On a successive property of strongly starlikeness for multivalent functions

Abstract

For f analytic in the unit disk $\mathbb {D}$, of the form $f(z)=z^p+\cdots$, we consider some consequences of strongly starlikeness of $f^{(p-1)}(z)/p!$.

1 Introduction

We denote by $\mathcal {H}$ the class of functions f(z) which are holomorphic in the open unit disc ${\mathbb {D}}=\{z\in \mathbb {C}: |z|<1\}$. Denote by ${\mathcal {A}}_p$, $p\in \mathbb {N}=\{1,2,\ldots \}$, the class of functions $f(z)\in \mathcal {H}$ given by

$$\begin{aligned} f(z)=z^p+ \sum _{n=p+1}^{\infty }a_{n}z^{n},\quad (z\in {\mathbb {D}}). \end{aligned}$$

(1.1)

Lemma 1.1

[2, Theorem 5] If $f(z)\in {\mathcal {A}}_p$, then for all $z\in {\mathbb {D}}$, we have

$$\begin{aligned} {\mathfrak {Re}}\left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\}>0\quad \Rightarrow \quad \forall k\in \{1,\ldots ,p-1\}:\quad {\mathfrak {Re}}\left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} >0. \end{aligned}$$

(1.2)

In this paper we consider a generalization of the above result. In Lemma 1.1 we have assumed that $zf^{(p)}(z)/f^{(p-1)}(z)$ lies in the right half-plane while in this paper we work with a sector. The problem we solve here is: for what values of $\alpha ,\beta$ does an analytic function of the form (1.1) satisfy

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \alpha }{2}\quad \Rightarrow \quad \forall k\in \{1,\ldots ,p-1\}:\quad \left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| <\frac{\pi \beta }{2}? \end{aligned}$$

Recall that if $f(z)\in {\mathcal {A}}_p$ and

$$\begin{aligned} {\mathfrak {Re}}\left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} >0\quad (z\in {\mathbb {D}}), \end{aligned}$$

then $f^{(p-1)}(z)/p!\in {\mathcal {A}}_1$ is univalent in ${\mathbb {D}}$ and $f^{(p-1)}(z)/p!$ is called a starlike function. If $f(z)\in {\mathcal {A}}_p$, $\gamma \in (0,1]$, and

$$\begin{aligned} \arg \left| \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right| <\frac{\pi \gamma }{2},\quad z\in {\mathbb {D}}, \end{aligned}$$

(1.3)

then $f^{(p-1)}(z)/p!$ is called a strongly starlike function of order $\gamma$ and such functions we consider in the paper. This class for the case $p=1$ was introduced by Brannan and Kirwan [1]. Also, if $f(z)\in {\mathcal {A}}_p$ satisfies (1.3), then f(z) is called p-valently strongly starlike function of order $\gamma$. For the proof of main result we need the following lemma.

Lemma 1.2

[3] Let $q(z)=1+\sum _{n\ge m}^{\infty }c_nz^n$, $c_m\ne 0$ be analytic function in $|z|<1$ with $q(0)=1$, $q(z)\ne 0$. If there exists a point $z_0$, $|z_0|<1$, such that

$$\begin{aligned} |{\arg } \left\{ q(z)\right\} |<\frac{\pi \beta }{2}\ \ for\ \ |z|<|z_0| \end{aligned}$$

and

$$\begin{aligned} |{\arg }\left\{ q(z_0)\right\} |=\frac{\pi \beta }{2} \end{aligned}$$

for some $\beta >0$, then we have

$$\begin{aligned} \frac{z_0q'(z_0)}{q(z_0)}=\frac{2ik\arg \left\{ q(z_0)\right\} }{\pi }, \end{aligned}$$

for some $k\ge m(a+a^{-1})/2\ge m$, where

$$\begin{aligned} \left\{ q(z_0)\right\} ^{1/\beta }=\pm ia, \ \ and\ \ a>0. \end{aligned}$$

2 Main results

For given $0<\beta _{s-1}\le 1$ let us consider the number

$$\begin{aligned} \beta _s=\beta _{s-1}+\frac{2}{\pi }\tan ^{-1}\frac{\beta _{s-1} n(\beta _{s-1})\sin [\pi (1-\beta _{s-1})/2]}{sm(\beta _{s-1})+\beta _{s-1} n(\beta _{s-1})\cos [\pi (1-\beta _{s-1})/2], }\quad s=2,3,\ldots ,p, \end{aligned}$$

(2.1)

where

$$\begin{aligned} m(\beta _{s-1})=(1+\beta _{s-1})^{(1+\beta _{s-1})/2},\quad {and} \quad n(\beta _{s-1})=(1-\beta _{s-1})^{(1-\beta _{s-1})/2}. \end{aligned}$$

(2.2)

Notice that if $0<\beta _{s-1}\le 1$, then $0<\beta _s\le 1$ too because from (2.1), (2.2), we have

$$\begin{aligned} \beta _s= & {} \beta _{s-1}+\frac{2}{\pi }\tan ^{-1}\frac{\beta _{s-1} n(\beta _{s-1})\sin [\pi (1-\beta _{s-1})/2]}{sm(\beta _{s-1})+\beta _{s-1} n(\beta _{s-1})\cos [\pi (1-\beta _{s-1})/2], }\\\le & {} \beta _{s-1}+\frac{2}{\pi }\tan ^{-1}\frac{\beta _{s-1} n(\beta _{s-1})\sin [\pi (1-\beta _{s-1})/2]}{\beta _{s-1} n(\beta _{s-1})\cos [\pi (1-\beta _{s-1})/2], }\\= & {} 1. \end{aligned}$$

Therefore, if we have a number $\beta _1\in (0,1]$, then from (2.1), we can find a sequence $\beta _p,\beta _{p-1},\ldots ,\beta _2,\beta _1$, such that

$$\begin{aligned} 0<\beta _1\le \beta _{2}\le \cdots \le \beta _{p-1}\le \beta _p\le 1. \end{aligned}$$

(2.3)

Theorem 2.1

Let $f(z)\in {\mathcal {A}}_p$, $p\ge 2$. For given $\beta _{p-1}\in (0,1]$ there exists $\beta _{p}\in (0,1]$ of the form (2.1) such that for all $z\in {\mathbb {D}}$, we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \beta _{p}}{2}\quad \Rightarrow \quad \left| \arg \left\{ \frac{zf^{(p-1)}(z)}{f^{(p-2)}(z)}\right\} \right| <\frac{\pi \beta _{p-1}}{2}. \end{aligned}$$

(2.4)

Proof

Let us put

$$\begin{aligned} q_1(z)=\frac{zf^{(p-1)}(z)}{2f^{(p-2)}z},\quad q_1(0)=1. \end{aligned}$$

Then it follows that

$$\begin{aligned} \frac{zq_1'(z)}{q_1(z)}=1+\frac{zf^{(p)}(z)}{f^{(p-1)}(z)} -\frac{zf^{(p-1)}(z)}{f^{(p-2)}(z)} \end{aligned}$$

and

$$\begin{aligned} 2q_1(z)+\frac{zq_1'(z)}{q_1(z)}=1+\frac{zf^{(p)}(z)}{f^{(p-1)}(z)} \end{aligned}$$

and so

$$\begin{aligned} \arg \left\{ q_1(z)\right\} +\arg \left\{ 2+\frac{zq_1'(z)}{q_1^2(z)}\right\} =\arg \left\{ 1+\frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} . \end{aligned}$$

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

$$\begin{aligned}&|\arg \{q_1(z)\}|<\pi \beta _{p-1}/2\quad {\mathrm{for}}\quad |z|<|z_0|,\quad |\arg \{q_1(z_0)\}|=\pi \beta _{p-1}/2,\\&\quad \left\{ q_1(z_0)\right\} ^{1/\beta _{p-1}}=\pm ia,\quad {\mathrm{and}}\quad a>0, \end{aligned}$$

then from Lemma 1.2, we have

$$\begin{aligned} \frac{z_0 q_1'(z_0)}{q_1(z_0)}=\frac{2ik\arg \left\{ q_1(z_0)\right\} }{\pi } \end{aligned}$$

(2.5)

for some real k with $k\ge (a+a^{-1})/2\ge 1$. For the case $\arg \{q_1(z_0)\}=\pi \beta _{p-1}/2$, we have

$$\begin{aligned} \arg \left\{ \frac{z_0f^{(p)}(z_0)}{f^{(p-1)}(z_0)}\right\}\ge & {} \arg \left\{ 1+\frac{z_0f^{(p)}(z_0)}{f^{(p-1)}(z_0)}\right\} \\= & {} \arg \left\{ q_1(z_0)\right\} + \arg \left\{ 2+\frac{z_0 q_1'(z_0)}{q_1(z_0)}\frac{1}{q_1(z_0)}\right\} \\\ge & {} \frac{\pi \beta _{p-1}}{2} +\arg \left\{ 2+e^{i\pi (1-\beta _{p-1})/2}\frac{1}{(ia)^{\beta _{p-1}}}\right\} , \end{aligned}$$

where $\left\{ q_1(z_0)\right\} ^{1/\beta _{p-1}}=ia$ and a is a positive real number. Applying Lemma 1.2 we obtain

$$\begin{aligned}&\arg \left\{ \frac{z_0f^{(p)}(z_0)}{f^{(p-1)}(z_0)}\right\} \\&\quad \ge \frac{\pi \beta _{p-1}}{2}+\arg \left\{ e^{i\pi (1-\beta _{p-1})/2}\left( \left( \frac{1+\beta _{p-1}}{1-\beta _{p-1}}\right) ^{(1-\beta _{p-1})/2}\right. \right. \\&\qquad \left. \left. +\,\left( \frac{1+\beta _{p-1}}{1-\beta _{p-1}}\right) ^{-(1+\beta _{p-1})/2}\right) +2\right\} \\&\quad = \frac{\pi \beta _{p-1}}{2}+\tan ^{-1}\frac{\frac{\beta _{p-1}}{1-\beta _{p-1}}\left( \frac{1-\beta _{p-1}}{1+\beta _{p-1}}\right) ^{(1+\beta _{p-1})/2}\sin \frac{\pi (1-\beta _{p-1})}{2}}{2+\frac{\beta _{p-1}}{1-\beta _{p-1}}\left( \frac{1-\beta _{p-1}}{1+\beta _{p-1}}\right) ^{(1+\beta _{p-1})/2}\cos \frac{\pi (1-\beta _{p-1})}{2}} \\&\quad = \frac{\pi \beta _{p-1}}{2}+\tan ^{-1}\frac{\beta _{p-1} n(\beta _{p-1})\sin \frac{\pi (1-\beta _{p-1})}{2}}{2m(\beta _{p-1})+\beta _{p-1} n(\beta _{p-1})\cos \frac{\pi (1-\beta _{p-1})}{2}}. \end{aligned}$$

From (2.1), we can see that

$$\begin{aligned} \arg \left\{ \frac{z_0f^{(p)}(z_0)}{f^{(p-1)}(z_0)}\right\} \ge \frac{\pi \beta _{p}}{2}. \end{aligned}$$

(2.6)

This contradicts hypothesis in (2.4).

For the case $\arg \{q_1(z_0)\}=-\pi \beta _{p-1}/2$, applying the same method as the above, gives

$$\begin{aligned} \arg \left\{ \frac{z_0f^{(p)}(z_0)}{f^{(p-1)}(z_0)}\right\} \le -\frac{\pi \beta _{p}}{2}. \end{aligned}$$

(2.7)

This also contradicts hypothesis in (2.4) and therefore, we have

$$\begin{aligned} |\arg \{q_1(z)\}|<\pi \beta _{p-1}/2\quad {\mathrm{for}}\quad |z|<|1|. \end{aligned}$$

This completes the proof. $\square$

Let us go to next step and define the function

$$\begin{aligned} q_2(z)=\frac{zf^{(p-2)}(z)}{3f^{(p-3)}(z)},\quad q_2(0)=1 \end{aligned}$$

and applying the same method as the above, we have the following theorem.

Theorem 2.2

Let $f(z)\in {\mathcal {A}}_p$, $p\ge 2$, $0<\beta _2\le 1$ and suppose that

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right| <\frac{\pi \beta _{p}}{2},\quad z\in {\mathbb {D}}. \end{aligned}$$

(2.8)

Then we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p-2)}(z)}{f^{(p-3)}(z)}\right\} \right| <\frac{\pi \beta _{p-2}}{2},\quad z\in {\mathbb {D}}. \end{aligned}$$

(2.9)

where $\beta _{p-2}$ we obtain from $\beta _{p-1}$ using formula (2.1). Furthermore,

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p-1)}(z)}{f^{(p-2)}(z)}\right\} \right| <\frac{\pi \beta _{p-1}}{2},\quad z\in {\mathbb {D}}, \end{aligned}$$

(2.10)

and where $\beta _{p-1}$ we obtain from $\beta _{p}$ using formula (2.1) too.

Applying the same step as the above and under the hypothesis of Theorem 2.1, we have the following theorem

Theorem 2.3

Let $f(z)\in {\mathcal {A}}_p$, $p\ge 2$. For given $\beta _{1}\in (0,1]$ there exist $\beta _{k}\in (0,1]$, $k=2,\ldots ,p$, of the form (2.1) such that for all $z\in {\mathbb {D}}$, we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \beta _{p}}{2}\quad \Rightarrow \quad \forall k\in \{1,\ldots ,p-1\}:\quad \left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| <\frac{\pi \beta _{k}}{2}. \end{aligned}$$

Furthermore

$$\begin{aligned} 0<\beta _1\le \beta _{2}\le \cdots \le \beta _{p-1}\le \beta _p\le 1. \end{aligned}$$

It is easy to see that Theorem 2.3 holds for the case $\beta _{p}=\beta _{p-1}=\cdots =\beta _{1}=1$ and then Theorem 2.3 becomes Lemma 1.1 and in this sense Theorem 2.3 improves Lemma 1.1.

Corollary 2.4

Let $f(z)\in {\mathcal {A}}_p$, $p\ge 2$. If $\beta _1\in (0,1]$ and $\beta _{k}\in (0,1]$, $k=2,\ldots ,p$ are of the form (2.1), then for all $k=1,\ldots ,p-1$ and for all $z\in {\mathbb {D}}$, we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \beta _k}{2}\quad \Rightarrow \quad \left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| <\frac{\pi \beta _k}{2}. \end{aligned}$$

Proof

For given $\beta _1\in (0,1]$ there exist $\beta _{k}\in (0,1]$, $k=2,\ldots ,p$, of the form (2.1) such that

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \beta _{p}}{2}\quad \Rightarrow \quad \forall k\in \{1,\ldots ,p-1\}:\quad \left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| <\frac{\pi \beta _{k}}{2}. \end{aligned}$$

where

$$\begin{aligned} 0<\beta _1\le \beta _{2}\le \cdots \le \beta _{p-1}\le \beta _p\le 1. \end{aligned}$$

(2.11)

Therefore, from (2.11), we have

$$\begin{aligned}&\left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \beta _k}{2}\quad \Rightarrow \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|<\frac{\pi \beta _{p}}{2}\\&\quad \Rightarrow \quad \left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| <\frac{\pi \beta _{k}}{2}. \end{aligned}$$

$\square$

Corollary 2.5

If $f(z)\in {\mathcal {A}}_p$, $p\ge 2$, then for all $\gamma \in (0,1]$ and for all $k\in \{1,\ldots ,p\}$ and for all $s\in \{k,\ldots ,p-1\}$, and for all $z\in {\mathbb {D}}$, we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(s)}(z)}{f^{(s-1)}(z)}\right\} \right|<\frac{\pi \gamma }{2}\quad \Rightarrow \quad \left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| <\frac{\pi \gamma }{2}. \end{aligned}$$

From the properties of the sequence (2.1) we have following corollary.

Corollary 2.6

Let $f(z)\in {\mathcal {A}}_p$, $p\ge 2$, $0<\beta _{p}\le 1$ and suppose that

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right| <\frac{\pi \beta _{p}}{2},\quad z\in {\mathbb {D}}. \end{aligned}$$

Then we have

$$\begin{aligned} \sup _{z\in {\mathbb {D}}}\left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|> & {} \sup _{z\in {\mathbb {D}}}\left| \arg \left\{ \frac{zf^{(p-1)}(z)}{f^{(p-2)}(z)}\right\} \right|>\cdots>\sup _{z\in {\mathbb {D}}}\left| \arg \left\{ \frac{zf^{(k)}(z)}{f^{(k-1)}(z)}\right\} \right| \\> & {} \sup _{z\in {\mathbb {D}}}\left| \arg \left\{ \frac{zf^{(k-1)}(z)}{f^{(k-2)}(z)}\right\} \right|>\cdots >\sup _{z\in {\mathbb {D}}}\left| \arg \left\{ \frac{zf'(z)}{f(z)}\right\} \right| . \end{aligned}$$

Theorem 2.7

Let $\beta =\alpha +(2/\pi )\tan ^{-1}\alpha$ and $f(z)\in {\mathcal {A}}_p$, $p\ge 2$. Suppose also that

$$\begin{aligned} \left| \arg \left\{ f^{(p)}(z)\right\} \right| <\frac{\pi \beta }{2},\quad z\in {\mathbb {D}}. \end{aligned}$$

(2.12)

Then we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf'(z)}{f(z)}\right\} \right| <\frac{\pi \beta _{1}}{2},\quad z\in {\mathbb {D}}, \end{aligned}$$

(2.13)

where $\beta _{1}$ is described in (2.1) with $\beta _{p}=\alpha +\beta$.

Proof

If

$$\begin{aligned} \left| \arg \left\{ \frac{f^{(p-1)}(z)}{z}\right\} \right| <\frac{\pi \alpha }{2} \end{aligned}$$

(2.14)

in $|z<|z_0|$ and

$$\begin{aligned} \arg \left\{ \frac{f^{(p-1)}(z_0)}{z_0}\right\} =\frac{\pi \alpha }{2}\quad {\mathrm{or}}\quad \arg \left\{ \frac{f^{(p-1)}(z_0)}{z_0}\right\} =-\frac{\pi \alpha }{2}, \end{aligned}$$

(2.15)

then for the first case in (2.15), from Lemma 1.2, we have

$$\begin{aligned} \frac{z_0f^{(p)}(z_0)}{f^{(p-1)}(z_0)}-1=ik\alpha \end{aligned}$$

for some $k\ge 1$, This gives

$$\begin{aligned} \arg \left\{ f^{(p)}(z_0)\right\}= & {} \arg \left\{ \frac{f^{(p-1)}(z_0)}{z_0}(ik\alpha +1)\right\} \\= & {} \arg \left\{ \frac{f^{(p-1)}(z_0)}{z_0}\right\} +\arg \left\{ ik\alpha +1\right\} \\\ge & {} \frac{\pi }{2}\left\{ \alpha +(2/\pi )\tan ^{-1}\alpha \right\} =-\frac{\pi \beta }{2}. \end{aligned}$$

This contradicts hypothesis (2.12). In the second case in (2.15), applying the same method as in the first case, we obtain

$$\begin{aligned} \arg \left\{ f^{(p)}(z_0)\right\}\le & {} -\frac{\pi }{2}\left\{ \alpha +(2/\pi )\tan ^{-1}\alpha \right\} =\frac{\pi \beta }{2}. \end{aligned}$$

This also contradicts hypothesis (2.12). So (2.14) holds in the whole unit disc ${\mathbb {D}}$. From (2.12) and (2.14), we have

$$\begin{aligned} \left| \arg \left\{ \frac{zf^{(p)}(z)}{f^{(p-1)}(z)}\right\} \right|= & {} \left| \arg \left\{ f^{(p)}(z)\frac{z}{f^{(p-1)}(z)}\right\} \right| \\\le & {} \left| \arg \left\{ f^{(p)}(z)\right\} \right| +\left| \arg \left\{ \frac{f^{(p-1)}(z)}{z}\right\} \right| \\< & {} \frac{\pi (\alpha +\beta )}{2},\quad z\in {\mathbb {D}}. \end{aligned}$$

Applying Theorem 2.3, we obtain (2.13). $\square$