On the multivalency of certain analytic functions

Abstract

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

MSC:30C45, 30C80.

1 Introduction

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

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

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

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

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

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

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

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

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

and

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

For $p=1$ the classes ${\mathcal{S}}_{\alpha }^{\ast }(p)$ and ${\mathcal{C}}_{\alpha }(p)$ 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 [1] and [2] 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 [3]. In this paper, we need the following lemma.

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

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

(1.1)

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

$$|arg\left\{\frac{z{f}^{\prime }(z)}{f(z)}\right\}|\le |arg\{f^{\prime }(z)\}|,z\in \mathbb{D}.$$

Proof First we note that from

$$0\le arg\{z_1\}\le arg\{z_2\}\le \pi \Rightarrow arg\{z_1\}\le arg\{z_1+z_2\}\le arg\{z_2\},$$

the implication

$$0\le arg\{z_1\}\le \cdots \le arg\{z_n\}\le \pi \Rightarrow arg\{z_1\}\le arg\left\{\sum _{k=1}^n{z}_k\right\}\le arg\{z_n\}$$

(1.2)

follows by mathematical induction.

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

$$\begin{array}{rcl}arg\left\{\frac{f(z)}{z}\right\}& =& arg\left\{\frac{1}{r{e}^{i\theta }}{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)e^{i\theta }\mathrm{d}\rho \right\}\\ =& arg\left\{{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)\mathrm{d}\rho \right\}.\end{array}$$

(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) $arg\{f^{\prime }(\rho e^{i\theta })\}$ 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

$$\begin{array}{rcl}arg\left\{\frac{f(z)}{z}\right\}& =& arg\left\{{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)\mathrm{d}\rho \right\}\\ =& arg\left\{\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^n{f}^{\prime }\left({\rho }_k{e}^{i\theta }\right)\mathrm{\Delta }{\rho }_k\right\}\\ =& \underset{n\to \mathrm{\infty }}{lim}arg\left\{\sum _{k=1}^n{f}^{\prime }\left({\rho }_k{e}^{i\theta }\right)\mathrm{\Delta }{\rho }_k\right\}\\ \le & arg\left\{f^{\prime }\left(r{e}^{i\theta }\right)\right\}\end{array}$$

(1.6)

or we have

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

(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 $arg\{f^{\prime }(\rho e^{i\theta })\}$ 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

$$\begin{array}{rcl}arg\left\{\frac{f(z)}{z}\right\}& =& arg\left\{{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)\mathrm{d}\rho \right\}\\ \ge & arg\left\{f^{\prime }\left(r{e}^{i\theta }\right)\right\}\end{array}$$

and we also have

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

(1.8)

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

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

It completes the proof of Lemma 1.1. □

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

$$\left(\mathfrak{Im}\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right)\mathfrak{Im}\{z\}\ge 0,z\in \mathbb{D},$$

(1.9)

then we have

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

(1.10)

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

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

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

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

In both cases, we have (1.10). □

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

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

(1.11)

Recall that if $f(z)$ is analytic in $\mathbb{D}.$ and $(\mathfrak{Im}\{f(z)\})(\mathfrak{Im}\{z\})\ge 0$ in $\mathbb{D}.$, then f is called typically real function; see [[4], Chapter 10]. Therefore, Corollary 1.2 says that if $z{f}^{″}(z)/f^{\prime }(z)$ is a typically real function, then $z{f}^{\prime }(z)/f(z)$ is a typically real function, too.

2 Main result

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

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

(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

$$|arg\{f^{\prime }(z)\}|\le \frac{\pi }{2},z\in \mathbb{D}.$$

(2.2)

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

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

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

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

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

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

$$|arg\left\{\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right\}|\le |arg\{f^{″}(z)\}|,z\in \mathbb{D}.$$

Applying Lemma 2.2, we have the following theorem.

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

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

(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

$$|arg\{f^{″}(z)\}|<\frac{\pi }{2},z\in \mathbb{D}.$$

(2.4)

Then we have $f(z)\in \mathcal{C}(2)={\mathcal{C}}_1(2)$ or $f(z)$ is 2-valently convex in $\mathbb{D}.$.

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

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

Therefore, we have

$$1+\mathfrak{Re}\frac{z{f}^{″}(z)}{f^{\prime }(z)}>0,z\in \mathbb{D}.$$

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

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

(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

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

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

$$\begin{array}{rcl}arg\left\{\frac{f(z)}{z^p}\right\}& =& arg\left\{\frac{1}{r^p{e}^{ip\theta }}{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)e^{i\theta }\mathrm{d}\rho \right\}\\ =& arg\left\{{\int }_0^r\right|f^{\prime }\left(\rho e^{i\theta }\right)\left|e^{i(arg\{f^{\prime }(\rho e^{i\theta })\}-(p-1)\theta )}\mathrm{d}\rho \right\}\\ \le & arg\left\{f^{\prime }\left(r{e}^{i\theta }\right)\right\}-(p-1)\theta \\ =& arg\left\{\frac{f^{\prime }(z)}{z^{p-1}}\right\}\end{array}$$

and therefore we have

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

(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

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

(2.7)

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

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

It completes the proof of Lemma 2.4. □

Thus, we have the following theorems.

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

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

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

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

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

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

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

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

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

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

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

$$\mathfrak{Re}\left\{\frac{z{f}^{″}(z)}{f^{\prime }(z)}(\overline{z{e}^{-i\alpha }}-z{e}^{-i\alpha })\right\}\ge 0\mathit{\text{in }}|z|\le 0,$$

(2.8)

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

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

(2.9)

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

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

(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

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

This proves (2.9) and (2.10) and it shows that the function $arg{f}^{\prime }(\rho e^{i\theta })$ is an increasing function with respect to ρ, $0\le \rho \le 1$, and $\alpha \le \theta \le \alpha +\pi$, and that the function $arg{f}^{\prime }(\rho e^{i\theta })$ 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 $f(z)=z+{\sum }_{n=2}^{\mathrm{\infty }}{a}_n{z}^n$ be analytic in $\mathbb{D}.$ and suppose that it satisfies the following condition:

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

(2.11)

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

$$|arg\{f^{\prime }(z)\}|\le \frac{\pi }{2},z\in \mathbb{D}.$$

(2.12)

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

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

$$\begin{array}{rcl}0& =& {(arg\frac{f(z)}{z})}_{z=0}\\ \le & arg\left\{\frac{1}{\rho e^{i\theta }}{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)e^{i\theta }\mathrm{d}\rho \right\}\\ =& arg\frac{f(z)}{z}\\ =& arg{\int }_0^r{f}^{\prime }\left(\rho e^{i\theta }\right)\mathrm{d}\rho \\ =& arg{\int }_0^r\left|f^{\prime }\left(\rho e^{i\theta }\right)\right|e^{iarg\{f^{\prime }(\rho e^{i\theta })\}}\mathrm{d}\rho \\ \le & arg\{f^{\prime }(z)\}.\end{array}$$

This shows that

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

(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

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

(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

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

This completes the proof. □

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