New extensions concerned with results by Ponnusamy and Karunakaran

Abstract

A subclass $\mathcal{A}(n,k)$ of analytic functions $f(z)$ in the open unit disk $\mathbb{U}$ is introduced. By means of the result due to Fukui and Sakaguchi (Bull. Fac. Edu. Wakayama Univ. Natur. Sci. 30:1-3, 1980), some interesting properties of $f(z)$ in $\mathcal{A}(n,k)$ concerned with Ponnusamy and Karunakaran (Complex Var. Theory Appl. 11:79-86, 1989) are discussed.

MSC:30C45.

1 Introduction

Let $\mathcal{A}(n,k)$ be a class of functions $f(z)$ of the form

$$f(z)=z^n+\sum _{m=n+k}^{\mathrm{\infty }}{a}_m{z}^m(n\geqq 1,k\geqq 1)$$

(1.1)

which are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$. For two functions $f(z)$ and $g(z)$ belonging to the class $\mathcal{A}(1,1)$, Sakaguchi [1] proved the following result.

Theorem A Let $f(z)\in \mathcal{A}(1,1)$ and $g(z)\in \mathcal{A}(1,1)$ be starlike in $\mathbb{U}$. If $f(z)$ and $g(z)$ satisfy

$$Re\left(\frac{f^{\prime }(z)}{g^{\prime }(z)}\right)>0(z\in \mathbb{U}),$$

(1.2)

then

$$Re\left(\frac{f(z)}{g(z)}\right)>0(z\in \mathbb{U}).$$

(1.3)

After Theorem A, many mathematicians studying this field have applied this theorem to get some results (see [2]). In 1989, Ponnusamy and Karunakaran [3] improved Theorem A as follows.

Theorem B Let α be a complex number with $Re\alpha >0$ and $\beta <1$. Further, let $f(z)\in \mathcal{A}(n,k)$ and $g(z)\in \mathcal{A}(n,j)$ ($j\geqq 1$) satisfy

$$Re\left(\frac{\alpha g(z)}{z{g}^{\prime }(z)}\right)>\delta (z\in \mathbb{U})$$

(1.4)

with $0\leqq \delta <\frac{Re\alpha }{n}$. If $f(z)$ and $g(z)$ satisfy

$$Re\{(1-\alpha )\frac{f(z)}{g(z)}+\alpha \frac{f^{\prime }(z)}{g^{\prime }(z)}\}>\beta (z\in \mathbb{U}),$$

(1.5)

then

$$Re\left(\frac{f(z)}{g(z)}\right)>\frac{2\beta +\delta k}{2+\delta k}(z\in \mathbb{U}).$$

(1.6)

It is the purpose of the present paper to discuss Theorem B applying the lemma due to Fukui and Sakaguchi [4]. To discuss our problems, we need the following lemmas.

Lemma 1 Let $w(z)={\sum }_{n=k}^{\mathrm{\infty }}{a}_n{z}^n$ ($a_k\ne 0$, $k\geqq 1$) be analytic in $\mathbb{U}$. If the maximum value of $|w(z)|$ on the circle $|z|=r<1$ is attained at $z=z_0$, then we have

$$\frac{z_0{w}^{\prime }(z_0)}{w(z_0)}=\ell \geqq k,$$

(1.7)

which shows that $\frac{z_0{w}^{\prime }(z_0)}{w(z_0)}$ is a positive real number.

The proof of Lemma 1 can be found in [4], and we see that Lemma 1 is a generalization of Jack’s lemma given by Jack [5]. Applying Lemma 1, we derive the following.

Lemma 2 Let $p(z)=1+{\sum }_{n=k}^{\mathrm{\infty }}{c}_n{z}^n$ ($c_k\ne 0$, $k\geqq 1$) be analytic in $\mathbb{U}$ with $p(z)\ne 0$ ($z\in \mathbb{U}$). If there exists a point $z_0\in \mathbb{U}$ such that

$$Rep(z)>0(|z|<|z_0|)$$

and

$$Rep(z_0)=0,$$

then we have

$$-z_0{p}^{\prime }(z_0)\geqq \frac{\ell }{2}(1+{|p(z_0)|}^2),$$

(1.8)

and so

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}=i\ell ,$$

(1.9)

where

$$k\leqq \frac{k}{2}(a+\frac{1}{a})\leqq \ell (argp(z_0)=\frac{\pi }{2})$$

(1.10)

and

$$-k\geqq -\frac{k}{2}(a+\frac{1}{a})\geqq \ell (argp(z_0)=-\frac{\pi }{2})$$

(1.11)

with $p(z_0)=\pm ia$ ($a>0$).

Proof Let us consider

$$\varphi (z)=\frac{1-p(z)}{1+p(z)}=\frac{c_k}{2}{z}^k+\cdots$$

(1.12)

for $p(z)$. Then, it follows that $\varphi (0)={\varphi }^{\prime }(0)=\cdots ={\varphi }^{(k-1)}(0)=0$, $|\varphi (z)|<1$ ($|z|<|z_0|$) and $|\varphi (z_0)|=1$. Therefore, applying Lemma 1, we have that

$$\frac{z_0{\varphi }^{\prime }(z_0)}{\varphi (z_0)}=\frac{-2z_0{p}^{\prime }(z_0)}{1-{(p(z_0))}^2}=\frac{-2z_0{p}^{\prime }(z_0)}{1+{|p(z_0)|}^2}=\ell \geqq k.$$

(1.13)

This implies that $z_0{p}^{\prime }(z_0)$ is a negative real number and

$$-z_0{p}^{\prime }(z_0)\geqq \frac{k}{2}(1+{|p(z_0)|}^2).$$

(1.14)

Let us use the same method by Nunokawa [6]. If $argp(z_0)=\frac{\pi }{2}$, then we write $p(z_0)=ia$ ($a>0$). This gives us that

$$Im\left(\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)=Im(-\frac{i{z}_0{p}^{\prime }(z_0)}{a})\geqq \frac{k}{2}(a+\frac{1}{a}).$$

If $argp(z_0)=-\frac{\pi }{2}$, then we write $p(z_0)=-ia$ ($a>0$). Thus we have that

$$Im\left(\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)=Im\left(\frac{i{z}_0{p}^{\prime }(z_0)}{a}\right)\leqq -\frac{k}{2}(a+\frac{1}{a}).$$

This completes the proof of Lemma 2. □

2 Main results

With the help of Lemma 2, we derive the following theorem.

Theorem 1 Let α be a complex number with $Re\alpha >0$ and $\beta <1$. Further, let $f(z)\in \mathcal{A}(n,k)$ and $g(z)\in \mathcal{A}(n,j)$ ($j\geqq 1$) satisfy

$$Re\left(\frac{\alpha g(z)}{z{g}^{\prime }(z)}\right)>\delta (z\in \mathbb{U})$$

(2.1)

with $0\leqq \delta <\frac{Re\alpha }{n}$. If $f(z)$ and $g(z)$ satisfy

$$Re\{(1-\alpha )\frac{f(z)}{g(z)}+\alpha \frac{f^{\prime }(z)}{g^{\prime }(z)}\}+\frac{\delta k}{2(1-{\beta }_1)}|\frac{f(z)}{g(z)}-{\beta }_1{|}^2>\beta (z\in \mathbb{U}),$$

(2.2)

then

$$Re\left(\frac{f(z)}{g(z)}\right)>{\beta }_1(z\in \mathbb{U}),$$

(2.3)

where ${\beta }_1=\frac{2\beta +\delta k}{2+\delta k}$.

Proof Defining the function $p(z)$ by

$$p(z)=\frac{\frac{f(z)}{g(z)}-{\beta }_1}{1-{\beta }_1},$$

(2.4)

we see that $p(0)=1$ and

$$\begin{array}{r}Re\{(1-\alpha )\frac{f(z)}{g(z)}+\alpha \frac{f^{\prime }(z)}{g^{\prime }(z)}-\beta \}\\ =Re\{({\beta }_1-\beta )+(1-{\beta }_1)(p(z)+\frac{\alpha g(z)}{z{g}^{\prime }(z)}z{p}^{\prime }(z))\}\\ >-\frac{\delta k}{2(1-{\beta }_1)}|\frac{f(z)}{g(z)}-{\beta }_1{|}^2\end{array}$$

(2.5)

for all $z\in \mathbb{U}$. Let us suppose that there exists a point $z_0\in \mathbb{U}$ such that

$$|argp(z_0)|<\frac{\pi }{2}(|z|<|z_0|)$$

and

$$|argp(z_0)|=\frac{\pi }{2}.$$

Then, by means of Lemma 2, we have that

$$-z_0{p}^{\prime }(z_0)\geqq \frac{k}{2}(1+{|p(z_0)|}^2).$$

(2.6)

If follows from the above that

$$\begin{array}{r}Re\{(1-\alpha )\frac{f(z_0)}{g(z_0)}+\alpha \frac{f^{\prime }(z_0)}{g^{\prime }(z_0)}-\beta \}\\ =({\beta }_1-\beta )+(1-{\beta }_1)Re\{p(z_0)+\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}{z}_0{p}^{\prime }(z_0)\}\\ =({\beta }_1-\beta )-(1-{\beta }_1)Re\left\{\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}(-z_0{p}^{\prime }(z_0))\right\}\\ \leqq ({\beta }_1-\beta )-(1-{\beta }_1)\frac{\delta k}{2}(1+{|p(z_0)|}^2)\\ =-\frac{\delta k}{2(1-{\beta }_1)}|\frac{f(z_0)}{g(z_0)}-{\beta }_1{|}^2,\end{array}$$

which contradicts (2.5). This completes the proof of the theorem. □

Remark 1 If $f(z)$ and $g(z)$ satisfy $f(z)={\beta }_{1}g(z)$ in Theorem 1, then Theorem 1 becomes Theorem B given by Ponnusamy and Karunakaran [3]. We also have the following theorem.

Theorem 2 Let α be a complex number with $Re\alpha >0$ and $\beta <1$. Further, let $f(z)\in \mathcal{A}(n,k)$ and $g(z)\in \mathcal{A}(n,j)$ ($j\geqq 1$) satisfy the condition (2.1) with $0\leqq \delta <\frac{Re\alpha }{n}\leqq 1+\delta$. If $f(z)$ and $g(z)$ satisfy

$$|arg\{(1-\alpha )\frac{f(z)}{g(z)}+\alpha \frac{f^{\prime }(z)}{g^{\prime }(z)}-\beta \}|<\frac{\pi }{2}+{Tan}^{-1}\left(\frac{\delta k|p(z)|}{2(\frac{2r}{1-r^2}+\frac{|Im\alpha |}{n}+1)}\right)$$

(2.7)

for $|z|=r<1$, then

$$|arg(\frac{f(z)}{g(z)}-{\beta }_1)|<\frac{\pi }{2}(z\in \mathbb{U})$$

(2.8)

or

$$Re\left(\frac{f(z)}{g(z)}\right)>{\beta }_1(z\in \mathbb{U}),$$

(2.9)

where ${\beta }_1=\frac{2\beta +\delta k}{2+\delta k}$ and

$$p(z)=\frac{\frac{f(z)}{g(z)}-{\beta }_1}{1-{\beta }_1}.$$

Proof Note that the function $p(z)$ is analytic in $\mathbb{U}$ and $p(0)=1$. It follows that

$$\begin{array}{rl}|arg\{(1-\alpha )\frac{f(z)}{g(z)}+\alpha \frac{f^{\prime }(z)}{g^{\prime }(z)}-\beta \}|& =|arg\{({\beta }_1-\beta )+(1-{\beta }_1)(p(z)+\frac{\alpha g(z)}{z{g}^{\prime }(z)}z{p}^{\prime }(z))\}|\\ <\frac{\pi }{2}+{Tan}^{-1}\left(\frac{\delta k|p(z)|}{2(\frac{2r}{1-r^2}+\frac{|Im\alpha |}{n}+1)}\right)\end{array}$$

for $|z|=r<1$. If there exists a point $z_0\in \mathbb{U}$ such that

$$|argp(z_0)|<\frac{\pi }{2}(|z|<|z_0|)$$

and

$$|argp(z_0)|=\frac{\pi }{2},$$

then, by Lemma 2, we have that

$$\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}=i\ell ,$$

where

$$\frac{k}{2}(a+\frac{1}{a})\leqq \ell (argp(z_0)=\frac{\pi }{2})$$

and

$$-\frac{k}{2}(a+\frac{1}{a})\geqq \ell (argp(z_0)=-\frac{\pi }{2})$$

with $p(z_0)=\pm ia$ ($a>0$). If $argp(z_0)=\frac{\pi }{2}$, then it follows that

$$\begin{array}{r}arg\{(1-\alpha )\frac{f(z_0)}{g(z_0)}+\alpha \frac{f^{\prime }(z_0)}{g^{\prime }(z_0)}-\beta \}\\ =argp(z_0)\{\frac{{\beta }_1-\beta }{p(z_0)}+(1-{\beta }_1)(1+\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}\frac{z_0{p}^{\prime }(z_0)}{p(z_0)})\}\\ =\frac{\pi }{2}+arg\{-\left(\frac{{\beta }_1-\beta }{a}\right)i+(1-{\beta }_1)(1+i\ell \frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)})\}\\ =\frac{\pi }{2}+argI(z_0),\end{array}$$

where

$$I(z_0)=-\left(\frac{{\beta }_1-\beta }{a}\right)i+(1-{\beta }_1)(1+i\ell \frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}).$$

(2.10)

Note that

$$\begin{array}{rl}ImI(z_0)& =\frac{\beta -{\beta }_1}{a}+(1-{\beta }_1)\ell Re\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}\\ \geqq (1-{\beta }_1)\delta \ell +\frac{\beta -{\beta }_1}{a}\\ \geqq \frac{\delta k}{2}(1-{\beta }_1)(a+\frac{1}{a})+\frac{\beta -{\beta }_1}{a}\\ =\frac{\delta k}{2}(1-{\beta }_1)a>0\end{array}$$

(2.11)

and

$$ReI(z_0)=(1-{\beta }_1)(1-\ell Im\left(\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}\right))\leqq (1-{\beta }_1)(1+\ell |Im\left(\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}\right)\left|\right).$$

(2.12)

Letting

$$q(z)=\frac{\alpha g(z)}{z{g}^{\prime }(z)}+1-\frac{\alpha }{n},$$

(2.13)

we know that $q(z)$ is analytic in $\mathbb{U}$ with $q(0)=1$ and $Req(z)>0$ ($z\in \mathbb{U}$). Therefore, applying the subordinations, we can write that

$$q(z)=\frac{1-w(z)}{1+w(z)}$$

with the Schwarz function $w(z)$ analytic in $\mathbb{U}$, $w(0)=0$ and $|w(z)|\leqq |z|$. This leads us to

$$|w(z)|=\left|\frac{1-q(z)}{1+q(z)}\right|\leqq r(|z|\leqq r<1),$$

which is equivalent to

$$|q(z)-\frac{1+r^2}{1-r^2}|\leqq \frac{2r}{1-r^2}.$$

This gives us that

$$|Imq(z)|=|Im(\frac{\alpha g(z)}{z{g}^{\prime }(z)}+1-\frac{\alpha }{n})|\leqq \frac{2r}{1-r^2}$$

(2.14)

for $|z|=r<1$. Thus we have that

$$|Im\left(\frac{\alpha g(z_0)}{z_0{g}^{\prime }(z_0)}\right)|\leqq \frac{2r}{1-r^2}+\frac{|Im\alpha |}{n}(|z|=r<1).$$

(2.15)

Using (2.12) and (2.15), we obtain that

$$argI(z_0)={Tan}^{-1}\left(\frac{ImI(z_0)}{ReI(z_0)}\right)\geqq {Tan}^{-1}\left(\frac{\delta ka}{2(\frac{2r}{1-r^2}+\frac{|Im\alpha |}{n}+1)}\right),$$

which contradicts our condition (2.7).

If $argp(z_0)=-\frac{\pi }{2}$, using the same way, we also have that

$$arg\{(1-\alpha )\frac{f(z_0)}{g(z_0)}+\alpha \frac{f^{\prime }(z_0)}{g^{\prime }(z_0)}-\beta \}\leqq -\{\frac{\pi }{2}+{Tan}^{-1}\left(\frac{\delta ka}{2(\frac{2r}{1-r^2}+\frac{|Im\alpha |}{n}+1)}\right)\},$$

which contradicts (2.7). □