On analytic multivalent functions associated with lemniscate of Bernoulli

1.   Introduction and definitions

To understand in a clear way the notions used in our main results, we need to add here some basic literature of Geometric function theory. For this we start first with the notation $\mathcal{A}$ which denotes the class of holomorphic or analytic functions in the region $\mathfrak{D} = \left \{ z\in \mathbb{C} :\left \vert z\right \vert < 1\right \}$ and if a function $g\in \mathcal{A}$, then the relations $g\left(0\right) = g^{\prime }\left(0\right) -1 = 0$ must hold. Also, all univalent functions will be in a subfamily $\mathcal{S}$ of $\mathcal{A}$. Next we consider to define the idea of subordinations between analytic functions $g_1$ and $g_2,$ indicated by $g_1\left(z\right) \prec g_2\left(z\right),$ as; the functions $g_1, g_2\in \mathcal{A}$ are connected by the relation of subordination, if there exists an analytic function $w$ with the restrictions $w\left(0\right) = 0$ and $\left \vert w\left(z\right) \right \vert < 1$ such that $g_1(z) = g_2(w(z)).$ Moreover, if the function $g_2\in \mathcal{S}$ in $\mathfrak{D},$ then we obtain:

In 1992, Ma and Minda ^[16] considered a holomorphic function $\varphi$ normalized by the conditions $\varphi (0) = 1$ and $\varphi ^{\prime }(0) > 0$ with $Re\varphi > 0$ in $\mathfrak{D}$. The function $\varphi$ maps the disc $\mathfrak{D}$ onto region which is star-shaped about $1$ and symmetric along the real axis. In particular, the function $\varphi (z) = (1+Az)/(1+Bz), \ (-1\leq B < A\leq 1)$ maps $\mathfrak{D}$ onto the disc on the right-half plane with centre on the real axis and diameter end points $\frac{1-A}{1-B}$ and $\frac{1+A}{1+B}$. This interesting familiar function is named as Janowski function ^[10]. The image of the function $\varphi (z) = \sqrt{1+z}$ shows that the image domain is bounded by the right-half of the Bernoulli lemniscate given by $|w^{2}-1| < 1, \ $^[25]. The function $\varphi (z) = 1+\frac{4}{3}z+\frac{2}{3}z^{2}$ maps $\mathfrak{D}$ into the image set bounded by the cardioid given by $(9x^{2}+9y^{2}-18x+5)^{2}-16(9x^{2}+9y^{2}-6x+1) = 0,$ ^[21] and further studied in ^[23]. The function $\varphi (z) = 1+\sin z$ was examined by Cho and his coauthors in ^[3] while $\varphi (z) = e^{z}$ is recently studied in ^[17] and ^[24]. Further, by choosing particular $\varphi, \ $several subclasses of starlike functions have been studied. See the details in ^{[2,4,5,11,12,14,19]}.

Recently, Ali et al. ^[1] have obtained sufficient conditions on $\alpha$ such that

Similar implications have been studied by various authors, for example see the works of Halim and Omar ^[6], Haq et al ^[7], Kumar et al ^[13,15], Paprocki and Sokól ^[18], Raza et al ^[20] and Sharma et al ^[22].

In 1994, Hayman ^[8] studied multivalent ($p$-valent) functions which is a generalization of univalent functions and is defined as: an analytic function $g$ in an arbitrary domain $\mathcal{D}\subset \mathbb{C}$ is said to be $p$-valent, if for every complex number $\omega$, the equation $g(z) = \omega$ has maximum $p$ roots in $\mathcal{D}$ and for a complex number $\omega _{0}$ the equation $g(z) = \omega _{0}$ has exactly $p$ roots in $\mathcal{D}$. Let $\mathcal{A}_{p}$ $\left(p\in \mathbb{N} = \left \{ 1, 2, \ldots \right \} \right)$ denote the class of functions, say $g\in \mathcal{A}_{p}$, that are multivalent holomorphic in the unit disc $\mathfrak{D}$ and which have the following series expansion:

Using the idea of multivalent functions, we now introduce the class $\mathcal{SL}_{p}^{\ast }$ of multivalent starlike functions associated with lemniscate of Bernoulli and as given below:

In this article, we determine conditions on $\alpha$ such that for each

are subordinated to Janowski functions implies $\frac{g\left(z\right) }{ z^{p}}\prec \sqrt{1+z}, \ \left(z\in \mathfrak{D}\right)$. These results are then utilized to show that $g$ are in the class $\mathcal{SL}_{p}^{\ast }.$

1.1.   Lemma

Let $w$ be analytic non-constant function in $\mathfrak{D}$ with $w\left(0\right) = 0.$ If

then there exists a real number $m\ \left(m\geq 1\right)$ such that $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right).$

This Lemma is known as Jack's Lemma and it has been proved in ^[9].

2.   Main results

2.1.   Theorem

Let $g\in \mathcal{A}_{p}$ and satisfying

with the restriction on $\alpha$ is

Then

Proof

Let us define a function

where the function $p$ is analytic in $\mathfrak{D}$ with $p(0) = 1.$ Also consider

Now to prove our result we will only require to prove that $\left \vert w\left(z\right) \right \vert < 1.$ Logarithmically differentiating $\left(2.3\right)$ and then using $\left(2.2\right),$ we get

and so

Now, we suppose that a point $z_{0}\in \mathfrak{D}$ occurs such that

Also by Lemma 1.1, a number $m\geq 1$ exists with $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right)$. In addition, we also suppose that $w\left(z_{0}\right) = e^{i\theta }$ for $\theta \in \left[ -\pi, \pi \right].$ Then we have

Now if

then

which illustrates that the function $\phi \left(m\right)$ is increasing and hence $\phi \left(m\right) \geq \phi \left(1\right)$ for $m\geq 1,$ so

Now, by using $\left(2.1\right),$ we have

which contradicts the fact that $\begin{array}{c} p\left(z\right) \prec \frac{1+Az}{1+Bz}. \end{array}$ Thus $\left \vert w\left(z\right) \right \vert < 1$ and so we get the desired result.

Taking $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in the last result, we obtain the following Corollary:

2.2.   Corollary

Let $f\in \mathcal{A}_{p}$ and satisfying

with the condition on $\alpha$ is

Then $f\in \mathcal{SL}_{p}^{\ast }.$

2.3.   Theorem

If $g\in \mathcal{A}_{p}$ such that

with

then

Proof

Let us choose a function $p$ by

in such a way that $p$ is analytic in $\mathfrak{D}$ with $p(0) = 1.$ Also consider

Using some simple calculations, we obtain

and so

Let a point $z_{0}\in \mathfrak{D}$ exists in such a way

Then, by virtue of Lemma 1.1, a number $m\geq 1$ occurs such that$\ z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right)$. In addition, we set $w\left(z_{0}\right) = e^{i\theta },$ so we have

Now let

it implies

which illustrates that the function $\phi \left(m\right)$ is increasing and so $\phi \left(m\right) \geq \phi \left(1\right)$ for $m\geq 1,$ hence

Now, by using $\left(2.6\right),$ we have

which contradicts $\left(2.5\right).$ Thus $\left \vert w\left(z\right) \right \vert < 1$ and so the desired proof is completed.

Putting $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in last Theorem, we get the following Corollary:

2.4.   Corollary

If $f\in \mathcal{A}_{p}$ and satisfying

with

then $f\in \mathcal{SL}_{p}^{\ast }.$

2.5.   Theorem

If $g\in \mathcal{A}_{p}$ and satisfy the subordination relation

with the condition on $\alpha$

is true, then

Proof

Let us define a function

Then $p$ is analytic in $\mathfrak{D}$ with $p(0) = 1.$ Also let us consider

Using some simplification, we obtain

and so

Let us choose a point $z_{0}\in \mathfrak{D}$ such a way that

Then, by the consequences of Lemma 1.1, a number $m\geq 1$ occurs such that $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right)$ and also put $w\left(z_{0}\right) = e^{i\theta },$for $\theta \in \left[ -\pi, \pi \right],$ we have

Then

which demonstrates that the function $\phi \left(m\right)$ is increasing and thus $\phi \left(m\right) \geq \phi \left(1\right)$ for $m\geq 1,$ hence

Now, using $\left(2.8\right),$ we have

which contradicts $\left(2.7\right)$. Thus $\left \vert w\left(z\right) \right \vert < 1$ and so we get the required proof.

If we set $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in last theorem, we easily have the following Corollary:

2.6.   Corollary

Assume that

and if $f\in \mathcal{A}_{p}$ satisfy

then $f\in \mathcal{SL}_{p}^{\ast }.$

2.7.   Theorem

If $g\in \mathcal{A}_{p}$ satisfy the subordination

with restriction on $\alpha$ is

then

Proof. Let us define a function

where $p$ is analytic in $\mathfrak{D}$ with $p(0)-1 = 0.$ Also let

Using some simple calculations, we obtain

and so

Let us pick a point $z_{0}\in \mathfrak{D}$ in such a way that

Then, by using Lemma 1.1$,$ a number $m\geq 1$ exists such that $z_{0}w^{\prime }\left(z_{0}\right) = mw\left(z_{0}\right) \ $and put $w\left(z_{0}\right) = e^{i\theta },$ for $\theta \in \left[ -\pi, \pi \right]$, we have

Now let

then

which shows that $\phi \left(m\right)$ is an increasing function and hence it will have its minimum value at $m = 1$, so

Using $\left(2.9\right),$ we easily obtain

which is a contradiction to the fact that $p\left(z\right) \prec \frac{1+Az }{1+Bz},$ and so $\left \vert w\left(z\right) \right \vert < 1.$ Hence we get the desired result.

If we put $g\left(z\right) = \frac{z^{p+1}f^{\prime }\left(z\right) }{pf(z)}$ in last Theorem, we achieve the following result:

2.8.   Corollary

If $f\in \mathcal{A}_{p}$ and satisfy the condition

and

then $f\in \mathcal{SL}_{p}^{\ast }.$

Conflict of interest

All authors declare no conflict of interest in this paper.