)\right\}-arg\left\{\frac{f\left(z\right)}{z}\right\}|\\ \le & |arg\left\{{f}^{\prime }\left(z\right)\right\}|.\end{array}$

It completes the proof of Lemma 1.1. □

Corollary 1.2 Assume that $f\in \mathcal{A}$ with $f\left(z\right)/z\ne 0$ in $\mathbb{D}.$. Assume also that $f\left(z\right)$ satisfies the following condition:

$\left(\mathfrak{Im}\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)\mathfrak{Im}\left\{z\right\}\ge 0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(1.9)

then we have

$arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}arg\left\{z\right\}\ge 0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(1.10)

Proof The conditions (1.1) and (1.9) are equivalent. If $arg\left\{z\right\}\ge 0$, then $z=r{e}^{i\theta }\in \mathbb{D}$ with $0\le \theta \le \pi$. By (1.7), we have also

$arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}\ge 0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

If $arg\left\{z\right\}\le 0$, then $z=r{e}^{i\theta }\in \mathbb{D}$ with $\pi <\theta \le 2\pi$. By (1.8), we have also

$arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}\le 0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

In both cases, we have (1.10). □

The inequality (1.10) can be written in the equivalent form

$\left(\mathfrak{Im}\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)\mathfrak{Im}\left\{z\right\}\ge 0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(1.11)

Recall that if $f\left(z\right)$ is analytic in $\mathbb{D}.$ and $\left(\mathfrak{Im}\left\{f\left(z\right)\right\}\right)\left(\mathfrak{Im}\left\{z\right\}\right)\ge 0$ in $\mathbb{D}.$, then f is called typically real function; see [, Chapter 10]. Therefore, Corollary 1.2 says that if $z{f}^{″}\left(z\right)/{f}^{\prime }\left(z\right)$ is a typically real function, then $z{f}^{\prime }\left(z\right)/f\left(z\right)$ is a typically real function, too.

## 2 Main result

Theorem 2.1 Let $f\left(z\right)\in \mathcal{A}$. Assume that for all θ, $0\le \theta <2\pi$, $f\left(z\right)$ satisfies the following condition:

$\begin{array}{rcl}\left(\mathfrak{Im}\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)sin\theta & =& \rho \left(\frac{\mathrm{d}arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}}{\mathrm{d}\rho }\right)sin\theta \\ \ge & 0,\end{array}$
(2.1)

where $z=\rho {e}^{i\theta }$, $0\le \rho \le r<1$, moves on the segment from $z=0$ to $z=r{e}^{i\theta }$ and

$|arg\left\{{f}^{\prime }\left(z\right)\right\}|\le \frac{\pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.2)

Then $f\left(z\right)$ is starlike in $\mathbb{D}.$ or $f\left(z\right)\in {\mathcal{S}}^{\ast }$.

Proof From the hypothesis (2.1) and the hypothesis (2.2) and applying Lemma 1.1, we have

$|arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|\le |arg\left\{{f}^{\prime }\left(z\right)\right\}|\le \frac{\pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

This shows that $f\left(z\right)$ is starlike in $\mathbb{D}.$. □

Applying the same method as in the proof of Lemma 1.1, we have the following lemma.

Lemma 2.2 Let $f\left(z\right)\in \mathcal{A}\left(2\right)$. Assume that for all θ, $0\le \theta <2\pi$, $f\left(z\right)$ satisfies the following condition:

$\begin{array}{rcl}\left(\mathfrak{Im}\frac{z{f}^{‴}\left(z\right)}{{f}^{″}\left(z\right)}\right)sin\theta & =& \rho \left(\frac{\mathrm{d}arg\left\{{f}^{″}\left(\rho {e}^{i\theta }\right)\right\}}{\mathrm{d}\rho }\right)sin\theta \\ \ge & 0,\end{array}$

where $z=\rho {e}^{i\theta }$, $0\le \rho \le r<1$ moves on the segment from $z=0$ to $z=r{e}^{i\theta }$. Then we have

$|arg\left\{\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}|\le |arg\left\{{f}^{″}\left(z\right)\right\}|,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Applying Lemma 2.2, we have the following theorem.

Theorem 2.3 Let $f\left(z\right)\in \mathcal{A}\left(2\right)$. Suppose that for all θ, $0\le \theta <2\pi$, $f\left(z\right)$ satisfies the following condition:

$\begin{array}{rcl}\left(\mathfrak{Im}\frac{z{f}^{‴}\left(z\right)}{{f}^{″}\left(z\right)}\right)sin\theta & =& \rho \left(\frac{\mathrm{d}arg\left\{{f}^{″}\left(\rho {e}^{i\theta }\right)\right\}}{\mathrm{d}\rho }\right)sin\theta \\ \ge & 0,\end{array}$
(2.3)

where $z=\rho {e}^{i\theta }$, $0\le \rho \le r<1$, moves on the segment from $z=0$ to $z=r{e}^{i\theta }$ and

$|arg\left\{{f}^{″}\left(z\right)\right\}|<\frac{\pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.4)

Then we have $f\left(z\right)\in \mathcal{C}\left(2\right)={\mathcal{C}}_{1}\left(2\right)$ or $f\left(z\right)$ is 2-valently convex in $\mathbb{D}.$.

Proof From the hypothesis (2.3) and (2.4) and applying Lemma 2.2, we have

$|arg\left\{\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}|\le |arg\left\{{f}^{″}\left(z\right)\right\}|<\frac{\pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Therefore, we have

$1+\mathfrak{Re}\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

It completes the proof. □

Applying the same method as in the proof of Lemma 1.1 and Lemma 2.2, we can generalize Theorem 2.1 and Theorem 2.3 as follows.

Lemma 2.4 Let $f\left(z\right)\in \mathcal{A}\left(p\right)$. Suppose that for all θ, $0\le \theta <2\pi$, $f\left(z\right)$ satisfies the following condition:

$\begin{array}{rl}\left(\mathfrak{Im}\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)sin\theta & =\rho \left(\frac{\mathrm{d}\left(arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}-\left(p-1\right)\theta \right)}{\mathrm{d}\rho }\right)sin\theta \\ =\rho \left(\frac{\mathrm{d}arg\left\{{f}^{\prime }\left(z\right)/{z}^{p-1}\right\}}{\mathrm{d}\rho }\right)sin\theta \\ \ge 0,\end{array}$
(2.5)

where $z=\rho {e}^{i\theta }$, $0\le \rho \le r<1$ moves on the segment from $z=0$ to $z=r{e}^{i\theta }$. Then we have

$|arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|\le |arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-1}}\right\}|,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Proof For the case $0\le \theta \le \pi$, from the hypothesis (2.5), we have

$\begin{array}{rcl}arg\left\{\frac{f\left(z\right)}{{z}^{p}}\right\}& =& arg\left\{\frac{1}{{r}^{p}{e}^{ip\theta }}{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right){e}^{i\theta }\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}\\ =& arg\left\{{\int }_{0}^{r}|{f}^{\prime }\left(\rho {e}^{i\theta }\right)|{e}^{i\left(arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}-\left(p-1\right)\theta \right)}\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}\\ \le & arg\left\{{f}^{\prime }\left(r{e}^{i\theta }\right)\right\}-\left(p-1\right)\theta \\ =& arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-1}}\right\}\end{array}$

and therefore we have

$0\le arg\left\{\frac{f\left(z\right)}{{z}^{p}}\right\}\le arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-1}}\right\}$
(2.6)

for $z=r{e}^{i\theta }$ and $0\le \theta \le \pi$.

For the case $\pi <\theta <2\pi$, applying the same method as above and in the proof of Lemma 1.1 and Lemma 2.2, we have

$0\ge arg\left\{\frac{f\left(z\right)}{z}\right\}\ge arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-1}}\right\}$
(2.7)

for $z=r{e}^{i\theta }$ and $\pi <\theta <2\pi$. From (2.6) and (2.7), we have

$|arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|=|arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p}}\right\}-arg\left\{\frac{f\left(z\right)}{{z}^{p}}\right\}|\le |arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-1}}\right\}|.$

It completes the proof of Lemma 2.4. □

Thus, we have the following theorems.

Theorem 2.5 Let $f\left(z\right)\in \mathcal{A}\left(p\right)$. Assume that for all θ, $0\le \theta <2\pi$, $f\left(z\right)$ satisfies the following condition:

$\begin{array}{rl}\left(\frac{\mathrm{d}\left(arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}-\left(p-1\right)\theta \right)}{\mathrm{d}\rho }\right)sin\theta & =\left(\frac{\mathrm{d}arg\left\{{f}^{\prime }\left(z\right)/{z}^{p-1}\right\}}{\mathrm{d}\rho }\right)sin\theta \\ \ge 0,\end{array}$

where $z=\rho {e}^{i\theta }$, $0\le \theta <2\pi$, $0\le \rho \le r<1$, moves on the segment from $z=0$ to $z=r{e}^{i\theta }$ and suppose that

$|arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-1}}\right\}|\le \frac{\alpha \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$

where $0<\alpha \le 1$. Then we have $f\left(z\right)\in {\mathcal{S}}_{\alpha }^{\ast }\left(p\right)$ or $f\left(z\right)$ is p-valently and strongly starlike of order α in $\mathbb{D}.$.

Theorem 2.6 Let $f\left(z\right)\in \mathcal{A}\left(p\right)$, $p\ge 2$. Assume that for all θ, $0\le \theta <2\pi$, $f\left(z\right)$ satisfies the following condition:

$\begin{array}{rcl}\left(\mathfrak{Im}\frac{z{f}^{‴}z\right)}{{f}^{″}\left(z\right)}\right)sin\theta & =& \rho \left(\frac{\mathrm{d}\left(arg\left\{{f}^{″}\left(\rho {e}^{i\theta }\right)\right\}-\left(p-2\right)\theta \right)}{\mathrm{d}\rho }\right)sin\theta \\ =& \rho \left(\frac{\mathrm{d}arg\left\{{f}^{″}\left(z\right)/{z}^{p-2}\right\}}{\mathrm{d}\rho }\right)sin\theta \\ \ge & 0,\end{array}$

where $z=\rho {e}^{i\theta }$, $0\le \rho \le r<1$ moves on the segment from $z=0$ to $z=r{e}^{i\theta }$ and suppose that

$|arg\left\{\frac{{f}^{\prime }\left(z\right)}{{z}^{p-2}}\right\}|\le \frac{\alpha \pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$

where $0<\alpha \le 1$. Then we have $f\left(z\right)\in {\mathcal{C}}_{\alpha }\left(p\right)$ or $f\left(z\right)$ is p-valently and strongly convex of order α.

Lemma 2.7 Let $f\left(z\right)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in $|z|\le 1$ and suppose that it satisfies the following condition:

(2.8)

where $0\le \alpha \le \pi$. Then for $\alpha \le \theta \le \alpha +\pi$ we have

$\begin{array}{rcl}\rho \left(\frac{\mathrm{d}\left(arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}\right)}{\mathrm{d}\rho }\right)& =& \mathfrak{Im}\left\{\frac{z{f}^{\prime }z\right)}{f\left(z\right)}\right\}\\ \ge & 0,\end{array}$
(2.9)

while for $\alpha +\pi \le \theta \le \alpha +2\pi$ we have

$\begin{array}{rcl}\rho \left(\frac{\mathrm{d}\left(arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}\right)}{\mathrm{d}\rho }\right)& =& \mathfrak{Im}\left\{\frac{z{f}^{\prime }z\right)}{f\left(z\right)}\right\}\\ \le & 0,\end{array}$
(2.10)

where $z=\rho {e}^{i\theta }$, $0\le \rho \le |z|\le 1$.

Proof Let $z=\rho {e}^{i\theta }$, $0\le \rho \le |z|\le 1$. Then it follows that

$\begin{array}{r}\mathfrak{Re}\left\{\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\left(\overline{z{e}^{-i\alpha }}-z{e}^{-i\alpha }\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\mathfrak{Re}\left\{\frac{\mathrm{d}log{f}^{\prime }\left(z\right)}{\mathrm{d}z}\left(\rho {e}^{-i\left(\theta -\alpha \right)}-\rho {e}^{i\left(\theta -\alpha \right)}\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\mathfrak{Re}\left\{\rho \left(\frac{\mathrm{d}log|{f}^{\prime }\left(\rho {e}^{i\theta }\right)|}{\mathrm{d}\rho }+i\frac{\mathrm{d}arg{f}^{\prime }\left(\rho {e}^{i\theta }\right)}{\mathrm{d}\rho }\right)\left(-2i\right)\right\}sin\left(\theta -\alpha \right)\\ \phantom{\rule{1em}{0ex}}=2\rho \frac{\mathrm{d}arg{f}^{\prime }\left(\rho {e}^{i\theta }\right)}{\mathrm{d}\rho }sin\left(\theta -\alpha \right)\\ \phantom{\rule{1em}{0ex}}\ge 0.\end{array}$

This proves (2.9) and (2.10) and it shows that the function $arg{f}^{\prime }\left(\rho {e}^{i\theta }\right)$ is an increasing function with respect to ρ, $0\le \rho \le 1$, and $\alpha \le \theta \le \alpha +\pi$, and that the function $arg{f}^{\prime }\left(\rho {e}^{i\theta }\right)$ is a decreasing function with respect to ρ, $0\le \rho \le 1$, and $\alpha +\pi \le \theta \le \alpha +2\pi$. □

Theorem 2.8 Let $f\left(z\right)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in $\mathbb{D}.$ and suppose that it satisfies the following condition:

$\mathfrak{Re}\left\{\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\left(\overline{z{e}^{-i\alpha }}-z{e}^{-i\alpha }\right)\right\}\ge 0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.11)

where $0\le \alpha \le \pi$ and

$|arg\left\{{f}^{\prime }\left(z\right)\right\}|\le \frac{\pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.12)

Then $f\left(z\right)$ is starlike in $\mathbb{D}.$.

Proof From Lemma 2.7 and (2.12), for the case $0\le \theta \le \pi$, we have

$\begin{array}{rcl}0& =& {\left(arg\frac{f\left(z\right)}{z}\right)}_{z=0}\\ \le & arg\left\{\frac{1}{\rho {e}^{i\theta }}{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right){e}^{i\theta }\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}\\ =& arg\frac{f\left(z\right)}{z}\\ =& arg{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \\ =& arg{\int }_{0}^{r}|{f}^{\prime }\left(\rho {e}^{i\theta }\right)|{e}^{iarg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \\ \le & arg\left\{{f}^{\prime }\left(z\right)\right\}.\end{array}$

This shows that

$0\le arg\left\{\frac{f\left(z\right)}{z}\right\}\le arg\left\{{f}^{\prime }\left(z\right)\right\},$
(2.13)

where $z=r{e}^{i\theta }$, $0\le r<1$, and $\alpha \le \theta \le \alpha +\pi$.

For the case $\pi \le \theta \le 2\pi$, applying the same method as above, we have

$0\ge arg\left\{\frac{f\left(z\right)}{z}\right\}\ge arg\left\{{f}^{\prime }\left(z\right)\right\},$
(2.14)

where $z=r{e}^{i\theta }$, $0\le r<1$, and $\pi +\alpha \le \theta \le 2\pi +\alpha$. Applying (2.12), (2.13), and (2.14), we have

$\begin{array}{rl}|arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|& =|arg\left\{{f}^{\prime }\left(z\right)\right\}-arg\left\{\frac{f\left(z\right)}{z}\right\}|\le |arg\left\{{f}^{\prime }\left(z\right)\right\}|\\ <\frac{\pi }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.\end{array}$

This completes the proof. □

Remark 2.9 The functions $f\left(z\right)=z+\alpha {z}^{2}/2$ satisfy the conditions of Theorem 2.8 whenever $|\alpha |\le 1/2$.

## References

1. Stankiewicz J: Quelques problèmes extrêmaux dans les classes des fonctions α -angulairement étoilées. Ann. Univ. Mariae Curie-Skłodowska, Sect. A 1966, 20: 59-75.

2. Brannan DA, Kirwan WE: On some classes of bounded univalent functions. J. Lond. Math. Soc. 1969,1(2):431-443.

3. Duren PL: Univalent Functions. Springer, New York; 1983.

4. Goodman AW: Univalent Functions. Vols. I and II. Mariner Publishing Co., Tampa; 1983.

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Correspondence to Janusz Sokół.

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