Growth and covering theorems for biholomorphic mappings are important parts in geometric function theory of several complex variables. Especially as the biholomorphic mappings were restricted with the geometric characteristic, many scholars started to discuss the problem. From the geometric characteristic of spirallike mappings of type $\beta$, Feng Shuxia [1] gave the definitions of almost spirallike mappings of type $\beta$ and order $\alpha$, spirallike mappings of type $\beta$ and order $\alpha$ and strongly spirallike mappings of type $\beta$ and order $\alpha$, and obtained their growth and covering theorems. Liu X S [2] considered the order of zero and aiso obtained the growth and covering theorems for spirallike mappings of type $\beta$, almost starllike mappings of order $\alpha$ and starllike mappings of order $\alpha$. And also there are many refining growth and covering theorems for other biholomorphic mappings (see [3-5]).

In 2001, Hamada and Kohr [6] firstly gave the definition of strongly spirallike mappings of type $\alpha$ on the unit ball $B^{n}$ in $\mathbb{C}^{n}$, and later Xu Q H [4] generalized the definition on the unit ball $B$ in complex Banach spaces. In this paper, we mainly discuss the growth and covering theorems for strongly spirallike mappings of type $\beta$ on $B$, and where $D$ is the unit disc.

Definition 1.1 [4] Suppose that $f$ is a normalized locally biholomorphic mapping on $B$. If

$|e^{-i\beta}\displaystyle\frac{1}{\|x\|}T_{x}[(Df(x))^{-1}f(x)]-\displaystyle\frac{1+c^{2}}{1-c^{2}}\cos\beta+i\sin\beta|<\displaystyle\frac{2c}{1-c^{2}}\cos\beta,c\in(0,1),\beta\in(-\frac{\pi}{2},\frac{\pi}{2}),$

then $f$ is said to be a strongly spirallike mapping of type $\beta$, and if $\beta=0$, then $f$ is said to be a strongly starllike mapping.

Remark From Definition 1.1 we know that the image of the unit ball under the mapping $e^{-i\beta}\displaystyle\frac{1}{\|x\|}T_{x}[(Df(x))^{-1}f(x)]$ is in the circle where the center is $\displaystyle\frac{1+c^{2}}{1-c^{2}}\cos\beta-i\sin\beta$ and the radius is $\displaystyle\frac{2c}{1-c^{2}}\cos\beta$. Yet

$\displaystyle\frac{1+c^{2}}{1-c^{2}}\cos\beta-\displaystyle\frac{2c}{1-c^{2}}\cos\beta=\displaystyle\frac{1-c}{1+c}\cos\beta>0,$

so the circle must be in the right half-plain, and thus strongly spirallike mappings of type $\beta$ must be spirallike mappings of type $\beta$.

Lemma 1.2 (see [7]) Suppose that $f:D\rightarrow D$ is holomorphic and $f(0)=0$, then $|f'(0)|<1$, and $|f(z)|\leq |z|,\forall z\in D$.

Lemma 1.3 (see [8]) Suppose that $f:D\rightarrow D$ is holomorphic and $z=0$ is the zero of $f(z)$ of order $k(k\in\mathbb{N})$, then $|f(z)|\leq |z|^{k},\forall z\in D$.

Lemma 1.4 (see [1]) Suppose that $f:B\rightarrow X$ is a normalized biholomorphic spirallike mapping of type $\beta,\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$. If we denote $x(t)=f^{-1}(\exp(-e{-i\beta}t)f(x)) (0\leq t<+\infty)$ for $x\in B\setminus{0}$, then

(1) $\|x(t)\|$ strictly decreases monotonically for $t\in[0,+\infty)$;

(2) $\lim\displaystyle\frac{\|f(x(t))\|}{\|x(t)\|}=1,\displaystyle\frac{dx}{dt}(t)=-e^{-i\beta}[(Df(x(t)))^{-1}f(x(t))]$ for arbitrarily $t\in(0,+\infty)$;

(3) $\displaystyle\frac{d\|f(x(t))\|}{dt}=-\cos\beta\|f(x(t))\|,t\in(0,+\infty)$.

Lemma 1.5 (see [1]) Suppose that $x(t):[0,+\infty)\rightarrow X$ is differentiable in the point of $x\in(0,+\infty)$, and $\|x(t)\|$ is also differentiable in the point of $s$, then

${\rm Re} T_{x(t)}[\displaystyle\frac{dx}{dt}(s)]=\displaystyle\frac{\|dx(s)\|}{dt},s\in[0,+\infty).$

Theorem 2.1 Suppose that $f$ is a normalized biholomorphic strongly spirallike mapping of type $\beta$ on $B$, $c\in(0,1)$, $\beta\in(-\displaystyle\frac{\pi}{2},\frac{\pi}{2})$, then

$\displaystyle\frac{\|x\|}{(1+c\|x\|)^{2}}\leq\|f(x)\|\leq\displaystyle\frac{\|x\|}{(1-c\|x\|)^{2}},f(B)\supset\displaystyle\frac{1}{4}B.$

When $\beta=0$, i.e., $f$ is a strongly starllike mapping on $B$, we have the same result.

Proof Fix $x\in B\setminus\{0\}$, and let $x_{0}=\displaystyle\frac{x}{\|x\|}$, then $x_{0}\in\partial B$.Let

$\begin{equation} \label{eq:1}g(\zeta)= \left\{ \begin{aligned} &e^{-i\beta}\displaystyle\frac{1}{\zeta}T_{x_{0}}[(Df(\zeta x_{0}))^{-1}f(\zeta x_{0})],\zeta\in D\setminus\{0\},\nonumber\\ &e^{-i\beta},\zeta=0. \end{aligned} \right.\end{equation}$

(1)

Since $f$ is a normalized biholomorphic strongly spirallike mapping of type $\beta$ on $B$, then by Definition 1.1, we have

$|e^{-i\beta}\displaystyle\frac{1}{\|x\|}T_{x}[(Df(x))^{-1}f(x)]-(\displaystyle\frac{1+c^{2}}{1-c^{2}}\cos\beta-i\sin\beta)|<\displaystyle\frac{2c}{1-c^{2}}\cos\beta,$

Suppose $x=\zeta x_{0}=|\zeta|e^{i\theta}x_{0},\zeta\neq 0$, then we have

$|e^{-i\beta}\displaystyle\frac{1}{\zeta}T_{x_{0}}[(Df(\zeta x_{0}))^{-1}f(\zeta x_{0})]-(\displaystyle\frac{1+c^{2}}{1-c^{2}}\cos\beta-i\sin\beta)| <\displaystyle\frac{2c}{1-c^{2}}\cos\beta,$

i.e.,

$|g(\zeta)-(\displaystyle\frac{1+c^{2}}{1-c^{2}}\cos\beta-i\sin\beta)|<\displaystyle\frac{2c}{1-c^{2}}\cos\beta,\zeta\in D\setminus\{0\}.$

Furthermore, for $\zeta=0$, the above inequality holds by eq. (1), thus the above inequality holds for arbitrarily $\zeta\in D$, so we have

$|\displaystyle\frac{1-c^{2}}{2c\cos\beta}[g(\zeta)+i\sin\beta)]-\displaystyle\frac{1+c^{2}}{2c}|<1,\zeta\in D.$

Suppose

$p(\zeta)=\displaystyle\frac{1-c^{2}}{2c\cos\beta}[g(\zeta)+i\sin\beta)]-\displaystyle\frac{1+c^{2}}{2c},$

then we have $|p(\zeta)|<1$ and $p(0)=-c$, Let $\varphi(\zeta)=\displaystyle\frac{p(\zeta)-p(0)}{1-p(0)p(\zeta)}$, it is clear that $|\varphi(\zeta)|<1$ and $\varphi(0)=0$, so we have $|\varphi(\zeta)|<|\zeta|$ by Lemma 1.2, i.e.,

$|\displaystyle\frac{\displaystyle\frac{1-c^{2}}{2c\cos\beta}[g(\zeta)+i\sin\beta)]-\displaystyle\frac{1+c^{2}}{2c}+c} {1+c[\displaystyle\frac{1-c^{2}}{2c\cos\beta}[g(\zeta)+i\sin\beta)]-\displaystyle\frac{1+c^{2}}{2c}]}|<|\zeta|,$

i.e.,

$|\displaystyle\frac{(g(\zeta)+i\sin\beta))-\cos\beta} {(g(\zeta)+i\sin\beta))+\cos\beta}|<c|\zeta|.$

Let $g(\zeta)+i\sin\beta=A+Bi(A,B\in\mathbb{R})$, and from the above inequality, then we have

$(A-\displaystyle\frac{1+c^{2}|\zeta|^{2}}{1-c^{2}|\zeta|^{2}}\cos\beta)^{2}+B^{2} <\displaystyle\frac{4c^{2}|\zeta|^{2}}{(1-c^{2}|\zeta|^{2})^{2}}\cos^{2}\beta,$

so $g(\zeta)+i\sin\beta$ is in the circle taking $a$ as the center and $r$ as the radius in the complex plain, and here

$a=\displaystyle\frac{1+c^{2}|\zeta|^{2}}{1-c^{2}|\zeta|^{2}}\cos\beta,r=\displaystyle\frac{2c|\zeta|}{1-c^{2}|\zeta|^{2}}\cos\beta.$

Then Re$a-r\leq {\rm Re} (g(\zeta)+i\sin\beta)\leq {\rm Re} a+r,$ i.e.,

$\displaystyle\frac{1-c|\zeta|}{1+c|\zeta|}\cos\beta\leq {\rm Re} g(\zeta)\leq\displaystyle\frac{1+c|\zeta|}{1-c|\zeta|}\cos\beta.$

According to (1), and let $\zeta=\|x\|$, we obtain

$\displaystyle\frac{\|x\|(1-c\|x\|)}{1+c\|x\|}\cos\beta\leq {\rm Re}\{e^{-i\beta}T_{x}[(Df(x))^{-1}f(x)]\} \leq\displaystyle\frac{\|x\|(1+c\|x\|)}{1-c\|x\|}\cos\beta.$

(2)

Let

$x(t)=f^{-1}(\exp(-e^{-i\beta}t)f(x))(0\leq t<+\infty),$

by Lemma 1.4, then $\|x(t)\|$ is differentiable almost everywhere in $[0,+\infty)$, and

$\displaystyle\frac{dx}{dt}(t)=-e^{-i\beta}[(Df(x(t)))^{-1}f(x(t))].$

(3)

From (2), (3) and Lemma 1.5, we obtain

$-\displaystyle\frac{\|x(t)\|(1+c\|x(t)\|)}{1-c\|x(t)\|}\cos\beta\leq {\rm Re} T_{x(t)}(\displaystyle\frac{dx(t)}{dt})=\displaystyle\frac{d\|x(t)\|}{dt} \leq-\displaystyle\frac{\|x(t)\|(1-c\|x(t)\|)}{1+c\|x(t)\|}\cos\beta.$

(4)

Also from Lemma 1.4, we have $\displaystyle\frac{d\|f(x(t))\|}{dt}=-\cos\beta\|f(x(t))\|$, then from (4) we obtain

$\begin{aligned} &\displaystyle\frac{1}{\|f(x(t))\|}\displaystyle\frac{d\|f(x(t))\|}{dt}\displaystyle\frac{\|x(t)\|(1+c\|x(t)\|)}{1-c\|x(t)\|} \leq\displaystyle\frac{d\|x(t)\|}{dt} \\ \leq&\displaystyle\frac{1}{\|f(x(t))\|}\displaystyle\frac{d\|f(x(t))\|}{dt}\displaystyle\frac{\|x(t)\|(1-c\|x(t)\|)}{1+c\|x(t)\|}.\end{aligned}$

(5)

For $\tau\geq 0$, from the left of (5) we obtain

$\int^{\tau}_{0}\displaystyle\frac{1}{\|f(x(t))\|}\displaystyle\frac{d\|f(x(t))\|}{dt}dt \leq\int^{\tau}_{0}\displaystyle\frac{1-c\|x(t)\|}{\|x(t)\|(1+c\|x(t)\|)}\displaystyle\frac{d\|x(t)\|}{dt}dt,$

thus

$\log\|f(x(\tau))\|-\log\|f(x)\|\leq \log\displaystyle\frac{\|x(\tau)\|}{(1+c\|x(\tau)\|)^{2}}-\log\displaystyle\frac{\|x\|}{(1+c\|x\|)^{2}},$

i.e.,

$\displaystyle\frac{\|f(x(\tau))\|}{\|f(x)\|}\leq\displaystyle\frac{\|x(\tau)\|}{(1+c\|x(\tau)\|)^{2}}\displaystyle\frac{(1+c\|x\|)^{2}}{\|x\|}.$

Let $\tau\rightarrow+\infty$, from Lemma 1.4 we obtain $\|f(x)\|\geq\displaystyle\frac{\|x\|}{(1+c\|x\|)^{2}}$.Also from the right of (5) we can obtain $\|f(x)\|\leq\displaystyle\frac{\|x\|}{(1-c\|x\|)^{2}}$.Hence

$\displaystyle\frac{\|x\|}{(1+c\|x\|)^{2}}\leq\|f(x)\|\leq\displaystyle\frac{\|x\|}{(1-c\|x\|)^{2}},$

and so we have $f(B)\supset\displaystyle\frac{1}{4}B$, this completes the proof.

Theorem 2.2 Suppose that $f$ is a normalized biholomorphic strongly spirallike mapping of type $\beta$ on $B$, $c\in(0,1)$, $\beta\in(-\displaystyle\frac{\pi}{2},\frac{\pi}{2})$, and $x=0$ is the zero of $f(x)-x$ of order $k+1(k\in\mathbb{N})$, then

$\displaystyle\frac{\|x\|}{(1+c\|x\|^{k})^{\frac{2}{k}}}\leq\|f(x)\|\leq\displaystyle\frac{\|x\|}{(1-c\|x\|^{k})^{\frac{2}{k}}},f(B)\supset\displaystyle\frac{1}{2^{\frac{2}{k}}}B.$

When $\beta=0$, that is, when $f$ is a strongly starllike mapping on $B$, we have the same result.

Proof With the same method of the proof of Theorem 2.1, and from Lemma 1.3 we can obtain $|\varphi(\zeta)|<|\zeta|^{k}$.Replacing $|\zeta|$ with $|\zeta|^{k}$ in the proof of Theorem 2.1, so we can obtain

$\begin{aligned}&\displaystyle\frac{1}{\|f(x(t))\|}\displaystyle\frac{d\|f(x(t))\|}{dt}\displaystyle\frac{\|x(t)\|(1+c\|x(t)\|^{k})}{1-c\|x(t)\|^{k}} \leq\displaystyle\frac{d\|x(t)\|}{dt} \\ \leq& \displaystyle\frac{1}{\|f(x(t))\|}\displaystyle\frac{d\|f(x(t))\|}{dt}\displaystyle\frac{\|x(t)\|(1-c\|x(t)\|^{k})}{1+c\|x(t)\|^{k}}.\end{aligned}$

(6)

From the left of (6), we have

$\int^{\tau}_{0}\displaystyle\frac{1}{\|f(x(t))\|}\displaystyle\frac{d\|f(x(t))\|}{dt}dt \leq\int^{\tau}_{0}\displaystyle\frac{1-c\|x(t)\|^{k}}{\|x(t)\|(1+c\|x(t)\|^{k})}\displaystyle\frac{d\|x(t)\|}{dt}dt,$

thus

$\begin{eqnarray*} &&\log\|f(x(\tau))\|-\log\|f(x)\|\leq \log\displaystyle\frac{\|x(\tau)\|}{(1+c\|x(\tau)\|)^{2}}-\log\displaystyle\frac{\|x\|}{(1+c\|x\|)^{2}}\nonumber\\ &=&\log\{\|x(\tau)\|\exp\int^{\|x(\tau)\|}_{0}[\displaystyle\frac{1-cw^{k}}{1+cw^{k}}-1]\displaystyle\frac{dw}{w}\}- \log\{\|x\|\exp\int^{\|x\|}_{0}[\displaystyle\frac{1-cw^{k}}{1+cw^{k}}-1]\displaystyle\frac{dw}{w}\}\nonumber\\ &=&\log\{\|x(\tau)\|(1+c\|x(\tau)\|^{k})^{-\frac{2}{k}}\}-\log\{\|x\|(1+c\|x\|^{k})^{-\frac{2}{k}}\},\end{eqnarray*}$

So we have

$\displaystyle\frac{\|f(x(\tau))\|}{\|f(x)\|}\leq\displaystyle\frac{\|x(\tau)\|}{(1+c\|x(\tau)\|^{k})^{\frac{2}{k}}}\displaystyle\frac{(1+c\|x\|^{k})^{\frac{2}{k}}}{\|x\|}.$

Thus we obtain

$\|f(x)\|\geq\displaystyle\frac{\|x\|}{(1+c\|x\|^{k})^{\frac{2}{k}}}.$

Also from the right of(6) we can obtain

$\|f(x)\|\leq\displaystyle\frac{\|x\|}{(1-c\|x\|^{k})^{\frac{2}{k}}},$

this completes the proof.