On the improvement of Mocanu’s conditions

Abstract

We estimate $|Arg\{p(z)\}|$ for functions of the form $p(z)=1+a_{1}z+a_2{z}^2+a_3{z}^3+\cdots$ in the unit disc $\mathbb{D}=\{z:|z|<1\}$ 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 ℋ be the class of functions analytic in the unit disk $\mathbb{D}=\{z\in \mathbb{C}:|z|<1\}$, and denote by the class of analytic functions in and usually normalized, i.e., $\mathcal{A}=\{f\in \mathcal{H}:f(0)=0,f^{\prime }(0)=1\}$.

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

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

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

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 $|Arg\{p(z)\}|$ for functions of the form $p(z)=1+a_{1}z+a_2{z}^2+a_3{z}^3+\cdots$ in the unit disc , under several assumptions, and then we use this estimation for the case $p(z)=z{f}^{\prime }(z)/f(z)$. 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 [7–11].

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

$$|Arg\{f^{\prime }(z)\}|<\frac{\alpha \pi }{2}\approx 1.0076658,z\in \mathbb{D},$$

(2.1)

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

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

(2.2)

or f is starlike in .

Proof By (2.1), we have

$${\{f^{\prime }(z)\}}^{1/\alpha }\prec \frac{1+z}{1-z},z\in \mathbb{D}.$$

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

$${\left\{f^{\prime }\left(x{e}^{i\theta }\right)\right\}}^{1/\alpha }\in D(C,R)\text{for all }x\in [0,\rho ),\theta \in (-\pi ,\pi ].$$

(2.3)

A simple geometric observation yields to

$$|Arg\left\{{\left(f^{\prime }\left(\rho e^{i\theta }\right)\right)}^{1/\alpha }\right\}|\le {sin}^{-1}\frac{R}{C}={sin}^{-1}\frac{2\rho }{1+{\rho }^2}\text{for all }\rho \in [0,1),\theta \in (-\pi ,\pi ].$$

(2.4)

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

$$\begin{array}{rcl}|Arg\left\{\frac{f(z)}{z}\right\}|& =& |Arg\left\{{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)\mathrm{d}\rho \right\}|\\ \le & {\int }_0^r|Arg\left\{f^{\prime }\left(\rho e^{i\theta }\right)\right\}|\mathrm{d}\rho \\ =& \alpha {\int }_0^r|Arg\left\{{\left(f^{\prime }\left(\rho e^{i\theta }\right)\right)}^{1/\alpha }\right\}|\mathrm{d}\rho \\ \le & \alpha {\int }_0^r{sin}^{-1}\frac{2\rho }{1+{\rho }^2}\mathrm{d}\rho \\ =& \alpha \{\rho {sin}^{-1}\frac{2\rho }{1+{\rho }^2}-log(1+{\rho }^2)\}{|}_{\rho =0}^{\rho =r}\\ =& \alpha \{r{sin}^{-1}\frac{2r}{1+r^2}-log(1+r^2)\}.\end{array}$$

The function

$$h(r)=r{sin}^{-1}\frac{2r}{1+r^2}-log(1+r^2),r\in [0,1)$$

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

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

Using this and (2.1), we obtain

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

It completes the proof. □

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

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

$$\mathfrak{Re}\{p(z)+z{p}^{\prime }(z)\}>0,z\in \mathbb{D},$$

(2.5)

then

$$|Arg\{p(z)\}|<\frac{\pi }{2}-log2=0.877649\dots ,z\in \mathbb{D}.$$

(2.6)

Proof By (2.5), we have

$$p(z)+z{p}^{\prime }(z)\prec \frac{1+z}{1-z},z\in \mathbb{D}.$$

(2.7)

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

$$|p\left(x{e}^{i\theta }\right)+x{e}^{i\theta }{p}^{\prime }\left(x{e}^{i\theta }\right)-\frac{1+{\rho }^2}{1-{\rho }^2}|<\frac{2\rho }{1-{\rho }^2}\text{for all }x\in [0,\rho ),\theta \in (-\pi ,\pi ].$$

(2.8)

A simple geometric observation yields to

$$|Arg\{p\left(\rho e^{i\theta }\right)+\rho e^{i\theta }{p}^{\prime }\left(\rho e^{i\theta }\right)\}|\le {sin}^{-1}\frac{2\rho }{1+{\rho }^2}\text{for all }\rho \in [0,1),\theta \in (-\pi ,\pi ].$$

(2.9)

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

$$\begin{array}{rl}|Arg\{p(z)\}|& =|Arg\left\{\frac{zp(z)}{z}\right\}|\\ =|Arg\left\{\frac{{\int }_0^z{(tp(t))}^{\prime }\mathrm{d}t}{z}\right\}|\\ =|Arg\left\{\frac{{\int }_0^z(p(t)+t{p}^{\prime }(t))\mathrm{d}t}{z}\right\}|\\ =|Arg\left\{\frac{{\int }_0^r(p(\rho e^{i\theta })+\rho e^{i\theta }{p}^{\prime }(\rho e^{i\theta }))e^{i\theta }\mathrm{d}\rho }{r{e}^{i\theta }}\right\}|\\ =|Arg\left\{{\int }_0^r(p\left(\rho e^{i\theta }\right)+\rho e^{i\theta }{p}^{\prime }\left(\rho e^{i\theta }\right))\mathrm{d}\rho \right\}-Arg\{r\}|\\ \le {\int }_0^r|Arg\{p\left(\rho e^{i\theta }\right)+\rho e^{i\theta }{p}^{\prime }\left(\rho e^{i\theta }\right)\}|\mathrm{d}\rho .\end{array}$$

Therefore, by using (2.9), we have

$$\begin{array}{rcl}|Arg\{p(z)\}|& \le & \alpha {\int }_0^r{sin}^{-1}\frac{2\rho }{1+{\rho }^2}\mathrm{d}\rho \\ =& \alpha \{\rho {sin}^{-1}\frac{2\rho }{1+{\rho }^2}-log(1+{\rho }^2)\}{|}_{\rho =0}^{\rho =r}\\ <& \alpha \{\rho {sin}^{-1}\frac{2\rho }{1+{\rho }^2}-log(1+{\rho }^2)\}{|}_{\rho =0}^{\rho =1}\\ =& \frac{\pi }{2}-log2=0.8776491464\dots ,z\in \mathbb{D}.\end{array}$$

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

$${\theta }_1=max\{\theta :|Arg\{\frac{2}{e^{i\theta }}log(1+e^{i\theta })-1\}\left|\right\}=0.91106219\dots .$$

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

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

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

then

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

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

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

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

then

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

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

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

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

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

then

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

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

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

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

(2.10)

Then we have

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

(2.11)

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

$$\mathfrak{Re}\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}>0\mathit{\text{for}}|z|<r_0,$$

(2.12)

where $0.902<r_0<0.903$ is the positive root of the equation

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

(2.13)

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

$$|Arg\{f^{\prime }(z)\}|\le {sin}^{-1}|z|,z\in \mathbb{D}$$

(2.14)

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

$$\begin{array}{rl}|Arg\left\{\frac{f(z)}{z}\right\}|& =|Arg\left\{\frac{1}{r{e}^{i\theta }}{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)e^{i\theta }\mathrm{d}\rho \right\}|\\ =|Arg\left\{{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)\mathrm{d}\rho \right\}|\\ \le {\int }_0^r|Arg\left\{f^{\prime }\left(\rho e^{i\theta }\right)\right\}|\mathrm{d}\rho \\ <{\int }_0^r{sin}^{-1}\rho \mathrm{d}\rho .\end{array}$$

Then we have

$${\int }_0^r{sin}^{-1}\rho \mathrm{d}\rho =r{sin}^{-1}r+\sqrt{1-r^2}-1.$$

Therefore, and from (2.14), we have

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

The function

$$\begin{array}{rcl}G(r)& =& (1+r){sin}^{-1}r+\sqrt{1-r^2}-1\\ =& {sin}^{-1}r+{\int }_0^r{sin}^{-1}\rho \mathrm{d}\rho \end{array}$$

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

$$G(0.902)=1.57030\dots <\frac{\pi }{2}=1.5707963\dots <G(0.903)=1.573753\dots .$$

Therefore, the equation (2.13) has the solution $r_0$, $0.902<r_0<0.903$, and

$$\mathfrak{Re}\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}>0\text{for }|z|<r_0\approx 0.903.$$

This completes the proof. □

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

$$|f^{\prime }(z)-1|<\alpha ,z\in \mathbb{D},$$

(2.15)

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

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

(2.16)

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

$$|Arg\{f^{\prime }(z)\}|\le {sin}^{-1}\alpha |z|,z\in \mathbb{D},$$

(2.17)

and

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

(2.18)

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

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

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

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

The function

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

increases in the segment $[0,2/\sqrt{5}]$ as the sum of two increasing functions. Moreover, $H(0)=0$, $H(2/\sqrt{5})=\pi /2$, so the equation (2.16) has in $(0,1]$ 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]:

$$[|f^{\prime }(z)-1|<2/\sqrt{5},z\in \mathbb{D}]\Rightarrow [\mathfrak{Re}\{z{f}^{\prime }(z)/f(z)\}>0,z\in \mathbb{D}].$$

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

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

(2.19)

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

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

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

Proof Suppose that there exists a point $z_0\in \mathbb{D}$ such that

$$|Arg\{p(z)\}|<\frac{\pi \beta }{2}\text{for }|z|<|z_0|$$

(2.20)

and

$$|Arg\{p(z_0)\}|=\frac{\pi \beta }{2},$$

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

$${\{p(z_0)\}}^{1/\beta }=\pm ia,a>0\text{and}\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}=ik\beta ,$$

where

$$k\ge \frac{1}{2}(a+\frac{1}{a})\text{when }Arg\{p(z_0)\}=\frac{\pi \beta }{2}$$

and

$$k\le -\frac{1}{2}(a+\frac{1}{a})\text{when }Arg\{p(z_0)\}=-\frac{\pi \beta }{2},$$

moreover,

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

(2.21)

For the case $Arg\{p(z_0)\}=\frac{\pi \beta }{2}$, we have from (2.21),

$$\begin{array}{rcl}Arg\{p(z_0)+\alpha \left(\frac{z{p}^{\prime }(z_0)}{p(z_0)}\right)\}& =& Arg\{p(z_0)\}+Arg\{1+\alpha \left(\frac{z{p}^{\prime }(z_0)}{p^2(z_0)}\right)\}\\ =& \frac{\pi \beta }{2}+Arg\{1+\frac{|\alpha |\beta k}{a^{\beta }}{e}^{-i\pi (1+\beta )/2}\}\\ \le & -\{{tan}^{-1}\left(\frac{|\alpha |\delta (\beta )sin\frac{\pi (1+\beta )}{2}}{1+|\alpha |\delta (\beta )cos\frac{\pi (1+\beta )}{2}}\right)-\frac{\pi \beta }{2}\}.\end{array}$$

This contradicts (2.19), and for the case $Arg\{p(z_0)\}=-\frac{\pi \beta }{2}$, applying the same method as above, we have

$$Arg\{p(z_0)+\alpha \left(\frac{z{p}^{\prime }(z_0)}{p(z_0)}\right)\}\ge \{{tan}^{-1}\left(\frac{|\alpha |\delta (\beta )sin\frac{\pi (1+\beta )}{2}}{1+|\alpha |\delta (\beta )cos\frac{\pi (1+\beta )}{2}}\right)-\frac{\pi \beta }{2}\}.$$

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

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

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

(2.22)

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

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

Then $Arg\{p(z)\}<\frac{\pi \beta }{2}$ 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

$$|Arg\{p(z)\}|<\frac{\pi \beta }{2}\text{for }|z|<|z_0|$$

(2.23)

and

$$|Arg\{p(z_0)\}|=\frac{\pi \beta }{2},$$

then by Nunokawa’s lemma [12], we have for the case $Arg\{p(z_0)\}=\frac{\pi \beta }{2}$,

$$\begin{array}{rcl}Arg\{p(z_0)+\alpha \left(\frac{z{p}^{\prime }(z_0)}{p(z_0)}\right)\}& =& Arg\{p(z_0)\}+Arg\{1+\alpha \left(\frac{z{p}^{\prime }(z_0)}{p^2(z_0)}\right)\}\\ =& \frac{\pi \beta }{2}+Arg\{1+\frac{\alpha \beta k}{a^{\beta }}{e}^{i\pi (1+\beta )/2}\}\\ \le & \frac{\pi \beta }{2}+{tan}^{-1}\left(\frac{|\alpha |\delta (\beta )sin\frac{\pi (1+\beta )}{2}}{1+|\alpha |\delta (\beta )cos\frac{\pi (1+\beta )}{2}}\right)\frac{\pi \beta }{2}.\end{array}$$

This contradicts (2.22), and for the case $Arg\{p(z_0)\}=-\frac{\pi \beta }{2}$, applying the same method as above, we have

$$Arg\{p(z_0)+\alpha \left(\frac{z{p}^{\prime }(z_0)}{p(z_0)}\right)\}\le -\{\frac{\pi \beta }{2}+{tan}^{-1}\left(\frac{|\alpha |\delta (\beta )sin\frac{\pi (1+\beta )}{2}}{1+|\alpha |\delta (\beta )cos\frac{\pi (1+\beta )}{2}}\right)\frac{\pi \beta }{2}\}.$$

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