# On the improvement of Mocanu’s conditions

## Abstract

We estimate $|Arg\left\{p\left(z\right)\right\}|$ for functions of the form $p\left(z\right)=1+{a}_{1}z+{a}_{2}{z}^{2}+{a}_{3}{z}^{3}+\cdots$ in the unit disc $\mathbb{D}=\left\{z:|z|<1\right\}$ under several assumptions. By using Nunokawa’s lemma, we improve a few of Mocanu’s results obtained by differential subordinations. Some applications for strongly starlikeness and convexity are formulated.

MSC:30C45, 30C80.

## 1 Introduction

Let be the class of functions analytic in the unit disk $\mathbb{D}=\left\{z\in \mathbb{C}:|z|<1\right\}$, and denote by the class of analytic functions in and usually normalized, i.e., $\mathcal{A}=\left\{f\in \mathcal{H}:f\left(0\right)=0,{f}^{\prime }\left(0\right)=1\right\}$.

Let ${\mathcal{SS}}^{\ast }\left(\beta \right)$ denote the class of strongly starlike functions of order β, $0<\beta \le 1$,

${\mathcal{SS}}^{\ast }\left(\beta \right):=\left\{f\in \mathcal{A}:|Arg\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}|<\frac{\beta \pi }{2},z\in \mathbb{D}\right\},$

which was introduced in [1] and [2]. We say that $f\in \mathcal{A}$ is in the class ${\mathcal{SC}}^{\ast }\left(\beta \right)$ of strongly convex functions of order β when $z{f}^{\prime }\left(z\right)\in {\mathcal{SS}}^{\ast }\left(\beta \right)$. We say that $f\in \mathcal{H}$ is subordinate to $g\in \mathcal{H}$ in the unit disc , written $f\prec g$ if and only if there exists an analytic function $w\in \mathcal{H}$ such that $w\left(0\right)=0$, $|w\left(z\right)|<1$ and $f\left(z\right)=g\left[w\left(z\right)\right]$ for $z\in \mathbb{D}\subseteq g\left(\mathbb{D}\right)$. In particular, if g is univalent in then the subordination principle says that $f\prec g$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(|z| for all $r\in \left(0,1\right)$.

## 2 Main result

In this section, we investigate conditions, under which a function $f\in \mathcal{A}$ is strongly starlike or strongly convex. We also estimate $|Arg\left\{p\left(z\right)\right\}|$ for functions of the form $p\left(z\right)=1+{a}_{1}z+{a}_{2}{z}^{2}+{a}_{3}{z}^{3}+\cdots$ in the unit disc , under several assumptions, and then we use this estimation for the case $p\left(z\right)=z{f}^{\prime }\left(z\right)/f\left(z\right)$. By using Nunokawa’s lemma [3], we improve a few Mocanu’s [4, 5] results obtained by differential subordinations. Some sufficient conditions for functions to be in several subclasses of strongly starlike functions can also be found in the recent papers [6] and [711].

Theorem 2.1 Let $f\left(z\right)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in the unit disc . If

$|Arg\left\{{f}^{\prime }\left(z\right)\right\}|<\frac{\alpha \pi }{2}\approx 1.0076658,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.1)

where $\alpha =1/\left(1+\beta \right)=1/\left(2-\left(log4\right)/\pi \right)\approx 0.641548$, $\beta =1-\left(log4\right)/\pi \approx 0.5587$, then

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

or f is starlike in .

Proof By (2.1), we have

${\left\{{f}^{\prime }\left(z\right)\right\}}^{1/\alpha }\prec \frac{1+z}{1-z},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Let $z=\rho {e}^{i\theta }$, $\rho \in \left[0,1\right)$, $\theta \in \left(-\pi ,\pi \right]$. The function $w\left(z\right)=\left(1+z\right)/\left(1-z\right)$ is univalent in and maps $|z|<\rho <1$ onto the open disc $D\left(C,R\right)$ with the center $C=\left(1+{\rho }^{2}\right)/\left(1-{\rho }^{2}\right)$ and the radius $R=\left(2\rho \right)/\left(1-{\rho }^{2}\right)$. Then by the subordination principle under univalent function,

(2.3)

A simple geometric observation yields to

(2.4)

Therefore, applying the same idea as [[3], pp.1292-1293] for $z=r{e}^{i\theta }$, $r\in \left[0,1\right)$, $\theta \in \left(-\pi ,\pi \right]$, we have

$\begin{array}{rcl}|Arg\left\{\frac{f\left(z\right)}{z}\right\}|& =& |Arg\left\{{\int }_{0}^{r}{f}^{\prime }\left(\rho {e}^{i\theta }\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}|\\ \le & {\int }_{0}^{r}|Arg\left\{{f}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}|\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \\ =& \alpha {\int }_{0}^{r}|Arg\left\{{\left({f}^{\prime }\left(\rho {e}^{i\theta }\right)\right)}^{1/\alpha }\right\}|\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \\ \le & \alpha {\int }_{0}^{r}{sin}^{-1}\frac{2\rho }{1+{\rho }^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \\ =& \alpha \left\{\rho {sin}^{-1}\frac{2\rho }{1+{\rho }^{2}}-log\left(1+{\rho }^{2}\right)\right\}{|}_{\rho =0}^{\rho =r}\\ =& \alpha \left\{r{sin}^{-1}\frac{2r}{1+{r}^{2}}-log\left(1+{r}^{2}\right)\right\}.\end{array}$

The function

$h\left(r\right)=r{sin}^{-1}\frac{2r}{1+{r}^{2}}-log\left(1+{r}^{2}\right),\phantom{\rule{1em}{0ex}}r\in \left[0,1\right)$

is increasing because ${h}^{\prime }\left(r\right)={sin}^{-1}\left\{2r/\left(1+{r}^{2}\right)\right\}>0$. Now, letting $r\to {1}^{-}$, we obtain

$\begin{array}{rcl}|Arg\left\{\frac{f\left(z\right)}{z}\right\}|& \le & \alpha \left(\pi /2-log2\right)=\frac{\alpha \pi }{2}\left\{1-\frac{log4}{\pi }\right\}\\ =& \frac{\pi }{2}\alpha \beta ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.\end{array}$

Using this and (2.1), we obtain

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

It completes the proof. □

Remark 2.2 Theorem 2.1 is an improvement of Mocanu’s result in [4].

Theorem 2.3 Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in the unit disc . If

$\mathfrak{Re}\left\{p\left(z\right)+z{p}^{\prime }\left(z\right)\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.5)

then

$|Arg\left\{p\left(z\right)\right\}|<\frac{\pi }{2}-log2=0.877649\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.6)

Proof By (2.5), we have

$p\left(z\right)+z{p}^{\prime }\left(z\right)\prec \frac{1+z}{1-z},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.7)

Let $z=\rho {e}^{i\theta }$, $\rho \in \left[0,1\right)$, $\theta \in \left(-\pi ,\pi \right]$. The subordination principle used for (2.7) gives

(2.8)

A simple geometric observation yields to

(2.9)

Therefore, for $z=r{e}^{i\theta }$, $r\in \left[0,1\right)$, $\theta \in \left(-\pi ,\pi \right]$, we have

$\begin{array}{rl}|Arg\left\{p\left(z\right)\right\}|& =|Arg\left\{\frac{zp\left(z\right)}{z}\right\}|\\ =|Arg\left\{\frac{{\int }_{0}^{z}{\left(tp\left(t\right)\right)}^{\prime }\phantom{\rule{0.2em}{0ex}}\mathrm{d}t}{z}\right\}|\\ =|Arg\left\{\frac{{\int }_{0}^{z}\left(p\left(t\right)+t{p}^{\prime }\left(t\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t}{z}\right\}|\\ =|Arg\left\{\frac{{\int }_{0}^{r}\left(p\left(\rho {e}^{i\theta }\right)+\rho {e}^{i\theta }{p}^{\prime }\left(\rho {e}^{i\theta }\right)\right){e}^{i\theta }\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho }{r{e}^{i\theta }}\right\}|\\ =|Arg\left\{{\int }_{0}^{r}\left(p\left(\rho {e}^{i\theta }\right)+\rho {e}^{i\theta }{p}^{\prime }\left(\rho {e}^{i\theta }\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \right\}-Arg\left\{r\right\}|\\ \le {\int }_{0}^{r}|Arg\left\{p\left(\rho {e}^{i\theta }\right)+\rho {e}^{i\theta }{p}^{\prime }\left(\rho {e}^{i\theta }\right)\right\}|\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho .\end{array}$

Therefore, by using (2.9), we have

$\begin{array}{rcl}|Arg\left\{p\left(z\right)\right\}|& \le & \alpha {\int }_{0}^{r}{sin}^{-1}\frac{2\rho }{1+{\rho }^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \\ =& \alpha \left\{\rho {sin}^{-1}\frac{2\rho }{1+{\rho }^{2}}-log\left(1+{\rho }^{2}\right)\right\}{|}_{\rho =0}^{\rho =r}\\ <& \alpha \left\{\rho {sin}^{-1}\frac{2\rho }{1+{\rho }^{2}}-log\left(1+{\rho }^{2}\right)\right\}{|}_{\rho =0}^{\rho =1}\\ =& \frac{\pi }{2}-log2=0.8776491464\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.\end{array}$

It leads to the desired conclusion. □

Remark 2.4 Theorem 2.3 is an improvement of Mocanu’s result in [5], where instead of ${\gamma }_{0}=\frac{\pi }{2}-log2=0.8776491464\dots$ is

${\theta }_{1}=max\left\{\theta :|Arg\left\{\frac{2}{{e}^{i\theta }}log\left(1+{e}^{i\theta }\right)-1\right\}|\right\}=0.91106219\dots .$

Substituting $p\left(z\right)=f\left(z\right)/z$, $f\in \mathcal{A}$, in Theorem 2.3 leads to the following corollary.

Corollary 2.5 If $f\in \mathcal{A}$ and it satisfies

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

then

$|Arg\left\{\frac{f\left(z\right)}{z}\right\}|<\frac{\pi }{2}-log2=0.877649\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

Substituting $p\left(z\right)=z{f}^{\prime }\left(z\right)/f\left(z\right)$, $f\in \mathcal{A}$, in Theorem 2.3 gives the following corollary.

Corollary 2.6 If $f\in \mathcal{A}$ and it satisfies

$\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\left(2+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)-{\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)}^{2}\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$

then

$|Arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|<\frac{\pi }{2}-log2=0.877649\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

This means that f is strongly starlike of order $1-\left(log4\right)/\pi =0.558728799\dots$ .

Substituting $p\left(z\right)=1+z{f}^{″}\left(z\right)/{f}^{\prime }\left(z\right)$, $f\in \mathcal{A}$, in Theorem 2.3 gives the following corollary.

Corollary 2.7 If $f\in \mathcal{A}$ and it satisfies

$\mathfrak{Re}\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\left(2+\frac{z{f}^{‴}\left(z\right)}{{f}^{″}\left(z\right)}\right)-{\left(\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right)}^{2}\right\}>0,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$

then

$|Arg\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}|<\frac{\pi }{2}-log2=0.877649\dots ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$

This means that f is strongly convex of order $1-\left(log4\right)/\pi =0.558728799\dots$ .

Theorem 2.8 Let $f\left(z\right)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in the unit disc , and suppose that

$|{f}^{\prime }\left(z\right)-1|<1,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.10)

Then we have

$|Arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|<\left(1+r\right){sin}^{-1}r+\sqrt{1-{r}^{2}}-1,$
(2.11)

where $r=|z|<1$, and, therefore, we have

$\mathfrak{Re}\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}>0\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.25em}{0ex}}|z|<{r}_{0},$
(2.12)

where $0.902<{r}_{0}<0.903$ is the positive root of the equation

${sin}^{-1}r=\frac{\pi -2\left(\sqrt{1-{r}^{2}}-1\right)}{2\left(1+r\right)}.$
(2.13)

Proof From (2.10), we have ${f}^{\prime }\left(z\right)\prec 1+z$, so the subordination principle gives

$|Arg\left\{{f}^{\prime }\left(z\right)\right\}|\le {sin}^{-1}|z|,\phantom{\rule{1em}{0ex}}z\in \mathbb{D}$
(2.14)

and for $z=r{e}^{i\theta }$,

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

Then we have

${\int }_{0}^{r}{sin}^{-1}\rho \phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho =r{sin}^{-1}r+\sqrt{1-{r}^{2}}-1.$

Therefore, and from (2.14), we have

$\begin{array}{c}|Arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}\le |Arg\left\{{f}^{\prime }\left(z\right)\right\}|+|Arg\left\{\frac{f\left(z\right)}{z}\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}<{sin}^{-1}r+r{sin}^{-1}r+\sqrt{1-{r}^{2}}-1,\phantom{\rule{1em}{0ex}}|z|=r<1.\hfill \end{array}$

The function

$\begin{array}{rcl}G\left(r\right)& =& \left(1+r\right){sin}^{-1}r+\sqrt{1-{r}^{2}}-1\\ =& {sin}^{-1}r+{\int }_{0}^{r}{sin}^{-1}\rho \phantom{\rule{0.2em}{0ex}}\mathrm{d}\rho \end{array}$

increases in $\left[0,1\right]$ as the sum of two increasing functions. Moreover, $G\left(0\right)=0$, $G\left(1\right)=\pi -1$, and it satisfies

$G\left(0.902\right)=1.57030\dots <\frac{\pi }{2}=1.5707963\dots

Therefore, the equation (2.13) has the solution ${r}_{0}$, $0.902<{r}_{0}<0.903$, and

This completes the proof. □

Theorem 2.9 Let $f\left(z\right)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ be analytic in the unit disc , and suppose that

$|{f}^{\prime }\left(z\right)-1|<\alpha ,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.15)

with $\alpha \in \left(0,2/\sqrt{5}\right]$. Then f is strongly starlike of order β, where $\beta \in \left(0,1\right]$ is the positive root of the equation

${sin}^{-1}\left\{\alpha \sqrt{1-{\alpha }^{2}/4}+\frac{\alpha }{2}\sqrt{1-{\alpha }^{2}}\right\}=\frac{\pi \beta }{2}.$
(2.16)

Proof We have ${f}^{\prime }\left(z\right)\prec 1+\alpha z$. Applying the result from [[4], p.118] we have also that $f\left(z\right)/z\prec 1+\alpha z/2$ in . This shows that

$|Arg\left\{{f}^{\prime }\left(z\right)\right\}|\le {sin}^{-1}\alpha |z|,\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.17)

and

$|Arg\left\{\frac{f\left(z\right)}{z}\right\}|\le {sin}^{-1}\frac{\alpha |z|}{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.$
(2.18)

Therefore, using (2.17) and (2.18), we have

$\begin{array}{c}|Arg\left\{\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}\le |Arg\left\{{f}^{\prime }\left(z\right)\right\}|+|Arg\left\{\frac{f\left(z\right)}{z}\right\}|\hfill \\ \phantom{\rule{1em}{0ex}}<{sin}^{-1}\alpha +{sin}^{-1}\frac{\alpha }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D}.\hfill \end{array}$

For $\alpha \in \left(0,2/\sqrt{5}\right]$, we have ${\alpha }^{2}+{\left(\alpha /2\right)}^{2}\le 1$, so we can use the formula

${sin}^{-1}\alpha +{sin}^{-1}\frac{\alpha }{2}={sin}^{-1}\left\{\alpha \sqrt{1-{\alpha }^{2}/4}+\frac{\alpha }{2}\sqrt{1-{\alpha }^{2}}\right\}.$

The function

$\begin{array}{rcl}H\left(\alpha \right)& =& {sin}^{-1}\left\{\alpha \sqrt{1-{\alpha }^{2}/4}+\frac{\alpha }{2}\sqrt{1-{\alpha }^{2}}\right\}\\ =& {sin}^{-1}\alpha +{sin}^{-1}\frac{\alpha }{2}\end{array}$

increases in the segment $\left[0,2/\sqrt{5}\right]$ as the sum of two increasing functions. Moreover, $H\left(0\right)=0$, $H\left(2/\sqrt{5}\right)=\pi /2$, so the equation (2.16) has in $\left(0,1\right]$ the solution β. This completes the proof. □

Putting $\alpha =2/\sqrt{5}$, we get $\beta =1$ and Theorem 2.15 becomes the result from [[4], p.118]:

$\left[|{f}^{\prime }\left(z\right)-1|<2/\sqrt{5},z\in \mathbb{D}\right]\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}\left[\mathfrak{Re}\left\{z{f}^{\prime }\left(z\right)/f\left(z\right)\right\}>0,z\in \mathbb{D}\right].$

Theorem 2.10 Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}{z}^{n}$ be analytic in the unit disc , and suppose that

$|Arg\left\{p\left(z\right)+\alpha \left(\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}\right)\right\}|<{tan}^{-1}\frac{|\alpha |\delta \left(\beta \right)sin\left(\left(1+\beta \right)\pi /2\right)}{1+|\alpha |\delta \left(\beta \right)cos\left(\left(1+\beta \right)\pi /2\right)}-\frac{\pi \beta }{2},\phantom{\rule{1em}{0ex}}z\in \mathbb{D},$
(2.19)

where $\alpha <0$, $0<\beta <1$, and

$\delta \left(\beta \right)=\frac{\beta }{2}\left({\left(\frac{1-\beta }{1+\beta }\right)}^{\beta +1}+{\left(\frac{1-\beta }{1+\beta }\right)}^{\beta -1}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}|\alpha |>\frac{sin\left(\pi \beta /2\right)}{\delta \left(\beta \right)}.$

Then $Arg\left\{p\left(z\right)\right\}<\frac{\beta \pi }{2}$ in .

Proof Suppose that there exists a point ${z}_{0}\in \mathbb{D}$ such that

(2.20)

and

$|Arg\left\{p\left({z}_{0}\right)\right\}|=\frac{\pi \beta }{2},$

then by Nunokawa’s lemma [12], we have

${\left\{p\left({z}_{0}\right)\right\}}^{1/\beta }=±ia,\phantom{\rule{1em}{0ex}}a>0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{{z}_{0}{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}=ik\beta ,$

where

and

moreover,

$\frac{\beta k}{{a}^{\beta }}\ge \delta \left(\beta \right).$
(2.21)

For the case $Arg\left\{p\left({z}_{0}\right)\right\}=\frac{\pi \beta }{2}$, we have from (2.21),

$\begin{array}{rcl}Arg\left\{p\left({z}_{0}\right)+\alpha \left(\frac{z{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)\right\}& =& Arg\left\{p\left({z}_{0}\right)\right\}+Arg\left\{1+\alpha \left(\frac{z{p}^{\prime }\left({z}_{0}\right)}{{p}^{2}\left({z}_{0}\right)}\right)\right\}\\ =& \frac{\pi \beta }{2}+Arg\left\{1+\frac{|\alpha |\beta k}{{a}^{\beta }}{e}^{-i\pi \left(1+\beta \right)/2}\right\}\\ \le & -\left\{{tan}^{-1}\left(\frac{|\alpha |\delta \left(\beta \right)sin\frac{\pi \left(1+\beta \right)}{2}}{1+|\alpha |\delta \left(\beta \right)cos\frac{\pi \left(1+\beta \right)}{2}}\right)-\frac{\pi \beta }{2}\right\}.\end{array}$

This contradicts (2.19), and for the case $Arg\left\{p\left({z}_{0}\right)\right\}=-\frac{\pi \beta }{2}$, applying the same method as above, we have

$Arg\left\{p\left({z}_{0}\right)+\alpha \left(\frac{z{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)\right\}\ge \left\{{tan}^{-1}\left(\frac{|\alpha |\delta \left(\beta \right)sin\frac{\pi \left(1+\beta \right)}{2}}{1+|\alpha |\delta \left(\beta \right)cos\frac{\pi \left(1+\beta \right)}{2}}\right)-\frac{\pi \beta }{2}\right\}.$

This contradicts also (2.19), and, therefore, it completes the proof. □

Theorem 2.11 Let $p\left(z\right)=1+{\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}{z}^{n}$ be analytic in the unit disc , and suppose that

$\begin{array}{r}|Arg\left\{p\left(z\right)+\alpha \left(\frac{z{p}^{\prime }\left(z\right)}{p\left(z\right)}\right)\right\}|\\ \phantom{\rule{1em}{0ex}}<\frac{\pi }{2}\left\{\beta +\frac{2}{\pi }{tan}^{-1}\frac{|\alpha |\delta \left(\beta \right)sin\left(\left(1-\beta \right)\pi /2\right)}{1+|\alpha |\delta \left(\beta \right)cos\left(\left(1-\beta \right)\pi /2\right)}\right\}\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.25em}{0ex}}z\in \mathbb{D},\end{array}$
(2.22)

where $0<\alpha$, $0<\beta <1$, and

$\delta \left(\beta \right)=\frac{\beta }{2}\left({\left(\frac{1-\beta }{1+\beta }\right)}^{\beta +1}+{\left(\frac{1-\beta }{1+\beta }\right)}^{\beta -1}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\alpha >\frac{sin\left(\pi \beta /2\right)}{\delta \left(\beta \right)}.$

Then $Arg\left\{p\left(z\right)\right\}<\frac{\pi \beta }{2}$ in .

Proof The proof runs as the previous proof, take $\alpha >0$ into account. Suppose that there exists a point ${z}_{0}\in \mathbb{D}$ such that

(2.23)

and

$|Arg\left\{p\left({z}_{0}\right)\right\}|=\frac{\pi \beta }{2},$

then by Nunokawa’s lemma [12], we have for the case $Arg\left\{p\left({z}_{0}\right)\right\}=\frac{\pi \beta }{2}$,

$\begin{array}{rcl}Arg\left\{p\left({z}_{0}\right)+\alpha \left(\frac{z{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)\right\}& =& Arg\left\{p\left({z}_{0}\right)\right\}+Arg\left\{1+\alpha \left(\frac{z{p}^{\prime }\left({z}_{0}\right)}{{p}^{2}\left({z}_{0}\right)}\right)\right\}\\ =& \frac{\pi \beta }{2}+Arg\left\{1+\frac{\alpha \beta k}{{a}^{\beta }}{e}^{i\pi \left(1+\beta \right)/2}\right\}\\ \le & \frac{\pi \beta }{2}+{tan}^{-1}\left(\frac{|\alpha |\delta \left(\beta \right)sin\frac{\pi \left(1+\beta \right)}{2}}{1+|\alpha |\delta \left(\beta \right)cos\frac{\pi \left(1+\beta \right)}{2}}\right)\frac{\pi \beta }{2}.\end{array}$

This contradicts (2.22), and for the case $Arg\left\{p\left({z}_{0}\right)\right\}=-\frac{\pi \beta }{2}$, applying the same method as above, we have

$Arg\left\{p\left({z}_{0}\right)+\alpha \left(\frac{z{p}^{\prime }\left({z}_{0}\right)}{p\left({z}_{0}\right)}\right)\right\}\le -\left\{\frac{\pi \beta }{2}+{tan}^{-1}\left(\frac{|\alpha |\delta \left(\beta \right)sin\frac{\pi \left(1+\beta \right)}{2}}{1+|\alpha |\delta \left(\beta \right)cos\frac{\pi \left(1+\beta \right)}{2}}\right)\frac{\pi \beta }{2}\right\}.$

This contradicts also (2.22), and, therefore, it completes the proof. □

## References

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## Acknowledgements

This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2011-0007037).

Dedicated to Professor Hari M Srivastava.

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Correspondence to NE Cho.

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The authors declare that they have no competing interests.

### Authors’ contributions

All authors jointly worked on the results, and they read and approved the final manuscript.

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Nunokawa, M., Owa, S., Cho, N. et al. On the improvement of Mocanu’s conditions. J Inequal Appl 2013, 426 (2013). https://doi.org/10.1186/1029-242X-2013-426