# New subclasses of analytic functions

## Abstract

For analytic functions f (z) in the open unit disk $\mathcal{U}$, subclasses $\mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3}:\lambda \right),\mathcal{P}\left(\theta ,\alpha \right)$, and $\mathcal{K}\left(\theta ,\alpha \right)$ are introduced. The object of the present article is to discuss some interesting properties of functions f (z) associated with classes $\mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3}:\lambda \right),\mathcal{P}\left(\theta ,\alpha \right)$, and $\mathcal{K}\left(\theta ,\alpha \right)$.

Mathematics Subject Classification (2010): 30C45.

## 1. Introduction and Definitions

Let $\mathcal{A}$ denotes the class of the normalized functions of the form

$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_{n}{z}^{n},$
(1.1)

which are analytic in the open unit disk $\mathcal{U}=\left\{z\in ℂ:|z|<1\right\}$. Also, a function f (z) belonging to $\mathcal{A}$ is said to be convex of order α if it satisfies

$\text{Re}\phantom{\rule{1em}{0ex}}\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}>\alpha \phantom{\rule{1em}{0ex}}\left(z\in \mathcal{U}\right)$
(1.2)

for some α(0 ≤ α < 1). We denote by $\mathcal{K}\left(\alpha \right)$ the subclass of $\mathcal{A}$ consisting of functions which are convex of order α in $\mathcal{U}$ (see, [1, 2]). Further, a function f (z) belonging to $\mathcal{A}$ is said to be in the class $\mathcal{P}\left(\alpha \right)$ iff

$\text{Re}\phantom{\rule{1em}{0ex}}\left(z{f}^{″}\left(z\right)+{f}^{\prime }\left(z\right)\right)>\alpha ,\phantom{\rule{1em}{0ex}}\left(z\in \mathcal{U}\right).$
(1.3)

for some α(0 ≤ α < 1).

For analytic functions f (z), Uyanik and Owa , obtained some interesting properties for analytic functions in the subclass $\mathcal{A}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ defined by

$\begin{array}{c}\left|{\beta }_{1}z{\left(\frac{f\left(z\right)}{z}\right)}^{\prime }+{\beta }_{2}z{\left(\frac{f\left(z\right)}{z}\right)}^{\prime \prime }+{\beta }_{3}z{\left(\frac{f\left(z\right)}{z}\right)}^{\prime \prime \prime }\right|\le \lambda \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left({\beta }_{1},{\beta }_{2},{\beta }_{3}\in ℂ;\lambda >0;z\in \mathcal{U}\right),\end{array}$

associated with close-to-convex functions and starlike functions of order α.

Definition 1.1. A function f (z) belonging to $\mathcal{A}$ is said to be in the class $\mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$, if it satisfies

$\left|{\beta }_{1}z{f}^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}^{\left(4\right)}\left(z\right)\right|\le \lambda \phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(z\in \mathcal{U}\right),$
(1.4)

for some complex numbers β1, β2, β3, and for some real λ > 0.

Example 1.2. Let us consider the function f γ (z), γ , given by

${f}_{\gamma }\left(z\right)=z{\left(1+z\right)}^{\gamma }.$

Then, we observe that

$\begin{array}{l}|{\beta }_{1}z{f}_{\gamma }^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}_{\gamma }^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}_{\gamma }^{\left(4\right)}\left(z\right)|\\ =|\sum _{n=2}^{\infty }n\left(n-1\right)\left(\begin{array}{c}\gamma \\ n-1\end{array}\right)\left({\beta }_{1}+\left(n-2\right){\beta }_{2}+\left(n-2\right)\left(n-3\right){\beta }_{3}\right){z}^{n-1}|,\end{array}$

where

$\left(\begin{array}{c}\hfill \gamma \hfill \\ \hfill n-1\hfill \end{array}\right)=\frac{\gamma \left(\gamma -1\right)\left(\gamma -2\right)\cdots \left(\gamma -n+2\right)}{\left(n-1\right)!}.$

Therefore, if γ = 1, then

$\left|{\beta }_{1}z{f}_{1}^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}_{1}^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}_{1}^{\left(4\right)}\left(z\right)\right|=\left|2{\beta }_{1}z\right|\le 2\left|{\beta }_{1}\right|.$

This implies that ${f}_{1}\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ for λ ≥ 2 |β1|. If γ = 2, then

$\left|{\beta }_{1}z{f}_{2}^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}_{2}^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}_{2}^{\left(4\right)}\left(z\right)\right|=\left|4{\beta }_{1}z+6\left({\beta }_{1}+{\beta }_{2}\right){z}^{2}\right|\le 10\left|{\beta }_{1}\right|+6\left|{\beta }_{2}\right|.$

Therefore, ${f}_{2}\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ for λ ≥ 10 |β1| + 6 |β2|. Further, if γ = 3; then we have

$\begin{array}{c}\left|{\beta }_{1}z{f}_{3}^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}_{3}^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}_{3}^{\left(4\right)}\left(z\right)\right|\\ \phantom{\rule{1em}{0ex}}=\left|6{\beta }_{1}z+18\left({\beta }_{1}+{\beta }_{2}\right){z}^{2}+12\left({\beta }_{1}+2{\beta }_{2}+2{\beta }_{3}\right){z}^{3}\right|\\ \phantom{\rule{1em}{0ex}}\le 36\left|{\beta }_{1}\right|+42\left|{\beta }_{2}\right|+24\left|{\beta }_{3}\right|.\end{array}$

Thus, ${f}_{3}\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ for λ ≥ 36 |β1| + 42 |β2| + 24 |β3|.

Now, let ${\mathcal{A}}_{\theta }$ denotes the subclass of $\mathcal{A}$ consisting of functions f (z) with

${a}_{n}=\left|{a}_{n}\right|{e}^{i\left(\left(n-1\right)\theta +\pi \right)}\phantom{\rule{1em}{0ex}}\left(n=2,3,...\right).$

Also, we introduce the subclasses $\mathcal{P}\left(\theta ,\alpha \right)$ and $\mathcal{K}\left(\theta ,\alpha \right)$ of ${\mathcal{A}}_{\theta }$ as follows:

$\mathcal{P}\left(\theta ,\alpha \right)={\mathcal{A}}_{\theta }\cap \mathcal{P}\left(\alpha \right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{2.77695pt}{0ex}}\mathcal{K}\left(\theta ,\alpha \right)={\mathcal{A}}_{\theta }\cap \mathcal{K}\left(\alpha \right).$

## 2. Properties of the class $\mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$

We first prove

Theorem 2.1. If $f\left(z\right)\in \mathcal{A}$ satisfies

$\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\le \lambda$
(2.1)

for some complex numbers β1, β2, β3 and for some real λ > 0, then $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$.

Proof. We observe that

$\begin{array}{c}\left|{\beta }_{1}z{f}_{3}^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}_{3}^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}_{3}^{\left(4\right)}\left(z\right)\right|\\ \phantom{\rule{1em}{0ex}}=\left|\sum _{n=2}^{\infty }n\left(n-1\right)\left({\beta }_{1}+\left(n-2\right){\beta }_{2}+\left(n-2\right)\left(n-3\right){\beta }_{3}\right){a}_{n}{z}^{n-1}\right|\\ \phantom{\rule{1em}{0ex}}\le \sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|{\left|z\right|}^{n-1}\\ \phantom{\rule{1em}{0ex}}<\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|.\end{array}$

Therefore, if f (z) satisfies the inequality (2.1), then $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$.

Next, we prove

Theorem 2.2. if $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ with arg β1 = arg β2 = arg β3 = ϕ and a n = |a n |ei((n-1)θ-ϕ)(n = 2, 3,...), then we have

$\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\le \lambda .$

Proof. For $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$, we see that

$\begin{array}{c}\left|{\beta }_{1}z{f}_{3}^{″}\left(z\right)+{\beta }_{2}{z}^{2}{f}_{3}^{‴}\left(z\right)+{\beta }_{3}{z}^{3}{f}_{3}^{\left(4\right)}\left(z\right)\right|\\ \phantom{\rule{1em}{0ex}}=\left|\sum _{n=2}^{\infty }n\left(n-1\right)\left({\beta }_{1}+\left(n-2\right){\beta }_{2}+\left(n-2\right)\left(n-3\right){\beta }_{3}\right){a}_{n}{z}^{n-1}\right|\\ \phantom{\rule{1em}{0ex}}=\left|\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|{e}^{i\left(n-1\right)\theta }{z}^{n-1}\right|\\ \phantom{\rule{1em}{0ex}}\le \lambda .\end{array}$

for all $z\in \mathcal{U}$. Let us consider a point $z\in \mathcal{U}$ such that z = |z| e-.

Then we have

$\left|\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|{\left|z\right|}^{n-1}\right|\le \lambda .$

Letting |z| → 1-, we obtain

$\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\le \lambda .$

Corollary 2.3. If $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ with arg β1 = arg β2 = arg β3 = ϕ and a n = |a n | ei((n-1)θ-ϕ)(n = 2, 3,...), then we have

$\left|{a}_{n}\right|\le \frac{\lambda }{n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(n=2,3,...\right).$

Example 2.4. Let us consider the function $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ with arg β1 = arg β2 = arg β3 = ϕ and

${a}_{n}=\frac{\lambda {e}^{i\left(\left(n-1\right)\theta -\varphi \right)}}{{n}^{2}{\left(n-1\right)}^{2}\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left(n=2,3...\right).$

Then, we see that

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\\ =\lambda \sum _{n=2}^{\infty }\frac{1}{n\left(n-1\right)}=\lambda \sum _{n=2}^{\infty }\left(\frac{1}{n-1}-\frac{1}{n}\right)=\lambda .\end{array}$

Corollary 2.5. If $f\left(z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ with arg β1 = arg β2 = arg β3 = ϕ and a n = |a n | ei((n-1)θ-ϕ)(n = 2, 3,...), then we have

$\left|z\right|-\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n}-{A}_{j}{\left|z\right|}^{j+1}\le \left|f\left(z\right)\right|\le \left|z\right|+\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n}+{A}_{j}{\left|z\right|}^{j+1}$

with

${A}_{j}=\frac{\left(\lambda -\sum _{n=2}^{j}n\left(n-1\right)\left(|{\beta }_{1}|+\left(n-2\right)|{\beta }_{2}|+\left(n-2\right)\left(n-3\right)|{\beta }_{3}|\right)|{a}_{n}|\right)}{j\left(j+1\right)\left(|{\beta }_{1}|\right)+\left(j-1\right)|{\beta }_{2}|+\left(j-1\right)\left(j-2\right)|{\beta }_{3}|\right)}$

and

$1-\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n-1}-{B}_{j}{\left|z\right|}^{j}\le \left|{f}^{\prime }\left(z\right)\right|\le 1+\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n-1}+{B}_{j}{\left|z\right|}^{j}$

with

${B}_{j}=\frac{\left(\lambda -\sum _{n=2}^{j}n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\right)}{j\left(\left|{\beta }_{1}\right|+\left(j-1\right)\left|{\beta }_{2}\right|+\left(j-1\right)\left(j-2\right)\left|{\beta }_{3}\right|\right)}$

Proof. In view of Theorem 2.1, we know that

$\begin{array}{c}\sum _{n=j+1}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\\ \phantom{\rule{1em}{0ex}}\le \lambda -\sum _{n=2}^{j}n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|.\end{array}$

Further, we note that

$\begin{array}{c}j\left(j+1\right)\left(\left|{\beta }_{1}\right|+\left(j-1\right)\left|{\beta }_{2}\right|+\left(j-1\right)\left(j-2\right)\left|{\beta }_{3}\right|\right)\sum _{n=j+1}^{\infty }\left|{a}_{n}\right|\\ \phantom{\rule{1em}{0ex}}\le \sum _{n=j+1}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|,\end{array}$

which is equivalent to

$\begin{array}{ll}\hfill \sum _{n=j+1}^{\infty }\left|{a}_{n}\right|& \le \frac{\left(\lambda -\sum _{n=2}^{j}n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\right)}{j\left(j+1\right)\left(\left|{\beta }_{1}\right|+\left(j-1\right)\left|{\beta }_{2}\right|+\left(j-1\right)\left(j-2\right)\left|{\beta }_{3}\right|\right)}\phantom{\rule{2em}{0ex}}\\ ={A}_{j}.\phantom{\rule{2em}{0ex}}\end{array}$

Thus, we have

$\left|f\left(z\right)\right|\le \left|z\right|+\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n}+\sum _{n=j+1}^{\infty }\left|{a}_{n}\right|{\left|z\right|}^{n}\le \left|z\right|+\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n}+{A}_{j}{\left|z\right|}^{j+1}$

and

$\left|f\left(z\right)\right|\ge \left|z\right|-\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n}-\sum _{n=j+1}^{\infty }\left|{a}_{n}\right|{\left|z\right|}^{n}\ge \left|z\right|-\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n}-{A}_{j}{\left|z\right|}^{j+1}.$

Next, we observe that

$\begin{array}{c}j\left(\left|{\beta }_{1}\right|+\left(j-1\right)\left|{\beta }_{2}\right|+\left(j-1\right)\left(j-2\right)\left|{\beta }_{3}\right|\right)\sum _{n=j+1}^{\infty }n\left|{a}_{n}\right|\\ \phantom{\rule{1em}{0ex}}\le \sum _{n=j+1}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\\ \phantom{\rule{1em}{0ex}}\le \lambda \sum _{n=2}^{j}n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-1\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|,\end{array}$

which implies that

$\begin{array}{ll}\hfill \sum _{n=j+1}^{\infty }n\left|{a}_{n}\right|& \le \frac{\left(\lambda -\sum _{n=2}^{j}n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right)\left|{a}_{n}\right|\right)}{j\left(\left|{\beta }_{1}\right|+\left(j-1\right)\left|{\beta }_{2}\right|+\left(j-1\right)\left(j-2\right)\left|{\beta }_{3}\right|\right)}\phantom{\rule{2em}{0ex}}\\ ={B}_{j}.\phantom{\rule{2em}{0ex}}\end{array}$

Therefore, we obtain that

$\left|{f}^{\prime }\left(z\right)\right|\le 1+\sum _{n=2}^{j}n\left|{a}_{n}\right|{\left|z\right|}^{n-1}+\sum _{n=j+1}^{\infty }n\left|{a}_{n}\right|{\left|z\right|}^{n-1}\le 1+\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n-1}+{B}_{j}{\left|z\right|}^{j}$

and

$\left|{f}^{\prime }\left(z\right)\right|\ge 1-\sum _{n=2}^{j}n\left|{a}_{n}\right|{\left|z\right|}^{n-1}-\sum _{n=j+1}^{\infty }n\left|{a}_{n}\right|{\left|z\right|}^{n-1}\ge 1-\sum _{n=2}^{j}\left|{a}_{n}\right|{\left|z\right|}^{n-1}-{B}_{j}{\left|z\right|}^{j}.$

## 3. Radius problem for the class $\mathcal{P}\left(\theta ,\alpha \right)$

To obtain the radius problem for the class $\mathcal{P}\left(\theta ,\alpha \right)$, we need the following lemma.

Lemma 3.1. If $f\left(z\right)\in \mathcal{P}\left(\theta ,\alpha \right)$, then

$\sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|\le 1-\alpha .$
(3.1)

Proof. Let $f\left(z\right)\in \mathcal{P}\left(\theta ,\alpha \right)$. Then, we have

$\begin{array}{ll}\hfill \text{Re}\phantom{\rule{1em}{0ex}}\left\{\left(z{f}^{″}\left(z\right)+{f}^{\prime }\left(z\right)\right)\right\}& =\text{Re}\phantom{\rule{1em}{0ex}}\left\{1+\sum _{n=2}^{\infty }{n}^{2}{a}_{n}{z}^{n-1}\right\}\phantom{\rule{2em}{0ex}}\\ =\text{Re}\phantom{\rule{1em}{0ex}}\left\{1+\sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|{e}^{i\left(\left(n-1\right)\theta +\pi \right)}{z}^{n-1}\right\}\phantom{\rule{2em}{0ex}}\\ =\text{Re}\phantom{\rule{1em}{0ex}}\left\{1-\sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|{e}^{i\left(\left(n-1\right)\theta \right)}{z}^{n-1}\right\}>\alpha \phantom{\rule{2em}{0ex}}\end{array}$

for all $z\in \mathcal{U}$. Let us consider a point $z\in \mathcal{U}$ such that z = |z| e-.

Then we have

$1-\sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|{\left|z\right|}^{n-1}>\alpha$

Letting |z| → 1-, we obtain the inequality (3.1).

Corollary 3.2. If $f\left(z\right)\in \mathcal{P}\left(\theta ,\alpha \right)$, then

$\left|{a}_{n}\right|\le \frac{1-\alpha }{{n}^{2}}\phantom{\rule{1em}{0ex}}\left(n=2,3,...\right).$

Remark 3.3. By Lemma 3.1, we observe that if $f\left(z\right)\in \mathcal{P}\left(\theta ,\alpha \right)$, then

$\sum _{n=2}^{\infty }n\left(n-1\right)\left|{a}_{n}\right|\le \sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|\le 1-\alpha .$

Applying Theorem 2.1 and Lemma 3.1, we derive

Theorem 3.4. Let $f\left(z\right)\in \mathcal{P}\left(\theta ,\alpha \right)$, and δ (0 < |δ| < 1). Then the function $\frac{1}{\delta }f\left(\delta z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ for (0 < |δ| ≤ |δ0(λ)|, where |δ0(λ)| is the smallest positive root of the equation

$\begin{array}{l}|{\beta }_{1}|\frac{|\delta |\sqrt{2\left(1-\alpha \right)}}{{\left(1-{|\delta |}^{2}\right)}^{3/2}}+|{\beta }_{2}|\frac{{|\delta |}^{2}\sqrt{\left(6+18{|\delta |}^{2}\right)\left(1-\alpha -2{|{a}_{2}|}^{2}\right)}}{{\left(1-{|\delta |}^{2}\right)}^{5/2}}\\ +|{\beta }_{3}|\frac{4\sqrt{3}{|\delta |}^{3}\sqrt{\left(1+8{|\delta |}^{2}+6{|\delta |}^{4}\right)\left(1-\alpha -2\right){|{a}_{2}|}^{2}-6{|{a}_{3}|}^{2}\right)}}{{\left(1-{|\delta |}^{2}\right)}^{7/2}}=\lambda \end{array}$
(3.2)

in 0 < |δ| < 1.

Proof. For $f\left(z\right)\in \mathcal{P}\left(\theta ,\alpha \right)$, we see that

$\frac{1}{\delta }f\left(\delta z\right)=z+\sum _{n=2}^{\infty }{\delta }^{n-1}{a}_{n}{z}^{n}$

and

$\sum _{n=2}^{\infty }n\left(n-1\right){\left|{a}_{n}\right|}^{2}\le 1-\alpha .$

Thus, to show that $\frac{1}{\delta }f\left(\delta z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$, from Theorem 2.1, it is sufficient to prove that

$\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right){\left|\delta \right|}^{n-1}\left|{a}_{n}\right|\le \lambda .$

Applying Cauchy-Schwarz inequality, we note that

$\begin{array}{c}\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right){\left|\delta \right|}^{n-1}\left|{a}_{n}\right|\\ \phantom{\rule{1em}{0ex}}\le \frac{\left|{\beta }_{1}\right|}{\left|\delta \right|}{\left(\sum _{n=2}^{\infty }n\left(n-1\right){\left|\delta \right|}^{2n}\right)}^{\frac{1}{2}}{\left(\sum _{n=2}^{\infty }n\left(n-1\right){\left|{a}_{n}\right|}^{2}\right)}^{\frac{1}{2}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left|{\beta }_{2}\right|}{\left|\delta \right|}{\left(\sum _{n=3}^{\infty }n\left(n-1\right){\left(n-2\right)}^{2}{\left|\delta \right|}^{2n}\right)}^{\frac{1}{2}}{\left(\sum _{n=3}^{\infty }n\left(n-1\right){\left|{a}_{n}\right|}^{2}\right)}^{\frac{1}{2}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left|{\beta }_{3}\right|}{\left|\delta \right|}{\left(\sum _{n=4}^{\infty }n\left(n-1\right){\left(n-2\right)}^{2}{\left(n-3\right)}^{2}{\left|\delta \right|}^{2n}\right)}^{\frac{1}{2}}{\left(\sum _{n=4}^{\infty }n\left(n-1\right){\left|{a}_{n}\right|}^{2}\right)}^{\frac{1}{2}}\\ \phantom{\rule{1em}{0ex}}\le \frac{\left|{\beta }_{1}\right|}{\left|\delta \right|}{\left(\sum _{n=2}^{\infty }n\left(n-1\right){\left|\delta \right|}^{2n}\right)}^{\frac{1}{2}}\sqrt{1-\alpha }\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left|{\beta }_{2}\right|}{\left|\delta \right|}{\left(\sum _{n=3}^{\infty }n\left(n-1\right){\left(n-2\right)}^{2}{\left|\delta \right|}^{2n}\right)}^{\frac{1}{2}}\sqrt{1-\alpha -2{\left|{a}_{2}\right|}^{2}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{\left|{\beta }_{3}\right|}{\left|\delta \right|}{\left(\sum _{n=4}^{\infty }n\left(n-1\right){\left(n-2\right)}^{2}{\left(n-3\right)}^{2}{\left|\delta \right|}^{2n}\right)}^{\frac{1}{2}}\sqrt{1-\alpha -2{\left|{\alpha }_{2}\right|}^{2}-6{\left|{a}_{3}\right|}^{2}}.\end{array}$
(3.3)

We note that

$\sum _{n=0}^{\infty }{x}^{n}=\frac{1}{1-x},\phantom{\rule{1em}{0ex}}\left(\left|x\right|<1\right),$

thus, we have

$\sum _{n=2}^{\infty }n\left(n-1\right){x}^{n}=\frac{2{x}^{2}}{{\left(1-x\right)}^{3}}.$
(3.4)

Since

$\sum _{n=3}^{\infty }\left(n-2\right){x}^{n-1}={x}^{2}\left(\sum _{n=3}^{\infty }\left(n-2\right){x}^{n-3}\right)={x}^{2}{\left(\sum _{n=3}^{\infty }{x}^{n-2}\right)}^{\prime }=\frac{{x}^{2}}{{\left(1-x\right)}^{2}},$

we see that

$\sum _{n=3}^{\infty }\left(n-1\right){\left(n-2\right)}^{2}{x}^{n}={x}^{3}{\left(\frac{{x}^{2}}{{\left(1-x\right)}^{2}}\right)}^{″}=\frac{2{x}^{3}+4{x}^{4}}{{\left(1-x\right)}^{4}}.$

and thus, we obtain

$\sum _{n=3}^{\infty }n\left(n-1\right){\left(n-2\right)}^{2}{x}^{n}=\frac{6{x}^{3}+18{x}^{4}}{{\left(1-x\right)}^{5}}.$
(3.5)

Furthermore, we have

$\begin{array}{ll}\hfill \sum _{n=4}^{\infty }\left(n-1\right){\left(n-2\right)}^{2}{\left(n-3\right)}^{2}{x}^{n}& ={x}^{4}\left(\sum _{n=4}^{\infty }\left(n-1\right){\left(n-2\right)}^{2}{\left(n-3\right)}^{2}{x}^{n-4}\right)\phantom{\rule{2em}{0ex}}\\ ={x}^{4}{\left(\sum _{n=4}^{\infty }\left(n-2\right)\left(n-3\right){x}^{n-1}\right)}^{‴},\phantom{\rule{2em}{0ex}}\end{array}$

but

$\sum _{n=4}^{\infty }\left(n-2\right)\left(n-3\right){x}^{n-1}={x}^{3}\left(\sum _{n=4}^{\infty }\left(n-2\right)\left(n-3\right){x}^{n-4}\right)=\frac{2{x}^{3}}{{\left(1-x\right)}^{3}}$

thus, we have

$\sum _{n=4}^{\infty }\left(n-1\right){\left(n-2\right)}^{2}{\left(n-3\right)}^{2}{x}^{n}=\frac{12{x}^{4}+72{x}^{5}+36{x}^{6}}{{\left(1-x\right)}^{6}},$

which yields

$\sum _{n=4}^{\infty }n\left(n-1\right){\left(n-2\right)}^{2}{\left(n-3\right)}^{2}{x}^{n}=\frac{48{x}^{4}\left(1+8x+6{x}^{2}\right)}{{\left(1-x\right)}^{7}}.$
(3.6)

Therefore, from (3.3)-(3.6) with |δ|2 = x, we obtain

$\begin{array}{c}\sum _{n=2}^{\infty }n\left(n-1\right)\left(\left|{\beta }_{1}\right|+\left(n-2\right)\left|{\beta }_{2}\right|+\left(n-2\right)\left(n-3\right)\left|{\beta }_{3}\right|\right){\left|\delta \right|}^{n-1}\left|{a}_{n}\right|\\ \le \left|{\beta }_{1}\right|\frac{\left|\delta \right|\sqrt{2\left(1-\alpha \right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{3/2}}+\left|{\beta }_{2}\right|\frac{{\left|\delta \right|}^{2}\sqrt{\left(6+18{\left|\delta \right|}^{2}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}\right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{5/2}}\\ \left|{\beta }_{3}\right|\frac{4\sqrt{3}{\left|\delta \right|}^{3}\sqrt{\left(1+8{\left|\delta \right|}^{2}+6{\left|\delta \right|}^{4}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}-6{\left|{a}_{3}\right|}^{2}\right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{7/2}}\end{array}$

Now, let us consider the complex number δ (0 < |δ| < 1) such that

$\begin{array}{c}\left|{\beta }_{1}\right|\frac{\left|\delta \right|\sqrt{2\left(1-\alpha \right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{3/2}}+\left|{\beta }_{2}\right|\frac{{\left|\delta \right|}^{2}\sqrt{\left(6+18{\left|\delta \right|}^{2}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}\right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{5/2}}\\ \phantom{\rule{1em}{0ex}}\left|{\beta }_{3}\right|\frac{4\sqrt{3}{\left|\delta \right|}^{3}\sqrt{\left(1+8{\left|\delta \right|}^{2}+6{\left|\delta \right|}^{4}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}-6{\left|{a}_{3}\right|}^{2}\right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{7/2}}=\lambda .\end{array}$

If we define the function h(|δ|) by

$\begin{array}{ll}\hfill h\left(\left|\delta \right|\right)& =\left|{\beta }_{1}\right|\left|\delta \right|{\left(1-{\left|\delta \right|}^{2}\right)}^{2}\sqrt{2\left(1-\alpha \right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left|{\beta }_{2}\right|{\left|\delta \right|}^{2}\left(1-{\left|\delta \right|}^{2}\right)\sqrt{\left(6+18{\left|\delta \right|}^{2}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+4\sqrt{3}\left|{\beta }_{3}\right|{\left|\delta \right|}^{3}\sqrt{\left(1+8{\left|\delta \right|}^{2}+6{\left|\delta \right|}^{4}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}-6{\left|{a}_{3}\right|}^{2}\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\lambda {\left(1-{\left|\delta \right|}^{2}\right)}^{7/2},\phantom{\rule{2em}{0ex}}\end{array}$

then we have h(0) = -λ < 0 and $h\left(1\right)=12\sqrt{5}\left|{\beta }_{3}\right|\sqrt{1-\alpha -2{\left|{a}_{2}\right|}^{2}-6{\left|{a}_{3}\right|}^{2}}>0$. This means that there exists some δ0 such that h(|δ0|) = 0 (0 < |δ0| < 1). This completes the proof of the theorem.

## 4. Radius problem for the class $\mathcal{K}\left(\theta ,\alpha \right)$

For the class $\mathcal{K}\left(\theta ,\alpha \right)$, we prove the following lemma.

Lemma 4.1. If $f\left(z\right)\in \mathcal{K}\left(\theta ,\alpha \right)$, then

$\sum _{n=2}^{\infty }n\left(n-\alpha \right)\left|{a}_{n}\right|\le 1-\alpha .$
(4.1)

Proof. Let $f\left(z\right)\in \mathcal{K}\left(\theta ,\alpha \right)$. Then, we have

$\begin{array}{ll}\hfill \text{Re}\phantom{\rule{1em}{0ex}}\left\{1+\frac{z{f}^{″}\left(z\right)}{{f}^{\prime }\left(z\right)}\right\}& =\text{Re}\phantom{\rule{1em}{0ex}}\left\{\frac{1+\sum _{n=2}^{\infty }{n}^{2}{a}_{n}{z}^{n-1}}{1+\sum _{n=2}^{\infty }n{a}_{n}{z}^{n-1}}\right\}\phantom{\rule{2em}{0ex}}\\ =\text{Re}\phantom{\rule{1em}{0ex}}\left\{\frac{1-\sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|{e}^{i\left(n-1\right)\theta }{z}^{n-1}}{1-\sum _{n=2}^{\infty }n\left|{a}_{n}\right|{e}^{i\left(n-1\right)\theta }{z}^{n-1}}\right\}>\alpha \phantom{\rule{2em}{0ex}}\end{array}$

for all $z\in \mathcal{U}$. Let us consider a point $z\in \mathcal{U}$. such that z = |z|e-.

Then we have

$\frac{1-\sum _{n=2}^{\infty }{n}^{2}\left|{a}_{n}\right|{\left|z\right|}^{n-1}}{1-\sum _{n=2}^{\infty }n\left|{a}_{n}\right|{\left|z\right|}^{n-1}}>\alpha$

Letting |z| → 1-, we obtain the inequality (4.1).

Corollary 4.2. If $f\left(z\right)\in \mathcal{K}\left(\theta ,\alpha \right)$, then

$\left|{a}_{n}\right|\le \frac{1-\alpha }{n\left(n-\alpha \right)}\phantom{\rule{1em}{0ex}}\left(n=2,3,\dots \right).$

Remark 4.3. If $f\left(z\right)\in \mathcal{K}\left(\theta ,\alpha \right)$, then

$\sum _{n=2}^{\infty }n\left(n-1\right)\left|{a}_{n}\right|\le \sum _{n=2}^{\infty }n\left(n-\alpha \right)\left|{a}_{n}\right|\le 1-\alpha .$

Applying Theorem 2.1, Lemma 4.1 and using the same technique as in the proof of Theorem 3.4, we derive

Theorem 4.4. Let $f\left(z\right)\in \mathcal{K}\left(\theta ,\alpha \right)$, and δ (0 < |δ| < 1). Then the function $\frac{1}{\delta }f\left(\delta z\right)\in \mathcal{T}\left({\beta }_{1},{\beta }_{2},{\beta }_{3};\lambda \right)$ for (0 < |δ| ≤ |δ0(λ)|, where |δ0(λ)| is the smallest positive root of the equation

$\begin{array}{c}\left|{\beta }_{1}\right|\frac{\left|\delta \right|\sqrt{2\left(1-\alpha \right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{3/2}}+\left|{\beta }_{2}\right|\frac{{\left|\delta \right|}^{2}\sqrt{\left(6+18{\left|\delta \right|}^{2}\right)\left(1-\alpha -2{\left|{a}_{2}\right|}^{2}\right)}}{{\left(1-{\left|\delta \right|}^{2}\right)}^{5/2}}\\ \phantom{\rule{1em}{0ex}}+\left|{\beta }_{3}\right|\frac{4\sqrt{3}{\left|\delta \right|}^{3}\sqrt{\left(1+8{\left|\delta \right|}^{2}+6{\left|\delta \right|}^{4}\right)\left(1-\alpha -2{\left|{\alpha }_{2}\right|}^{2}-6{\left|{a}_{3}\right|}^{2}\right)}}{{\left(1-|\delta {|}^{2}\right)}^{7/2}}=\lambda \end{array}$
(4.2)

in 0 < |δ| < 1.

## References

1. Duren PL: Univalent Functions. Springer-Verlag, Berlin; 1983.

2. Goodman AW: Univalent Functions. Volume 1–2. Mariner, Tampa; 1983.

3. Uyanik N, Owa S: New extensions for classes of analytic functions associated with close-to-convex and starlike of order α. Math Comput Model 2011, 54: 359–366. 10.1016/j.mcm.2011.02.020

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Correspondence to Basem Aref Frasin.

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Frasin, B.A. New subclasses of analytic functions. J Inequal Appl 2012, 24 (2012). https://doi.org/10.1186/1029-242X-2012-24 