On an extension of Nunokawa’s lemma

Abstract

Jack’s Lemma says that if f(z) is regular in the disc $|z|\le r$, $f(0)=0$, and |f(z)| assumes its maximum at $z_0$ on the circle $|z|=r$, then $z_0f'(z)_0/f(z_0)\ge 1$. This Lemma was generalized in several directions. In this paper we consider an improvement of some first author’s results of this type.

1 Introduction

Let $\mathcal {H}$ be the family of analytic functions in the region $\mathbb {D}=\left\{ z:\left| z\right| <1\right\}$ on the complex plane $\mathbb {C}$. Let $\mathcal {A}\subset \mathcal {H}$ denote the set of all functions f(z) that are analytic in $\mathbb {D}$ with the series representation

$$\begin{aligned} f(z)=z+\sum _{n=2}^{\infty }a_{n}\text { }z^{n},\quad z\in \mathbb {D} \text {.} \end{aligned}$$

(1.1)

Also by $\mathcal {S}$ we means a subfamily of the set $\mathcal {A}$ which contains univalent functions. We next denote by $\mathcal {P}$ the class of analytic functions p(z) which are normalized by

$$\begin{aligned} p\left( z\right) =1+\sum _{n=1}^{\infty }p_{n}z^{n}, \end{aligned}$$

(1.2)

such that $\mathfrak {Re} \{p(z)\} >0,\ \ z\in \mathbb {D}.$ Furthermore, by using the set $\mathcal {P}$, let $\mathcal {S}^{*}$ and $\mathcal {C}$ denote the families of starlike and convex functions in $\mathbb {D}$ which are defined as

$$\begin{aligned} \mathcal {S}^{*}= & {} \left\{ f:f\in \mathcal {S}\text { and }\frac{ zf^{\prime }\left( z\right) }{f\left( z\right) }\in \mathcal {P},\text { }z\in \mathbb {D}\right\} , \nonumber \\&\nonumber \\ \mathcal {C}= & {} \left\{ f:f\in \mathcal {S}\text { and }\frac{(zf^{\prime }\left( z\right) )^{\prime }}{f^{\prime }\left( z\right) }\in \mathcal {P}, \text { }z\in \mathbb {D}\right\} . \end{aligned}$$

(1.3)

The above subfamilies of the set $\mathcal {S}$ are among the most studied families. Recall also, that if $f\in \mathcal A$ satisfies

$$\begin{aligned} {\mathfrak {Re}} \left\{ \frac{zf'(z)}{e^{i\alpha }g(z)}\right\} >0 ,\quad z\in \mathbb D \end{aligned}$$

(1.4)

for some $g\in \mathcal S^*$ and some $\alpha \in (-\pi /2,\pi /2)$, then f is said to be close-to-convex in $\mathbb D$ and denoted by $f\in \mathcal K$. An univalent function $f\in \mathcal A$ belongs to $\mathcal K$ if and only if the complement E of the image-region $F=\left\{ f(z): |z|<1\right\}$ is the union of rays that are disjoint (except that the origin of one ray may lie on another one of the rays). We note that if $g\left( z\right) =z$, then the class $\mathcal {K}$ reduces to the set $\mathcal {R}$ of bounded turning functions.

Jack’s Lemma [2], says that if f(z) is regular in the disc $|z|\le r$, $f(0)=0$, and |f(z)| assumes its maximum at $z_0$ on the circle $|z|=r$, then $z_0f'(z)_0/f(z_0)\ge 1$. Now we will consider a lemma, which is a small extension of Jack’s Lemma.

We will test behavior of $z_0f'(z)_0/f(z_0)$ when |f(z)| assumes its local maximum at $z_0$ on the following neighborhood

$$\begin{aligned} N_{\varepsilon }(z_0)=\{z:|z|\le |z_0|,\quad |z-z_0|\le \varepsilon \}. \end{aligned}$$

Lemma 1.1

[1] Assume that $p(z)=a_kz^k+a_{k+1}z^{k+1}+\cdots$, $a_k\ne 0$, $k\ge 1$ is analytic in $\mathbb D$. If there exists $z_0\in \mathbb D$ such that

$$\begin{aligned} \max _{z\in N_{\varepsilon }(z_0)}|p(z)|=|p(z_0)| \end{aligned}$$

(1.5)

for some $\varepsilon >0$, then $z_0p'(z_0)/p(z_0)$ is a real number and

$$\begin{aligned} \frac{ z_0p'(z_0)}{p(z_0)}\ge k\ge 0. \end{aligned}$$

(1.6)

Proof

Let us put

$$\begin{aligned} p(z)=z^k\varphi (z). \end{aligned}$$

Then $\varphi (z)$ is analytic in $\mathbb D$ and $|\varphi (z)|$ takes its local maximum value at the point $z=z_0$ on the circular arc $z=|z_0|e^{i\theta }$, $\arg z_0-\varepsilon<\theta <\arg z_0+\varepsilon$. When z moves on the circle $|z|=|z_0|$ with positive direction, then $\arg \varphi (z)$ is increasing at $z=z_0$. Therefore, we have

$$\begin{aligned} \left[ \frac{\mathrm{d}\arg \varphi (|z_0|e^{i\theta })}{\mathrm{d\theta }}\right] _{\theta =\theta _0} =\mathfrak {Re}\frac{ z_0\varphi '(z_0)}{\varphi (z_0)}\ge 0 \end{aligned}$$

and so

$$\begin{aligned} \mathfrak {Re}\frac{z_0p'(z_0)}{p(z_0)} =\mathfrak {Re}\left( k+\frac{ z_0\varphi '(z_0)}{\varphi (z_0)}\right) \ge k\ge 0. \end{aligned}$$

(1.7)

From the hypothesis, we have

$$\begin{aligned} \left[ \frac{\mathrm{d}|p(|z_0|e^{i\theta })|}{\mathrm{d\theta }}\right] _{\theta =\theta _0} = 0, \end{aligned}$$

so

$$\begin{aligned} \left[ \frac{\mathrm{d}\log |p(|z_0|e^{i\theta })|}{\mathrm{d\theta }}\right] _{\theta =\theta _0} = 0 \end{aligned}$$

too. Therefore, we have

$$\begin{aligned} \left[ \mathfrak {Im}\frac{\mathrm{d}\log p(z)}{\mathrm{d} \log z}\right] _{z=z_0}= & {} \left[ -\frac{\mathrm{d}\log |p(|z_0|e^{i\theta })|}{\mathrm{d\theta }}\right] _{\theta =\theta _0}\\= & {} \left[ -\left( \frac{\mathrm{d}|p(z)|}{\mathrm{d}\theta }\right) \frac{1}{|p(z)|}\right] _{\theta =\theta _0}\\= & {} 0=\mathfrak {Im}\frac{ z_0p'(z_0)}{p(z_0)}. \end{aligned}$$

This shows

$$\begin{aligned} \frac{ z_0p'(z_0)}{p(z_0)}\ge k\ge 0. \end{aligned}$$

(1.8)

$\square$

Theorem 1.2

Assume that $p(z)=1+a_kz^k+a_{k+1}z^{k+1}+\cdots$, $a_k\ne 0$, $k\ge 1$ is analytic in $\mathbb D$. Assume there exist two points $z_1,z_2\in \mathbb D$ such that $|z_1|=|z_2|$ and

$$\begin{aligned} \arg p(z_2)= -\frac{\pi \beta }{2},\quad \arg p(z_1)=\frac{\pi \alpha }{2} \end{aligned}$$

(1.9)

for some $\alpha$, $\beta$, $\alpha ,\beta \in (0,2)$. If

$$\begin{aligned} -\frac{\pi \beta }{2}<\arg p(z)<\frac{\pi \alpha }{2},\quad |z|<|z_1|=|z_2| \end{aligned}$$

(1.10)

and

$$\begin{aligned} \max _{z\in N_{\varepsilon }(z_1)}|\arg \{p(z)\}|= & {} |\arg \{p(z_1)\}|\end{aligned}$$

(1.11)

$$\begin{aligned} \max _{z\in N_{\varepsilon }(z_2)}|\arg \{ p(z)\}|= & {} |\arg \{p(z_2)\}| \end{aligned}$$

(1.12)

for some $\varepsilon >0$, then we have

$$\begin{aligned} \frac{ z_1p'(z_1)}{p(z_1)}=\frac{i\alpha k}{2}\left( a+\frac{1}{a}\right) , \quad \frac{\alpha k}{2}\left( a+\frac{1}{a}\right) \ge \alpha k, \end{aligned}$$

(1.13)

where $p^{1/\alpha }(z_1)=ia$, $a>0$, and

$$\begin{aligned} \frac{ z_2p'(z_2)}{p(z_2)}=-\frac{i\beta k}{2}\left( b+\frac{1}{b}\right) , \quad -\frac{\beta k}{2}\left( b+\frac{1}{b}\right) \le -\beta k, \end{aligned}$$

(1.14)

where $p^{1/\beta }(z_2)=-ib$, $b>0$.

Proof

Let us check the first case $\arg p(z_1)=\pi \alpha /2$ and put

$$\begin{aligned} q(z)= [p(z)]^{1/\alpha },\quad \varphi (z)=\frac{1-q(z)}{1+q(z)}, \end{aligned}$$

then it follows that

$$\begin{aligned} q(0)=1,q'(0)=q''(0)=\cdots =q^{(k-1)}(0)=0\quad \mathrm{and}\quad q^{(k)}(0)\ne 0 \end{aligned}$$

and so

$$\begin{aligned} \varphi (0)=\varphi '(0)=\varphi ''(0)=\cdots =\varphi ^{(k-1)}(0)=0\quad \mathrm{and}\quad \varphi ^{(k)}(0)\ne 0. \end{aligned}$$

From the hypothesis, we have

$$\begin{aligned} \arg q(z)<\frac{\pi }{2},\quad z\in D_\varepsilon (z_1)=N_\varepsilon (z_1){\setminus }\{z_1\} \end{aligned}$$

for some $\varepsilon >0$ and

$$\begin{aligned} \arg q(z_1)=\frac{\pi }{2}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} |\varphi (z)|<1,\quad z\in D_\varepsilon (z_1) \end{aligned}$$

and

$$\begin{aligned} |\varphi (z_1)|=1. \end{aligned}$$

This shows that $|\varphi (z)|$ takes its local maximum value at the point $z=z_1$ in the the domain $N_\varepsilon (z_1)$ and applying Lemma 1.1, we have

$$\begin{aligned} \frac{z_1\varphi '(z_1)}{\varphi (z_1)}= & {} -\frac{z_1q'(z_1)}{1-q(z_1)}-\frac{z_1q'(z_1)}{1+q(z_1)}\\= & {} -\frac{2z_1q'(z_1)}{(1-q(z_1))(1+q(z_1))}\\= & {} -\frac{2z_1q'(z_1)}{(1-ia)(1+ia)}\\= & {} -\frac{2z_1q'(z_1)}{1+a^2}\\\ge & {} k \end{aligned}$$

where $k\ge 1$ and $q(z_1)=ia$ and $a>0$. This shows that $z_1q'(z_1)$ is a negative real number. This gives

$$\begin{aligned} \mathfrak {Im} \frac{ z_1q'(z_1)}{q(z_1)} =\mathfrak {Im} \frac{ z_1q'(z_1)}{ia} \ge \frac{k}{2}\left( a+\frac{1}{a}\right) \ge k. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \frac{ z_1q'(z_1)}{q(z_1)}=\frac{1}{\alpha }\frac{ z_1p'(z_1)}{p(z_1)} \end{aligned}$$

and putting $z=|z_1|e^{i\theta }$ in the section $\arg z_1-\varepsilon<\theta <\arg z_1+\varepsilon$, then $\arg \{q(z_1)\}$ takes its local maximum $\pi /2$ at the point $\theta =\arg \{z_1\}$ in this section and so, we have

$$\begin{aligned} \mathfrak {Re} \frac{ z_1q'(z_1)}{q(z_1)} =\left[ \mathfrak {Re} \frac{\mathrm{d}\log q(z)}{\mathrm{d} z}\right] _{z=z_1}=\left[ \frac{\mathrm{d}\arg q(z)}{\mathrm{d} \theta }\right] _{\theta =\arg z_1}=0. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \frac{ z_1p'(z_1)}{p(z_1)}=i\frac{\alpha k}{2}\left( a+\frac{1}{a}\right) , \quad \frac{\alpha k}{2}\left( a+\frac{1}{a}\right) \ge \alpha k \end{aligned}$$

where $p^{1/\alpha }(z_1)=ia$, $a>0$. This completes the proof of (1.13).

For the next case $\arg p(z_2)=-\pi \beta /2$ let us put

$$\begin{aligned} \rho (z)= [p(z)]^{1/\beta },\quad \psi (z)=\frac{1-\rho (z)}{1+\rho (z)}, \end{aligned}$$

then it follows that

$$\begin{aligned} \rho (0)=1,\rho '(0)=\rho ''(0)=\cdots =\rho ^{(k-1)}(0)=0\quad \mathrm{and}\quad \rho ^{(k)}(0)\ne 0 \end{aligned}$$

and so

$$\begin{aligned} \psi (0)=\psi '(0)=\psi ''(0)=\cdots =\psi ^{(k-1)}(0)=0\quad \mathrm{and}\quad \psi ^{(k)}(0)\ne 0. \end{aligned}$$

From the hypothesis, we have

$$\begin{aligned} \arg \rho (z)>-\frac{\pi }{2},\quad z\in D_\varepsilon (z_2)=N_\varepsilon (z_2){\setminus }\{z_2\} \end{aligned}$$

for some $\varepsilon >0$ and

$$\begin{aligned} \arg \rho (z_2)=-\frac{\pi }{2}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} |\psi (z)|<1,\quad z\in D_\varepsilon (z_2) \end{aligned}$$

and

$$\begin{aligned} |\psi (z_2)|=1. \end{aligned}$$

This shows that $|\psi (z)|$ takes its local maximum value at the point $z=z_2$ in the the domain $N_\varepsilon (z_2)$ and applying Lemma 1.1, we have

$$\begin{aligned} \frac{z_2\psi '(z_2)}{\psi (z_2)}= & {} -\frac{z_2\rho '(z_2)}{1-\rho (z_2)}-\frac{z_2\rho '(z_2)}{1+\rho (z_2)}\\= & {} -\frac{2z_2\rho '(z_2)}{(1-\rho (z_2))(1+\rho (z_2))}\\= & {} -\frac{2z_2\rho '(z_2)}{(1-ib)(1+ib)}\\= & {} -\frac{2z_2\rho '(z_2)}{1+b^2}\\\ge & {} k, \end{aligned}$$

where $k\ge 1$ and $\rho (z_2)=-ib$ with $b>0$. This shows that $z_2\rho '(z_2)$ is a negative real number. This gives

$$\begin{aligned} \mathfrak {Im} \frac{ z_2\rho '(z_2)}{\rho (z_2)} =\mathfrak {Im} \frac{ z_2\rho '(z_2)}{-ib} \le \frac{k}{2}\left( b+\frac{1}{b}\right) \ge k. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \frac{ z_2\rho '(z_2)}{\rho (z_2)}=\frac{1}{\beta }\frac{ z_2p'(z_2)}{p(z_2)} \end{aligned}$$

and putting $z=|z_2|e^{i\theta }$ in the section $\arg z_2-\varepsilon<\theta <\arg z_2+\varepsilon$, then $\arg \{\rho (z_2)\}$ takes its local minimum $-\pi /2$ at the point $\theta =\arg \{z_2\}$ in this section and so, we have

$$\begin{aligned} \mathfrak {Re} \frac{ z_2\rho '(z_2)}{\rho (z_2)} =\left[ \mathfrak {Re} \frac{\mathrm{d}\log \rho (z)}{\mathrm{d} z}\right] _{z=z_2}=\left[ \frac{\mathrm{d}\arg \rho (z)}{\mathrm{d} \theta }\right] _{\theta =\arg z_2}=0. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \frac{ z_2p'(z_2)}{p(z_2)}=-i\frac{\beta k}{2}\left( b+\frac{1}{b}\right) \end{aligned}$$

is a pure imaginary number, where $p^{1/\beta }(z_2)=-ib$, $b>0$ so

$$\begin{aligned} -\frac{\beta k}{2}\left( b+\frac{1}{b}\right) \le -\beta k. \end{aligned}$$

This completes the proof of (1.14). $\square$

Theorem 1.2 is a generalization of a result in [4, 5].

Corollary 1.3

Under the assumptions of Theorem 1.2 we have

$$\begin{aligned} \int _{z_2}^{z_1}\left\{ \frac{zp'(z)}{p(z)}\right\} '\mathrm{d}z\ge (\alpha +\beta )k. \end{aligned}$$

Corollary 1.3 implies that the length of image curve under $zp'(z)/p(z)$ of the circular arc $|z|=|z_1|=|z_2|$ from $z_2$ to $z_1$ is great or equal to $(\alpha +\beta )k$.

Corollary 1.4

Under the assumptions of Theorem 1.2 and if $\arg z_2<\arg z_1$ and

$$\begin{aligned} \left| \mathfrak {Re}\left[ \left\{ \frac{p'(z)}{p(z)}+ \frac{zp''(z)}{p'(z)}-z\left( \frac{p'(z)}{p(z)}\right) ^2\right\} iz\right] \right| <\frac{1}{2\pi },\quad z\in \mathbb D, \end{aligned}$$

(1.15)

then we have

$$\begin{aligned} \mathfrak {Re}\{p(z)\}>0,\quad z\in \mathbb D. \end{aligned}$$

Proof

From Lemma 1.1 we have

$$\begin{aligned} 0<\alpha +\beta \le (\alpha +\beta )k\ge \alpha +\beta \le \left| \int _{z_2}^{z_1}\left\{ \frac{zp'(z)}{p(z)}\right\} '\mathrm{d}z\right| . \end{aligned}$$

On the other hand and applying (1.15), we have

$$\begin{aligned} 0<\alpha +\beta \le \left| \int _{z_2}^{z_1}\left\{ \frac{zp'(z)}{p(z)}\right\} '\mathrm{d}z\right|= & {} \left| \mathfrak {Re}\int _{\theta _2}^{\theta _1}\left\{ \frac{p'(z)}{p(z)}+ \frac{zp''(z)}{p'(z)}-z\left( \frac{p'(z)}{p(z)}\right) ^2\right\} iz\mathrm{d}\theta \right| \\\le & {} \int _{\theta _2}^{\theta _1}\left| \mathfrak {Re}\left[ \left\{ \frac{p'(z)}{p(z)}+ \frac{zp''(z)}{p'(z)}-z\left( \frac{p'(z)}{p(z)}\right) ^2\right\} iz\right] \right| \mathrm{d}\theta \\\le & {} \frac{\theta _1-\theta _2}{2\pi }<1, \end{aligned}$$

where $\arg z_1=\theta _1$, $\arg z_2=\theta _2$ and $\theta _2<\theta _1$. This shows that

$$\begin{aligned} 0<\alpha +\beta <1 \end{aligned}$$

which implies $0<\alpha <1$ and $0<\beta <1$ or

$$\begin{aligned} \mathfrak {Re}\{p(z)\}>0,\quad z\in \mathbb D. \end{aligned}$$

$\square$

In [8, p.54] we can find the following result for the convolution of power series

$$\begin{aligned} \sum _{n=0}^{\infty }a_nz^n*\sum _{n=0}^{\infty }b_nz^n=\sum _{n=0}^{\infty }a_nb_nz^n. \end{aligned}$$

Lemma 1.5

If $g(z)\in {\mathcal C}$, $f(z)\in {\mathcal S}^*$ and F(z) is analytic in $\mathbb D$, then for all $z\in \mathbb D$

$$\begin{aligned} \frac{g(z) * F(z)f(z)}{g(z) * f(z)}\in \overline{co}~F(\mathbb D), \end{aligned}$$

where $\overline{co}A$ denotes the closed convex hull of A.

Theorem 1.6

If $f(z)=a_0+a_1z+a_2z^2+\cdots , h(z)=b_0+b_1z+b_2z^{2}+\cdots \in \mathcal H$, $zh''(z)/(2b_2)\in \mathcal S^*$ and

$$\begin{aligned} q(z)=\frac{f'(z)-a_1}{h'(z)-b_1},\quad p(z)=\frac{f''(z)}{h''(z)},\quad z\in \mathbb D, \end{aligned}$$

are analytic functions in $\mathbb D$, then

$$\begin{aligned} q(\mathbb D)\subseteq \overline{co}~p(\mathbb D). \end{aligned}$$

(1.16)

Proof

Denote

$$\begin{aligned} s(z)=\sum _{n=2}^{\infty } \frac{z^{n-1}}{n-1}=-\log (1-z),\quad z\in \mathbb D, \end{aligned}$$

then $s(z)\in \mathcal C$.

Applying Lemma 1.5, we have

$$\begin{aligned} \frac{f'(z)-a_1}{h'(z)-b_1}= & {} \frac{\sum _{n=2}^{\infty } na_nz^{n-1}}{\sum _{n=2}^{\infty } nb_nz^{n-1}}\\= & {} \frac{\sum _{n=2}^{\infty } na_nz^{n-2}}{\sum _{n=2}^{\infty } nb_nz^{n-2}}\\= & {} \frac{\sum _{n=2}^{\infty } \frac{z^{n-2}}{n-1} * \sum _{n=2}^{\infty } n(n-1)a_nz^{n-2}}{\sum _{n=2}^{\infty } \frac{z^{n-2}}{n-1} * \sum _{n=2}^{\infty } n(n-1)b_nz^{n-2}}\\= & {} \frac{\frac{s(z)}{z}*f''(z)}{\frac{s(z)}{z}*h''(z)} \\= & {} \frac{\frac{s(z)}{z}*p(z)h''(z)}{\frac{s(z)}{z}*h''(z)} \\= & {} \frac{s(z)*p(z)\frac{zh''(z)}{2b_2}}{s(z)*\frac{zh''(z)}{2b_2}} \\\in & {} \overline{co}~p(\mathbb D) \end{aligned}$$

because $s(z)\in \mathcal C$ and $zh''(z)/(2b_2)\in \mathcal S^*$. $\square$

The differential subordination theory provides another method od proving this type of results. For two functions f(z) and g(z) analytic in $\mathbb D$, we say that ff(z) is subordinate to g(z), written by $f\left( z\right) \prec g\left( z\right)$, if there exists a function w(z), analytic in $\mathbb D$, with $w(0) =0$ and $|w(z)| <1$ such that $f(z)=g( w(z))$.

In [3, p.70] we can find the following result

Lemma 1.7

Let H(z) be convex univalent, $P(z)\in \mathcal H$ and $\mathfrak {Re}\{P(z)\}>0$ in $\mathbb D$. If p(z) is amalytic in $\mathbb D$ with $p(0)=H(0)$, then

$$\begin{aligned} p(z)+P(z)zp'(z) \prec H(z)\quad \Rightarrow \quad p(z) \prec H(z). \end{aligned}$$

(1.17)

Theorem 1.8

Let H(z) be convex univalent in $\mathbb D$. Assume that

$$\begin{aligned} \frac{f^{(s)}(z)}{g^{(s)}(z)},\quad \frac{f^{(s-1)}(z)}{g^{(s-1)}(z)},\quad \frac{g^{(s-1)}(z)}{zg^{(s)}(z)} \end{aligned}$$

are analytic in $\mathbb D$ for some positive integer s and

$$\begin{aligned} \mathfrak {Re}\left\{ \frac{g^{(s-1)}(z)}{zg^{(s)}(z)}\right\} >0,\quad z\in \mathbb D. \end{aligned}$$

(1.18)

If

$$\begin{aligned} \frac{f^{(s)}(z)}{g^{(s)}(z)}\prec H(z), \end{aligned}$$

and with $f^{(s-1)}(0)/g^{(s-1)}(0)=H(0)$ then

$$\begin{aligned} \frac{f^{(s-1)}(z)}{g^{(s-1)}(z)}\prec H(z). \end{aligned}$$

(1.19)

Proof

Denote

$$\begin{aligned} p(z)=\frac{f^{(s-1)}(z)}{h^{(s-1)}(z)}. \end{aligned}$$

Then we have $p(z)h^{(s-1)}(z)=f^{(s-1)}(z)$, by differentation, we obtain

$$\begin{aligned} p(z)+ \frac{h^{(s-1)}(z)}{zh^{(s)}(z)}zp'(z)=\frac{f^{(s)}(z)}{h^{(s)}(z)}\prec H(z). \end{aligned}$$

Applying (1.17) and (1.18) give $p(z) \prec H(z)$, i. e. (1.19).

Corollary 1.9

Assume that $\alpha \in (0,1]$ and

$$\begin{aligned} \frac{f''(z)}{h''(z)},\quad \frac{f'(z)}{h'(z)},\quad \frac{h(z)}{zh'(z)},\quad \frac{h'(z)}{zh''(z)} \end{aligned}$$

are analytic in $\mathbb D$ Then

$$\begin{aligned} \frac{f'(z)}{h'(z)}\prec \left\{ \frac{1+z}{1-z}\right\} ^{\alpha }~and~\ ~ \mathfrak {Re}\left\{ \frac{h(z)}{zh'(z)}\right\} >0,\quad z\in \mathbb D,\quad \Rightarrow \quad \frac{f(z)}{h(z)}\prec \left\{ \frac{1+z}{1-z}\right\} ^{\alpha } \end{aligned}$$

(1.20)

and

$$\begin{aligned} \frac{f''(z)}{h''(z)}\prec \left\{ \frac{1+z}{1-z}\right\} ^{\alpha }~and~\ ~ \mathfrak {Re}\left\{ \frac{h'(z)}{zh''(z)}\right\} >0,\quad z\in \mathbb D,\quad \Rightarrow \frac{f'(z)}{h'(z)}\prec \left\{ \frac{1+z}{1-z}\right\} ^{\alpha }. \end{aligned}$$

(1.21)

Consider, $h(z)=e^z-1$, which satisfies

$$\begin{aligned} \mathfrak {Re}\frac{zh''(z)}{h'(z)}=\mathfrak {Re}\{z\}>-1,\quad z\in \mathbb D. \end{aligned}$$

Therefore, $h(z)\in \mathcal C$ and so $h(z)\in \mathcal S^*$ which gives

$$\begin{aligned} \mathfrak {Re}\left\{ \frac{h(z)}{zh'(z)}\right\} >0,\quad z\in \mathbb D. \end{aligned}$$

Applying (1.20) we obtain

$$\begin{aligned} \frac{f'(z)}{e^z}\prec \left\{ \frac{1+z}{1-z}\right\} ^{\alpha }~ \Rightarrow \frac{f(z)}{e^z-1}\prec \left\{ \frac{1+z}{1-z}\right\} ^{\alpha }. \end{aligned}$$

Corollary 1.10

Assume that $\alpha \in [0,1)$ and

$$\begin{aligned} \frac{f''(z)}{h''(z)},\quad \frac{f'(z)}{h'(z)},\quad \frac{h(z)}{zh'(z)},\quad \frac{h'(z)}{zh''(z)} \end{aligned}$$

are analytic in $\mathbb D$. Then

$$\begin{aligned} \mathfrak {Re}\left\{ \frac{h(z)}{zh'(z)}\right\}>0~and~\ ~ \mathfrak {Re}\frac{f'(z)}{h'(z)}>\alpha ,\quad z\in \mathbb D\quad \Rightarrow \mathfrak {Re}\frac{f(z)}{h(z)}>\alpha ,\quad z\in \mathbb D \end{aligned}$$

(1.22)

and

$$\begin{aligned} \mathfrak {Re}\left\{ \frac{h'(z)}{zh''(z)}\right\}>0~and~\ ~ \mathfrak {Re}\frac{f''(z)}{h''(z)}>\alpha ,\quad z\in \mathbb D\quad \Rightarrow \mathfrak {Re}\frac{f'(z)}{h'(z)}>\alpha ,\quad z\in \mathbb D. \end{aligned}$$

(1.23)

Property (1.22) was known earlier, see Theorem 2.6f in [3, p.63].

Theorem 1.11

Assume that $h(z)\in \mathcal H$ and

$$\begin{aligned} \mathfrak {Re}\frac{h'(z)-\gamma -z}{z(h''(z)-1)}>0,\quad z\in \mathbb D \end{aligned}$$

(1.24)

for some $\gamma \in (1,\infty ]$ and

$$\begin{aligned} \left| \arg \{h'(z)-\gamma -z\}\right| <\frac{\beta \pi }{2},\quad z\in \mathbb D, \end{aligned}$$

(1.25)

for some $\beta \in (0,1]$. Assume that $f(z)\in \mathcal H$ and

$$\begin{aligned} \left| \arg \frac{f''(z)-1}{h''(z)-1}\right| <\frac{\alpha \pi }{2},\quad z\in \mathbb D, \end{aligned}$$

(1.26)

for some $\alpha \in (0,1]$. Furthermore, assume that

$$\begin{aligned} F(z)=\frac{f'(z)-\gamma -z}{h'(z)-\gamma -z} \end{aligned}$$

is analytic in $\mathbb D$ with $F(0)=1$. Then

$$\begin{aligned} \left| \arg \{f'(z)\}\right| <\frac{(\alpha +\beta )\pi }{2},\quad z\in \mathbb D. \end{aligned}$$

(1.27)

Proof

Consider Theorem 1.8 with $s=2$, $H(z)=\left( \frac{1+z}{1-z}\right) ^{\alpha }$ and $g'(z)= h'(z)-\gamma -z$. Then H(z) is convex univalent and condition (1.18) becomes (1.24). Applying this for (1.26) gives us

$$\begin{aligned} \left| \arg \left\{ \frac{f'(z)-\gamma -z}{h'(z)-\gamma -z}\right\} \right| < \frac{\alpha \pi }{2},\quad z\in \mathbb D. \end{aligned}$$

(1.28)

Now, applying (1.25), we have

$$\begin{aligned}&~&\left| \arg \{f'(z)-\gamma -z\}\right| - \left| \arg \{h'(z)-\gamma -z\}\right| \\\le & {} \left| \arg \{f'(z)-\gamma -z\}-\arg \{h'(z)-\gamma -z\}\right| \\= & {} \left| \arg \left\{ \frac{f'(z)-\gamma -z}{h'(z)-\gamma -z}\right\} \right| \\< & {} \frac{\alpha \pi }{2},\quad z\in \mathbb D \end{aligned}$$

because of (1.28). So we have

$$\begin{aligned} \left| \arg \{f'(z)-\gamma -z\}\right|\le & {} \frac{\alpha \pi }{2}+ \left| \arg \{h'(z)-\gamma -z\}\right| \\< & {} \frac{(\alpha +\beta )\pi }{2},\quad z\in \mathbb D. \end{aligned}$$

A simple geometric observation shows that

$$\begin{aligned} \left| \arg \{f'(z)\}\right| < \left| \arg \{f'(z)-\gamma -z\}\right| ,\quad z\in \mathbb D,\quad \gamma \in (1,\infty ]. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| \arg \{f'(z)\}\right|< \left| \arg \{f'(z)-\gamma -z\}\right| <\frac{(\alpha +\beta )\pi }{2},\quad z\in \mathbb D. \end{aligned}$$

$\square$

Condition (1.27) implies that f(z) is strongly close-to-convex of order $\alpha +\beta$ with respect convex function $g(z)=z$. We obtain more when $\alpha +\beta \le 1$ in the following corollary.

Corollary 1.12

Assume that $h(z)\in \mathcal H$ and

$$\begin{aligned} \mathfrak {Re}\frac{h'(z)-\gamma -z}{z(h''(z)-1)}>0,\quad z\in \mathbb D \end{aligned}$$

(1.29)

for some $\gamma \in (1,\infty ]$ and

$$\begin{aligned} \left| \arg \{h'(z)-\gamma -z\}\right| <\frac{\beta \pi }{2},\quad z\in \mathbb D, \end{aligned}$$

(1.30)

for some $\beta \in (0,1]$. Assume that $f(z)\in \mathcal H$ and

$$\begin{aligned} \left| \arg \frac{f''(z)-1}{h''(z)-1}\right| <\frac{\alpha \pi }{2},\quad z\in \mathbb D, \end{aligned}$$

(1.31)

for some $\alpha \in (0,1-\beta ]$. Furthermore, assume that

$$\begin{aligned} F(z)=\frac{f'(z)-\gamma -z}{h'(z)-\gamma -z} \end{aligned}$$

is analytic in $\mathbb D$ with $F(0)=1$. Then

$$\begin{aligned} \mathfrak {Re}\{f'(z)\}>0,\quad z\in \mathbb D. \end{aligned}$$

(1.32)

Proof

We have $\alpha +\beta \le 1$, then by Theorem 1.11, we have

$$\begin{aligned} \left| \arg \{f'(z)\}\right| < \frac{(\alpha +\beta )\pi }{2}\le \frac{\pi }{2},\quad z\in \mathbb D. \end{aligned}$$

This gives (1.32). $\square$

Condition (1.32) implies the univelence f(z) in $\mathbb D$ by Noshiro–Warshawski’s theorem.