数理解析研究所講究録
第 1717 巻 2010 年 46-54

46

Subordination relations for certain analytic
functions missing some coefficients
Kazuo Kuroki and Shigeyoshi Owa
Abstract
For a positive integer , applying Schwarz’s lemma related to analytic functions$w(z)=c_{\tau 1}z^{n}+\cdots$ in the open unit disk
, some properties concerning with the$f(z)=a+a_{1}z+\cdots$
subordinations for functioiis
alid $g(z)=a+b_{n}z^{n}+\cdots$ which
are analytic in are discussed, and an extension of some subordination relation
which was proven by T. J. Suffiidge (Duke Math. J. 37 (1970), 775-777) is given.
$n$

$U$

$U$

1

Introduction and preliminaries
Let$\mathcal{H}$denote the class of functions $f(z)$ which are analytic in the open unit disk
and

$U=\{z:z\in \mathbb{C}$For a positive integerof the form$n$$|z|<1\}$ .

and a complex number , let$a$$\mathcal{H}[a, n]$be the class of functions$f(z)\in \mathcal{H}$$f(z)=a+ \sum_{k=n}^{\infty}a_{k}z^{k}$.

Also, let $A$ denote the class of functions
normalized by $f(O)=0$ and $f’(O)=1$ .
$A$
The subclass of consisting of all univalent functions $f(z)$ inis denoted by $S$ . An
important member of the class $S$ is the Koebe function$f(z)\in \mathcal{H}$$U$$k(z)= \frac{z}{(1-z)^{2}}=z+\sum_{k=2}^{\infty}kz^{k}$.

A function
is said to be convex in if it is univalent in and $f(U)$ is a convex
domain (A domain
is said to be convex if the line segment joining any two points
of $D$ lies entirely in D). It is well-known that the function $f(z)$ is convex in if and only
if $f’(O)\neq 0$ and$f(z)\in \mathcal{H}$

$U$

$U$

$D\subset \mathbb{C}$

$U$

(1.1)

${\rm Re}(1+ \frac{zf’’(z)}{f(z)})>0$

The normalized class of convex functions denoted by$f(z)\in S$ for which $f(U)$ is convex.

$(z\in U)$$\mathcal{K}$.

consists of the set of all functions

2000 Mathematics Subject Classification: Primary $30C45$ .

Keywords and Phrases: Differential subordination, Schwarz’s lemma, Lindel\"of principle,
Jack lemma, convex function, starlike function, subordination relation.
$s$

47
Furthermore, a function
is said to be starlike in if it is univalent in and$f(U)$ is a starlike domain (A domain
is said to be starlike with respect to theorigin if $0\in D$ and the line segment joining and any point of lies entirely in D). It is
well-known that the function $f(z)$ is starlike in if and only if $f(O)=0,$ $f’(O)\neq 0$ and
$U$

$U$

$f(z)\in \mathcal{H}$$D\subset \mathbb{C}$

$D$

$0$

$U$

(1.2)

$(z\in U)$${\rm Re}( \frac{zf’(z)}{f(z)})>0$

The class of starlike functions denoted by
for which $f(U)$ is starlike.
The equivalent analytic descriptions ofconsists of the set of all functions $f(z)\in S$$S^{*}$

and$\mathcal{K}$.

$S^{*}$are given respectively as follows.

Remark 1.1 A necessary and suffcient condition for
the inequality (1.1). Also, $f(z)\in S^{*}$ if and only if$f(z)\in \mathcal{A}$

From the definitions of$\mathcal{K}$and$S^{*}$, we know thatsatisfies
is thatsatisfies the inequality (1.2).
$f(z)\in \mathcal{A}$$f(z)\in \mathcal{K}$$f(z)\in \mathcal{K}$if and only if $zf’(z)\in S^{*}$ .

Wc ncxt introduce thc familiar ) 1
between analyticof differential subol 1
functions. Let $f(z)$ and $g(z)$ be members of the class . Then the function $g(z)$ is said
to be subordinate to $f(z)$ in , written by$L$$ri$ $ici_{I^{J}}1c$$di_{1}$$\dot{\epsilon}\iota tio\iota 18$$\mathcal{H}$

$U$

(1.3)

$(z\in U)$ ,$g(z)\prec f(z)$if there exists a function
with $w(O)=0$ and $|w(z)|<1$ $(z\in U)$ , and such that$g(z)=f(w(z))$ $(z\in U)$ . From the definition of the subordinations, it is easy to showthat the subordination (1.3) implies that$w(z)\in \mathcal{H}$(1.4)$g(O)=f(O)$In particular, if $f(z)$ is univalent incondination (1. 4).

$U$

and

$g(U)\subset f(U)$ .

, then the subordination (1.3) is equivalent to the

In order to discuss our main results, we need the following lemma which is well-known
as Jack lemma [3] proven by Miller and Mocanu [5] (see also [4]). For $0<r_{0}<1$ , we let$s$$U_{r_{0}}=\{z:z\in \mathbb{C}$

and

Let

be continuous on,$\partial U_{r_{0}}=\{z:z\in \mathbb{C}$$|z|=r_{0}\}$be a positive integer, and let $z_{0}\in U$ with $|z_{0}|=r$ and $0<r<1$ .

$w(z)= \sum_{k=n}^{\infty}c_{k}z^{k}=c_{n}z^{n}+c_{n+1}z^{n+1}+\cdots$$\overline{U_{r}}$

and

$s$

$n$

(1.5)

$|z|<r_{0}\}$. Jack lemma is contained in Lemma 1.2.

$\overline{U_{r_{0}}}=U_{r_{0}}\cup\partial U_{r_{0}}$Lemma 1.2
Also, letandand analytic on$U_{r}\cup\{z_{0}\}$with$w(z)\not\equiv 0$$|w(z_{0})|=m_{\frac{ax}{r}}|w(z)|z\in$

,

. If

48
then there exists a number$k$with $k\geqq n$ , and such that$\frac{z_{0}w^{f}(z_{0})}{w(z_{0})}=k$.

Suffridge [6] independently discovered some particular case of $Jack^{:}s$ lemma from a
result of Julia [1], and deduced the following subordination relation for convex functions
by making good use of it.
Lemma 1.3

Let$g(z)\prec f(z)$$(z\in U)$$f(z)\in \mathcal{K}$.

and $g(z)\in \mathcal{H}[0,1]$ . If $zg’(z)\prec zf’(z)$$(z\in U)$ ,then

In the present paper, applying Schwarz lemma related to analytic function $w(z)$ which
has the form (1.5), we discuss some subordination properties containing the Lindel\"of (orsubordination) principle (cf. [2, Vol. I, Theorem 10]) concerning with the subordination(1.3) for $f(z)\in \mathcal{H}[a, 1]$ and $g(z)\in \mathcal{H}[a, n]$ . Further, by making use of those properties
and Lemma 1.2, we deduce the following subordination relation
$s$

(1.6)

$zg’(z)\prec nzf’(z)$for $f(z)\in \mathcal{H}[a, 1]$ and$(z\in U)$$g(z)\in \mathcal{H}[a, n]$

implies$g(z)\prec f(z)$$(z\in U)$, where $f(z)$ is convex in U.

Some properties for certain subordination2

To considering some subordination properties, we need Schwarz‘s lemma related to(see [2, Vol. I, Theorem 11]) as follows.

$w(z)\in \mathcal{H}[0, n]$Lemma 2.1

Let$w(z)= \sum_{k=n}^{\infty}c_{k}z^{k}\in \mathcal{H}[0,n]$(2.1). If $w(z)$ satisfies $|w(z)|<1$$(z\in U)$, then$|w(z)|\leqq|z|^{n}$for each point$z_{0}\in U\backslash \{0\}$$z\in$, thenU. Further, if equality occurs in the inequality (2.1) for one point(2.2)$w(z)=xz^{n}$for some complex number with $|x|=1$ , and the equality in the inequality (2.1) holds for$x$

all $z\in$ U. Finally, we have$|c_{\eta}|= \frac{1}{n!}|w^{(n)}(0)|\leqq 1$,

and $|c_{n}|=1$ if and only if $w(z)$ is given by the equation (2.2).

Proof.

If we define the function$\phi(z)$by$\phi(z)=\frac{w(z)}{z^{n}}$$(z\in U)$,

49
is analytic in with $\phi(0)=c_{n}$ . Suppose that $0<\rho<1$ . Then the modulus. Hence this factfortakes the maximum value at some point
combined with $|w(z)|<1$ $(z\in U)$ yields that

then

$U$

$\phi(z)$$|\phi(z)|$$z\in\overline{U_{\rho}}$$z_{\rho}\in\partial U_{\rho}$$(z\in\overline{U_{\rho}})$$| \phi(z)|\leqq|\phi(z_{/},)|=\frac{|w(z_{\rho})|}{|z_{\rho}|^{n}}<\frac{1}{\rho^{n}}$which is satisfied for every(2.3)$\rho$

with $0<\rho<1$ . Thus, by letting$|\phi(z)|\leqq 1$$(z\in U)$,$\rhoarrow 1^{-}$, we obtain,which implies that the inequality (2.1). Then since the inequality (2.3) is satisfied for all
interior points of , we may conclude that$U$$|\phi(0)|=|c_{n}|\leqq 1$.

Moreover, if there is a pointsuch that $|w(z_{0})|=|z_{0}|^{n}$ , then we have $|\phi(z_{0})|=1$ .
Also, it is clear that $|c_{n}|=1$ if and only if $|\phi(0)|=1$ . Thus, from the inequality (2.3), by, we see that
applying the maximum modulus principle to analytic function$|\phi(z)|=1$$\phi(z)=x$, and hence we must havefor some complex number
is constant andwith $|x|=1$ , which implies that equation (2.2).
$\sim 0\gamma\in U\backslash \{0\}$$\phi(z)$$\phi(z)$$\square$$x$

Applying Lemma 2.1. we can obtain the following coefficient estimation by the subordination (1.3) for $f(z)\in \mathcal{H}[a, 1]$ and $g(z)\in \mathcal{H}[a, n]$ .
Theorem 2.2$g(z)\prec f(z)$Let$(z\in U)$$f(z)=a+ \sum_{k=1}^{0C}a_{k}z^{k}\in \mathcal{H}[a, 1]$

and $g(z)=a+ \sum_{k=n}^{x}b_{k}z^{k}\in \mathcal{H}[a, n]$ . If

, then$|b_{n}|\leqq|a_{1}|$,and equality occurs if and only if $g(z)=f(xz^{n})$ for some complex number

Proof.

$x$with $|x|=1$ .

, since $g(z)\prec f(z)$ $(z\in U)$ , there exists
Forandan analytic function $w(z)\in \mathcal{H}[0, n]$ which has the form (1.5), with $|w(z)|<1$ $(z\in U)$ ,and such that $g(z)=f(w(z))$ $(z\in U)$ , where$f(z)\in \mathcal{H}[a, 1]$$g(z)\in \mathcal{H}[a, n]$$f(w(z))=a+a_{1}c_{n}z^{n}+\cdots+a_{1}c_{2n-1}z^{2n-1}+(a_{1}c_{2n}+a_{2}c_{n}^{2})z^{2n}+\cdots$

Then, by comparing the coefficient, of in the both sides of equality $g(z)=f(w(z))$ , we
, and hence by Lemma 2.1, we haveobtain$z^{n}$$b_{n}=a_{1}c_{n}$$|b_{n}|=|a_{1}c_{n}|\leqq|a_{1}|$.

Further, since $|c_{n}|=1$ if and only if equation (2.2) holds true, it is clear that $|b_{n}|=|a_{1}|$if and only if $g(z)=f(xz^{n})$ for some complex number with $|x|=1$ .
$x$$\square$Remark 2.3 Letting $n=1$ in Theorem 2.2, we know that $g(z)\prec f(z)$ $(z\in U)$ which
implies that $|g’(0)|\leqq|f’(0)|$ with equality for $g(z)=f(xz)$ , where $|x|=1$ .

50Moreover, Lemma 2.1 provides a slight extension of the Lindel\"of principle bellow.

Theorem 2.4

Let$f(z)\in \mathcal{H}[a, 1]$

and$g(z)\in \mathcal{H}[a, n]$. If $g(z)\prec f(z)$$(z\in U)$, then$g(U_{r})\subset f(U_{r^{n}})$is on the boundaw offor one point, then there is a complex number with $|x|=1$ such that $g(z)=f(xz^{n})$ , and$g(z)$ is on the boundary
.
point
for
of

for each with $0<r<1$ . Further, if$r$$z_{0}\in\partial U_{r}$$x$$f(U_{r^{\mathfrak{n}}})$

Proof.

$f(U_{r^{n}})$$g(z_{0})$If $g(z)\prec f(z)$such that $g(z)=f(w(z))$
by Lemma 2.1, since$z\in\partial U_{r}$$ever^{\vee}y$then there exists an analytic function $w(z)\in \mathcal{H}[0,n]$$(z\in U)$ and $w(z)$ satisfies the conditions of Lemma 2.1. Hencefor each with $0<r<1$ , we see that$(z\in U)$ ,$w(U_{r})\subset U_{r^{n}}$$r$$g(U_{r})=f(w(U_{r}))\subset f(U_{r^{n}})$for each with $0<r<1$ .
Further, if there is a point
since $w(z_{0})$ is on the boundary of$r$$z_{0}\in\partial U_{r}$such that, we have$g(z_{0})$is a boundary point of$f(U_{r^{n}})$, then$U_{r^{n}}$$|w(z_{0})|=r^{n}$,where $|z_{0}|=r$ , and this means that the equality for (2.1) holds for a point.
Thus, by Lemma 2.1, we have $w(z)=xz^{n}$ for some complex nmnber with $|x|=1$ , which
implies that$g(z)=f(xz^{n})$ ,$z_{0}\in\partial U_{r}$$x$where $|x|=1$ . And, from the above equation, it is easy to see that $g(z)$ is the boundary
point offor every point.
$f(U_{r^{n}})$Remark 2.5 Since$(z\in U)$ implies$z\in\partial U_{r}$$f(U_{r^{n}})\subset f(U_{r})$$\square$for each with $0<r<1$ , we know that $g(z)\prec f(z)$$r$$g(U_{r})\subset f(U_{r})$for each with $0<r<1$ by Theorem 2.4. This is the well-known property as the Lindel\"ofprinciple (cf. [2]). Also, Theorem 2.4 for $n=1$ is the Lindel\"of principle.
$r$

As an example of Theorem 2.4, we give the following.
Example 2.6 For a complex number such that $|a|<M$ with $M>0$ , let us consider
two functions $f(z)$ and $g(z)$ given respectively by
$a$

(2.4)

$f(z)=M \frac{a+Mz}{M+\overline{a}z}\in \mathcal{H}[a, 1]$

Then, since $f(z^{n})\prec f(z)$$(z\in U)$

and$g(z)=M \frac{a+Mz^{n}}{M+\overline{a}z^{n}}\in \mathcal{H}[a, n]$, we know that$g(z)=f(z^{n})\prec f(z)$$(z\in U)$

.

.

51
For a radius with $0<r<1$ , a simple check shows that $f(z)$ given by (2.4) maps
onto the interior of the circle with radius and center at , where
$r$

(2.5)

$R= \frac{r^{n}M(M^{2}-|a|^{2})}{M^{2}-r^{2n}|a|^{2}}$and.

$U_{r^{n}}$

$C$

$R$

$(|a|<M)$ .

$C= \frac{(1-r^{2n})M^{2}a}{M^{2}-r^{2n}|a|^{2}}$Further the calculation provides that $g(z)$ given by (2.4) maps
onto the open disk withradius and center at , where and are defined by (2.5), and hence we must have$U_{r}$

$R$

$C$

$C$

$R$

$g(U_{r})=f(U_{r^{n}})$for each$r$with $0<r<1$ .

An extension of some subordination relation

3

Applyilig some subordination properties which were discussed in the previous section,and by using Lemma 1.2, we will give the proof of the subordination relation (1.6).

Let $f(z)\in \mathcal{H}[a, 1]$ and $g(z)\in \mathcal{H}[a, n]$ , and suppose that $f(z)$ is convex
U. If $zg’(z)\prec nzf’(z)$ $(z\in U)$ , then $g(z)\prec f(\approx)$ $(z\in U)$ .

Theorem 3.1$in$

Proof.

If we let$F(z)=nzf^{f}(z)$then the assumption $z\mathscr{S}(z)\prec nzf’(z)$(3.1)

and $G(z)=zg’(z)$ ,$(z\in U)$$G(z)\prec F(z)$can be rewritten by$(z\in U)$.

Moreover, if we set$f(z)=a+ \sum_{k=1}^{\infty}a_{k}z^{k}$and$g(z)=a+ \sum_{k=n}^{\infty}b_{k}z^{k}$,then$F(z)=n \sum_{k=1}^{\infty}ka_{k}z^{k}$andIt follows that the subordination (3.1) implies
if and only if(3.2)$G(z)= \sum_{k=n}^{\infty}kb_{k}z^{k}$$|b_{n}|\leqq|a_{1}|$.

by Theorem 2.2. Since $|b_{n}|=|a_{1}|$$G(z)=F(xz^{n})$ ,where $|x|=1$ , we have $zg’(\approx)=nxz^{n}f’(xz^{n})$ which implies that $g(z)=f(xz^{n})$ , where$|x|=1$ , and this means that $g(z)\prec f(z)$ $(z\in U)$ . Therefore, we may continue theargument assuming that $|b_{n}|<|a_{1}|$ .

Suppose that $g(z)$ is not subordinate to $f(z)$ in U. If we let$f_{n}(z)=f(z^{n})=a+ \sum_{k=1}^{\infty}a_{k}z^{nk}=a+a_{1}z^{n}+a_{2}z^{2n}+\cdots$,

52
for all sufficiently smallthen since $|b_{n}|<|a_{1}|$ implies thatvalues of , there exists a radius with $0<r<1$ such that
. Since $f(z)\in \mathcal{H}[a, 1]$ isfor some real and , and thatconvex in which implies that $f(z)$ is univalent in , we know that the inverse
with $f^{-1}(a)=0$ . Then fromonto
is analytic in a domain $D=f(U)$ and maps
. Also, it follows fromis analytic on, we see that$g(z)$
maydefine a function $w(z)$ by
is analytic on . Therefore, wethat$g(U_{-})\subset f_{n}(U_{\epsilon})=f(U_{\overline{-\vee}n})$$g(re^{i\theta})=f_{n}(re^{i\varphi})=f(r^{n}e^{in\varphi})$$r$$\epsilon$$g(\overline{U_{r}})\subset f_{n}(\overline{U_{r}})=f(\overline{U_{r^{n}}})$$\theta$$\varphi$

$U$

$D$

$f^{-1}$$g(\overline{U_{r}})\subset f(\overline{U_{r^{\mathfrak{n}}}})\subset f(U)$$g(z)\in \mathcal{H}[a, n]$(3.3)$g(\overline{U_{r}})$$\overline{U_{r}}$$w(z)=f^{-1}(g(z))$$(z\in U_{r})$which has the form (1.5). Then $w(z)$ is analytic on, we have. Since$0$$f^{-1}$$U$$U$$\overline{U_{r}}$with $w(O)=f^{-1}(g(O))=f^{-1}(a)=$$w(\overline{U_{r}})=f^{-1}(g(\overline{U_{r}}))\subset\overline{U_{r^{\mathfrak{n}}}}$$|w(z)|\leqq r^{n}$$(z\in\overline{U_{r}})$.

Further, noting that $w(re^{i\theta})=f^{-1}(g(re^{i\theta}))=f^{-1}(f(r^{n}e^{tn\varphi}))=r^{n}e^{in\varphi}$, we find that$|w(re^{i\theta})|=|r^{n}e^{in\varphi}|=r^{n}=m_{\frac{ax}{r}}z\in|w(z)|$That is, the modulus $|w(z)|$ takes the maximum value
Thus, according to Lemma 1.2, there is a real number(3.4)$\frac{z_{0}w’(z_{0})}{w(z_{0})}=k$.

at a point
so that $k\geqq n\geqq 1$ and$z=re^{i\theta}\in\overline{U_{r}}$$r^{n}$$k$.

,. Equation (3.3) implies that $g(z)=f(w(z))$ and $g^{f}(z)=w’(z)f’(w(z))$ .
where
If we use these relations at $z=z_{0}$ and equation (3.4), then from $w(z_{0})=w(re^{i\ell})=r^{n}e^{in\varphi}$ ,we see that$z_{0}=re^{i\theta}$

(3.5)

.

$k= \frac{z_{0}w’(z_{0})}{w(z_{0})}=\frac{z_{0}g^{f}(z_{0})}{w(z_{0})f’(w(z_{0}))}=\frac{re^{i\theta}g^{f}(re^{i\theta})}{r^{n}e^{in\varphi}f^{f}(r^{n}e^{in\varphi})}\geqq n$

Since $F(z)=nzf^{f}(z)$ and $G(z)=zg’(z)$ , the inequality (3.5) is the same as(3.6)$\frac{G(re^{i\theta})}{F(r^{n}e^{i\iota\rho})}\geqq 1$.

In addition, it follows from the inequality (3.6) that
(3.7)

Now, as$\arg(G(re^{i\theta}))=\arg(F(r^{n}e^{in\varphi}))$$f(z)=a+ \sum_{k=1}^{\infty}a_{k}z^{k}$is convex in$\frac{f(z)-a}{a_{1}}\in \mathcal{K}$$U$, we have

if and only if.

$\frac{f(z)-a}{a_{1}}\in \mathcal{K}$. Moreover, since$z( \frac{f(z)-a}{a_{1}})’\in S^{*}$,

it is clear that $F(z)=nzf’(z)$ is starlike and univalent in U. By Theorem 2.4, we seethat $G(z)\prec F(z)$ $(z\in U)$ implies(3.8)$G(U_{r})\subset F(U_{r^{n}})$$(0<r<1)$ .

53Then, since $G(O)=F(O)=0$ and $F(U)$ is starlike with respect to the origin, we find that
(3.9)$|G(re^{i\theta})|\leqq|F(r^{n}e^{in\varphi})|$for some real and which satisfy the equality (3.7). Here, if $G(z_{0})$ is on the boundary, then by Theorem 2.4, we have $G(z)=F(xz^{n})$ , wherefor one pointof. From thisfor every point$|x|=1$ , and $G(z)$ is on the boundary of(3.9)
occurs forfact and the relation (3.8), we see that the equality in the inequality$G(z)=F(xz^{n})$ , where $|x|=1$ . But we now continue the argument assuming that$|b_{n}|<|a_{1}|$ , which is same as that $G(z)$ does not have the form (3.2). Therefore, we obtainthat $G(z)\prec F(z)$ $(z\in U)$ which implies that$\theta$$\varphi$$F(U_{r^{n}})$$\approx 0\in\partial U_{r}$$z\in\partial U_{r}$$F(U_{r^{n}})$(3.10)$|G(re^{i\theta})|<|F(r^{n}e^{in\rho})|$for some real and which satisfy the equality (3.7). Moreover, the inequality (3.10)
combined with the equality (3.7) yields that$\theta$$\varphi$(3.11)$\frac{G(re^{x\theta})}{F(r^{n}e^{l^{-}n\varphi})}<1$.

From the above-mentioned, since the inequality (3.6) contradicts the inequality (3.11),
we see that the inequality (3.6) contradicts the assumption (3.1) of the theorem, andhence we must have $g(z)\prec f(z)$ $(z\in U)$ . Therefore, we conclude that $zg’(z)\prec nzf^{f}(z)$$(z\in U)$ , which completes the proof of Theorem 3.1.
$(z\in U)$ implies $g(z)\prec f(z)$$\square$Remark 3.2 Letting $n=1,$ $a=0$ and $a_{1}=1$ in Theorem 3.1, we obtain Lemma 1.3
which was shown by Suffridge [6].
As an example of Theorem 3.1, we introduce the following.
Example 3.3 For a complex number$f(z)$ and $g(z)$ given respectively by(3.12)$f(z)= \frac{a+\overline{a}z}{1-z}\in \mathcal{H}[a, 1]$$a$such thatand${\rm Re} a\neq 0$, let us consider two functions$g(z)= \frac{a+\overline{a}z^{n}}{1-z^{n}}\in \mathcal{H}[a, n]$.

Then, it is easy to see that the above function $f(z)$ is convex and univalent in U.
A simple calculation yields that$zf’(z)= \frac{2({\rm Re} a)z}{(1-z)^{2}}=2({\rm Re} a)k(z)$and$zg^{f}(z)= \frac{2n({\rm Re} a)z^{n}}{(1-z^{n})^{2}}=2n({\rm Re} a)k(z^{n})$where $k(z)$ is the Koebe function. Since $k(z^{n})\prec k(z)$,$(z\in U)$$zg’(z)=2n({\rm Re} a)k(z^{n})\prec 2n({\rm Re} a)k(z)=nzf’(z)$, we have$(z\in U)$,

which implies that all the assumptions of Theorem 3.1 are satisfied. Hence by Theorem
3.1, we obtain the subordination relation (1.6). Actually, it is clear that$g(z)=f(z^{n})\prec f(z)$for $f(z)$ and $g(z)$ given by (3.12).

$(z\in U)$

54

References[1] C. Carath\’eodory, Funktionentheo nie E, Basel, 1950.

[2] A. W. Goodman, Univalent Functions, Vol.I and , Mariner, Tampa, Florida, 1983.
$\Pi$[3] I. S. Jack, Functions starlike and convex of order(1971), 469-474.

$\alpha$, J. London Math. Soc. (2), 3[4] S. S. Miller and P. T. Mocanu, Second order diffierential
plane, J. Math. Anal. Appl. 65 (1978), 289-305.

$\dot{r,}neqnal\iota’tie,,s$

in the complex[5] S. S. Miller and P. T. Mocanu, Differential Subordinations, Pure and Applied Mathematics 225, Marcel Dekker, 2000.
[6] T. J. Suffridge, Some remarks on convex maps of the unit disk, Duke Math. J. 37(1970), 775-777.

Kazuo Kuroki
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan
e-mail : freedom@sakai. $zaq$ . ne.jp
Shigeyoshi Owa
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan
e-mail: owa@math. kindai. $ac$ .jp