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\title{ Janowski subclasses of starlike\\ mappings
}


\author{Paula Curt}
\address{``Babe\c{s}-Bolyai'' University, \\ Faculty of Economics and Business Administration\\
58-60, Teodor Mihali Street,\\
400591 Cluj-Napoca,\\
Romania}
\email{paula.curt@econ.ubbcluj.ro}
%

\dedicatory{Dedicated to the memory of 	Professor Gabriela Kohr}

\subjclass{32H02, 30C45}
\keywords{biholomorphic mapping, locally biholomorphic mapping starlike mapping, 	almost starlike mapping.}


\begin{abstract}
In this paper, two subclasses of biholomorphic starlike mappings named Janowski starlike and Janowski almost starlike with complex parameters are introduced and studied.  We determine $M$ such that holomorphic mappings $f$ which satisfy the condition $\|Df(z)-I\|\le M$, $z\in B^n$, are Janowski starlike, respectively Janowski almost starlike. 
We also derive sufficient conditions for normalized holomorphic mappings
(expressed in terms of their coefficient bounds) to belong to one of the subclasses of mappings mentioned above.


  

\end{abstract}

\maketitle





\section{ Introduction and Preliminaries}

The main reason for studying properties of  subclasses of biholomorphic mappings in several variables is the fact  that many of the classical results regarding  the class of univalent functions in one complex variable cannot be extended (without imposing supplementary restrictions) to higher dimensions.

In this paper we generalize the Janowski starlike and almost starlike classes of biholomorphic mappings studied in \cite{Cutai}.

In \cite {Ja}, W. Janowski introduced the following class of univalent normalized
functions defined on the unit disk $U$ of the complex plane.

If $A,B\in \mathbb{R}$, $-1\le B<A<1\le 1$, then
%\begin{equation}
%\label{1.1}
$$S^*[A,B]=\left\{f\in {\mathcal H}(U),\ f(0)=0,\ f'(0)=1,\
\f{zf'(z)}{f(z)}\prec \f{1+Az}{1+Bz}\right\}.$$
%\end{equation}

Closely related to Janowski starlike class of functions is the following
class of univalent functions \cite{Sil, SilSi}.

If $a,b\in \mathbb{R}$, $a\ge b$ then
%\begin{equation}
%\label{1.2}
$$S^*(a,b)=\left\{f\in {\mathcal H}(U),\ f(0)=0,\ f'(0)=1,\
\left|\f{zf'(z)}{f(z)}-a\right|<b\right\}.$$
%\end{equation}


Various results concerning the classes $S^*[A,B]$ and $S^*(a,b)$
can be found in \cite{Ja, Sil, SilSi}
(and the references therein).

Later, in \cite{KuOS} the authors introduced and studied the class $S^*[A,B]$ for some complex parameters $A$ and $B$, $A\ne B$ which satisfy one of the following two conditions:
\begin{align}
\label{1}
&|A|\le 1, |B|<1 \ \text{and} \ \Re(1-A\overline B)\ge |A-B| \\\label{2}
& |A|\le 1, |B|=1 \ \text{and} \ 1- A\overline B>0.
\end{align}

 

Many results related to the class $S^*[A,B]$ in the case when $A, B\in \mathbb C$ may be found in \cite{AHMK, KuOS}.

The main goal of the present paper is to generalize the Janowski mappings studied in \cite{Cutai} by introducing the $n$-dimensional version of the class $S^*[A,B]$ with complex parameters $A$ and $B$ that satisfy equivalent conditions to (\ref{1}) and (\ref{2}).

To this end, we recall some notions of  function theory in several complex variables  that will be used throughout the paper.

We denote by $\mathbb{C}^n$  the space of $n$ complex variables
$z=(z_1,z_2,\ldots ,z_n)$
with the Euclidean inner product
$\la z,w\ra=\sum_{j=1}^n z_j \ov{w_j}$
and the Euclidean norm
$\|z\|=\la z,z\ra^{1/2}$.
The open unit ball
$\{z\in \mathbb{C}^n:\ \|z\|<1\}$
is denoted by $B^n$. In
the case of one complex variable $B^1$ is denoted by $U$.
By  ${\mathcal L}(\mathbb{C}^n)$ we denote the the space of linear continuous
operators from $\mathbb{C}^n$ into $\mathbb{C}^n$ with the standard operator norm,
$\|A\|=\sup\{\|A(z)\|:\ \|z\|=1\}$. 
$I_n=I$
is the identity in ${\mathcal L}(\mathbb{C}^n)$.

Let ${\mathcal H}(B^n)$ be the set of holomorphic mappings from $B^n$ into $\mathbb{C}^n$.
If $f\in {\mathcal H}(B^n)$ we say that $f$ is normalized if $f(0)=0$ and the complex Jacobian
matrix of $f$ at $z=0$, $Df(0)$, is the identity operator $I$.
It is well known that if  $f\in {\mathcal H}(B^n)$ is a normalized mapping, then $f$ has the following series
expansion:
%\begin{equation}
%\label{2.1}
$f(z)=z+\sum_{k=2}^\infty A_k(z^k),\ z\in B^n,$
%\end{equation}
where
$A_k(w^k)=\f{1}{k!}D^kf(0)(w^k)$,
and $D^kf(0)(w^k)$ is the $k^{th}$ order Fr\'echet derivative of $f$ at $z=0$.

A holomorphic mapping $f:B^n\longrightarrow \mathbb{C}^n$  is said to be biholomorphic if the inverse
$f^{-1}$ exists and is biholomorphic on the open set $f(B^n)$.
It is known that any holomorphic and injective mapping on $B^n$
is biholomorphic on $B^n$.


Let $S(B^n)$ be the set of normalized biholomorphic mappings on $B^n$.
If $f\in {\mathcal H}(B^n)$, we say that $f$ is locally biholomorphic on $B^n$
if
$\det Df(z)\ne 0$, $z\in B^n$.
Let ${\mathcal L}S(B^n)$ be the set of normalized locally biholomorphic mappings
on $B^n$.

It is well-known \cite{Suff} that a locally biholomorphic mapping $f\in {\mathcal H}(B^n)$
such that $f(0)=0$ is starlike if and only if
\begin{equation}
\nonumber
\R \la (Df(z))^{-1}f(z),z\ra>0,\ z\in B^n\setminus \{0\}.
\end{equation}

We denote by $S^*(B^n)$  the set of biholomorphic normalized starlike
mappings.

We present next the notions of starlikeness of order $\alpha \in [0,1)$
 (see \cite{Cu1, Ko2}) and almost
 starlikeness of order $\alpha \in [0,1)$ (see \cite{Ko1, GraKocarte}). We denote by $S_\alpha ^*(B^n)$ the class of mappings that are starlike of order $\alpha $
 on $B^n$ and by
 ${\mathcal A}S_\alpha ^*(B^n)$ the class of almost starlike mappings of order $\alpha $ on $B^n$.
\begin{definition}
	\label{d2.1}
  Let $\alpha \in [0,1)$.
		\begin{equation}
		\nonumber
		S_\alpha ^*(B^n)=\left\{f\in {\mathcal L}S(B^n):
		\R \f{\|z\|^2}{\la (Df(z))^{-1}f(z),z\ra}>\alpha, z\in B^n\setminus \{0\}\right\}
		\end{equation}
		\begin{equation}
		\nonumber
		{\mathcal A}S_\alpha ^*(B^n)=\left\{f\in {\mathcal L}S(B^n):
	\R \f{\la (Df(z))^{-1}f(z),z\ra}{\|z\|^2}>\alpha, z\in B^n\setminus \{0\}\!\right\}.
		\end{equation}
		
	
\end{definition}





It is obvious that $f$ is starlike of order 0 on $B^n$ if and only if $f$ is almost
starlike of order 0 on $B^n$.
Moreover, we have that
$S_0^*(B^n)={\mathcal A}S_0^*(B^n)=S^*(B^n)$.

In \cite{Cutai} we introduced and studied the following classes of biholomorphic starlike
mappings on $B^n$.

\begin{definition}
	\label{d3.1}
	Let $a,b\in \mathbb{R}$ such that $|a-1|<b\le a$.
	\begin{equation}
	\nonumber
	S^*(a,b,B^n)=\left\{f\in {\mathcal L}S(B^n):\
	\left|\f{\|z\|^2}{\la (Df(z))^{-1}f(z),z\ra}-a\right|<b,\ z\in B^n\setminus \{0\}\right\}
	\end{equation}
	\begin{equation}
	\nonumber
	{\mathcal A}S^*(a,b,B^n)=\left\{f\in {\mathcal L}S(B^n):\
	\left|\f{\la (Df(z))^{-1}f(z),z\ra}{\|z\|^2}-a\right|<b,\ z\in B^n\setminus \{0\}\!\right\}.
	\end{equation}
	
	
\end{definition}

If  in the previous definition we take $a=b$,  then  $S^*(a,a,B^n)={\mathcal A}S_{1/2a}^*(B^n)$ and ${\mathcal A}S^*(a,a,B^n)=S_{1/2a}^*(B^n)$.

Various results concerning the classes $S^*(a,b,B^n)$ and ${\mathcal A}S^*(a,b,B^n)$ can be found in \cite{ Cutai, LiZhang, Manu1, Manu2}.

The following set of normalized mappings is the generalization to $n$-complex
variables of the well-known Carath\'eodory class of one variable
holomorphic functions with positive real part on the unit disk of complex plane.
\begin{equation}
\nonumber
{\mathcal M}=\{h\in {\mathcal H}(B^n):\ h(0)=0,\ Dh(0)=I,\
\R \la h(z),z\ra>0,\ z\in B^n\setminus \{0\}\}.
\end{equation}

The class ${\mathcal M}$, introduced in \cite{Pfa} plays a fundamental role in the study
of the Loewner differential equation (see for example \cite {Cucarte, Grahako, GraKocarte} and the references therein).
Also, it is closely related to certain subclasses of biholomorphic mappings
on $B$, such as starlike mappings, mappings with parametric representation \cite{Grahako}, etc.

Some subclasses of the class ${\mathcal M}$ are presented next.



 Let $g\in {\mathcal H}(U)$ be an univalent function, such that
		$g(0)=1$
		and $\R g(\zeta )>0$ on $U$.
		
		Let ${\mathcal M}_g$ be the class of holomorphic mappings given by
		\begin{equation}
		\nonumber
		%\label{2.7}
		{\mathcal M}_g\!=\!\left\{\!h\in {\mathcal H}(B^n): h(0)\!=\!0,\ Dh(0)\!=\!I,\
		\f{1}{\|z\|^2}\la h(z),z\ra\!\in\! g(U),\ z\!\in\! B\setminus \{0\}\!\right\}.
		\end{equation}
		


\noindent
It is clear that
${\mathcal M}_g\subseteq {\mathcal M}$ and for
$g(\zeta )=\f{1-\zeta }{1+\zeta }$
it follows that ${\mathcal M}_g={\mathcal M}$.
Particular choices of the function $g$
provide various subclasses of class ${\mathcal M}$.

\begin{remark}
	\label{r2.4}
	\begin{enumerate}
		\item [a)] Let $0\le \alpha <1$, let $g:U\to \mathbb{C}$ be the function defined by
		$g(\zeta )=\f{1-\zeta }{1-(2\alpha -1)\zeta }$
		and let $f\in {\mathcal L}S(B^n)$.
		Then 
		\begin{equation}
		\nonumber
		f\in S_\alpha ^*(B^n) \iff [Df(z)]^{-1}f(z)\in {\mathcal M}_g.
		\end{equation}
		\item [b)]Let $0\le \alpha <1$, let $g:U\to \mathbb{C}$ be the function defined by
		$g(\zeta )=\f{1+(2\alpha -1)\zeta }{1+\zeta }$
		and let $f\in {\mathcal L}S(B^n)$.
		Then
		\begin{equation}
		\nonumber
		f\in {\mathcal A}S_\alpha ^*(B^n) \iff
		[Df(z)]^{-1}f(z)\in {\mathcal M}_g.
		\end{equation}	
		\item [c)]	 Let $a,b\in \mathbb{R}$ such that $|a-1|<b\le a$, let $g:U\to \mathbb{C}$ be the function defined by
		$g(\zeta )=\f{1+(a-a^2+b^2)/b\zeta }{1+(1-a)/b\zeta }$
		and let $f\in {\mathcal L}S(B^n)$.
				Then
		\begin{equation}
	\nonumber
		f\in {\mathcal A}S^*(a,b,B^n) \iff 	[Df(z)]^{-1}f(z)\in {\mathcal M}_g.
		\end{equation}
	\item [d)]  Let $a,b\in \mathbb{R}$ such that $|a-1|<b\le a$, let $g:U\to \mathbb{C}$ be the function defined by
	$g(\zeta )=\f{1+(a-1)/b\zeta }{1+(a^2-a-b^2)/b\zeta }$
	and let $f\in {\mathcal L}S(B^n)$.
	Then
		\begin{equation}
		\nonumber
		f\in S^*(a,b,B^n) \iff [Df(z)]^{-1}f(z)\in {\mathcal M}_g.,
		\end{equation}
	
	\end{enumerate}
	  
		
	\end{remark}
For the first  assertion in the previous remark see \cite{Cu1, Ko2}, for the second see \cite{Ko1, GraKocarte} and  for the last two see \cite{Cutai}.

If $g$ is a univalent function with $g(0)=1$ and positive real part on  $U$ 
we denote by $S_g^*(B^n)$ the subset of $S^*(B^n)$ consisting of the normalized
locally biholomorphic mappings $f$ such that
$(Df(z))^{-1}f(z)\in {\mathcal M}_g$.

In this paper, our main concern is the case when the function  $g\in {\mathcal H}(U)$ has positive real part on $U$ and is of the following
particular form:
$g(\zeta)=\f{1+A\zeta }{1+B\zeta},
$ with  $A, B\in \mathbb C$, $A\ne B$.





\section{Janowski subclasses of starlike mappings}

In this section we introduce and study two subclasses of biholomorphic mappings named Janowski starlike and Janowski almost starlike with complex parameters.

Let $a\in \mathbb C, b\in \mathbb{R}$ such that $|a-1|<b\le \Re a$. We denote by $S^*(a,b,B^n)$ the class of Janowski starlike mappings 
on $B^n$ and by
${\mathcal A}S^*(a,b,B^n)$ the class of Janowski almost starlike mappings  on $B^n$.
\begin{definition}
	\label{d5.1}
	 Let $a\in \mathbb C, b\in \mathbb{R}$ such that $|a-1|<b\le \Re a$.
		\begin{equation}
		\nonumber
		S^*(a,b,B^n)=\left\{f\in {\mathcal L}S(B^n):\
		\left|\f{\|z\|^2}{\la (Df(z))^{-1}f(z),z\ra}-a\right|<b,\ z\in B^n\setminus \{0\}\right\}
		\end{equation}
		\begin{equation}
		\nonumber
		{\mathcal A}S^*(a,b,B^n)=\left\{f\in {\mathcal L}S(B^n):\
		\left|\f{\la (Df(z))^{-1}f(z),z\ra}{\|z\|^2}-a\right|<b,\ z\in B^n\setminus \{0\}\!\right\}.
		\end{equation}
		
	
\end{definition}

We  mention that for $a\in \mathbb R$ Definition \ref{d5.1} reduces to the Definition \ref{d3.1}. We also mention that the mappings in $S^*(a,b,B^n)$ or in ${\mathcal A}S^*(a,b,B^n)$
are starlike since the conditions satisfied by  $a$ and $b$ in Definition \ref{d5.1}
imply that the disk of center $a$ and radius $b$ is a subset of the right half- plane.

In the next remark we present the relationships between the two subclasses of starlike mappings introduced above.

\begin{remark}
	\label{r3.2}
Let $a\in \mathbb{C},b\in \mathbb{R}$ such that $|a-1|<b\le \Re a$.
		
		Then the following assertions are true:
		
		(i) $S^*(a,b,B^n)={\mathcal A}S^*\left(\frac{\overline a}{|a|^2-b^2},\frac{b}{|a|^2-b^2},B^n\right)$,
		if $b<|a|$;
		
		(ii) $S^*(a,a,B^n)={\mathcal A}S_{\frac{1}{2a}}^*(B^n)$, if $b=|a|$;
		
		(iii) ${\mathcal A}S^*(a,b,B^n)=S^*\left(\frac{\overline a}{|a|^2-b^2},\frac{b}{|a|^2-b^2},B^n\right)$, if $b<|a|$;
		
		(iv) ${\mathcal A}S^*(a,a,B^n)=S_{\frac{1}{2a}}^*(B^n)$, if $b=|a|$.
	
	\end{remark}

\begin{proof}
	The assertions (i) and (iii) are immediate consequences of the fact
	that the disk of center $a$ and radius $b$ is mapped by the function
	$1/\zeta $ onto the disk of center $\frac{\overline a}{|a|^2-b^2}$
	and radius $\frac{b}{|a|^2-b^2}$.
	
	The assertions (ii) and (iv) are immediate consequences of the fact
	that  the disk of center $a$ and radius $a$ is mapped by the function $1/\zeta $
	onto the half-plane
	$\left\{\zeta \mid \Re \zeta >\frac{1}{2a}\right\}$.
\end{proof}

The following remark (see also \cite{KuOS}) presents the conditions satisfied by the complex parameters $A, B$, $A\ne B$ such that  $g(\zeta)=\f{1+A\zeta }{1+B\zeta}, \ \zeta \in U,$ is a holomorphic function with  positive real part on $U$.
\begin{remark}
	\label{r3.0}
	Let $A, B\in \mathbb C$, $A\ne B$ and let $g\in {\mathcal H}(U)$ be the function defined by $g(\zeta)=\f{1+A\zeta }{1+B\zeta}$. If  $g$ has  positive real part on $U$, then the complex parameters $A$ and $B$ satisfy one of the following conditions: 
	\begin{align}
	\label{1.1}
	& |B|<1 \ \text{and} \ \Re(1-A\overline B)\ge |A-B| \\\label{1.2}
	& |B|=1 \ \text{and} \ -1\le A\overline B<1
	\end{align}
\end{remark}

\begin{proof}
	The fact that $g\in {\mathcal H}(U)$ immediately implies that $|B|\le 1$.
	
	The function $g$, as a homographic function, maps the unit disk $U$ either onto an open disk (when $|B|<1$) or onto a half-plane (when $|B|=1$). It remains for us to determine the conditions satisfied by $A$ and $B$ such that the image $g(U)$ to be situated in the right half-plane.
		
	When $|B|<1$,
	the function $g$ maps the unit disk $U$ onto the open unit disk given by
	\begin{equation}
	\nonumber
	\left|g(\zeta)-\f{1-A\overline B}{1-|B|^2}\right|<\f{|A-B|}{1-|B|^2}, \ \zeta \in U.
	\end{equation}
The above disk is in the right half-plane if 
	\begin{equation}
	\nonumber
	\Re g(\zeta)>\f{\Re (1-A\overline B)-|A-B|}{1-|B|^2}\ge 0,\  \zeta \in U,
	\end{equation}
	hence (\ref{1.1}) is fulfilled.
	
	When $|B|=1$, the function $g$ maps the unit disk $U$ onto a half-plane, which has to be situated in the right half-plane, hence the image of the unit circle is a vertical line. Therefore, $\Re g(\overline B)$=$\Re g(i\overline B)$, wherefrom we obtain that $A\overline B\in \mathbb R$ and $g(\overline B)=\f{1+A\overline B}{2}$.
	In this case $\Re g(\zeta)>0, \zeta \in U$,  if and only if
	\begin{equation}
	\nonumber
	1=\Re g(0)>\f{ 1+A\overline B}{2}\ge 0,\ 
	\end{equation}
hence 	(\ref{1.2}) is fulfilled. This completes the proof.
	\end{proof}
		We remark that each of the conditions (\ref{1.1}) or  (\ref{1.1}) easily implies that $|A|\le 1$. 

We also remark that we have proved that the condition $|A|\le 1$ in (\ref{1}) is redundant. The condition (\ref{1.1}) is  equivalent to  (\ref{1}).

We next present the appropriate univalent function $g$ such that
the Janowski class  ${\mathcal A}S^*(a,b,B^n)$
can be rewritten as $S_g^*(B^n)$.
\begin{remark}
	\label{r3.1}
	  	(i) Let $a\in \mathbb{C} \mbox{ and} \  b\in \mathbb{R} $ such that $|a-1|<b\le \Re a$. 	Then
	  \begin{equation}
	  \label{3.5}
	  {\mathcal A}S^*(a,b,B^n)=S_g^*(B^n),
	  \mbox {where} \	g(\zeta )=\dfrac{1+\frac {a-|a|^2+b^2}{b}\zeta }{1+\frac {1-\overline a}{b}\zeta },\ \zeta \in U
	  \end{equation}
	  
	  
	  (ii) Let $A, B \in \mathbb C$, $ A\neq B$,  and let
		$g:U\longrightarrow \mathbb C$, 
		$g(\zeta )=\frac{1+A\zeta }{1+B\zeta }$ be a holomorphic function with positive real part on $U$.
		Then:
		\begin{align}
		\label{3.3}
		& S_g^*(B^n)={\mathcal A}S^*\left(\frac{1-A\overline B}{1-|B|^2},\frac{|A-B|}{1-|B|^2},B^n\right)
		\mbox{ if } |B|<1\mbox{ and} \ \Re (1-A\overline B)\geq |A-B|\\
		& S_g^*(B^n)={\mathcal A}S_{\frac {1+A\overline B}{2}}^*(B^n)\mbox{ if } |B|= 1 \mbox{ and} -1<A\overline B<1.\nonumber
		\end{align}
	
		
	
\end{remark}
\begin{proof}
	
		To prove (\ref{3.5}) we have to determine $A,B\in \mathbb C$, $A\ne B,|B|<1, \Re (1-A\overline B)\geq |A-B|$ such that
	$a=\f{1-A\overline B}{1-|B|^2}$
	and
	$b=\f{|A-B|}{1-|B|^2}$.
	
	Straightforward computations lead to the following values:
	$B=\f{1-\overline a}{b}\frac{|A-B|}{\overline A-\overline B}=\f{1-\overline a}{b}e^{i\phi}, \phi=\arg{(A-B)}$, and $A=\frac{b^2+a-|a|^2}{b}e^{i\phi}$.
		Since the image of $U$ under the function $g$ is invariant to the rotations of the unit disk, without loss of generality we can assume that  $B=\f{1-\overline a}{b}$
	and hence
	$A=\f{b^2+a-|a|^2}{b}$. 
	
	By direct computations we obtain that $1-A\overline B=\frac{\ov a(b^2-|a-1|^2)}{b^2}$ and $B=\frac {b^2-|a-1|^2}{b}$. Therefore the desired inequality $\Re (1-A\overline B)\geq |A-B|$ is equivalent to $\Re \ov a\ge b$, which is true. The inequality $|B|<1$ results immediately from $|a-1|<b$. This completes the proof.
	
	The equalities in (\ref{3.3}) easily follow from the fact that the unit disk
	is mapped by the function
	$g(\zeta )=\f{1+A\zeta }{1+B\zeta }$
	onto the disk centered at
	$a=\f{1-A\overline B}{1-|B|^2}$
	with radius
	$b=\f{|A-B|}{1-|B|^2}$
	in the case when $|B|<1$,
	respectively onto the half-plane
	$\left\{\zeta \mid \R \zeta >\f{1+A\overline B}{2}\right\}$
	in the case when $|B|=1$.
	
	
	

	
\end{proof}
We next present the appropriate univalent function $g$ such that
the Janowski class $S^*(a,b,B^n)$ 
can be rewritten as $S_g^*(B^n)$.
\begin{remark}
	\label{r3.11}
	  		(i) Let $a\in \mathbb{C} \mbox{ and} \  b\in \mathbb{R} $ such that $|a-1|<b\le \Re a$.
		Then
			\begin{equation}
		\label{3.6}
		S^*(a,b,B^n)=S_g^*(B^n),	 
		\mbox {where} \	g(\zeta )=\dfrac{1+\frac {\overline a-1}{b}\zeta }{1+\frac {|a|^2-a-b^2}{b}\zeta }, \zeta \in U.
		\end{equation}
	(ii) Let $A, B \in \mathbb C$, $ A\neq B$,  and let
	$g:U\longrightarrow \mathbb C$, 
	$g(\zeta )=\frac{1+A\zeta }{1+B\zeta }$ be a holomorphic function with positive real part on $U$.
	Then:
	
	\begin{align}
	\label{3.4}
	& S_g^*(B^n)=S^*\left(\frac{1-A\overline B}{1-|A|^2},\frac{|A-B|}{1-|A|^2},B^n\right)\mbox{ for } |A|<1\mbox{ and} \ \Re (1-A\overline B)\geq |A-B|\\
	& S_g^*(B^n)=S_{\frac {1+A\overline B}{2}}^*(B^n)\mbox{ if } |A|= 1 \mbox{ and} -1<A\overline B<1.\nonumber
	\end{align}
\end{remark}
\begin{proof}
	
	The equality in (\ref{3.6}) is an immediate consequence of (\ref{3.5})
	and assertion (iii) from Remark \ref{r3.2}.
	
	The equalities in (\ref{3.4}) can be  justified by using similar arguments to those presented in the proof of (ii), Remark \ref{r3.1}. 
	

	\end{proof}
	

\section{Sufficient Conditions for Janowski Starlikeness}

In this section we obtain sufficient conditions for  normalized holomorphic
mappings to belong to $S^*(a,b,B^n)$,
respectively ${\mathcal A}S^*(a,b,B^n)$, where $a\in \mathbb C, b\in \mathbb{R}$,
$|a-1|<b\le \Re a$.

We deduce first a sufficient condition for  normalized holomorphic
mappings to belong to 
$S_g^*(B^n)$, where $g:U\to \mathbb{C}$ is defined by
$g(\zeta )=\f{1+A\zeta }{1+B\zeta }$ and
$A, B \in \mathbb C$, $ A\neq B,  |B|<1$ and $\ \Re (1-A\overline B)\geq |A-B|$.
\begin{theorem}
	\label{t4.0}
		Let
	$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
	be a holomorphic mapping on $B^n$ and let
	$g:U\to \mathbb{C}$ be the function defined by
	$g(\zeta )=\frac{1+A\zeta }{1+B\zeta }$, $\zeta \in U$, where
	$A, B \in \mathbb C$, $ A\neq B,  |B|<1$ and $\ \Re (1-A\overline B)\geq |A-B|$.

	If
	\begin{equation}
	\label{4.00}
	\|Df(z)-I\|\le M, \  \ z\in B^n
	\end{equation}
	where $M$ is defined by
	\begin{equation}
	\label{4.0}
	M=\frac{2|A-B|(1-|B|)}{2|1-A\overline B|+2|A-B|+(1-|B|^2)}
	\end{equation}
		then $f\in S_g^*(B^n)$.
\end{theorem}
\begin{proof}
In order that $f$ to be in $ S_g^*(B^n)$ it is sufficient to show that  
\begin{equation}
\nonumber
\left \|\frac{1-|B|^2}{|A-B|}
(Df(z))^{-1}f(z)-\f{1-A\overline B}{|A-B|}z\right\|<\|z\|, \forall z\in B^n\setminus \{0\}.
\end{equation}

	
	If $h$ is the holomorphic function defined by 
	
	\begin{equation}
	\nonumber
h(z)=\frac{1-|B|^2}{|A-B|}
(Df(z))^{-1}f(z)-\f{1-A\overline B}{|A-B|}z, z\in B^n
	\end{equation}
	 then $h(0)=0$, $\|Dh(0)\|=|B|<1$ hence $\lim_{z\rightarrow 0}\f {\|h(z)\|}{\|z\|}<1$. 
	
	Therefore, it suffices to prove that $\|h(z)\|<\|z\|$ for all $  z\in B^n\setminus\{0\}.$
	
	If the previous inequality is not true, then there exists a point $z_0 \in B^n\setminus\{0\}$ such that $\|h(z_0)\|=\|z_0\|$. By denoting $w_0=(Df(z_0))^{-1}f(z_0)$, after direct computations we obtain 
	\begin{equation} 
\label{4.01}
\|w_0\|\le \frac{|1-A\overline B|+|A-B|}{1-|B|^2}\|z_0\|.
\end{equation}
Since $\|Df(z_0)-I\|\le M$, we have first $\|Df(z_0)w_0-w_0\|\le M\|w_0\|$ and hence 
\begin{equation}
\label{4.02}
\|f(z_0)-w_0\|\le M\|w_0\|
\end{equation}

Now we prove that 
\begin{equation}
\label{4.03}
\|f(z)-z\|<\f{M}{2}, \forall z\in B^n.
\end{equation}
 If not, then there exists a point $z_1\in B^n\setminus \{0\}$ such that 
 \begin{equation}
 \label{4.000}
\max_{\|z\|\le \|z_1\|}\|f(z)-z\|=\|f(z_1)-z_1\|=\f{M}{2}
 \end{equation}

 
 According to Lemma1 \cite{LiuZhu}, there exists a real number $t\ge 2$ such that 
 \begin{equation}
 \nonumber
 \la Df(z_1)(z_1)-z_1, f(z_1)-z_1\ra=t\|f(z_1)-z_1\|^2.
 \end{equation}
 In view of  the relations (\ref{4.00}) and (\ref {4.000}),  the previous equality implies 
 \begin{equation}
 t\frac {M^2}{4}\le \|Df(z_1)(z_1)-z_1\|\cdot \|f(z_1)-z_1\|\le \frac{M^2}{2}\|z_1\|<\frac{M^2}{2},
 \end{equation} 
hence $t<2 $, which is a  contradiction.

Therefore, the relation (\ref{4.03}) is true and by applying the Schwarz's Lemma we obtain
\begin{equation}
\label{4.04}
\|f(z)-z\|\le \f{M}{2}\|z\|^2, \ \forall z \in B^n.
\end{equation}
By using the relations (\ref{4.02}) and (\ref{4.04}) we obtain first that
\begin{equation}
\nonumber
M\|w_0\|\ge \|z_0-w_0\|-\|f(z_0)-z_0\|>\|z_0-w_0\|-\f{M}{2}\|z_0\|,
\end{equation}
and then
\begin{equation}
\label{4.05}
M<\f {\|z_0-w_0\|}{\f{\|z_0\|}{2}+\|w_0\|}.
\end{equation}
On the other hand, by using the fact that $\|h(z_0)\|=\|z_0\|$, (\ref{4.01}) and (\ref{4.0}) we get
\begin{align}
\nonumber
\f {\|z_0-w_0\|}{\f{\|z_0\|}{2}+\|w_0\|} & = \f{\left\|w_0-\frac{|1-A\overline B|}{1-|B|^2}z_0+\frac{|1-A\overline B|}{1-|B|^2}z_0-z_0\right\|}{\f{\|z_0\|}{2}+\|w_0\|}  \\
&\ge\f{\frac{|A-B|-|B(A-B)|}{1-|B|^2}\|z_0\|}{\f{\|z_0\|}{2}+\|w_0\|}=\frac{\frac{|A-B|}{1-|B|^2}}{\f{1}{2}+\f{\|w_0\|}{\|z_0\|}}\nonumber\\
&\ge \frac{\frac{|A-B|}{1+|B|}}{\f{1}{2}+\f{|1-A\overline B|+|A-B|}{1-|B|^2}}=M \nonumber
\end{align}
Because  the previous inequality contradicts (\ref{4.05}), the assumption that there exists a point $z_0 \in B^n\setminus\{0\}$ such that $\|h(z_0)\|=\|z_0\|$ is false, hence $\|h(z)\|<\|z\|$ for all $  z\in B^n\setminus\{0\}$. This completes the proof.

\end{proof}

	Let $a\in \mathbb{C} \mbox{ and} \  b\in \mathbb{R} $ such that $|a-1|<b\le \Re a$.

By taking $A=\frac{a-|a|^2+b^2}{b}$ and $B=\frac{1-\overline a}{b}$
in Theorem \ref{t4.0} and by using Remark \ref{r3.1} we obtain
the following sufficient condition for a holomorphic mapping
to belong to ${\mathcal A}S^*(a,b,B^n)$

\begin{theorem}
	\label{t4.02}
	Let
	$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
	be a holomorphic mapping on $B^n$ and let  $a\in \mathbb{C},b\in \mathbb{R}$ such that
	$|a-1|<b\le\Re a$.
	
	If
	$$\|Df(z)-I\|\le \f {2(b-|1-a|)}{2(|a|+b)+1}, \ \ z\in B^n$$
	then $f\in {\mathcal A}S^*(a,b,B)$.
\end{theorem}
	Let $a\in \mathbb{C} \mbox{ and} \  b\in \mathbb{R} $ such that $|a-1|<b\le \Re a$. 
	
	If we take
	$A=\frac{\overline a-1}{b}$ and $B=\frac{|a|^2-a-b^2}{b}$
	in Theorem \ref{t4.0} and use Remark \ref{r3.11} we get the following sufficient condition
	for a holomorphic mapping to be in $S^*(a,b,B^n)$.
	
	\begin{theorem}
		\label{t4.01}
		Let
		$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
		be a holomorphic mapping on $B^n$ and let $a\in \mathbb{C},b\in \mathbb{R}$ such that
		$|a-1|<b\le\Re a$ and $b<|a|$.
		
		If
		$$\|Df(z)-I\|\le \f {2(b-||a|^2-a+b^2|)}{(|a|+b)(|a|-b+2)}, \ \ z\in B^n$$
		then $f\in S^*(a,b,B^n)$.
	\end{theorem}
Next theorems present sufficient conditions that are expressed in terms of coefficients
bounds of the considered mappings.
\begin{theorem}
	\label{t4.1}
	Let
	$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
	be a holomorphic mapping on $B^n$ and let
	$g:U\to \mathbb{C}$ be the univalent function defined by
	$g(\zeta )=\dfrac{1+A\zeta }{1+B\zeta }$, $\zeta \in U$, where
	$A, B \in \mathbb C$, $ A\neq B,  |B|<1$ and $\ \Re (1-A\overline B)\geq |A-B|$.
	
	If
	\begin{equation}
	\label{4.1}
	\sum_{k=2}^\infty \left(\left|k\f{1-A\overline B}{|A-B|}-\f{1-|B|^2}{|A-B|}\right|+k\right)\|A_k\|
	\le 1-|B|
	\end{equation}
	then $f\in S_g^*(B^n)$. Moreover, for all $z\in B^n$
	\begin{equation}
	\label{4.06}
	\|z\|-\frac {1-|B|}{\left|\frac{2(1-A\overline B)-(1-|B|^2)}{|A-B|}\right|+2}\|z\|^2\le \|f(z)\|\le \|z\|+\frac {1-|B|}{\left|\frac{2(1-A\overline B)-(1-|B|^2)}{|A-B|}\right|+2}\|z\|^2.
	\end{equation}
	
\end{theorem}


\begin{proof}

From the inequality (\ref{4.1}) it follows that
\begin{align}
\sum_{k=2}^\infty k\|A_k\|& \le \f{2}{2\frac{|1-A\overline B|}{|A-B|}-\frac {1-|B|^2}{|A-B|}+2}
\sum_{k=2}^\infty \left(\left|k\f{1-A\overline B}{|A-B|}-\f{1-|B|^2}{|A-B|}\right|+k\right)\|A_k\|\nonumber\\
&\le \f{2(1-|B|)}{2\frac{|1-A\overline B|}{|A-B|}-\frac {1-|B|^2}{|A-B|}+2}<1.\nonumber
\end{align}
By direct computation of Fr\'echet derivatives of $f$ we obtain
\begin{align}
\|Df(z)-I\|=\left\|\sum_{k=2}^\infty kA_k(z^{k-1},\cdot )\right\|
\le \sum_{k=2}^\infty k\|A_k\|<1,\ z\in B^n.\nonumber
\end{align}
Hence we obtain that
$Df(z)=I-(I-Df(z))$
is an invertible linear operator and
\begin{equation}
\label{4.2}
\|(Df(z))^{-1}\|\le \f{1}{1-\|I-Df(z)\|}
\le \f{1}{1-\sum_{k=2}^\infty k\|A_k\|},\ z\in B^n.
\end{equation}
For every $z\in B^n\setminus \{0\}$, 
we have
\begin{align}
&\left\|\f{1-|B|^2}{|A-B|}f(z)-
\f{1-A\overline B}{|A-B|}Df(z)(z)\right\|=
 \nonumber \\
&\left\|\f{A\overline B-|B|^2}{|A-B|}z+\sum_{k=2}^\infty
\!\left (\f{1-|B|^2}{|A-B|}-\f{1-A\overline B}{|A-B|}k\!\right )\!
A_k(z^k)\right\|<\nonumber \\ 
&\|z\|\left(|B|+\sum_{k=2}^\infty
\left|k\f{|1-A\overline B|}{|A-B|}-\f{1-|B|^2}{|A-B|}\right| \|A_k\|\right).\nonumber
\end{align}
By using the inequality (\ref{4.2}) and the previous inequality we obtain
\begin{align}
&\f{1}{\|z\|^2}\left|\left\la \f{1-|B|^2}{|A-B|}
(Df(z))^{-1}f(z),z\right\ra-\f{1-A\overline B}{|A-B|}\|z\|^2\right|\le \nonumber \\
&\f{1}{\|z\|}\|(Df(z))^{-1}\|\cdot
\left\|\frac {1-|B|^2}{|A-B|}f(z)-\f{1-A\overline B}{|A-B|}Df(z)(z)\right\|< \nonumber \\
&\f{|B|+\sum_{k=2}^\infty
	\left|k\f{|1-A\overline B|}{|A-B|}-\f{1-|B|^2}{|A-B|}\right| \|A_k\|}
{1-\sum_{k=2}^\infty \|A_k\|k}\le 1,\nonumber
\end{align}
where for the last inequality we have used the relation (\ref{4.1}).

In conclusion
$$\left|\left\la \f{(Df(z))^{-1}f(z)}{\|z\|^2},z\right\ra
-\f{1-A\overline B}{1-|B|^2}\right|<\f{|A-B|}{1-|B|^2},\
z\in B^n\setminus \{0\},$$
which implies that $f\in S_g^*(B^n)$ as desired.

From the inequality (\ref{4.1}) it follows that
\begin{equation}
\nonumber
\min_{k\ge 2}\left(\left|k\f{1-A\overline B}{|A-B|}-\f{1-|B|^2}{|A-B|}\right|+2\right)\sum_{k=2}^\infty \|A_k\|\le 
1-|B|.
\end{equation}
We can easily see that the minimum is attained for $k=2$, hence the following inequality holds
\begin{equation}
\nonumber
\sum_{k=2}^\infty \|A_k\|\le \frac {1-|B|}{\left|\frac{2(1-A\overline B)-(1-|B|^2)}{|A-B|}\right|+2}.
\end{equation}
Combining the previous inequality with the following ones
\begin{equation}
\nonumber 
\|z\|-\sum_{k=2}^\infty \|A_k\|\|z\|^2\le \|f(z)\|\le  \|z\|+\sum_{k=2}^\infty \|A_k\|\|z\|^2,
\end{equation}
we get (\ref{4.06}) as desired.
\end{proof}
Let $a\in \mathbb{C} \mbox{ and} \  b\in \mathbb{R} $ such that $|a-1|<b\le \Re a$.

By taking $A=\frac{a-|a|^2+b^2}{b}$ and $B=\frac{1-\overline a}{b}$
in Theorem \ref{t4.1}  we obtain
the following sufficient condition for a holomorphic mapping
to be in ${\mathcal A}S^*(a,b,B^n)$.

\begin{theorem}
	\label{t4.2}
	Let
	$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
	be a holomorphic mapping on $B^n$ and let  $a\in \mathbb{C},b\in \mathbb{R}$ such that
	$|a-1|<b\le\Re a$.
	
	If
	$$\sum_{k=2}^\infty \left(|ka-1|+kb\right)\|A_k\|\le b-|1-a|$$
	then $f\in {\mathcal A}S^*(a,b,B)$.  Moreover, for all $z\in B^n$
	\begin{equation}
	\label{4.08}
	\|z\|-\frac {b-|1-a|}{|2a-1|+2b}\|z\|^2\le \|f(z)\|\le \|z\|+\frac {b-|1-a|}{|2a-1|+2b}\|z\|^2.
	\end{equation}
\end{theorem}

If in the previous theorem we take
$a=b=\f{1}{2\alpha }$, $0<\alpha <1$,
we get the next result \cite{LiuZhu1}.

\begin{corollary}
	\label{c4.3}
	Let
	$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
	be a locally biholomorphic mapping on $B^n$ and let $\alpha \in \mathbb{R}$
	be such that $0<\alpha <1$.
	
	If
	$$\sum_{k=2}^\infty (k-\alpha )\|A_k\|\le \f{1-|2\alpha -1|}{2}$$
	then $f\in S_\alpha ^*(B^n)$.
\end{corollary}

Let $a\in \mathbb{C} \mbox{ and} \  b\in \mathbb{R} $ such that $|a-1|<b\le \Re a$. 

If we take $A=\frac{\overline a-1}{b}$ and $B=\frac{|a|^2-a-b^2}{b}$
in Theorem \ref{t4.1} we get the following sufficient condition
for a holomorphic mapping to be in $S^*(a,b,B^n)$.

\begin{theorem}
	\label{t4.4}
	Let
	$f(z)=z+\sum_{k=2}^\infty A_k(z^k)$
	be a holomorphic mapping on $B^n$ and let $a\in \mathbb{C},b\in \mathbb{R}$ such that
	$|a-1|<b\le \Re a$ and $b<|a|$.
	
	If
	$$\sum_{k=2}^\infty\left( \left| k\overline a-(|a|^2-b^2)\right|+kb\right)\|A_k\|\le b-||a|^2-a+b^2|$$
	then $f\in S^*(a,b,B^n)$.  Moreover, for all $z\in B^n$
	\begin{equation}
	\label{4.07}
	\|z\|-\frac {b(b-||a|^2-a-b^2)}{|2\overline a-(|a|^2-b^2)|+2b^2}\|z\|^2\le \|f(z)\|\le \|z\|+\frac {b(b-||a|^2-a-b^2)}{|2\overline a-(|a|^2-b^2)|+2b^2}\|z\|^2.
	\end{equation}
\end{theorem}



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\end{document}