1. Introduction
For 
$r \gt 0,$ let 
$\mathbb{D}_r:=\{z\in\mathbb{C}: |z| \lt r \}$, 
$\mathbb{D}:=\mathbb{D}_1$, 
$\overline{\mathbb{D}}:=\{z\in\mathbb{C}:|z|\leqslant 1\}$ and let 
$\mathbb{R} T:=\{z\in\mathbb{C}:|z|=1\}.$ Let 
${\mathcal H}(\mathbb{D}_r)$ denote the class of all analytic functions f in 
$\mathbb{D}_r$ and let 
$\mathcal{H}:=\mathcal{H}(\mathbb{D}).$ Then 
$f\in {\mathcal H}(\mathbb{D}_r)$ has the following representation
\begin{equation}
f(z) = \sum_{n=0}^{\infty}a_n(f) z^n,\quad z\in\mathbb{D}_r.
\end{equation} Let 
$\mathcal {A}(\mathbb{D}_r)$ be the subclass of 
${\mathcal H}(\mathbb{D}_r)$ of all f normalized by 
$f(0)=0=f'(0)-1$ and let 
${\mathcal A}:={\mathcal A}(\mathbb{D}).$ By 
${\mathcal S}$ we denote the subclass of all univalent (i.e. analytic and injective in 
$\mathbb{D}$) functions in 
${\mathcal A}$.
Given 
$\alpha\in (0,1],$ let 
${\mathcal S}^*_{\alpha}$ denote class of all functions 
$f\in {\mathcal A}$ such that
\begin{equation}
\left|\operatorname{Arg} \frac{zf'(z)}{f(z)} \right| \lt \alpha\frac{\pi}{2}, \quad z\in \mathbb{D},
\end{equation} and the so-called strongly starlike of order 
$\alpha.$ For 
$\alpha:=1$ the class 
$\mathcal{S}_1^*=:\mathcal{S}^*$ is the well-known class of starlike functions, i.e. functions f which map univalently 
$\mathbb{D}$ onto a set which is star-shaped with respect to the origin. Then, the condition (1.2) can be written as
\begin{equation*}
\operatorname{Re} \frac{zf'(z)}{f(z)} \gt 0,\quad z\in\mathbb{D}.
\end{equation*} The class of strongly starlike functions was introduced by Stankiewicz [Reference Stankiewicz18] and [Reference Stankiewicz19] and independently by Brannan and Kirwan [Reference Brannan and Kirwan1] (see also [Reference Goodman6, Vol. I, pp. 137–142]). Stankiewicz [Reference Stankiewicz19] presented an external geometrical characterization of strongly starlike functions. Brannan and Kirwan found a geometrical condition called δ-visibility which is sufficient for functions to be strongly starlike. In turn, Ma and Minda [Reference Ma and Minda15] gave the internal characterization of functions in 
$\mathcal{S}_\alpha^*$ basing on the concept of k-starlike domains. Further results regarding the geometry of strongly starlike functions were presented in [Reference Lecko13, Chapter IV], [Reference Lecko14] and [Reference Sugawa20]. Since 
$\mathcal{S}^*\subset \mathcal{S}$ (cf. [Reference Duren5, pp. 40–41]) and 
${\mathcal S}^*_{\alpha}\subset \mathcal{S}^*$ for every 
$\alpha\in(0,1]$, it follows that 
${\mathcal S}^*_{\alpha}\subset \mathcal{S}$ for every 
$\alpha\in(0,1].$
If 
$f\in\mathcal{S},$ then the inverse function 
$F:=f^{-1}$ is well-defined and analytic in 
$\mathbb{D}_{r(f)},$ where 
$r(f):=\sup(\{r \gt 0:\mathbb{D}_r\subset f(\mathbb{D})\}).$ Thus
\begin{equation}
F(w)=w+\sum_{n=2}^\infty A_nw^n,\quad w\in \mathbb{D}_{r(f)},
\end{equation} where 
$A_n:=a_n(F).$ By Koebe one-quarter theorem (e.g. [Reference Duren5, p. 31]), it follows that 
$r(f) \geqslant 1/4$ for every 
$f\in\mathcal{S}.$
For 
$f\in\mathcal{S}$ define
\begin{equation*}
F_f(z) := \frac12\log\frac{f(z)}{z}=\sum_{n=1}^\infty\gamma_n z^n,\quad z\in\mathbb{D},
\end{equation*} a logarithmic function associated with 
$f.$ The numbers 
$\gamma_n:=a_n(F_f)$ are called the logarithmic coefficients of 
$f.$ It is well-known that the logarithmic coefficients play a crucial role in Milin’s conjecture (see [Reference Milin16], [Reference Duren5, p. 155]).
Referring to the above idea, for 
$f\in\mathcal{S}$, there exists the unique function 
$F_{f^{-1}}$ analytic in 
$\mathbb{D}_{r(f)}$ such that
\begin{equation}
F_{f^{-1}}(w):=\frac12\log\frac{f^{-1}(w)}{w}=\sum_{n=1}^\infty \varGamma_n w^n,\quad w\in\mathbb{D}_{r(f)},
\end{equation} where 
$\varGamma_n:=a_n\left(F_{f^{-1}}\right)$ are logarithmic coefficients of the inverse function 
$f^{-1}.$
It follows from Equation (1.3) that (e.g. [Reference Goodman6, Vol. I, p. 57])
\begin{equation}
\begin{aligned}
&A_2=-a_2,\quad A_3=-a_3+2a_2^2\quad \mathrm{and}\quad A_4=-a_4+5a_2a_3-5a_2^3,
\end{aligned}
\end{equation} where 
$a_n:=a_n(f).$ Thus from Equation (1.4) we derive that
\begin{equation*}
\varGamma_1=\frac12A_2,\quad \varGamma_2=\frac12A_3-\frac14A_2^2,\quad \varGamma_3=\frac12A_4-\frac12A_2A_3+\frac16A_2^3,
\end{equation*}and next using Equation (1.5) we obtain
\begin{equation}
\begin{aligned}
&\varGamma_1=-\frac12a_2,\quad \varGamma_2=-\frac12a_3+\frac34a_2^2\quad \mathrm{and}\quad \varGamma_3=-\frac12a_4+2a_2a_3-\frac53a_2^3.
\end{aligned}
\end{equation} For 
$q,n\in\mathbb{N},$ the Hankel matrix 
$H_{q,n}(f)$ of 
$f\in\mathcal{A}$ of the form (1.1) is defined as
\begin{equation}
H_{q,n}(f) :=
\left[\begin{matrix}
a_{n} & a_{n+1}& \cdots & a_{n+q-1} \\
a_{n+1} & a_{n+2} & \cdots & a_{n+q}\\
\vdots & \vdots & \vdots & \vdots \\
a_{n+q-1} & a_{n+q} &\cdots & a_{n+2(q-1)}
\end{matrix}\right].
\end{equation} In recent years, there has been a great deal of attention devoted to finding bounds for the modulus of the second and third Hankel determinants 
$\det H_{2,2}(f)$ and 
$\det H_{3,1}(f)$, when f belongs to various subclasses of 
$\mathcal{A}$ (see [Reference Cho, Kowalczyk, Kwon, Lecko and Sim2, Reference Kowalczyk, Lecko and Sim10, Reference Kowalczyk, Lecko and Thomas11] for further references).
Based on these ideas, in [Reference Kowalczyk and Lecko8] and [Reference Kowalczyk and Lecko9], the authors started the study the Hankel determinant 
$\det H_{q,n}(F_f)$ whose entries are logarithmic coefficients of 
$f\in\mathcal{S},$ that is, an in Equation (1.7) are replaced by 
$\gamma_n.$ In this paper, we continue analogous research considering the Hankel determinant 
$\det H_{q,n}(F_{f^{-1}})$ whose entries are logarithmic coefficients of inverse functions, i.e. an in Equation (1.7) are now replaced by 
$\varGamma_n.$ We demonstrate the sharp estimates of
\begin{equation*}
\left|\det H_{2,1}\left(F_{f^{-1}}\right)\right|=\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|=\frac{1}{48}\left|13a_2^4-12a_3^2+12a_2a_4-12a_2^2a_3\right|
\end{equation*} in the classes 
$\mathcal{S}_\alpha^*$.
2. Preliminary lemmas
Denote by 
${\mathcal P}$ the class of analytic functions 
$p\in\mathcal{H}$ with positive real part given by
\begin{equation}
p(z)=1+\sum_{n=1}^{\infty}c_n z^n,\quad z\in \mathbb{D},
\end{equation} where 
$c_n:=a_n(p).$
In the proof of the main result, we will use the following lemma which contains the well-known formula for c 2 (see, e.g. [Reference Pommerenke17, p. 166]) and the formula for c 3 (see [Reference Cho, Kowalczyk and Lecko3, Lemma 2.4] with further remarks related to extremal functions).
Lemma 1. If 
$p \in {\mathcal P}$ is of the form (2.1), then
\begin{equation}
c_1 = 2\zeta_1,
\end{equation}
\begin{equation}
c_2 = 2\zeta_1^2 + 2(1-|\zeta_1|^2)\zeta_2
\end{equation}and
\begin{equation}
c_3 = 2\zeta_1^3+2(1-|\zeta_1|^2)(2\zeta_1-\overline{\zeta_1}\zeta_2)\zeta_2 + 2(1-|\zeta_1|^2)(1-|\zeta_2|^2)\zeta_3
\end{equation} for some 
$\zeta_1,\zeta_2, \zeta_3 \in \overline{\mathbb{D}}.$
For 
$\zeta_1 \in \mathbb{T}$, there is a unique function 
$p \in {\mathcal P}$ with c 1 as in Equation (2.2), namely,
\begin{equation*}
p(z) = \frac{1+\zeta_1 z}{1-\zeta_1 z}, \quad z\in\mathbb{D}.
\end{equation*} For 
$\zeta_1\in\mathbb{D}$ and 
$\zeta_2 \in \mathbb{T}$, there is a unique function 
$p \in {\mathcal P}$ with c 1 and c 2 as in Equations (2.2) and (2.3), namely,
\begin{equation}
p(z) = \frac{1+( \overline{\zeta}_1 \zeta_2 +\zeta_1 )z + \zeta_2 z^2}{1+( \overline{\zeta}_1 \zeta_2 -\zeta_1 )z - \zeta_2 z^2}, \quad z\in\mathbb{D}.
\end{equation}Lemma 2. ([Reference Choi, Kim and Sugawa4])
For real numbers A, B, C, let
\begin{equation*}
Y(A,B,C) := \max\left(\left\{|A+Bz+Cz^2|+1-|z|^2: z\in\overline{\mathbb{D}}\right\}\right).
\end{equation*}I. If
$AC\geqslant 0,$ then
\begin{equation*}
Y(A,B,C)=\left\{
\begin{array}{ll}
|A|+|B|+|C|, & |B|\geqslant 2(1-|C|),\\
1+|A|+\dfrac{B^2}{4(1-|C|)}, & |B| \lt 2(1-|C|).
\end{array}
\right.
\end{equation*}II. If
$AC \lt 0,$ then
\begin{equation*}
Y(A,B,C)=\left\{
\begin{array}{lll}
1-|A|+\dfrac{B^2}{4(1-|C|)}, & -4AC(C^{-2}-1)\leqslant B^2 \wedge |B| \lt 2(1-|C|), \\
1+|A|+\dfrac{B^2}{4(1+|C|)}, & B^2 \lt \min\left\{4(1+|C|)^2,-4AC(C^{-2}-1)\right\}, \\
R(A,B,C), & {\rm otherwise} ,
\end{array}
\right.
\end{equation*}where
\begin{equation*}
R(A,B,C):=\left\{
\begin{array}{lll}
|A|+|B|-|C|, & |C|(|B|+4|A|)\leqslant |AB|,\\
-|A|+|B|+|C|, & |AB|\leqslant |C|(|B|-4|A|), \\
(|C|+|A|)\sqrt{1-\dfrac{B^2}{4AC}}, & {\rm otherwise}.
\end{array}
\right.
\end{equation*}
We recall now Laguerre’s rule of counting zeros of polynomials in an interval (see [Reference Jameson7], [Reference Laguerre12], [Reference Turowicz21, pp. 19–20]). We will apply Laguerre’s algorithm in the proof of the main theorem. Given a real polynomial
\begin{equation*}
Q(u) := d_0u^n + d_1u^{n-1} + \cdots+ d_{n-1}u + d_n,\quad u\in\mathbb{R} ,\ d_0,\dots,d_n\in\mathbb{R} ,
\end{equation*} consider a finite sequence 
$(q_k), k = 0, 1,\dots, n,$ of polynomials of the form
\begin{equation*}
q_k(u) =\sum_{j=0}^k d_ju^{k-j},\quad u\in\mathbb{R} .
\end{equation*} For each 
$u_0\in\mathbb{R},$ let 
$N(Q; u_0)$ denote the number of sign changes in the sequence 
$(q_k(u_0)), k = 0, 1,\dots, n.$ Given an interval 
$I \subset \mathbb{R},$ denote by 
$Z(Q; I)$ the number of zeros of Q in I counted with their orders. Then the following theorem due to Laguerre holds.
Theorem 1. If a < b and 
$Q(a)Q(b)\neq 0,$ then
\begin{equation*}
Z(Q; (a, b)) = N(Q; a) - N(Q; b)
\end{equation*}or
\begin{equation*}N(Q; a) - N(Q; b) - Z(Q; (a, b))\end{equation*}is an even positive integer.
Note that
\begin{equation*}
q_k(0) = d_k,\quad q_k(1) =\sum_{j=0}^k d_j.
\end{equation*} Thus, when 
$[a, b] := [0, 1],$ Theorem 1 reduces to the following useful corollary.
Corollary 1. If 
$Q(0)Q(1)\neq 0,$ then
\begin{equation*}
Z(Q; (0, 1)) = N(Q; 0) - N(Q; 1)
\end{equation*}or
\begin{equation*}
N(Q; 0) - N(Q; 1) - Z(Q; (0, 1))
\end{equation*} is an even positive integer, where 
$N(Q; 0)$ and 
$N(Q; 1)$ are the numbers of sign changes in the sequence of polynomial coefficients 
$(d_k)$ and in the sequence of sums 
$(\sum_{j=0}^k d_j ),$ where 
$k = 0, 1,\dots , n,$ respectively.
3. Main result
The main result of this paper is the following.
Theorem 2. Let 
$\alpha\in(0,1].$ If 
$f\in {\mathcal{S}}^*_{\alpha},$ then
\begin{equation}
\left|\det H_{2,1}\left(F_{f^{-1}}\right)\right|=\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|\leqslant
\begin{cases}
\dfrac{1}{4}\alpha^2,&0 \lt \alpha \lt \dfrac{1}{5},\\
\dfrac{\alpha^2(15\alpha^2+5\alpha+2)}{(35\alpha^2+30\alpha+7)},& \dfrac{1}{5}\leqslant \alpha\leqslant\alpha_0,\\
\dfrac{1}{36}\alpha^2(35\alpha^2+4),& \alpha_0 \lt \alpha\leqslant 1,\\
\end{cases}
\end{equation} where 
$\alpha_0\approx 0.39059 $ is the unique root in 
$(0,1]$ of the equation
\begin{equation}
1225\alpha^4+1050\alpha^3-155\alpha^2-60\alpha-44=0.
\end{equation}All inequalities are sharp.
Proof. Let 
$f\in{\mathcal{S}}^*_{\alpha}$ be of the form (1.1). Then by Equation (1.2), there exists 
$p\in\mathcal{P}$ of the form (2.1) such that
\begin{equation}
\frac{zf'(z)}{f(z)}=(p(z))^{\alpha},\quad z\in \mathbb{D}.
\end{equation}Putting the series (1.1) and (2.1) into (3.3), by equating the coefficients we get
\begin{equation}
\begin{aligned}
a_2&=\alpha c_1,\quad a_3=\frac{1}{2}\alpha\left(c_2+\frac{3\alpha-1}{2}c_1^2\right),\\
a_4&=\frac{1}{3}\alpha \left(c_3+\frac{5\alpha-2}{2}c_1c_2+\frac{17\alpha^2-15\alpha+4}{12}c_1^3\right).
\end{aligned}
\end{equation}Hence and from Equation (1.6) we obtain
\begin{equation*}
\begin{aligned}
&\varGamma_1 =-\frac{1}{2}\alpha c_1, \quad \varGamma_2 =-\frac{1}{8}\alpha(2c_2-(3\alpha+1) c_1^2),\\
&\varGamma_3 =-\frac{1}{72}\alpha\left(12c_3-(42\alpha+12)c_1c_2+ (29\alpha^2+21\alpha+4)c_1^3\right),
\end{aligned}
\end{equation*}and therefore
\begin{equation}
\varGamma_1\varGamma_3-\varGamma_2^2
= \frac{1}{576}\alpha^2(c_1^4(35\alpha^2+30\alpha+7)-12(5\alpha+1)c_1^2c_2+48c_1c_3-36c_2^2).
\end{equation} Since both the class 
${\mathcal{S}}^*_{\alpha}$ and 
$|\det H_{2,1}\left(F_{f^{-1}}\right)|$ are rotationally invariant, without loss of generality we may assume that 
$a_2\geqslant 0,$ which in view of Equation (3.4) yields 
$ c_1 \geqslant 0,$ i.e. by Equation (2.2) that 
$\zeta_1\in[0,1].$ Thus substituting Equations (2.2)–(2.4) into Equation (3.5), we obtain
\begin{equation}
\begin{aligned}
\varGamma_1\varGamma_3-\varGamma_2^2 =
\frac{\alpha^2}{36}&\left((35\alpha^2+4)\zeta_1^4-30\alpha(1-\zeta_1^2)\zeta_1^2\zeta_2-3(1-\zeta_1^2)(\zeta_1^2+3)\zeta_2^2\right.\\
&\left.+12\zeta_1(1-\zeta_1^2)(1-|\zeta_2|^2)\zeta_3 \right)
\end{aligned}
\end{equation} for some 
$\zeta_1\in[0,1]$ and 
$\zeta_2,\zeta_3 \in \overline{\mathbb{D}}$.
A. Suppose that 
$\zeta_1=0.$ Then from Equation (3.6),
\begin{equation*}
\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|=\frac{\alpha^2}{4}|\zeta_2|^2\leqslant \frac{\alpha^2}{4}.
\end{equation*} B. Suppose that 
$\zeta_1=1.$ Then from Equation (3.6),
\begin{equation*}
\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|=\frac{1}{36}\alpha^2(35\alpha^2+4).
\end{equation*} C. Suppose that 
$\zeta_1\in(0,1).$ Since 
$\zeta_3 \in \overline{\mathbb{D}}$, from Equation (3.6) we get
\begin{equation*} \left|\varGamma_1\varGamma_3-\varGamma_2^2\right| \leqslant \frac{1}{3}\alpha^2\zeta_1(1-\zeta_1^2) \varPhi(A,B,C), \end{equation*}where
\begin{equation*} \varPhi(A,B,C) := \left| A +B\zeta_2 +C\zeta_2^2 \right| + 1 -|\zeta_2|^2, \end{equation*}with
\begin{equation*}
A:=\frac{\zeta_1^3(35\alpha^2+4)}{12(1-\zeta_1^2)},\quad B:=\frac{-5\alpha\zeta_1}{2},\quad C:=\frac{-(\zeta_1^2+3)}{4\zeta_1}.
\end{equation*}Observe that AC < 0 and therefore we apply only the part II of Lemma 2.
C1. Let’s consider the condition 
$|B| \lt 2(1-|C|),$ i.e.
\begin{equation*}
\frac{5\alpha\zeta_1}{2} \lt 2\left(1-\frac{\zeta_1^2+3}{4\zeta_1}\right).
\end{equation*}The above inequality is equivalent to
\begin{equation}
\frac{\zeta_1^2(5\alpha+1)-4\zeta_1+3}{2\zeta_1} \lt 0,
\end{equation} which is equivalent to 
$(5\alpha+1)\zeta_1^2-4\zeta_1+3 \lt 0.$ However
\begin{equation*}(5\alpha+1)\zeta_1^2-4\zeta_1+3=5\alpha\zeta_1^2+(1-\zeta_1)(3-\zeta_1) \gt 0\end{equation*} for 
$\zeta_1\in(0,1)$, which shows that the inequality (3.7) is false.
C2. Since
\begin{equation*}
-4AC\left(\frac{1}{C^2}-1\right)=-\frac{(9-\zeta_1^2)(35\alpha^2+4)\zeta_1^2}{12(\zeta_1^2+3)} \lt 0
\end{equation*} for 
$\zeta_1\in(0,1),$ we deduce that the condition 
$B^2 \lt \min\{4(1+|C|)^2,-4AC(C^{-2}-1)\}$ is equivalent to
\begin{equation}
\frac{\zeta_1^2[(10\alpha^2-1)\zeta_1^2+135\alpha^2+9]}{3(\zeta_1^2+3)} \lt 0,
\end{equation} which is equivalent to 
$(10\alpha^2-1)\zeta_1^2+135\alpha^2+9 \lt 0$ for 
$\zeta_1\in(0,1).$ However, in the case when 
$10\alpha^2-1\geqslant 0$ we have
\begin{equation*}
(10\alpha^2-1)\zeta_1^2+135\alpha^2+9\geqslant 135\alpha^2+9 \gt 0,
\end{equation*} and in the case when 
$10\alpha^2-1 \lt 0$ we have
\begin{equation*}
(10\alpha^2-1)\zeta_1^2+135\alpha^2+9\geqslant 145\alpha^2+8 \gt 0,
\end{equation*} for all 
$\zeta_1\in(0,1).$ Thus the inequality (3.8) is false.
C3. The inequality 
$|C|(|B|+4|A|)\leqslant|AB|$ is equivalent to
\begin{equation*}
\frac{(175\alpha^3-70\alpha^2+35\alpha-8)\zeta_1^4-6(35\alpha^2-5\alpha+4)\zeta_1^2-45\alpha}{24(1-\zeta_1^2)}\geqslant 0
\end{equation*}which is equivalent to
\begin{equation}
\varphi_\alpha(\zeta_1^2)\geqslant 0,
\end{equation} where for 
$t\in\mathbb{R} ,$
\begin{equation*}
\varphi_\alpha(t):=(175\alpha^3-70\alpha^2+35\alpha-8)t^2-6(35\alpha^2-5\alpha+4)t-45\alpha.
\end{equation*} Observe that the equation 
$175\alpha^3-70\alpha^2+35\alpha-8= 0$ has only one real root α 1 in 
$(0,1],$ where
\begin{equation*}
\alpha_1:=\frac{1}{105}(13769+882\sqrt{445})^{1/3}-\frac{77}{15(13769+882\sqrt{445})^{1/3}}+\frac{2}{15}\approx 0.2758,
\end{equation*} and that the inequality (3.9) is false for 
$\alpha:=\alpha_1.$ Let now 
$\alpha\in(0,1]\setminus\{\alpha_1\}.$ For φα, we have 
$\Delta:=144(525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4) \gt 0,$ which is true for all 
$\alpha\in(0,1]\setminus\{\alpha_1\}.$ Hence the square trinomial φα has two roots
\begin{equation*}
t_{1,2}:=\frac{3(35\alpha^2-5\alpha+4)\pm 6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4}}{175\alpha^3-70\alpha^2+35\alpha-8}.
\end{equation*} Note that for all 
$\alpha\in(0,1]\setminus\{\alpha_1\}$ we have 
$-6(35\alpha^2-5\alpha+4) \lt 0.$ Now for 
$\alpha\in(\alpha_1,1]$ we have 
$175\alpha^3-70\alpha^2+35\alpha-8 \gt 0.$ Hence 
$t_2 \lt 0$ because the inequality
\begin{equation*}
3(35\alpha^2-5\alpha+4)-6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4} \lt 0
\end{equation*}is equivalent to
\begin{equation*}
-45\alpha(175\alpha^3 - 70\alpha^2 + 35\alpha - 8) \lt 0,
\end{equation*} which is true for all 
$\alpha\in(\alpha_1,1].$ On the other hand, the inequality 
$t_1 \gt 1$ is equivalent to
\begin{equation}
6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4} \gt 5 (35\alpha^3 - 35 \alpha^2 + 10\alpha - 4),
\end{equation} which is evidently true for 
$\alpha\in(\alpha_1,\alpha_2],$ where 
$\alpha_2\approx 0.82155,$ since then the right hand side of Equation (3.10) is non-positive. For 
$\alpha\in(\alpha_2,1]$ by squaring both sides of Equation (3.10), we equivalently get the inequality
\begin{equation*}
(5\alpha - 8) (6125\alpha^5 - 2450\alpha^4 + 1925\alpha^3 - 560\alpha^2 + 140\alpha - 32) \lt 0
\end{equation*} which is true for 
$\alpha\in(\alpha_2,1].$ Thus we conclude that for 
$\alpha\in(\alpha_1,1]$ the inequality (3.9) is false.
Let 
$\alpha\in(0,\alpha_1).$ Then 
$175\alpha^3-70\alpha^2+35\alpha-8 \lt 0$ and therefore 
$t_1 \lt 0$ evidently. Moreover, the inequality 
$t_2 \lt 0$ is equivalent to
\begin{equation*}
3(35\alpha^2-5\alpha+4)-6\sqrt{525\alpha^4 - 175\alpha^3 + 120\alpha^2 - 20\alpha + 4} \gt 0
\end{equation*}which is equivalent to
\begin{equation*}
-45\alpha(175\alpha^3 - 70\alpha^2 + 35\alpha - 8) \gt 0,
\end{equation*} which is true for all 
$\alpha\in(0,\alpha_1).$ Thus we conclude that for 
$\alpha\in(0,\alpha_1)$ the inequality (3.9) is false.
C4. The inequality 
$|C|(|B|-4|A|)\geqslant|AB|$ is equivalent to
\begin{equation*}
\frac{(175\alpha^3+70\alpha^2+35\alpha+8)\zeta_1^4+6\zeta_1^2(35\alpha^2+5\alpha+4)-45\alpha}{24(1-\zeta_1^2)}\leqslant 0
\end{equation*}which is equivalent to
\begin{equation}
\gamma_\alpha(\zeta_1^2)\leqslant 0,
\end{equation} where for 
$t\in\mathbb{R} ,$
\begin{equation*}
\gamma_\alpha(t):=(175\alpha^3+70\alpha^2+35\alpha+8)t^2+6(35\alpha^2+5\alpha+4)t-45\alpha.
\end{equation*} For γα we have 
$\Delta:=144 (525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4) \gt 0$ for all 
$\alpha\in(0,1].$ Hence the square trinomial γα has two roots
\begin{equation*}
t_{3,4}:=\frac{-3(35\alpha^2+5\alpha+4)\pm6\sqrt{525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4}}{175\alpha^3+70\alpha^2+35\alpha+8}.
\end{equation*} Note that 
$t_4 \lt 0$ evidently. Observe now that 
$t_3 \gt 0.$ Indeed, this inequality is equivalent to
\begin{equation*}
-3(35\alpha^2+5\alpha+4)+6\sqrt{525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4} \gt 0
\end{equation*}which is equivalent to the evidently true inequality
\begin{equation*}
45\alpha(175\alpha^3 + 70\alpha^2 + 35\alpha + 8) \gt 0,\quad \alpha\in(0,1].
\end{equation*} Moreover, 
$t_3 \lt 1$ is equivalent to
\begin{equation*}
6\sqrt{525\alpha^4 + 175\alpha^3 + 120\alpha^2 + 20 \alpha + 4} \lt 5 (35\alpha^3 + 35\alpha^2 + 10\alpha + 4)
\end{equation*}that after squaring both sides is equivalent to
\begin{equation*}
(5\alpha + 8) (6125\alpha^5 + 2450\alpha^4 + 1925\alpha^3 + 560 \alpha^2 + 140\alpha + 32) \gt 0,
\end{equation*} which is true for all 
$\alpha\in(0,1].$ Therefore the inequality (3.11) is true for 
$\zeta_1\in(0,\zeta_1^0],$ where 
$\zeta_1^0:=\sqrt{t_3}.$
Applying Lemma 2 for 
$0 \lt \zeta_1\leqslant \zeta_1^0$, we get
\begin{equation*}
\left|\varGamma_1\varGamma_3-\varGamma_2^2\right| \leqslant\frac{1}{3}\alpha^2\zeta_1(1-\zeta_1^2)(-|A|+|B|+|C|)=\rho_\alpha(\zeta_1),
\end{equation*}where
\begin{equation*}
\rho_\alpha(t):= -\frac{1}{36}\alpha^2((35\alpha^2+30\alpha+7)t^4-6(5\alpha-1)t^2-9),\quad t\in\mathbb{R} .
\end{equation*}We have
\begin{equation*}
\rho_\alpha(0)=\frac{1}{4}\alpha^2
\end{equation*}and
\begin{equation*}
\begin{aligned}
\rho_\alpha(\zeta_1^0)=&\frac{2\alpha^2}{(175\alpha^3+70\alpha^2+35\alpha+8)^2}\\
&\times \left[-18375\alpha^6-16625\alpha^5-10150\alpha^4-3775\alpha^3 -1025\alpha^2-150\alpha-12\right.\\
&\left.+(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\right.\\
&\left.\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\right].
\end{aligned}
\end{equation*} Note that for 
$\alpha\in(0,1/5]$ the equation
\begin{equation}
\rho_\alpha'(t)=-\frac{1}{9}\alpha^2t((35\alpha^2+30\alpha+7)t^2-3(5\alpha-1))=0
\end{equation} has no root in 
$(0,\zeta_1^0)$ and then evidently
\begin{equation*}
\rho_\alpha(t)\leqslant \rho_\alpha(0)=\frac{1}{4}\alpha^2,\quad 0\leqslant t\leqslant \zeta_1^0.
\end{equation*} For 
$\alpha\in(1/5,1]$, Equation (3.12) has a unique positive root, namely
\begin{equation}
t_5:=\sqrt{\frac{3(5\alpha-1)}{35\alpha^2+30\alpha+7}}.
\end{equation} It remains to check the condition 
$t_5 \lt \zeta_1^0$ equivalently written as
\begin{equation*}
\frac{10(105\alpha^4+70\alpha^3+32\alpha^2+8\alpha+1)}{35\alpha^2+30\alpha+7} \lt \sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4},
\end{equation*}which is equivalent to
\begin{equation*}
\frac{(175\alpha^3+70\alpha^2+35\alpha+8)(2625\alpha^5-175\alpha^4-925\alpha^3-425\alpha^2-80\alpha-12)}{(35\alpha^2+30\alpha+7)^2} \lt 0.
\end{equation*} The last inequality is true for 
$\alpha\in(1/5,\alpha_3),$ where 
$\alpha_3\approx0.812678$ is the unique root in 
$(0,1)$ of the equation
\begin{equation*}
2625\alpha^5-175\alpha^4-925\alpha^3-425\alpha^2-80\alpha-12=0.
\end{equation*} Then ρα attains its maximum value on 
$(0,\zeta_1^0]$ at t 5 with
\begin{equation*}
\rho_\alpha(t_5)=\frac{\alpha^2(15\alpha^2+5\alpha+2)}{35\alpha^2+30\alpha+7}.
\end{equation*} If 
$\alpha\in[\alpha_3,1],$ then evidently,
\begin{equation*}
\rho_\alpha(t)\leqslant\max\left(\{\rho_\alpha(0),\rho_\alpha(\zeta_1^0)\}\right)=\rho_\alpha(\zeta_1^0),\quad 0\leqslant t\leqslant \zeta_1^0.
\end{equation*} C5. Applying Lemma 2 for 
$\zeta_1^0 \lt \zeta_1 \lt 1$, we get
\begin{equation*}
\left|\varGamma_1\varGamma_3-\varGamma_2^2\right| \leqslant\frac{1}{3}\alpha^2\zeta_1(1-\zeta_1^2)\left(|A|+|C| \right)\sqrt{1-\frac{B^2}{4AC}}=\psi_\alpha(\zeta_1),
\end{equation*} where for 
$t\in[0,1],$
\begin{equation*}
\begin{aligned}
\psi_\alpha(t):= \frac{1}{18}\alpha^2((35\alpha^2+1)t^4-6t^2+9)\sqrt{\frac{-(10\alpha^2-1)t^2+45\alpha^2+3}{(35\alpha^2+4)(t^2+3)}}.
\end{aligned}
\end{equation*}We have
\begin{equation*}
\psi_\alpha (\zeta_1^0)=-\frac{2\alpha^2(35\alpha^2+4)}{(175\alpha^3+70\alpha^2+35\alpha+8)^2}H(\alpha)\sqrt{-\frac{G(\alpha)}{K(\alpha)}},
\end{equation*}where
\begin{equation*}
\begin{aligned}
H(\alpha):= &-1050\alpha^4-525\alpha^3-270\alpha^2-65\alpha-10+(35\alpha^2+10\alpha+3)\times\\
&\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4},
\end{aligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
G(\alpha):=&-2625\alpha^5-1400\alpha^4-750\alpha^3-195\alpha^2-30\alpha-4+(20\alpha^2-2)\times\\
&\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}
\end{aligned}
\end{equation*}and
\begin{equation*}
K(\alpha):=(175\alpha^3+35\alpha^2+30\alpha+4+2\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4})(35\alpha^2+4).
\end{equation*}Note that
\begin{equation*}
\psi_\alpha(1)=\frac{1}{36}\alpha^2(35\alpha^2+4).
\end{equation*}Differentiating ψα leads to the equation
\begin{equation*}
\begin{aligned}
\psi_\alpha'(t)=-\frac{1}{18}t\alpha^2\frac{Q(t^2)}{(35\alpha^2+4)(t^2+3)^2\sqrt{\dfrac{-(10\alpha^2-1)t^2+45\alpha^2+3}{(35\alpha^2+4)(t^2+3)}}}=0,
\end{aligned}
\end{equation*} where for 
$s\in[0,1],$
\begin{equation*}
\begin{aligned}
Q(s):=&4(35\alpha^2+1)(10\alpha^2-1)s^3+3(175\alpha^4-315\alpha^2-4)s^2\\
&-18(1050\alpha^4+115\alpha^2-2)s+2295\alpha^2+108.
\end{aligned}
\end{equation*} Now we describe the number of zeros of Q in the interval 
$(0,1)$ by combining Descartes’ and Laguerre’s rules. To apply Descartes’ rule, we check the numbers of sign changes of coefficients of the polynomial 
$Q.$ We have:
•
$ d_0(\alpha):=q_0(0)=4(35\alpha^2+1)(10\alpha^2-1) \gt 0$ iff $\alpha\in\left(1/\sqrt{10},1\right),$•
$ d_1(\alpha):=q_1(0)=3(175\alpha^4-315\alpha^2-4) \lt 0$ iff $\alpha\in\left(0,1\right),$•
$ d_2(\alpha):=q_2(0)=-18(1050\alpha^4+115\alpha^2-2) \gt 0$ iff $\alpha\in\left(0,\alpha_4\right),$ where
\begin{equation*}\alpha_4:=\frac{1}{2}\sqrt{\frac{1}{105}(\sqrt{865}-23))}\approx 0.12355,\end{equation*}•
$d_3(\alpha):=q_3(0)= 2295\alpha^2+108 \gt 0$ iff $\alpha\in\left(0,1\right).$
Thus there is one change of signs in 
$\left(0,1/\sqrt{10}\right)$, i.e. 
$N(Q,0)=1,$ and two changes of signs in 
$\left[1/\sqrt{10},1\right),$ i.e. 
$N(Q,0)=2.$ According to Descartes’ rule of signs, the polynomial Q has one positive real root in 
$\left(0,1/\sqrt{10}\right)$ and zero or two positive real roots in 
$\left[1/\sqrt{10},1\right).$
To apply Laguerres’ rule, it remains to compute the number 
$N(Q,1)$ of sign changes in the sequence of sums 
$\sum_{j=0}^ku_j(\alpha)$, where 
$k=0,1,2,3.$ We have
•
$d_0(\alpha)=4(35\alpha^2+1)(10\alpha^2-1) \gt 0$ iff $\alpha\in\left(1/\sqrt{10},1\right),$•
$d_0(\alpha)+d_1(\alpha)=1925 \alpha^4 - 1045\alpha^2 - 16 \gt 0$ iff $\alpha\in\left(\alpha_5,1\right),$ where
\begin{equation*}
\alpha_5:=\sqrt{(209 + 3\sqrt{5401})/770}\approx 0.74683,
\end{equation*}•
$d_0(\alpha)+d_1(\alpha)+d_2(\alpha)= -5 (3395 \alpha^4 + 623\alpha^2 - 4) \gt 0$ iff $\alpha\in\left(0,\alpha_6\right),$ where
\begin{equation*}
\alpha_6:=\sqrt{(3\sqrt{49161} - 623)/6790}\approx 0.078806,
\end{equation*}•
$d_0(\alpha)+d_1(\alpha)+d_2(\alpha)+d_3(\alpha)= -(35\alpha^2 + 4) (485\alpha^2 - 32) \gt 0$ iff $\alpha\in\left(0,\alpha_7\right),$ where $\alpha_7:=4\sqrt{2/485}\approx 0.25686.$
Thus there are no changes of signs in 
$\left(\alpha_7,1/\sqrt{10}\right)$, i.e. 
$N(Q,1)=0$, and one change of sign in 
$\left(0,\alpha_7\right]\cup\left[1/\sqrt{10},1\right)$ i.e. 
$N(Q,1)=1$. According to Laguerre’s rule, the polynomial Q has one root in 
$[0,1]$ for 
$\alpha\in\left(\alpha_7,1\right)$, and no roots in 
$[0,1]$ for 
$\alpha\in\left(0,\alpha_7\right].$ Therefore, for 
$\alpha\in\left(0,\alpha_7\right]$, the function ψα is increasing for 
$\zeta_1^0 \lt t \lt 1$ and hence
\begin{equation*}
\psi_\alpha(t)\leqslant \psi_\alpha(1),\quad \zeta_1^0 \lt t \lt 1.
\end{equation*} In turn, for 
$\alpha\in\left(\alpha_7,1\right)$, the function ψα has a unique critical point in 
$[0,1],$ where by using jointly Descartes’ and Laguerre’s rules we state that ψα attains its minimum value. Thus
\begin{equation*}
\psi_\alpha(t)\leqslant \max\left(\{\psi_\alpha(\zeta_1^0),\psi_\alpha(1)\}\right),\quad \zeta_1^0 \lt t \lt 1.
\end{equation*}Now we summarize results of sections C4 and C5.
(i) For
$\alpha\in\left(0,1/5\right),$ we compare $\psi_\alpha(1)$ and $\varrho_\alpha(0).$ Note that then $\varrho_\alpha(0)\geqslant \psi_\alpha(1)$ since it is equivalent to
\begin{equation*}\frac14\alpha^2-\dfrac{1}{36}\alpha^2(35\alpha^2+4)=\frac{1}{36}\alpha^2(5-35\alpha^2)\geqslant 0.\end{equation*}(ii) For
$\alpha\in\left[1/5,\alpha_3\right),$ we compare $\psi_\alpha(1)$ and $\varrho_\alpha(t_5).$ Note that the inequality
\begin{equation*}
\frac{\alpha^2(15\alpha^2+5\alpha+2)}{35\alpha^2+30\alpha+7}\geqslant \frac{1}{36}\alpha^2(35\alpha^2+4)
\end{equation*}is equivalent to
\begin{equation*}
\frac{1}{36}\alpha^2(1225\alpha^4+1050\alpha^3-155\alpha^2-60\alpha-44)\leqslant 0
\end{equation*}which is true for
$\alpha\in[1/5,\alpha_0],$ where $\alpha_0\approx0.390595$ is the unique root in $(0,1]$ of Equation (3.2). Thus $\varrho_\alpha(t_5)\geqslant \psi_\alpha(1)$ for $\alpha\in[1/5,\alpha_0],$ and $\varrho_\alpha(t_5) \lt \psi_\alpha(1)$ for $\alpha\in(\alpha_0,\alpha_3).$(iii) For
$\alpha\in\left[\alpha_3,1\right),$ we compare $\psi_\alpha(1)$ and $\rho_\alpha(\zeta_1^0).$ Note that the inequality $\rho_\alpha(\zeta_1^0)\leqslant \psi_\alpha(1)$ is equivalent to
\begin{equation*}
\begin{aligned} &-18375\alpha^6-16625\alpha^5-10150\alpha^4-3775\alpha^3 -1025\alpha^2-150\alpha-12\\
&+(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\\
\leqslant& \frac{1}{72}(35\alpha^2+4)(175\alpha^3+70\alpha^2+35\alpha+8)^2,
\end{aligned}
\end{equation*}equivalently written as
\begin{equation*}
\begin{aligned} &(1050\alpha^4+700\alpha^3+320\alpha^2+80\alpha+10)\times\sqrt{525\alpha^4+175\alpha^3+120\alpha^2+20\alpha+4}\\
\leqslant& \frac{1}{72}\left(1071875\alpha^8+857500\alpha^7+2045750\alpha^6
+1564500\alpha^5+881475\alpha^4+322200\alpha^3\right.\\
&\left.+85420\alpha^2+13040\alpha+1120\right),
\end{aligned}
\end{equation*}which is equivalent to
\begin{equation*}
\begin{aligned}
\frac{25}{5184}&(35\alpha^2+4)(175\alpha^3+70\alpha^2+35\alpha+8)^2\times\\
&\times \left(42875\alpha^8+34300\alpha^7+134750\alpha^6
+110460\alpha^5+17835\alpha^4
-7344\alpha^3\right.\\
&\left.-5036\alpha^2-1120\alpha-128\right)\geqslant 0
\end{aligned}
\end{equation*}which is true for
$\alpha\in\left[\alpha_3,1\right).$
D. We now show sharpness of all inequalities by using the formula (3.5). In the first inequality in Equation (3.1), the equality is attained by the function 
$f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3) with
\begin{equation*}
p(z):=\frac{1-z^2}{1+z^2},\quad z\in \mathbb{D},
\end{equation*} for which 
$c_1=c_3=0$ and 
$c_2=-2.$
In the second inequality in Equation (3.1), the equality is attained by the function 
$f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3), where 
$p\in\mathcal{P}$ is defined by Equation (2.5) with 
$\zeta_1=t_5=:\tau$ and 
$\zeta_2=1,$ i.e.
\begin{equation*}
p(z):=\frac{1+2\tau z+z^2}{1-z^2},\quad z\in\mathbb{D}.
\end{equation*}Here t 5 is described by Equation (3.13).
In the third inequality in Equation (3.1), the equality is attained by the function 
$f\in{\mathcal{S}}^*_{\alpha}$ given by Equation (3.3), where 
$p\in\mathcal{P}$ is defined by
\begin{equation*}
p(z):=\frac{1+z}{1-z},\quad z\in \mathbb{D},
\end{equation*} for which 
$c_1=c_2=c_3=2.$
This ends the proof of the theorem.
For α = 1, we have the following result:
Corollary 2. If 
$f\in {\mathcal{S}}^*,$ then
\begin{equation*}
\left|\det H_{2,1}\left(F_{f^{-1}}\right)\right|=\left|\varGamma_1\varGamma_3-\varGamma_2^2\right|\leqslant\frac{13}{12}.
\end{equation*}The inequality is sharp.
