Radius Problems for Ratios of Janowski Starlike Functions with Their Derivatives

Abstract

Let $-1 \le B<A \le 1$ and $F= f/f'$ for normalized locally univalent functions f. For analytic function p on the open unit disc $\mathbb {D}$ with $p(0)=1$ and satisfying the subordination $p(z) \prec (1+A z)/(1+B z)$, we determine bounds for $|p(z)- z p'(z)-1|$ and $|z p'(z)|/{{\mathrm{Re\,}}}p(z)$. As an application, we investigate the radius problem for F to satisfy $|F'(z)(z/F(z))^2-1| <1$, where f is a Janowski starlike function and the radius of univalence of F when f is a starlike function of order $\alpha$. Also, we have discussed the radius of starlikeness of F when f is a Janowski starlike function with fixed second coefficient. Apart from the radius problems, we give the sharp coefficient bounds of F when f is a Janowski starlike function and a sufficient condition for starlikeness of f when F is a Janowski starlike function. Our results generalize some of the earlier known results.

1 Introduction and Preliminaries

Let $\mathcal {A}$ be the class of all analytic functions $f:\mathbb {D}\rightarrow \mathbb {C}$ of the form $f(z)= z+ a_2 z^2 + a_3 z^3 + \cdots$, where $\mathbb {D}$ denotes the unit disc in $\mathbb {C}$. Let $\mathcal {S}$ denote the subclass of $\mathcal {A}$ consisting of univalent functions. Let $\mathcal {P}$ denote the class of all analytic functions $p:\mathbb {D}\rightarrow \mathbb {C}$ with $p(0)=1$ and ${{\mathrm{Re\,}}}p(z) >0$. We denote the subclasses of $\mathcal {S}$ consisting of starlike and convex functions by $\mathcal {S}^*$ and $\mathcal {C}$, respectively. Let us recall that an analytic function f is said to be subordinate to an analytic function g, written $f \prec g$, if $f=g\circ w$ for some analytic function $w:\mathbb {D}\rightarrow \mathbb {D}$ with $w(0)=0$. If g is univalent, then $f \prec g$ is equivalent to $f(0)=g(0)$ and $f(\mathbb {D}) \subset g(\mathbb {D})$. For $-1 \le B < A \le 1$, Janowski [10] introduced the classes:

$$\begin{aligned} \mathcal {P}[A,B] =&\left\{ p \in \mathcal {P}: p(z) \prec \frac{1+ A z}{1+ B z}\right\} \quad \text {and}\\ \mathcal {S}^*[A,B]=&\left\{ f \in \mathcal {S}: \frac{ z f'(z)}{f(z) }\in \mathcal {P}[A,B]\right\} . \end{aligned}$$

Tuan and Anh [26] put some restriction on the initial coefficient of functions in these classes and considered the following subclasses for a fixed b, $0 \le b \le 1$:

$$\begin{aligned}&\mathcal {P}_b[A,B]= \big \{ p \in \mathcal {P}[A,B]: p'(0)= b(A-B)\big \} \end{aligned}$$

and

$$\begin{aligned}&\mathcal {S}^*_b[A,B]= \{ f \in \mathcal {S}^*[A,B]: f''(0)= 2b (A-B), z f'(z)/f(z) \in \mathcal {P}_b[A,B] \}. \end{aligned}$$

For $0 < \alpha \le 1$, Nunokawa et al. [15] considered the class $\mathcal {G}(\alpha )$ of functions $f \in \mathcal {A}$ satisfying

$$\begin{aligned} {{\mathrm{Re\,}}}\left( 1 + \dfrac{z f''(z)}{f'(z)} \right) < 1+ \dfrac{\alpha }{2} \quad (z \in \mathbb {D}). \end{aligned}$$

For $f \in \mathcal {G}(\alpha )$, Jovanović and Obradović [11] have shown that $\mathcal {G}(\alpha ) \subset \mathcal {S}^*[1,1/(1+ \alpha )]$. Also, Obradović et al. [16] gave the sharp coefficient bounds $|a_n| \le \alpha /\big (n(n-1)\big )$ for $f(z)= z + \sum _{n=2}^\infty a_n z^n \in \mathcal {G}(\alpha )$. Similarly, for $0 \le b \le 1$, one can study the subclass $\mathcal {G}_b(\alpha )$ of $\mathcal {G}(\alpha )$ consisting of functions $f \in \mathcal {G}(\alpha )$ such that $f''(0)= b \alpha$. Let $\mathcal {U}$ denote the subclass of $\mathcal {A}$ consisting of functions f satisfying $|U_f(z)| <1$ for $z \in \mathbb {D}$, where $U_f(z)= f'(z)(z/f(z))^2-1$. This class was introduced by Ozaki and Nunokawa [18] and the authors showed that $\mathcal {U} \subset \mathcal {S}$. For a property $\mathfrak {P}$ and a set of functions $\mathfrak {M}$, the radius problem is to determine the largest radius $R_\mathfrak {P}$ such that every $f \in \mathfrak {M}$ has the property $\mathfrak {P}$ in the disc $\mathbb {D}_r:=\{z \in \mathbb {C}: |z| < r\}$ for each $r < R_\mathfrak {P}$ and for $r > R_\mathfrak {P}$, at least one f in $\mathfrak {M}$ does not have the property $\mathfrak {P}$ in $\mathbb {D}_r$. Recently, several authors have studied the radius problems of different well-known classes, for example, see [1, 2, 14, 20, 23].

In this paper, we give the sharp upper bound of $|p(z)- z p'(z)-1|$ for $p \in \mathcal {P}[A,B]$, the radius of starlikeness and the $\mathcal {U}$-radius of the function $F(z):= f(z)/f'(z)$ for $f \in \mathcal {S}^*[A,B]$ or $f \in \mathcal {S}_b^*[A,B]$. As corollaries, we give the radius of starlikeness and the $\mathcal {U}$-radius in certain special cases. Also, we give an upper bound of $|z p'(z)|/{{\mathrm{Re\,}}}p(z)$ for $p \in \mathcal {P}[A,B]$ which coincides with the bound given in [7, Theorem 1, p. 113] for $A=1, B=-1$. Apart from the radius problems, we give the sharp coefficient bounds of F when f is a Janowski starlike function and a sufficient condition for starlikeness of f when F is a Janowski starlike function. Some of our results improve the corresponding results in [17].

2 Estimations over $\mathcal {P}[A,B]$

By [6, Theorem 5, p. 742], $|c_n| \le A-B$ for $p(z)=1+ \sum _{n=1}^{\infty } c_n z^n \in \mathcal {P}[A,B]$ and therefore, we have $|p(z) - z p'(z) -1| \le (A-B) |z|^2/(1-|z|)^2$. In general, this bound is not sharp but for $p \in \mathcal {P}(\alpha ):=\mathcal {P}[1-2 \alpha ,-1]$, we obtain the sharp upper bound $|p(z) - z p'(z) -1| \le 2(1- \alpha ) |z|^2/(1-|z|)^2$. By using Dieudonné’s Lemma [8, p. 25] that

$$\begin{aligned} |z w'(z) - w(z)| \le \dfrac{|z|^2 - |w(z)|^2}{1- |z|^2} \end{aligned}$$

for an analytic function w defined on $\mathbb {D}$ with $w(0)=0$, $|w(z)|<1$, and the technique employed by Anh and Tuan in [4], we prove the following sharp version of the above result.

Lemma 2.1

If $p \in \mathcal {P}[A,B]$ then for $0<r= |z| < 1$, we have

$$\begin{aligned} |p(z)- z p'(z) -1| \le {\left\{ \begin{array}{ll} M(r;A,B), &{}\text { either } |B| < 0.8, r \in [0,1),\\ &{}\text { or, }0.8 \le |B| < 1 , r \in (0,r_0) \cup (\tilde{r}_0,1),\\ \dfrac{|B|(A-B) r^2}{(1- |B| r)^2}, &{} 0.8 \le |B| \le 1, r \in [r_0,\tilde{r}_0],\\ \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned}&M(r;A,B):=\dfrac{\big (A-B\big ) r^2 \big (1-|B| + |B|(2-|B|) r^2 + B^2 r^4\big )}{(1- |B|) (1- r^2)(1+ |B| r^2)^2} \end{aligned}$$

(2.1)

and

$$\begin{aligned}&r_0:= \dfrac{2(1-|B|)}{|B|+ \sqrt{|B|(5 |B| -4)}}, \quad \tilde{r}_0:= \dfrac{|B| + \sqrt{|B|(5 |B| -4)}}{2 |B|}. \end{aligned}$$

(2.2)

The result is sharp.

Proof

Since $p \in \mathcal {P}[A,B]$, $p(z)= \big (1+ A w(z)\big )/\big (1+ B w(z)\big )$ for some analytic function $w: \mathbb {D} \rightarrow \mathbb {D}$ with $w(0)=0$. This gives

$$\begin{aligned} w(z)= \dfrac{p(z)-1}{A- B p(z)}. \end{aligned}$$

(2.3)

By using [8, Dieudonné’s lemma, p. 25] and Eq. (2.3), we have

$$\begin{aligned} | p(z)- z p'(z) -1|&\le \dfrac{|z|^2 |A- B p(z)|^2 - |p(z) -1|^2 \big (1- |B| (1- |z|^2)\big )}{(A-B)(1- |z|^2)}. \end{aligned}$$

(2.4)

Since $p \in \mathcal {P}[A,B]$, the subordination principle shows that the image of $|z| \le r$ under p(z) is contained in the disc $|p(z)-a| \le d$, where $a= (1- A B r^2)/(1- B^2 r^2)$, $d= (A-B) r/(1- B^2 r^2)$. Put $|z|=r$ and $p(z)= R e^{i \theta }$, $R \in [a-d,a+d]$, $\theta \in (-\pi /2,\pi /2)$, then the right-hand side of the inequality (2.4) becomes

$$\begin{aligned} S(R, \theta ):= \dfrac{g(R,\theta )}{(A-B)(1-r^2)}, \end{aligned}$$

(2.5)

where

$$\begin{aligned} g(R,\theta )&:= \big (r^2 A^2 -1 + |B| - |B| r^2 \big ) - (1- |B|)\big (1+ |B| r^2\big )R^2 \\&\qquad + \, 2 R\big (- A B r^2 + 1 - |B|+ |B| r^2\big )\cos {\theta }. \end{aligned}$$

Since $S(R, \theta )$ is an even function of $\theta$, it is enough to consider $\theta \in [0, \pi /2)$.

Since we have

$$\begin{aligned} \dfrac{\partial {S(R,\theta )}}{\partial {\theta }}= -\dfrac{2 R \big ((|B|- A B) r^2 + 1- |B| \big )\sin {\theta }}{(A-B)(1- r^2)} \end{aligned}$$

and $(|B|- A B) r^2 + 1- |B|>0$, the function $S(R, \theta )$ is a decreasing function in $[0,\pi /2)$ and hence it attains maximum at $\theta =0$. Thus we have $| p(z)- z p'(z) -1| \le S(R,0)$. Also,

$$\begin{aligned} \dfrac{\partial {S(R,0)}}{\partial {R}}&= \dfrac{ 2\big ( R (1+ |B| r^2)(-1+ |B|) + 1 - |B| + (|B|- A B) r^2\big )}{(A-B)(1-r^2)}, \end{aligned}$$

and

$$\begin{aligned} \dfrac{\partial ^2{S(R,0)}}{\partial {R}^2}&=-\dfrac{ 2(1+ |B| r^2)(1- |B|)}{(A-B) (1-r^2)} \le 0. \end{aligned}$$

Now, for $B \ne -1$, $\partial {S(R,0)}/ \partial {R}=0$ gives

$$\begin{aligned} R_0:=\dfrac{1- |B| + (|B| - A B )r^2}{(1- |B|)(1+ |B| r^2)}. \end{aligned}$$

(2.6)

Since S(R, 0) is a concave function, it attains its maximum M(r; A, B), given by (2.1), at $R_0$ if $R_0 \in (a-d,a+d)$ and otherwise the maximum is attained either at $a-d$ or $a+d$.

Let $B \ne -1$, then

$$\begin{aligned}&a-d< R_0 \text { and } R_0 < a+d \text { are equivalent to } \phi (r,B):= |B| r^2 - B r + 1- |B| {>}0 \nonumber \\&\quad \text { and } \psi (r,B):= - |B| r^2 - B r -1 + |B|<0, \ \text {respectively} . \end{aligned}$$

(2.7)

Note that for $B\ge 0$, $\psi (r,B) <0$ and for $B\le 0$, $\phi (r,B) >0$ holds true. Therefore, we have

$$\begin{aligned} R_0 < a+d \ \text {for}\ B\ge 0 \quad \text {and} \quad R_0 > a-d \ \text {for} \ B \le 0. \end{aligned}$$

(2.8)

• (i) For $0< B < 0.8$, $\phi (r,B)$ does not have any real zero and $\phi (0,B) >0$, and therefore $\phi (r,B) >0$ for $r \in (0,1)$ and hence in view of (2.7) and (2.8), $R_0 \in (a-d,a+d)$. For $-0.8 < B <0$, $\psi (r,B)$ does not have any real zero and $\psi (0,B)<0$, and therefore $\psi (r,B) <0$ for $r \in (0,1)$ and hence in view of (2.7) and (2.8), $R_0 \in (a-d,a+d)$. For $B=0$, $R_0 = 1$ and in view of (2.8), we have $R_0 \in (a-d, a+d)$.

  Thus, for $|B| < 0.8$, $S(R,0) \le S(R_0,0)=:M(r;A,B)$, where M(r; A, B) is given by (2.1).

  For $0 \le B < 0.8$ and $-0.8 < B \le 0$, the result is sharp for the functions $p_1(z) = \big (1+ A w_1(z)\big )/\big (1+ B w_1(z)\big )$ and $p_2(z) = \big (1+ A w_2(z)\big )/\big (1+ B w_2(z)\big )$, respectively, where $w_1(z) = - z(z- c_1)/(1- c_1 z)$ and $w_2(z) = z(z- c_2)/(1- c_2 z)$ with $c_i$ $(i=1,2)$ determined from the equation

  $$\begin{aligned} {{\mathrm{Re\,}}}\dfrac{1+ A w_i(z)}{1+ B w_i(z)}= \dfrac{1- |B| + (|B| - A B )r^2}{(1- |B|)(1+ |B| r^2)} \text { at } z=r\ (i=1,2). \end{aligned}$$

  (2.9)

• (ii) For $0.8 < B <1$, $\phi (r,B)$ has two distinct zeros $r_0$ and $\tilde{r}_0$, given by (2.2), in (0, 1) and $\phi (0,B) >0$, hence $\phi (r,B) >0$ in $(0,r_0) \cup (\tilde{r}_0,1)$. For $B =0.8$, $\phi (r,B)$ has a zero of multiplicity two, $r_0 = \tilde{r}_0= 1/2$, in (0, 1) and hence $\phi (r,B) >0$ in $(0,1/2) \cup (1/2,1)$. Therefore, in view of (2.7) and (2.8), for $r \in (0,r_0) \cup (\tilde{r}_0,1)$, $0.8 \le B <1$, we have $R_0 \in (a-d,a+d)$.

  Also, for $-1 < B < -0.8$, $\psi (r,B)$ has two distinct zeros $r_0$ and $\tilde{r}_0$, given by (2.2), in (0, 1) and $\psi (0,B) <0$, hence $\psi (r,B) <0$ in $(0,r_0) \cup (\tilde{r}_0,1)$. For $B = -0.8$, $\psi (r,B)$ has a zero of multiplicity two, $r_0 = \tilde{r}_0= 1/2$, in (0, 1) and hence $\psi (r,B)<0$ in $(0,1/2)\cup (1/2,1)$. Therefore, in view of (2.7) and (2.8), for $r \in (0,r_0) \cup (\tilde{r}_0,1)$, $-1<B \le -0.8$, we have $R_0 \in (a-d,a+d)$.

  Thus, for $0.8 \le |B| < 1$ and $r \in (0,r_0) \cup (\tilde{r}_0,1)$, we have $S(R,0) \le S(R_0,0)=:M(r;A,B)$, where M(r; A, B) is given by (2.1). For $0.8 \le B <1$ and $-1 < B \le -0.8$, the sharpness follows for the functions $p_1$ and $p_2$, respectively, defined in case (i).

• (iii) By case (ii), for $0.8 \le B <1$, $\phi (r,B) \le 0$ in $[r_0, \tilde{r}_0]$ and therefore, in view of (2.7) and (2.8), $a-d \ge R_0$. Thus, $S(R,0) \le S(a-d,0)= B(A-B)r^2/(1-B r)^2$ for $0.8 \le B <1$ and $r \in [r_0, \tilde{r}_0]$. The bound is sharp for the function $p_3(z)= (1+Az)/(1+Bz)$ at $z=-r$.

  For $-1 < B \le -0.8$, by case (ii), $\psi (r,B) \ge 0$ in $[r_0, \tilde{r}_0]$ and hence, in view of (2.7) and (2.8), $R_0 \ge a+d$. Also, for $B=-1$, $\partial {S(R,0)}/ \partial {R}= 2(1+A) r^2 >0$. Thus, $S(R,0) \le S(a+d,0)= -B(A-B)r^2/(1+B r)^2$ for $-1 \le B \le -0.8$ and $r \in [r_0, \tilde{r}_0]$. The bound is sharp for the function $p_3(z)= (1+Az)/(1+Bz)$ at $z=r$.$\square$

For a function with positive real part, it is known [7, Theorem 1, p. 113] that

$$\begin{aligned} \frac{|z p'(z)|}{ {{\mathrm{Re\,}}}p(z)}\le \frac{2|z|}{1-|z|^2} \end{aligned}$$

This result is extended to the class $\mathcal {P}[A,B]$ in the following lemma.

Lemma 2.2

Let $p \in \mathcal {P}[A,B]$ and $0<r=|z|<1$. Then the upper bound of $|z p'(z)|/ {{\mathrm{Re\,}}}p(z)$ is

• (a) P(r; A, B) if $-1< B < A < 1$ and $r>r_0$;

• (b) Q(r; A, B) if the following holds:

  • (i) $-1< B < A < 1$, $A+B >0$, $r\le r_0$,

  • (ii) $A=1$, $-1 \le B < 1$ and

• (c) L(r; A, B) if $-1 \le B < A < 1$, $A+B <0$, $r\le r_0$,

where

$$\begin{aligned} P(r;A,B)&:=\left( \dfrac{1- A B - \sqrt{(1-A^2)(1-B^2)}}{A-B} \right) \left( \dfrac{2r}{1-r^2}\right) , \end{aligned}$$

(2.10)

$$\begin{aligned} Q(r;A,B)&:= \dfrac{(A-B) r}{(1- A r)(1- B r)}, \quad L(r;A,B):=\dfrac{(A-B) r}{(1+ A r)(1+ B r)}, \end{aligned}$$

(2.11)

and

$$\begin{aligned} r_0&:= \dfrac{|A+B|}{1+ A B + \sqrt{(1-A^2)(1-B^2)}}. \end{aligned}$$

(2.12)

The result is sharp for the cases (b) and (c).

Proof

Since $p \in \mathcal {P}[A,B]$, $p(z) = \big (1+A w(z)\big )/\big (1+ B w(z)\big )$ for some analytic function $w: \mathbb {D} \rightarrow \mathbb {D}$ with $w(0)=0$. This gives

$$\begin{aligned} p'(z)&= \dfrac{(A-B) w'(z)}{\big (1+ B w(z)\big )^2},\quad \text {and} \quad {{\mathrm{Re\,}}}p(z)= \dfrac{1+ A B |w(z)|^2 + (A+B) {{\mathrm{Re\,}}}w(z)}{|1+ B w(z)|^2}. \end{aligned}$$

Therefore, by using Carathéodary’s inequality $|w'(z)| \le (1- |w(z)|^2)/(1-|z|^2)$ for an analytic function w defined on $\mathbb {D}$ with $|w(z)| \le 1$ and Eq. (2.3), we have

$$\begin{aligned}&\dfrac{|z p'(z)|}{{{\mathrm{Re\,}}}p(z)}\\&\le \dfrac{(A-B)|z| \big (|A{-}B p(z)|^2{-}|p(z)-1|^2\big )}{(1{-} |z|^2)\big (|A-B p(z)|^2 {+} A B |p(z) -1|^2 {+}(A{+}B) {{\mathrm{Re\,}}}\big ((p(z) {-}1)(A- B \overline{p(z)}\big )\big )}. \end{aligned}$$

Substituting $|z|=r$, $p(z) = R e^{i \theta }$ where $R \in [a-d,a+d]$, $\theta \in (-\pi /2,\pi /2)$, $a= (1- A B r^2)/(1- B^2 r^2)$ and $d= (A-B) r/(1- B^2 r^2)$, in the above inequality and proceeding as in the previous lemma, we find that the right-hand side of the above inequality

$$\begin{aligned} S(R, \theta ):= \dfrac{r\big (A^2-1 +(B^2-1)R^2 + 2 R \cos {\theta }(1-A B)\big )}{(1-r^2)(A-B) R \cos {\theta }} \end{aligned}$$

as a function of $\theta$ would attain its maximum at $\theta =0$ except for the case $A = 1$ and $B = -1$. Therefore, we have $S(R,\theta ) \le S(R,0)$ except for the case $A = 1$ and $B = -1$. For $A = 1$ and $B = -1$, $S(R, \theta )= 2 r/(1-r^2)$ which is the required bound and coincides with the bound given in [7, Theorem 1, p. 113]. In this case, the sharpness follows at $z=-r$ for the function $p_0:\mathbb {D} \rightarrow \mathbb {C}$ defined by

$$\begin{aligned} p_0(z):= (1+ A z)/(1+ B z). \end{aligned}$$

(2.13)

Now, it is left to maximize S(R, 0) except for the case $A = 1$ and $B = -1$. Clearly,

$$\begin{aligned} \dfrac{\partial {S(R,0)}}{\partial {R}}&= \dfrac{r(-(A^2-1)+(B^2-1)R^2)}{(1-r^2)(A-B)R^2} \end{aligned}$$

and

$$\begin{aligned} \dfrac{\partial ^2{S(R,0)}}{\partial {R}^2}&= \dfrac{2 r(A^2-1)}{(1-r^2)(A-B)R^3} \le 0. \end{aligned}$$

Also, for $-1 < B <A \le 1$,

$$\begin{aligned} \dfrac{\partial {S(R,0)}}{\partial {R}}=0 \quad \text {if} \quad R =:R_0= \sqrt{\dfrac{1-A^2}{1-B^2}}. \end{aligned}$$

Since S(R, 0) is a concave function, it attains its maximum P(r; A, B), given by (2.10), at $R_0$ if $R_0 \in (a-d,a+d)$; otherwise, the maximum is attained either at $a-d$ or $a+d$.

• (a) Let $-1 < B < A <1$. If $A+B=0$, then $R_0=1$ and since $a-d<1$ and $a+d >1$, $R_0 \in (a-d,a+d)$.

  Let $A+B >0$. Since $\psi (r;A,B):= -(A+B) - 2(1+A B )r - (A+B) r^2 <0$, $R_0 < a+d$. Also, we see that $\phi (r;A,B):= A+B - 2(1+ A B) r + (A+B) r^2$ has a unique zero $r_0$ in (0, 1) given by (2.12) and $\phi (0;A,B)>0$, so $\phi (r;A,B)<0$ for $r > r_0$. Therefore, $a-d < R_0$ for $r >r_0$ and hence for $r>r_0$, we have $R_0 \in (a-d,a+d)$.

  Further, if $A+B <0$, clearly $\phi (r;A,B):= A+B - 2(1+ A B) r + (A+B) r^2<0$ and so $a-d < R_0$. Since $\psi (r;A,B):= -(A+B) - 2(1+A B )r - (A+B) r^2$ has a unique zero $r_0$ in (0, 1) given by (2.12) and $\psi (0;A,B)>0$, $\psi (r;A,B)<0$ for $r >r_0$. Therefore, $R_0 < a+d$ for $r >r_0$ and hence for $r>r_0$, we have $R_0 \in (a-d,a+d)$.

  Thus, in this case, $S(R,0) \le S(R_0,0)=:P(r;A,B)$, where P(r; A, B) is given by (2.10).

• (b) For $A=1$, $-1 < B <1$, S(R, 0) is a decreasing function of R and hence $S(R,0) \le S(a-d,0)=: Q(r;A,B)$, where Q(r; A, B) is given by (2.11).

  Also, by case (a), for $A+B >0$, $-1 < B < A <1$, $\phi (r;A,B) \ge 0$ for $r \le r_0$ and so $a-d \ge R_0$ for $r \le r_0$. Therefore, $S(R,0) \le S(a-d,0)=:Q(r;A,B)$.

  The result is sharp for the function $p_0$, given by (2.13), at $z=-r$.

• (c) For $-1 < A <1$, $B=-1$, S(R, 0) is an increasing function of R and therefore $S(R,0) \le S(a+d,0)=:L(r;A,B)$, where L(r; A, B) is given by (2.11).

  By case (a), for $A+B <0$, $-1 < B < A <1$, $\psi (r;A,B) \ge 0$ for $r \le r_0$ and so $R_0 \ge a+d$ for $r \le r_0$. Therefore, $S(R,0) \le S(a+d,0)=:L(r;A,B)$.

  The result is sharp for the function $p_0$, given by (2.13), at $z=r$. $\square$

3 Radius of Starlikeness and $\mathcal {U}$-Radius of $F:=f/f'$

We observe that $\mathcal {U} \subset \mathcal {S}$, see [18, Theorem 2, p. 394] and the Koebe function $k(z)=z/(1-z)^2$ is starlike but the function $F(z)=k(z)/k'(z)=(z-z^2)/(1+z)$ is not even univalent in $\mathbb {D}$ as $F'(\sqrt{2}-1)=0$. This motivates us to investigate the radius of starlikeness and the $\mathcal {U}$-radius of $F(z)=f(z)/f'(z)$ when f belongs to some well-known class of functions. In the following theorems, we consider the function f to belong to the class of Janowski starlike functions.

Theorem 3.1

Let $f \in \mathcal {S}^*[A,B]$, $F(z):= f(z)/f'(z)$ $(z \in \mathbb {D})$ and $0<r=|z|<1$. Let $\hat{r}^{**}$ be the unique zero in (0, 1] of the polynomial $\hat{g}(r;A,B):= A B r^4- 2 A B r^3 -(2 A + 2 B + A B + 1)r^2 - 2 r +1$. Then

• (a) if either of the following holds:

  • (i) $0 \le B < 1$,

  • (ii) $-1 < B <0, -1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big ) \notin (0,\hat{r}^{**}]$, then

  the radius of starlikeness of F is

  $$\begin{aligned} R&:= \sqrt{\dfrac{2\sqrt{1+B}}{(1+A) \sqrt{1+ B} + \sqrt{(1+A)(1+A + A B - 3 B)}}}. \end{aligned}$$

  (3.1)

• (b) if $-1 \le B <0, -1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big ) \in (0,\hat{r}^{**}]$, then the radius of starlikeness of F is $-1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big )$.

The following lemma is used to prove the result.

Lemma 3.2

[10, Theorem 3, p. 314] For all $p \in \mathcal {P}[A,B]$ and $|z| = r$, $0 < r < 1$,

$$\begin{aligned} {{\mathrm{Re\,}}}\dfrac{z p'(z)}{p(z)} \le {\left\{ \begin{array}{ll} Y_1(r; -A,-B), &{}0 < r \le \hat{r}^{**}; \\ Y_2(r; -A,-B), &{}\hat{r}^{**} < r < 1, \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} \ Y_1(r; A, B)&= -\dfrac{(A-B)r}{(1- A r)(1- B r)},\\ Y_2(r; A, B)&= \dfrac{2 \big (\sqrt{(1-B)(1+B r^2)(1-A)(1+ A r^2)}-(1- A B r^2)\big )}{(A-B)(1- r^2)}+ \dfrac{A+B}{A-B}, \end{aligned}$$

and $\hat{r}^{**}$ is the unique zero of the polynomial $\hat{g}(r;A,B):= A B r^4- 2 A B r^3 -(2 A + 2 B + A B + 1)r^2 - 2 r +1$ in (0, 1]. The bound is sharp.

Proof of Theorem 3.1

Since $f \in \mathcal {S}^*[A,B]$, $p(z):= z f'(z)/f(z) \in \mathcal {P}[A,B]$. Since $F(z)= z/p(z)$, we have

$$\begin{aligned} \dfrac{z F'(z)}{F(z)}= 1- \dfrac{z p'(z)}{p(z)}. \end{aligned}$$

By Lemma 3.2, it follows that

$$\begin{aligned} {{\mathrm{Re\,}}}\dfrac{z F'(z)}{F(z)} \ge {\left\{ \begin{array}{ll} J_1(r; A, B), &{}0<r \le \hat{r}^{**};\\ J_2(r;A,B), &{} \hat{r}^{**} < r < 1, \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} J_1(r; A, B)&:= \dfrac{A B r^2 + 2 B r +1}{(1+ A r)(1+ B r)}, \\ J_2(r; A, B)&:= \dfrac{-2 B}{A-B}+ \dfrac{2\big (\sqrt{(1+B)(1- B r^2)(1+A)(1-Ar^2)}- (1- A B r^2)\big )}{(A-B)(1-r^2)}. \end{aligned}$$

Note that $J_1(r; A, B)= J_2(r; A, B)$ in (0, 1] only at $\hat{r}^{**}$ and $\hat{r}^{**}=1$ only for $B =-1$.

It is easy to see that, for $-1 < B <1$,

$$\begin{aligned} J_2(r; A, B)>0 \ \text {if and only if } \ \phi (r; A, B) >0, \end{aligned}$$

(3.2)

where $\phi (r;A,B):= (A-B)\big ((1+B) - (1+A)(1+B) r^2 +(1+A) B r^4\big ).$

• (a) Let $0\le B<1$. In this case, $J_1(r; A, B)>0$ for $0<r<1$. Also, if $-1 < B <0$ and $-1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big ) \notin (0,\hat{r}^{**}]$, then $J_1(r;A,B)\ne 0$ in $(0,\hat{r}^{**}]$; since $J_1(0;A,B)> 0$, therefore $J_1(r;A,B)>0$ in $(0,\hat{r}^{**}]$.

  Now, $J_2(\hat{r}^{**}; A, B)= J_1(\hat{r}^{**}; A, B)>0$, therefore by (3.2), $\phi (\hat{r}^{**}; A, B) >0$. Also, $\phi (1; A, B) = -(A-B)^2 <0$, therefore, the bi-quadratic polynomial $\phi (r; A, B)$ has a unique zero R in $(\hat{r}^{**},1)$ given by (3.1). So, we have $\phi (r; A, B)>0$ for $r<R$. Thus, in view of (3.2), $J_2(r; A, B)>0$ for $r < R$ and hence ${{\mathrm{Re\,}}}z F'(z)/F(z)>0$ in $|z| < R$. The sharpness follows for the function $f_0:\mathbb {D} \rightarrow \mathbb {C}$ defined by

  $$\begin{aligned} \dfrac{z f_0'(z)}{f_0(z)}= \dfrac{\big (1-(1+A) d z + A z^2\big )}{\big (1-(1+B)d z + B z^2\big )}, \end{aligned}$$

  where

  $$\begin{aligned}&d = \dfrac{(1+ B r^2)s_1 - (1+ A r^2)}{r\big ((1+ B)s_1 - (1+A)\big )}, \quad s_1 = \sqrt{\frac{(1+A)(1-A r^2)}{(1+B)(1-Br^2)}}. \end{aligned}$$

• (b) Let $-1 \le B <0$. If $-1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big ) \in (0,\hat{r}^{**}]$, then $J_1(r;A,B) >0$ if $r < -1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big )$. Therefore, ${{\mathrm{Re\,}}}z F'(z)/F(z)>0$ in $|z| < -1/\big (B\big (1 + \sqrt{-(A-B)/B}\big )\big )$. Define a function $f_1: \mathbb {D} \rightarrow \mathbb {C}$ by $z f_1'(z)/f_1(z) =(1+ A z)/(1+ B z)$. The result is sharp for the function $f_1$.$\square$

The following classes are well known (for example, see [9, 13, 19, 21, 22, 25]):

• (i) $\mathcal {P}[1,-1]=\mathcal {P}$, $\mathcal {S}^*[1,-1]= \mathcal {S}^*$, $\mathcal {P}_b[1,-1]=: \mathcal {P}_b$, $\mathcal {S}^*_b[1,-1] =: \mathcal {S}^*_b;$

• (ii) for $0 \le \alpha <1$, $\mathcal {P}[1-2 \alpha ,-1]=:\mathcal {P}(\alpha )$, $\mathcal {S}^*[1-2 \alpha ,-1]=: \mathcal {S}^*(\alpha )$, $\mathcal {P}_b[1-2 \alpha ,-1]=: \mathcal {P}_b(\alpha )$, $\mathcal {S}^*_b[1-2 \alpha ,-1] =: \mathcal {S}^*_b(\alpha );$

• (iii) for $M > 1/2$, $\mathcal {P}[1,1/M-1]=:\mathcal {P}(M)$, $\mathcal {S}^*[1,1/M-1]=: \mathcal {S}^*(M)$, $\mathcal {P}_b[1,1/M-1]=: \mathcal {P}_b(M)$, $\mathcal {S}^*_b[1,1/M-1] =: \mathcal {S}^*_b(M);$

• (iv) for $0 < \beta \le 1$, $\mathcal {P}[\beta , -\beta ]=:\mathcal {P}^{(\beta )}$, $\mathcal {S}^*[\beta ,-\beta ]=: \mathcal {S}^{*^{(\beta )}}$, $\mathcal {P}_b[\beta ,-\beta ]=: \mathcal {P}_b^{(\beta )}$, $\mathcal {S}^*_b[\beta ,-\beta ] =: \mathcal {S}^{*^{(\beta )}}_b;$ and

• (v) for $0 \le \beta < 1$, $\mathcal {P}[1- \beta ,0]=:\mathcal {P}_{1-\beta }$, $\mathcal {S}^*[1- \beta ,0]=: \mathcal {S}^{*}_{1-\beta }$, $\mathcal {P}_b[1-\beta ,0]=: \mathcal {P}_{b,{1-\beta }}$, $\mathcal {S}^*_b[1-\beta ,0] =: \mathcal {S}^{*}_{b,1-\beta }.$

For some of these classes, we have the following results.

Corollary 3.3

Let $F(z):= f(z)/f'(z), z \in \mathbb {D}$. Then,

• (a) the radius of starlikeness of F is

  • (i) $1/\big (1 + \sqrt{2(1-\alpha )}\big )$ if $f \in \mathcal {S}^*(\alpha )$, $(0 \le \alpha < 1)$

  • (ii) $1/\sqrt{2-\beta }$ if $f \in \mathcal {S}^*_{1-\beta }$, $(0 \le \beta < 1)$;

• (b) if $f \in \mathcal {S}^{*{(\beta )}}$, $(0< \beta \le 1)$, the radius of starlikeness of F is

  $$\begin{aligned} {\left\{ \begin{array}{ll} \sqrt{\dfrac{2 \sqrt{1- \beta }}{\sqrt{1- \beta } (1+ \beta ) + \sqrt{(1+ \beta )(1+ 4 \beta - {\beta }^2)}}}, &{}0 < \beta < 0.800199,\\ \dfrac{\sqrt{2}-1}{\beta }, &{}\beta \ge 0.800199; \end{array}\right. } \end{aligned}$$

• (c) if $f \in \mathcal {S}^{*}(M)$, $(M > 1/2)$, the radius of starlikeness of F is

  $$\begin{aligned} {\left\{ \begin{array}{ll} \sqrt{\dfrac{1}{1 + \sqrt{2 M -1}}}, &{}1/2 < M < 5,\\ \dfrac{M}{\sqrt{(2 M-1)(M-1)}+M-1 }, &{}M \ge 5 ~and \end{array}\right. } \end{aligned}$$

• (d) if $f \in \mathcal {G}(\alpha )$, $(0 < \alpha \le 1)$, then ${{\mathrm{Re\,}}}(zF'(z)/F(z))>0$ in

  $$\begin{aligned} |z| < \sqrt{\big (2 + \alpha - \sqrt{\alpha (2 + \alpha )}\big )/2}. \end{aligned}$$

Proof

Clearly, (a)–(c) follow from the Theorem 3.1 by substituting the values of A and B. By [11, Theorem 1, p. 70], $\mathcal {G}(\alpha ) \subset \mathcal {S}^*[1,1/(1+\alpha )]$, and hence part (d) follows. $\square$

The part (d) of the above corollary provides an improved version of [17, Theorem 3.10, p. 138].

Theorem 3.4

If $f \in \mathcal {S}^*[A,B]$, $F(z):= f(z)/f'(z)$ $(z \in \mathbb {D})$ and $0<r=|z|<1$, then the $\mathcal {U}$-radius of F is

• (a) $R_1$ if either of the following holds:

  • (i) $0 \le B < 0.8$,

  • (ii) $0.8 \le B < 1$, $R_1 \in (0,r_0]$,

  • (iii) $0.8 \le B < 1$, $R_1 \notin (0,r_0]$ and $G_3(r;A,B)$ has no zero in $[r_0,\tilde{r}_0]$;

• (b) $R_3$ if $0.8 \le B <1$, $R_1 \notin (0,r_0]$ and $G_3(r;A,B)$ has a zero in $[r_0,\tilde{r}_0]$;

• (c) $R_2$ if either of the following holds:

  • (i) $-0.8 < B \le 0$,

  • (ii) $-1 < B \le -0.8$, $R_2 \in (0,r_0]$,

  • (iii) $-1 < B \le -0.8$, $R_2 \notin (0,r_0]$ and $G_4(r;A,B)$ has no zero in $[r_0,\tilde{r}_0]$;

• (d) $R_4$ if either of the following holds:

  • (i) $-1 < B \le -0.8$, $R_2 \notin (0,r_0]$ and $G_4(r;A,B)$ has a zero in $[r_0,\tilde{r}_0]$,

  • (ii) $B=-1$,

where for $-1<B<1$,

$$\begin{aligned} R_1&:= \sqrt{\dfrac{2\sqrt{1-B}}{(1+A - 2 B) \sqrt{1-B} + \sqrt{(1+A -2 B)\big (1+ A(1-B) + B(1+ 2 B)\big )}}} \end{aligned}$$

(3.3)

and $R_2$ is given by (3.1); for $0.8 \le B <1$,

$$\begin{aligned} R_3&:= \dfrac{1}{\sqrt{B}\big (\sqrt{B}+ \sqrt{A-B}\big )} \end{aligned}$$

(3.4)

is the smallest zero in $[r_0,\tilde{r}_0]$ of the polynomial

$$\begin{aligned} G_3(r;A,B)&:= (A B - 2 B^2)r^2+ 2 B r -1. \end{aligned}$$

(3.5)

For $-1 \le B \le -0.8$,

$$\begin{aligned} R_4:= \dfrac{1}{\sqrt{-B}\big (\sqrt{-B}+ \sqrt{A-B}\big )} \end{aligned}$$

(3.6)

is the smallest zero in $[r_0,\tilde{r}_0]$ of the polynomial

$$\begin{aligned} G_4(r;A,B)&:=- A B r^2- 2 B r -1; \end{aligned}$$

(3.7)

and $r_0$, $\tilde{r}_0$ are given by the Eq. (2.2).

Proof

Let $p(z)=z f'(z)/f(z)$. Then $F(z)= z/p(z)$ and $F'(z)= (p(z)- z p'(z))/p^2(z)$. Now, $U_F(z)= F'(z)(z/F(z))^2-1=p(z)- z p'(z) -1$. By using Lemma 2.1, we have

$$\begin{aligned} |U_F(z)|\le {\left\{ \begin{array}{ll} M(r;A,B), &{}\text {either } |B| < 0.8, r \in [0,1),\\ &{}\text { or, }0.8 \le |B| <1, r \in (0, r_0) \cup (\tilde{r}_0,1),\\ \dfrac{|B|(A-B) r^2}{(1- |B|r)^2}, &{}0.8 \le |B| \le 1, r \in [r_0, \tilde{r}_0], \end{array}\right. } \end{aligned}$$

where M(r; A, B) and $r_0$, $\tilde{r_0}$ are given by (2.1) and (2.2), respectively. Now,

$$\begin{aligned} M(r;A,B) <1 \text { if and only if } {\left\{ \begin{array}{ll} G_1(r;A,B)<0,&{}0 \le B<1,\\ G_2(r;A,B)<0,&{}-1< B \le 0, \end{array}\right. } \end{aligned}$$

(3.8)

where

$$\begin{aligned} G_1(r;A,B)&:= -1 + B -(1+A -2 B)(-1 +B) r^2 + B(1 + A - 2 B) r^4 \end{aligned}$$

and

$$\begin{aligned} G_2(r;A,B)&:= -1 - B + (1+A)(1+B) r^2 - B(1 + A) r^4. \end{aligned}$$

Also,

$$\begin{aligned} \dfrac{|B|(A-B) r^2}{(1- |B|r)^2} <1 \text { if and only if } {\left\{ \begin{array}{ll} G_3(r;A,B)<0,&{}0<B<1,\\ G_4(r;A,B)<0,&{}-1 \le B<0, \end{array}\right. } \end{aligned}$$

(3.9)

where $G_3(r;A,B)$ and $G_4(r;A,B)$ are given by Eqs. (3.5) and (3.7), respectively.

Since for $0\le B<1$, $G_1(0;A,B)= -(1-B)<0$ and $G_1(1;A,B)= A-B >0$, the bi-quadratic polynomial $G_1(r;A,B)$ has a unique zero $R_1$ in (0, 1) given by (3.3). Similarly, for $-1<B\le 0$, the bi-quadratic polynomial $G_2(r;A,B)$ has a unique zero $R_2$ in (0, 1) given by (3.1).

• (a) Clearly, if either $0 \le B < 0.8$ or $0.8 \le B <1$, $R_1 \in (0,r_0]$, then in view of (3.8), $M(r;A,B) <1$ for $r<R_1$.

  Let $0.8 \le B <1$, $R_1 \notin (0,r_0]$ and $G_3(r;A,B)$ has no zero in $[r_0,\tilde{r}_0]$.

  In view of (2.7) and (2.8), for $B>0$, $R_0 =a-d$ if and only if $\phi (r,B):= B r^2 - B r + 1- B =0$. Therefore, for $B>0$, $S(R_0,0)=:M(r;A,B)$ and $S(a-d,0)= B(A-B)r^2/(1- B r)^2$ are equal at the zeros $r=r_0$ and $r=\tilde{r}_0$ in (0, 1) of $\phi (r,B)$, where S, $R_0$ and $r_0$, $\tilde{r}_0$ are given by (2.5), (2.6) and (2.2), respectively. Hence, in view of (3.8) and (3.9), we have $G_1(r_0;A,B)= G_3(r_0;A,B)$ and $G_1(\tilde{r}_0;A,B)= G_3(\tilde{r}_0;A,B)$.

  Since $G_1(r;A,B)$ has no zero in $(0,r_0]$ and $G_1(0;A,B)= -(1-B)<0$, therefore $G_1(r_0;A,B)<0$ which implies $G_3(r_0;A,B)<0$.

  Further, since $G_3(r;A,B)$ has no zero in $[r_0,\tilde{r}_0]$, we have $G_3(r;A,B)<0$ in $[r_0,\tilde{r}_0]$ and so $G_1(\tilde{r}_0;A,B) = G_3(\tilde{r}_0;A,B)<0$. Thus, $G_1(r;A,B)<0$ for $r < R_1 \in (\tilde{r}_0,1)$ and hence, in view of (3.8), we have $M(r;A,B) <1$ for $r<R_1$.

• (b) Let $0.8 \le B <1$, $R_1 \notin (0,r_0]$ and $G_3(r;A,B)$ has a zero in $[r_0,\tilde{r}_0]$. As in part (a), $G_3(r_0;A,B)<0$ and thus $G_3(r;A,B)<0$ for $r < R_3$, where $R_3$ is the smallest zero of $G_3(r;A,B)$ in $[r_0,\tilde{r}_0]$ given by (3.4).

• (c) If either $-0.8 < B \le 0$, or $-1 < B \le -0.8$, $R_2 \in (0,r_0]$, then in view of (3.8), $M(r;A,B) <1$ for $r<R_2$.

  Let $-1 < B \le -0.8$, $R_2 \notin (0,r_0]$ and $G_4(r;A,B)$ has no zero in $[r_0,\tilde{r}_0]$. Similarly as in part (a), in view of (2.7) and (2.8), for $B<0$, $R_0 =a+d$ if and only if $\psi (r,B):= B r^2 - B r -1 -B =0$. Therefore, for $B<0$, $M(r,A,B) = -B(A-B)r^2/(1+ B r)^2$ at the zeros $r=r_0$ and $r=\tilde{r}_0$ in (0, 1) of $\psi (r,B)$, where $r_0$, $\tilde{r}_0$ are given by (2.2) and hence, in view of (3.8) and (3.9), we have $G_2(r;A,B)= G_4(r;A,B)$ at $r=r_0$ and $r=\tilde{r}_0$. Since $G_2(r;A,B)$ has no zero in $(0,r_0]$ and $G_2(0;A,B)= -(1+B)<0$, therefore $G_2(r_0;A,B)<0$ and hence $G_4(r_0;A,B)<0$. Further, since $G_4(r;A,B)$ has no zero in $[r_0,\tilde{r}_0]$, we have $G_4(r;A,B)<0$ in $[r_0,\tilde{r}_0]$ which implies $G_2(\tilde{r}_0;A,B) = G_4(\tilde{r}_0;A,B)<0$. Thus, in view of (3.8), $M(r;A,B) <1$ for $r<R_2 \in (\tilde{r}_0,1)$.

• (d) Let $-1 < B \le -0.8$, $R_2 \notin (0,r_0]$ and $G_4(r;A,B)$ has a zero in $[r_0,\tilde{r}_0]$. As in part (c), $G_4(r_0;A,B)<0$ and hence $G_4(r;A,B)<0$ for $r < R_4$, where $R_4$ is the smallest zero of $G_4(r;A,B)$ in $[r_0,\tilde{r}_0]$, given by (3.6). Also, for $B=-1$, $G_4(r;A,B)$ has a unique zero in (0, 1) namely $1/\big (1+ \sqrt{1+A}\big )$ and hence $G_4(r;A,B)<0$ for $r < 1/\big (1+ \sqrt{1+A}\big )$.

In the cases (a) and (c), the sharpness follows for the functions $f_1: \mathbb {D} \rightarrow \mathbb {C}$ and $f_2: \mathbb {D} \rightarrow \mathbb {C}$, respectively defined by $z f_i'(z)/f_i(z)= \big (1+ A w_i(z)\big )/\big (1+B w_i(z)\big ), (i=1,2)$, where $w_i(z)= (-1)^i z (z-c_i)/(1- c_i z), (i=1,2)$ with $c_i$ being determined from the Eq. (2.9). In the cases (b) and (d), the result is sharp for the function $f_3: \mathbb {D} \rightarrow \mathbb {C}$ defined by $z f_3'(z)/f_3(z) = (1+ A z)/(1+ B z)$. $\square$

Corollary 3.5

Let $F(z):= f(z)/f'(z), z \in \mathbb {D}$. Then

• (a) the $\mathcal {U}$-radius of F is

  • (i) $1/\big (1+ \sqrt{2(1-\alpha )}\big )$ if $f \in \mathcal {S}^*(\alpha )$, $(0 \le \alpha < 1)$,

  • (ii) $1/\sqrt{2 -\beta }$ if $f \in \mathcal {S}^*_{1-\beta }$, $(0 \le \beta < 1)$.

• (b) if $f \in \mathcal {S}^{*{(\beta )}}$ $(0< \beta \le 1)$, the $\mathcal {U}$-radius of F is

  $$\begin{aligned} {\left\{ \begin{array}{ll} \sqrt{\dfrac{2 \sqrt{1- \beta }}{\sqrt{1- \beta } (1+ \beta ) + \sqrt{(1+ \beta )(1+ 4 \beta - {\beta }^2)}}}, &{}0 < \beta < 0.800199,\\ \dfrac{\sqrt{2}-1}{\beta }, &{}\beta \ge 0.800199. \end{array}\right. } \end{aligned}$$

• (c) if $f \in \mathcal {G}(\alpha )$, $(0 < \alpha \le 1)$, $F \in \mathcal {U}$ in

  $$\begin{aligned} |z| < {\left\{ \begin{array}{ll} \dfrac{(1+ \alpha )}{(1+ \sqrt{\alpha })}, &{}0 < \alpha \le 0.141441,\\ \sqrt{\dfrac{-\alpha + \sqrt{(1+ \alpha )^2 +1}}{2}}, &{}0.141441 < \alpha \le 1. \end{array}\right. } \end{aligned}$$

• (d) if $f \in \mathcal {S}^{*}(M)$, $(M > 1/2)$, the $\mathcal {U}$-radius of F is

  $$\begin{aligned} {\left\{ \begin{array}{ll} \dfrac{M}{\sqrt{(2 M-1)(1-M)} + 1- M}, &{} 0.5 < M \le 0.533025 ,\\ \sqrt{\dfrac{M}{2M-1+ \sqrt{1- 2 M +2 M^2}}}, &{} 0.533025 < M \le 1 ,\\ \sqrt{\dfrac{1}{1 + \sqrt{2 M -1}}}, &{}1 \le M \le 5,\\ \dfrac{ M }{\sqrt{(2 M -1)(M-1)} + M-1}, &{}M > 5. \end{array}\right. } \end{aligned}$$

Proof

Clearly, (a), (b) and (d) follow from the Theorem 3.4 by substituting the values of A and B. By [11, Theorem 1, p. 70], $\mathcal {G}(\alpha ) \subset \mathcal {S}^*[1,1/(1+\alpha )]$, and hence part (c) follows. $\square$

The part (c) of the above corollary provides an improved version of [17, Theorem 3.14, p. 139], while (i) of part (a) is the same as [17, Theorem 3.3, p. 135].

Corollary 3.6

Let $f \in \mathcal {S}^*(\alpha )$. Then the radius of univalence of $F(z):=f(z)/f'(z)$ $(z \in \mathbb {D})$ is $1/\big (1+ \sqrt{2(1-\alpha )}\big )$.

Proof

Since $f \in \mathcal {S}^*(\alpha )$, by (i) of part (a) of the Corollary 3.5, the $\mathcal {U}$-radius of F is $1/(1+ \sqrt{2(1-\alpha )})$. Also, [18, Theorem 2, p. 394] gives $\mathcal {U} \subset \mathcal {S}$, and therefore, $F \in \mathcal {S}$ in $|z| < 1/\big (1+ \sqrt{2(1-\alpha )}\big )$. The sharpness follows for the function $f_0: \mathbb {D} \rightarrow \mathbb {C}$ defined by

$$\begin{aligned} \dfrac{z f_0'(z)}{f_0(z)} = \dfrac{1 + (1- 2 \alpha ) z}{1- z}. \end{aligned}$$

Corresponding to the function $f_0$, we have $F_0(z)= f_0(z)/f_0'(z)= z(1- z)/(1+ (1- 2 \alpha ) z)$. Clearly, $F_0'(1/(1+ \sqrt{2(1-\alpha )}))= 0$, and thus $F_0$ is not univalent in $|z|<r$, $r > 1/(1+ \sqrt{2(1-\alpha )}).$

Note that the above corollary can also be obtained by an application of Lemma 2.2 and the fact that if $p \in \mathcal {P}[A,B]$ $(-1\le B < A \le 1)$, then $q(z):= 1/p(z) \in \mathcal {P}[-B,-A]$ $(-1\le -A < -B \le 1)$. Precisely, for $f \in \mathcal {S}^*(\alpha )$, we have $F(z)= z q(z)$, $q(z):= f(z)/(z f'(z)) \in \mathcal {P}[1,-(1-2 \alpha )]$ and by Lemma 2.2, for $r=|z|<1$,

$$\begin{aligned} {{{\mathrm{Re\,}}}}\! F'(z)\ge {{{\mathrm{Re\,}}}}\! q(z) - |z q'(z)| \ge {{{\mathrm{Re\,}}}}\! q(z) \left( \dfrac{1- 2 r -(1-2 \alpha )r^2}{(1-r)(1+(1-2 \alpha )r)}\right) >0 \end{aligned}$$

if $r < 1/\big (1+ \sqrt{2(1-\alpha )}\big )$.

Theorem 3.7

If $f \in \mathcal {A }$ satisfies $F:= f/f' \in \mathcal {S}^*[A,B]$, $-1 \le B < A \le 1$, then for $z \in \mathbb {D}$,

$$\begin{aligned} \dfrac{z f'(z)}{f(z)} \prec {\left\{ \begin{array}{ll} (1+ Bz)^{-(A-B)/B}, &{}B \ne 0,\\ e^{- A z}, &{}B=0. \end{array}\right. } \end{aligned}$$

Proof

Let $p(z):= zf'(z)/f(z)$. Then $F(z)= z/p(z)$ and $z F'(z)/F(z) = 1- zp'(z)/p(z)$. Since $F \in \mathcal {S}^*[A,B]$, we have

$$\begin{aligned} -\dfrac{z p'(z)}{(A-B)p(z)} \prec \dfrac{z}{1+ B z}. \end{aligned}$$

Also, since $-1 \le B < A \le 1$,

$$\begin{aligned} g(z):= \int _0^z \dfrac{1}{1+ B t} \, dt= {\left\{ \begin{array}{ll} \big (1/B\big ) \log {(1+ B z)}, &{}B \ne 0,\\ z, &{}B=0 \end{array}\right. } \end{aligned}$$

(3.10)

is analytic in $\mathbb {D}$ such that $g(0)=0$ and $g'(0)=1$. Clearly, for $B=0$, g is a convex function. Let $B \ne 0$. Then

$$\begin{aligned} 1+ \dfrac{z g''(z)}{g'(z)} = \dfrac{1}{1+ B z}. \end{aligned}$$

For $B =-1$, ${{\mathrm{Re\,}}}(1/(1+ B z)) = {{\mathrm{Re\,}}}( 1/(1-z)) >1/2 >0.$ For $-1 < B <1$ $(B \ne 0)$, the image of $\mathbb {D}$ under $w(z):=1/(1+ B z)$ is the disc $|w - 1/(1- B^2)| < |B|/(1- B^2)$, and therefore ${{\mathrm{Re\,}}}(1/(1+ B z)) > 1/(1+ |B|) >0$ and hence g is a convex function.

By using [24, Theorem 3, p. 777], we conclude that

$$\begin{aligned} \left( -\dfrac{1}{A-B}\right) \log {p(z)}= \left( -\dfrac{1}{A-B}\right) \int _0^z \dfrac{p'(t)}{p(t)}\,dt \prec \int _0^z \dfrac{1}{1+ B t} \, dt. \end{aligned}$$

By using (3.10), the required result follows. $\square$

Since, for $B \ne 0$,

$$\begin{aligned} {{\mathrm{Re\,}}}\left( \dfrac{1}{1+B z}\right) ^{(A-B)/B}= \left| \dfrac{1}{1+B z}\right| ^{(A-B)/B} \cos {\left( \dfrac{(A-B)}{B} \arg {\dfrac{1}{1+ B z}}\right) } \end{aligned}$$

and $\arg {(1/(1+Bz))} \in (-\pi /2,\pi /2)$, we have the following simple deduction:

Corollary 3.8

If $f \in \mathcal {A }$ satisfies $F:= f/f' \in \mathcal {S}^*[A,B]$, $-1 \le B < A \le 1$, and $0<A/B<2$, then $f \in \mathcal {S}^*$.

Note that if $p \in \mathcal {P}[A,B]$ $(-1\le B < A \le 1)$, then $q(z):= 1/p(z) \in \mathcal {P}[-B,-A]$ $(-1\le -A < -B \le 1)$. Using this fact, we prove the following:

Theorem 3.9

If $f \in \mathcal {S}^*[A,B]$ and $F(z):= f(z)/f'(z) = z + \sum _{n=2}^\infty A_n z^n$, $z \in \mathbb {D}$ then for all n, $|A_n| \le A-B$. The result is sharp.

Proof

Since $f \in \mathcal {S}^*[A,B]$, $p(z):= z f'(z) /f(z) \in \mathcal {P}[A,B]$. Now, $F(z) = z P(z)$, where $P(z) = 1/p(z) \in \mathcal {P}[-B,-A]$. Let $P(z) = 1+ c_1 z + c_2 z^2 + \cdots$, then for each $n \ge 2$, $A_n = c_{n-1}$. By using [6, Theorem 5, p. 742], $|c_n| \le A-B$ for all $n \ge 1$, and so $|A_n| \le A-B$. The result is sharp for the function $f_n: \mathbb {D} \rightarrow \mathbb {C}$ satisfying

$$\begin{aligned} \dfrac{z f_n'(z)}{f_n(z)} =\dfrac{1+ A z^{n-1}}{1+ B z^{n-1}}. \end{aligned}$$

Corresponding to the function $f_n$, the function $F_n(z)= f_n(z)/f_n'(z)$ is given by

$$\begin{aligned} F_n(z) = \dfrac{z(1+ B z^{n-1})}{1+A z^{n-1}} = z - (A-B) z^n + (A^2-A B)z^{2 n -1} + \cdots .\qquad \qquad \square \end{aligned}$$

4 Radius of Starlikeness of $F:=f/f'$ with Fixed Second Coefficient

The second coefficient of univalent functions gives information about the function; indeed, the bound $|a_2|\le 2$ for $f\in \mathcal {S}$ gives the growth and distortion estimates of these functions. Therefore, the study of functions with fixed initial coefficients plays an important role in the function theory. See [1] for a brief history on functions with fixed initial coefficients. In this section, we first generalize Theorem 3.1 for $f \in \mathcal {S}_b^*[A,B]$. In fact, for $b=1$, the following theorem reduces to Theorem 3.1 but we kept the proof of Theorem 3.1 for the sake of explicitness.

Theorem 4.1

Let $f \in \mathcal {S}_b^*[A,B]$, $F(z):= f(z)/f'(z) (z \in \mathbb {D})$ and $0 < r=|z| <1$. Then the radius of starlikeness of F is given by

$$\begin{aligned} r_{\mathcal {S}_b^*} = {\left\{ \begin{array}{ll} R_b, &{}\phi _b(r;A,B) \text { has a zero in } (0,\hat{r_b}],\\ R, &{}\phi _b(r;A,B) \text { has no zero in } (0,\hat{r_b}], \end{array}\right. } \end{aligned}$$

where $\hat{r_b}$ is the unique zero of the polynomial

$$\begin{aligned} G_b(r;A,B):=&-1+\big (3 + A + B + b^2 (1+A)(1+B)\big )r^2+ 4 b (1+A) (1+B) r^3 \nonumber \\&+ \big (A+ B+ 3A B + b^2 (1+A)(1+B)\big )r^4-A B r^6 \end{aligned}$$

(4.1)

in (0, 1]; R , given by (3.1), is the unique zero of the polynomial

$$\begin{aligned} \tilde{\phi }(r;A,B)&:= (1+B) -(1+A)(1+B) r^2 + (1+A)B r^4 \end{aligned}$$

(4.2)

in $( \hat{r_b},1)$; and $R_b$ is the smallest zero of

$$\begin{aligned} \phi _b(r;A,B)&:= 1+ 2b(1+B) r +\big (-A + 3 B + b^2(1+A)(1+B)\big )r^2 \nonumber \\&\qquad + 2 b (1+A)B r^3 + A B r^4 \end{aligned}$$

(4.3)

in $(0,\hat{r_b}]$.

The proof of our result requires the following lemma.

Lemma 4.2

[26, Theorem 2.1, p. 305] Let $\alpha \ge 0, \beta \ge 0$. If $p \in \mathcal {P}_b[A,B]$, then for $|z|=r <1$, we have

where $R_1 = (L_1/K_1)^{1/2}, R_2' = (1- A C)/(1- B C), L_1= \beta (1-A)(1+A r^2), K_1=\alpha (A-B)(1-r^2)+ \beta (1-B) (1+B r^2) \text { and } C = r(r+b)/(1+b r)$. The result is sharp.

Proof of Theorem 4.1

Since $f \in \mathcal {S}_b^*[A,B]$, $p(z) = z f'(z)/f(z) \in \mathcal {P}_b[A,B]$. Also, $F(z) = z/p(z)$ which gives

$$\begin{aligned} \dfrac{z F'(z)}{F(z)}=1- \dfrac{z p'(z)}{p(z)}. \end{aligned}$$

An application of Lemma 4.2 to $q(z):=1/p(z)$ and use of the fact that if $p \in \mathcal {P}[A,B]$ $(-1\le B < A \le 1)$ then $1/p \in \mathcal {P}[-B,-A]$ $(-1\le -A < -B \le 1)$ and $zq'(z)/q(z)= -z p'(z)/p(z)$ gives

$$\begin{aligned} {{\mathrm{Re\,}}}\dfrac{z F'(z)}{F(z)}\ge {\left\{ \begin{array}{ll} X_b(r;A,B), &{}G_b(r;A,B) \le 0,\\ Y(r;A,B), &{}G_b(r;A,B) \ge 0, \end{array}\right. } \end{aligned}$$

(4.4)

where

$$\begin{aligned} X_b(r;A,B)&= \dfrac{\phi _b(r;A,B)}{\big (1+ r(b + A b + A r)\big )\big (1+ r(b + B b + B r)\big )},\end{aligned}$$

(4.5)

$$\begin{aligned} Y(r;A,B)&= \dfrac{-2\big (1+B -(1+A) B r^2 - \sqrt{(1+A)(1+B)(1-A r^2)(1-Br^2)}\big )}{(A-B)(1-r^2)}, \end{aligned}$$

(4.6)

$G_b(r;A,B)$ and $\phi _b(r;A,B)$ are given by (4.1) and (4.3), respectively. We first show that for $B \ne -1$, $G_b(r;A,B)$ is a strictly increasing function of r in (0, 1). For this, it is sufficient to show that $G_b'(r;A,B) >0$ in (0, 1). Let $B \ne -1$. Consider

$$\begin{aligned} G_b'(r;A,B)&=2 r \Big (3 + A + B + b^2(1+A)(1+B) + 6 b(1+A)(1+B) r \nonumber \\&\quad +\, 2\big (A + B + 3 A B + b^2(1+A)(1+B)\big )r^2 - 3 A B r^4\Big ) \nonumber \\&\ge 2 r I_b(r;A,B), \end{aligned}$$

(4.7)

where $I_b(r;A,B):= 3 + A + B + b^2(1+A)(1+B)+ 2\big (A + B + 3 A B + b^2(1+A)(1+B)\big )r^2 - 3 A B r^4.$

If $B=0$ then $A>0$ and consequently, $I_b(r;A,0)= 3 + A + b^2(1+A) + 2\big (A+ b^2(1+A)\big )r^2 >0$, hence $G_b'(r;A,0)>0$. For $A \ne 0$, $B \ne 0$, $I_b'(r;A,B) =0$ implies $r=0, \pm r_0$, where

$$\begin{aligned} r_0=\sqrt{\dfrac{A+B+3 A B + b^2(1+B)(1+A)}{3 A B}}. \end{aligned}$$

Clearly,

$$\begin{aligned} I_b(0;A,B)&= 3+ A + B + b^2(1+A)(1+B)>0 \end{aligned}$$

(4.8)

and

$$\begin{aligned} I_b(1;A,B)&= 3(1+A)(1+B)(1+b^2)>0. \end{aligned}$$

(4.9)

If $B>0$ then $I_b'(r;A,B) = 4 r (A + B + b^2(1+A)(1+B)+ 3 A B(1- r^2))>0$. Therefore, in this case, $I_b(r;A,B)$ is an increasing function of r in (0, 1) and hence, in view of (4.7) and (4.8), we have $G_b'(r;A,B) \ge 2 r I_b(0;A,B)>0$.

If $-1 < B <0$ and $A<0$ then $AB>0$. In this case, $r_0 \in (0,1)$ if and only if $A+ B + 3 A B + b^2(1+A)(1+B)>0$ and $A+ B + b^2(1+A)(1+B) <0$. So, if $r_0 \in (0,1)$ then $I_b''(r_0;A,B)= -8(A+B+3 A B + b^2(1+B)(1+A))<0$ and therefore, either $I_b(r;A,B) \ge I_b(0;A,B)$ or $I_b(r;A,B) \ge I_b(1;A,B)$ and hence, in view of (4.7), (4.8) and (4.9), $G_b'(r;A,B) >0$. Also, if $r_0 \notin (0,1)$ then either $I_b(r;A,B) \ge I_b(0;A,B)$ or $I_b(r;A,B) \ge I_b(1;A,B)$ and hence in view of (4.7), (4.8) and (4.9), $G_b'(r;A,B) >0$.

If $-1< B<0$ and $A>0$, then $A B <0$. In this case, $r_0 \in (0,1)$ if and only if $A+B+3 A B + b^2(1+B)(1+A) <0$ and $A + B + b^2(1+A)(1+B) >0$. So, if $r_0 \in (0,1)$ then $I_b''(r_0;A,B)= -8(A+B+3 A B + b^2(1+B)(1+A))>0$ and hence $I_b(r;A,B) \ge \min \{I_b(r_0;A,B),I_b(0;A,B), I_b(1;A,B)\}$. Also,

$$\begin{aligned} I_b(r_0;A,B)= & {} 3 + A + B + b^2(1+A)(1+B) \nonumber \\&+ \dfrac{(A + B + 3 A B + b^2(1+A)(1+B))^2}{3 A B} \nonumber \\> & {} 2( A + B + b^2(1+A)(1+B)) + 3(1+ A B) >0. \end{aligned}$$

(4.10)

Therefore, in view of (4.7), (4.8), (4.9) and (4.10), $G_b'(r;A,B) >0$. If $r_0 \notin (0,1)$ then either $I_b(r;A,B) \ge I_b(0;A,B)$ or $I_b(r;A,B) \ge I_b(1;A,B)$ and hence, in view of (4.7), (4.8) and (4.9), $G_b'(r;A,B) >0$.

Now, let us consider the case when $A=0$ and $-1< B<0$. Then $I_b(r;0,B):= 3 + B + b^2(1+B)+ 2( B + b^2(1+B))r^2$ and $I'_b(r;0,B)= 4( B + b^2(1+B))r$ which implies that $I_b(r;0,B) \ge I_b(0;0,B) >0$ if $B + b^2(1+B) \ge 0$ and $I_b(r;0,B) \ge I_b(1;0,B) >0$ if $B + b^2(1+B) \le 0$. Therefore, in view of (4.7), $G_b'(r;0,B) >0$.

Thus, we have seen that for $B \ne -1$, $G_b(r;A,B)$ is a strictly increasing function of r in (0, 1). Also, $G_b(0;A,B)=-1<0$ and $G_b(1;A,B)=2(1+A)(1+B)(1+b^2)>0$. Therefore, $G_b(r;A,B)$ has a unique zero, say, $\hat{r_b}$ in (0, 1). Also, $G_b(r;A,-1)$ has a unique zero $r=1$ in (0, 1]. Therefore, the inequality (4.4) becomes

$$\begin{aligned} {{\mathrm{Re\,}}}\dfrac{z F'(z)}{F(z)}\ge {\left\{ \begin{array}{ll} X_b(r;A,B), &{}r \le \hat{r_b} ,\\ Y(r;A,B), &{}r \ge \hat{r_b} , \end{array}\right. } \end{aligned}$$

where $\hat{r_b}$ is the unique zero of $G_b(r;A,B)$ in (0, 1]; $X_b(r;A,B)$ and Y(r; A, B) are given by (4.5) and (4.6), respectively.

Note that $X_b( r;A,B)= Y( r;A,B)$ at $r= \hat{r_b}$ only and $\hat{r_b}=1$ only for $B=-1$.

For $B=-1$, $\phi _b'(r;A,-1)= -2 r(1+A + 2(1+Ar^2) + 3(1+A)br) <0$; $\phi _b(0;A,-1)=1>0$ and $\phi _b(1;A,-1)= -2(1+A)(1+b) <0$, therefore $\phi _b(r;A,-1)$ has a unique zero in (0, 1).

It is easy to see that, for $-1<B<1$,

$$\begin{aligned} Y(r;A,B) >0 \text { if and only if } \tilde{\phi }(r;A,B) >0, \end{aligned}$$

(4.11)

where $\tilde{\phi }(r;A,B)$ is given by (4.2).

• (i) If $\phi _b(r;A,B)$ has a zero in $(0,\hat{r_b}]$, then ${{\mathrm{Re\,}}}z F'(z)/F(z) >0$ in $|z| < R_b$ where $R_b$ is the smallest zero of $\phi _b(r;A,B)$ in $(0,\hat{r_b}]$.

• (ii) If $\phi _b(r;A,B)$ has no zero in $(0,\hat{r_b}]$, then we must have $-1<B<1$. Since $\phi _b(0;A,B) =1 >0$, therefore $\phi _b(r;A,B)>0$ in $(0,\hat{r_b}]$ and hence $X_b(r;A,B)>0$ in $(0,\hat{r_b}]$. Since $Y( \hat{r_b};A,B)= X_b( \hat{r_b};A,B)>0$, therefore by (4.11), $\tilde{\phi }(\hat{r_b};A,B)>0$. Also, $\tilde{\phi }(1;A,B)= -(A-B) <0$. Therefore, the bi-quadratic polynomial $\tilde{\phi }(r;A,B)$ has a unique zero R, given by (3.1), in $( \hat{r_b},1)$. Thus, $\tilde{\phi }(r;A,B) >0$ for $r < R$. In view of (4.11), we have $Y(r;A,B) >0$ and hence ${{\mathrm{Re\,}}}z F'(z)/F(z)>0$ in $|z| < R$.

The sharpness follows immediately from the sharpness of Lemma 4.2. $\square$

Corollary 4.3

Let $F(z):= f(z)/f'(z), z \in \mathbb {D}$. Then

• (a) the radius of starlikeness of F is

  • (i) $\sqrt{1/\big (2 - \alpha + \sqrt{5- 6 \alpha + \alpha ^2}\big )}$ if $f \in \mathcal {S}_0^*(\alpha )$, $(0 \le \alpha < 1)$,

  • (ii) $1/\sqrt{2- \beta }$ if $f \in \mathcal {S}^*_{0,1-\beta }$, $(0 \le \beta < 1)$;

• (b) if $f \in \mathcal {S}_0^{*{(\beta )}}$, $(0< \beta \le 1)$, the radius of starlikeness of F is

  $$\begin{aligned} {\left\{ \begin{array}{ll} \sqrt{\dfrac{2 \sqrt{1- \beta }}{\sqrt{1- \beta } (1+ \beta ) + \sqrt{(1+ \beta )(1+ 4 \beta - {\beta }^2)}}}, &{}0 < \beta < 0.618034,\\ \sqrt{\dfrac{\sqrt{5}-2}{\beta }}, &{}\beta \ge 0.618034; \end{array}\right. } \end{aligned}$$

• (c) if $f \in \mathcal {S}_0^{*}(M)$, $(M > 1/2)$, the radius of starlikeness of F is

  $$\begin{aligned} {\left\{ \begin{array}{ll} \sqrt{\dfrac{1}{1+ \sqrt{2 M-1}}}, &{} 0.5 < M < 2.5 ,\\ \sqrt{\dfrac{2 M}{ 4 M-3 + \sqrt{9 - 28 M + 20 M^2}}}, &{}M \ge 2.5 ~and \end{array}\right. } \end{aligned}$$

• (d) if $f \in \mathcal {G}_0(\alpha )$, $(0 < \alpha \le 1)$, then $F \in \mathcal {S}^*$ in $|z| < \sqrt{\big (2+ \alpha - \sqrt{\alpha (2 + \alpha )}\big )/2}$.

Proof

Clearly, (a)–(c) follow from the Theorem 4.1 by substituting the values of A and B. By using [11, Theorem 1, p. 70] and the definitions of $\mathcal {G}_b(\alpha )$ and $\mathcal {S}_b^*(\alpha )$, we can easily see that for $0 \le b \le 1$, $\mathcal {G}_b(\alpha ) \subset \mathcal {S}_{\frac{b(1+ \alpha )}{2}}^*[1,1/(1+\alpha )]$ and so $\mathcal {G}_0(\alpha ) \subset \mathcal {S}_0^*[1,1/(1+\alpha )]$, hence part (d) follows. $\square$

The part (i) of (a), for $\alpha =1/2$, of the above corollary provides the sharp version of [17, Corollary 3.7, p. 137], while part (d) provides an improved version of [17, Corollary 3.12, p. 139].

The following theorem gives the $\mathcal {S}^*_{b,{1-\beta }}-$radius of $F:= f/f'$ for $f \in \mathcal {S}_b^*(\alpha )$ where $0 \le b \le 1$, $0 \le \beta <1$. To prove our result, we need the following lemma.

Lemma 4.4

[13, Theorem 2, p. 213] If $p \in \mathcal {P}_b(\alpha )$, $0 \le b \le 1$, then

$$\begin{aligned} \left| \dfrac{p'(z)}{p(z)}\right| \le \dfrac{2(1- \alpha )\big (b |z|^2 + 2 |z| +b \big )}{(1- |z|^2)\big ((1-2 \alpha ) |z|^2 + 2 b(1-\alpha )|z| +1 \big )}. \end{aligned}$$

Theorem 4.5

If $f \in \mathcal {S}_b^*(\alpha )$ and $F(z):=f(z)/f'(z)$ $(z \in \mathbb {D})$, then for $r=|z|<1$, $0 \le \beta <1$,

$$\begin{aligned} \left| \frac{z F'(z)}{F(z)}-1 \right| < 1- \beta \ \text {in}\ |z|< R, \end{aligned}$$

where $0<R < 1$ is the smallest positive zero of the polynomial

$$\begin{aligned}&-1 + \beta + 2 b (1- \alpha ) \beta r + 2 (2 - \alpha - \alpha \beta ) r^2 \nonumber \\&\quad + 2 b (1- \alpha )(2- \beta ) r^3 + (1-2 \alpha ) (1-\beta ) r^4. \end{aligned}$$

(4.12)

The result is sharp.

Proof

Let $p(z) = zf'(z)/f(z)$. Then $F(z)=z/p(z)$. Since $f \in \mathcal {S}_b^*(\alpha )$, $p \in \mathcal {P}_b(\alpha )$ and thus, by using Lemma 4.4, we have

$$\begin{aligned} \left| \frac{z F'(z)}{F(z)}-1 \right| = \left| \frac{z p'(z)}{p(z)} \right| \le \frac{2(1-\alpha )(br^2 + 2 r + b) r}{(1-r^2)\big ((1-2 \alpha ) r^2 + 2(1- \alpha ) b r +1\big )} < 1- \beta \end{aligned}$$

in $|z| < R$.

Define a function $f_0$ by

$$\begin{aligned} \dfrac{z f_0'(z)}{f_0(z)}= \dfrac{1+ 2b(1-\alpha )z + (1-2 \alpha )z^2}{1-z^2}, \end{aligned}$$

then $f_0 \in \mathcal {S}_b^*(\alpha ).$ Corresponding to the function $f_0$, $F_0(z)= f_0(z)/f_0'(z)$ is given by

$$\begin{aligned} F_0(z)= \dfrac{z(1-z^2)}{1+ 2 b(1- \alpha )z + (1-2 \alpha )z^2}. \end{aligned}$$

A calculation shows that

$$\begin{aligned} \frac{z F_0'(z)}{F_0(z)}-1 = \frac{-2(1- \alpha )z(b+2z + bz^2)}{(1-z^2)(1+2b(1-\alpha )z + (1-2 \alpha ) z^2)}, \end{aligned}$$

and hence, by using (4.12) at $z=R$, we have

$$\begin{aligned} \left| \frac{z F_0'(z)}{F_0(z)}-1 \right| = \frac{2(1- \alpha )R(b+2R + bR^2)}{(1-R^2)(1+2b(1-\alpha )R + (1-2 \alpha ) R^2)}= 1- \beta . \end{aligned}$$

$\square$

Corollary 4.6

Let $F(z):=f(z)/f'(z)$, $z \in \mathbb {D}$. Then

• (i) for $f \in \mathcal {S}_0^*(\alpha )$, $|z F'(z)/F(z)-1| < 1$ in $|z| < \sqrt{1/(2- \alpha + \sqrt{5 - 6 \alpha + \alpha ^2})}.$

• (ii) for $f \in \mathcal {S}_1^*(\alpha )$, $|z F'(z)/F(z)-1| < 1$ in $|z| < 1/(1 + \sqrt{2(1- \alpha )}).$