Proof. By logarithmic differentiation of 3 we get that \[\left\{ \begin{align} &\frac{z f'_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)}(z)} = 1 + \frac{1}{ab} \frac{z \mathfrak{W}'_{(\mu,a),(\nu,b)}(z)}{\mathfrak{W}_{(\mu,a),(\nu,b)}(z)} = 1 - \frac{1}{ab} \sum_{n \geq 1} \frac{2z^2}{\psi^2_{(\mu,a),(\nu,b), n} - z^2} \\[2ex] &\frac{z g_{(\mu,a),(\nu,b)}'(z)}{g_{(\mu,a),(\nu,b)}(z)} = 1 + \frac{z \mathfrak{W}'_{(\mu,a),(\nu,b)}(z)}{\mathfrak{W}_{(\mu,a),(\nu,b)}(z)} = 1 - \sum_{n \geq 1} \frac{2z^2}{\psi^2_{(\mu,a),(\nu,b), n} - z^2} \\[2ex] &\frac{z h'_{(\mu,a),(\nu,b)}(z)}{h_{(\mu,a),(\nu,b)}(z)} = 1 + \frac{1}{2} \frac{\sqrt{z} \mathfrak{W}'_{(\mu,a),(\nu,b)}(\sqrt{z})}{\mathfrak{W}_{(\mu,a),(\nu,b)}(\sqrt{z})} = 1 - \sum_{n \geq 1} \frac{z}{\psi^2_{(\mu,a),(\nu,b), n} - z} \end{align} \right.\] Taking \[\frac{z f'_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)}(z)} = 1 + \frac{1}{ab} \frac{z \mathfrak{W}'_{(\mu,a),(\nu,b)}(z)}{\mathfrak{W}_{(\mu,a),(\nu,b)}(z)} = 1 - \frac{1}{ab} \sum_{n \geq 1} \frac{2z^2}{\psi^2_{(\mu,a),(\nu,b), n} - z^2}\] with \(ab > 0\). Consider the continuous function given by \(K^{f_{(\mu,a),(\nu,b)}}: (0,\psi_{(\mu,a),(\nu,b),1}) \rightarrow \mathbb{R}\) defined as \[\label{eq3w} K^{f_{(\mu,a),(\nu,b)}} (r) = \frac{1}{ab} \sum_{n \geq 1} \frac{2r^2}{\psi^2_{(\mu,a),(\nu,b), n} - r^2} - \beta\tag{4}\] then \[\dfrac{d}{dr}K^{f_{(\mu,a),(\nu,b)}} (r) = \frac{1}{ab} \sum_{n \geq 1} \frac{4r^2 \psi^2_{(\mu,a),(\nu,b), n}}{(\psi^2_{(\mu,a),(\nu,b), n} - r^2)^2} > 0, \; \text{ for all } ab>0\] and for \(r<\psi_{(\mu,a),(\nu,b),1}\).
Also, \(K^{f_{(\mu,a),(\nu,b)}} (0)= -\beta < 0\) and \(\lim_{r\rightarrow \psi_{(\mu,a),(\nu,b),1}} K^{f_{(\mu,a),(\nu,b)}} (r)= \infty\). Thus there exist a unique positive root say \(\rho_{\varphi}(f_{(\mu,a),(\nu,b)})\) of \(K^{f_{(\mu,a),(\nu,b)}} (r)=0\) in \((0,\psi_{(\mu,a),(\nu,b),1})\).
Now, let \(\{ w: |w-1|<\beta \} \subseteq \varphi (\mathbb{D})\) be such that \(\varphi (-1)=1-\beta\) therefore, it is known that (lemma 1) if \(z \in \mathbb{C}\) and \(\delta \in \mathbb{R}\) such that \(|z|<r<\delta\) then \[Re \left( \dfrac{z}{\delta - z}\right) \leq \left| \dfrac{z}{\delta - z}\right| \leq \dfrac{|z|}{\delta - |z|}\] so, the inequality \[\label{eq4w} \left| \dfrac{z f^{'}_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)} (z)} -1 \right| = \left| \frac{1}{ab} \sum_{n \geq 1} \dfrac{2z^2}{\psi^2_{(\mu,a),(\nu,b),n}-z^2} \right| = \frac{1}{ab} \sum_{n \geq 1} \dfrac{2r^2}{\psi^2_{(\mu,a),(\nu,b),n}-r^2} \subseteq \beta\tag{5}\] implies that \(f_{(\mu,a),(\nu,b)} \in S^{*}(\varphi)\) in \(|z|<\rho_{\varphi}(f_{(\mu,a),(\nu,b)})\).
For sharpness, take \(|z| \in \mathbb{D}\) such that \(z=r=-\rho_{\varphi}(f_{(\mu,a),(\nu,b)})\) then using 4 and 5 , we get that \[\left| \dfrac{z f^{'}_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)} (z)} -1 = \beta \right|\] which means that \(\dfrac{z f^{'}_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)} (z)} \notin \varphi(\mathbb{D}), \;\forall |z| \geq \rho_{\varphi}(f_{(\mu,a),(\nu,b)})\), it proves the sharpness part.
Same reasoning will work for other two functions, that can be easily computed. ◻

Proof. From the functions defined in 3 we have \[\frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} = 1 + \frac{z \mathcal{W}''_{(\mu,a),(\nu,b)}(z)}{\mathcal{W}'_{(\mu,a),(\nu,b)}(z)}+ (\frac{1}{ab}-1) \frac{z \mathcal{W}'_{(\mu,a),(\nu,b)}(z)}{\mathcal{W}_{(\mu,a),(\nu,b)}(z)}\] \[= 1 - \sum_{n \geq 1} \frac{2z^2}{\tilde{\psi}^2_{(\mu,a),(\nu,b), n} - z^2} - (\frac{1}{ab}-1) \sum_{n \geq 1} \frac{2z^2}{\psi^2_{(\mu,a),(\nu,b), n} - z^2}\] Using lemma 1 we get that \[Re \left( 1+ \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right) \leq \left| 1+ \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right|\] implies that \[Re \left( \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right) \leq \left| \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right| \leq\] \[\sum_{n \geq 1} \frac{2z^2}{\tilde{\psi}^2_{(\mu,a),(\nu,b), n} - z^2} + (\frac{1}{ab}-1) \sum_{n \geq 1} \frac{2z^2}{\psi^2_{(\mu,a),(\nu,b), n} - z^2} = \frac{- r f''_{(\mu,a),(\nu,b)}(r)}{f'_{(\mu,a),(\nu,b)}(r)}\] holds in \(|z| = r < \tilde{\psi}_{(\mu,a),(\nu,b), 1}\) for \(ab \leq 1\).
Observe the inequality \[\left| \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right| \leq \frac{-r f''_{(\mu,a),(\nu,b)}(r)}{f'_{(\mu,a),(\nu,b)}(r)}\] also holds for \(ab > 1\) in \(r < \tilde{\psi}_{(\mu,a),(\nu,b), 1}\). Consider \(\rho^{c}_{\phi}(f_{(\mu,a),(\nu,b)})\) to be the smallest positive root of the equation \[\frac{r f''_{(\mu,a),(\nu,b)}(r)}{f'_{(\mu,a),(\nu,b)}(r)}+\beta=0\].
Now assume that \(\{ w: |w-1|<\beta \} \subseteq \varphi (\mathbb{D})\) be such that \(\varphi(-1)=1-\beta\) then it follows that \[\left| \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right| \leq \frac{-r f''_{(\mu,a),(\nu,b)}(r)}{f'_{(\mu,a),(\nu,b)}(r)} \leq \beta\] holds for \(|z|=r<\rho^{c}_{\phi}(f_{(\mu,a),(\nu,b)})\) for \(ab > 0\) thus \(f_{(\mu,a),(\nu,b)} \in C(\phi)\).
Now take \(z=-\rho^{c}_{\phi}(f_{(\mu,a),(\nu,b)})\) then \[\left| \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \right| = \frac{-r f''_{(\mu,a),(\nu,b)}(r)}{f'_{(\mu,a),(\nu,b)}(r)} = \beta\] implies that \[1+ \frac{z f''_{(\mu,a),(\nu,b)}(z)}{f'_{(\mu,a),(\nu,b)}(z)} \notin \varphi (\mathbb{D}),\;\forall |z|=r \geq \rho^{c}_{\phi}(f_{(\mu,a),(\nu,b)})\] which proves the sharpness part. Other two parts can be proved similarly. ◻

Proof. For proving the results, we need to show that \[\left| \log \left( \frac{z f'_{(\mu,a), (\nu,b)}(z)}{f_{(\mu,a), (\nu,b)}(z)} \right) \right| < 1, \qquad \left| \log \left( \frac{z g'_{(\mu,a), (\nu,b)}(z)}{g_{(\mu,a), (\nu,b)}(z)} \right) \right| < 1 \quad \text{and} \quad \left| \log \left( \frac{z h'_{(\mu,a), (\nu,b)}(z)}{h_{(\mu,a), (\nu,b)}(z)} \right) \right| < 1\] respectively for all \(z\) in disks of some radius and these inequalities do not hold true in any bigger disk other than that. Since, we know that \[\frac{z f'_{(\mu,a), (\nu,b)}(z)}{f_{(\mu,a), (\nu,b)}(z)} =1+\frac{1}{ab}\frac{z\mathfrak{W}'_{(\mu,a), (\nu,b)}(z)}{\mathfrak{W}_{(\mu,a), (\nu,b)}(z)} =1-\frac{2}{ab}\sum_{n=1}^{\infty}\frac{z^{2}}{\psi_{\rho,\beta,n}^{2}-z^{2}} .\]

Using Lemma 1, we get the following inequality \[\label{eq3a} \left| \frac{z f'_{(\mu,a), (\nu,b)}(z)}{f_{(\mu,a), (\nu,b)}(z)} -1+\frac{2}{ab}\sum_{n=1}^{\infty}\frac{r^{4}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}} \right| \le \frac{2}{ab}\sum_{n=1}^{\infty}\frac{\psi_{(\mu,a), (\nu,b),n}^{2}r^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}}\qquad{(1)}\] for all \(|z|\le r<\psi_{(\mu,a), (\nu,b),1}\). Note that the equality holds in the above equation when \(z=ir\). Take \(I_1:=(0,\psi_{(\mu,a), (\nu,b),1})\). Then the function \(F_f:I_1\to\mathbb{R}\) defined by \[F_f(r)=1-\frac{2}{ab}\sum_{n=1}^{\infty}\frac{r^{4}}{\psi_{(\mu,a), (\nu,b), n}^{4}-r^{4}} =\frac{1}{2}\left( \frac{r f'_{(\mu,a), (\nu,b)}(r)}{f_{(\mu,a), (\nu,b)}(r)} + \frac{i r f'_{(\mu,a), (\nu,b)}(ir)}{f_{(\mu,a), (\nu,b)}(ir)} \right).\] is continuous on \(I_1\) also \(F_f'(r)<0\) \(\forall\) \(r\in I_1\), thus \(F_f\) is decreasing on the interval \(I_1\).
Clearly \(F_f(0)=1>0\) and the function \(F_f\) has the limit \(-\infty\) when \(r \nearrow \psi_{(\mu,a), (\nu,b),1}\). Implies that \(F_f(a_{(\mu,a), (\nu,b),3})=0\) for some \(t_{(\mu,a), (\nu,b),3}\in I_1\). Therefore by intermediate value theorem there exists a unique real number \(t_{(\mu,a), (\nu,b),2}\) in the interval \((0,t_{(\mu,a), (\nu,b),3})\) such that \[F_f(t_{(\mu,a), (\nu,b),2})=\frac{1}{2}\left(1+\frac{1}{e}\right).\] Note that if \(0\le r\le t_{(\mu,a), (\nu,b),2}\), then \[\frac{1}{2}\left(1+\frac{1}{e}\right)\le F_f(r)\le 1.\] Now, define function \(\xi_f\) as \[\xi_f(r):=\frac{1}{e}-\frac{r f'_{(\mu,a), (\nu,b)}(r)}{f_{(\mu,a), (\nu,b)}(r)}.\]

Then the limit of the function \(\xi_f\) is \(\frac{1}{e}-1<0\) when \(r\searrow0\) and \(\infty\) when \(r\nearrow\psi_{(\mu,a), (\nu,b),1}\). Since \(\psi_f'(r)>0\) \(\forall\) \(r\in I\) so \(\xi_f\) is an increasing function of \(r\). Therefore the function \(\xi_f\) cuts the x-axis exactly once in the interval \(I_1\). By the computation \[\xi_f(t_{(\mu,a), (\nu,b),3}) =\frac{1}{e}-1+\frac{2}{ab}\sum_{n=1}^{\infty} \frac{t_{(\mu,a), (\nu,b),3}^{2}}{\psi_{(\mu,a), (\nu,b),n}^{2}-t_{(\mu,a), (\nu,b),3}^{2}}\] \[\ge \frac{1}{e}-1+\frac{2}{ab}\sum_{n=1}^{\infty} \frac{t_{(\mu,a), (\nu,b),3}^{4}}{\psi_{(\mu,a), (\nu,b),n}^{4}-t_{(\mu,a), (\nu,b),3}^{4}} = \frac{1}{e}F_f(t_{(\mu,a), (\nu,b),3}) = \frac{1}{e}>0\] we sees that the function \(\xi_f\) has a zero in the interval \((0,t_{(\mu,a), (\nu,b),3})\). Let this zero be denoted by \(t_{(\mu,a), (\nu,b),1}\). If \(t_{(\mu,a), (\nu,b),1}\le t_{(\mu,a), (\nu,b),2}\), then \(\xi_f(r)\le0\) for all \(r\le t_{(\mu,a), (\nu,b),1}\), that is \[\frac{2}{ab}\sum_{n=1}^{\infty} \frac{\psi_{(\mu,a), (\nu,b),n}^{2}r^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}} \le F_f(r)-\frac{1}{e}\]

for all \(r\le t_{(\mu,a), (\nu,b),1}\). If \(t_{(\mu,a), (\nu,b),2}<t_{(\mu,a), (\nu,b),1}<t_{(\mu,a), (\nu,b),3}\), then \[0<F_f(r)<\frac{1}{2}\left(1+\frac{1}{e}\right)\] for \(t_{(\mu,a), (\nu,b),2}<r<t_{(\mu,a), (\nu,b),1}\) and in this case, \(1\) does not belong to the disk ?? which leads to a contradiction. Hence using Lemma 1, we see that the disk ?? lies inside the region \(\Omega_e\) for all \(r\le t_{(\mu,a), (\nu,b),1}\). This shows that \[R_e^*(f_{(\mu,a), (\nu,b)})\ge t_{(\mu,a), (\nu,b),1}.\] Also \[\left| \log\left( \frac{t_{(\mu,a), (\nu,b),1}\cdot f'_{(\mu,a), (\nu,b)}(t_{(\mu,a), (\nu,b),1})}{f_{(\mu,a), (\nu,b)}(t_{(\mu,a), (\nu,b),1})} \right) \right| = \left| \log\left( \frac{1}{e}-\psi_f(t_{(\mu,a), (\nu,b),1}) \right) \right| =1\] proves that \(R_e^*(f_{(\mu,a), (\nu,b)})=t_{(\mu,a), (\nu,b),1}\).
For part b, we know that \[\frac{z g'_{(\mu,a), (\nu,b)}(z)}{g_{(\mu,a), (\nu,b)}(z)} =1+\frac{z\mathfrak{W}'_{(\mu,a), (\nu,b)}(z)}{\mathfrak{W}_{(\mu,a), (\nu,b)}(z)} =1-\sum_{n=1}^{\infty}\frac{2z^{2}}{\psi_{(\mu,a), (\nu,b),n}^{2}-z^{2}}.\]

Therefore from lemma 1 , \[\label{eq3b} \left| \frac{z g'_{(\mu,a), (\nu,b)}(z)}{g_{(\mu,a), (\nu,b)}(z)} -1+ \sum_{n=1}^{\infty} \frac{2r^{4}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}} \right| \le \sum_{n=1}^{\infty} \frac{2\psi_{(\mu,a), (\nu,b),n}^{2}r^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}}\qquad{(2)}\]

for all \(|z|\le r<\psi_{(\mu,a), (\nu,b),1}\) with equality at \(z=ir\). Taking \(I_1:=(0,\psi_{(\mu,a), (\nu,b),1})\), it can be seen that the function \[F_g(r):= 1-\sum_{n=1}^{\infty}\frac{2r^{4}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}} = \frac{1}{2} \left( \frac{r g'_{(\mu,a), (\nu,b)}(r)}{g_{(\mu,a), (\nu,b)}(r)} + \frac{i r g'_{(\mu,a), (\nu,b)}(ir)}{g_{(\mu,a), (\nu,b)}(ir)} \right)\]

is continuously decreasing on \(I_1\). Following the similar way as part (a), we observe that \(F_g(0)=1\) and \(\lim_{r\nearrow\psi_{(\mu,a), (\nu,b),1}}F_g(r)=-\infty\). This confirms the existence of a root of \(C_g\) in the interval \(I_1\), we denote it by \(b1_{(\mu,a), (\nu,b),3}\), . Again using intermediate value theorem we get a unique real number \(b1_{(\mu,a), (\nu,b),2}\) in the interval \((0,b1_{(\mu,a), (\nu,b),3})\) satisfying \[F_g(b1_{(\mu,a), (\nu,b),2})=\frac{1}{2}\left(1+\frac{1}{e}\right).\]

Now for \(0<r\le b1_{(\mu,a), (\nu,b),2}\), observe that \[\frac{1}{2}\left(1+\frac{1}{e}\right)\le F_g(r)\le 1.\]

Consider a function defined by \[\xi_g(r):=\frac{1}{e}-\frac{r g'_{(\mu,a), (\nu,b)}(r)}{g_{(\mu,a), (\nu,b)}(r)}.\]

Then \(\lim_{r\searrow0}\xi_g(r)=\frac{1}{e}-1<0\) and \(\lim_{r\nearrow\psi_{(\mu,a), (\nu,b),1}}\xi_g(r)=+\infty\). Also \(\xi_g'(r)>0\) for all \(r\in I_1\), this implies \(\xi_g\) is an increasing function of \(r\). Therefore by intermediate value theorem, there exists a unique root \(b1_{(\mu,a), (\nu,b),1}\) of the function \(\xi_g\) in the interval \(I_1\), or more precisely in the interval \((0,b1_{(\mu,a), (\nu,b),3})\) since

\[\xi_g(b1_{(\mu,a), (\nu,b),3}) = \frac{1}{e}-1+ \sum_{n=1}^{\infty} \frac{2(b1)_{(\mu,a), (\nu,b),3}^{2}}{\psi_{(\mu,a), (\nu,b),n}^{2}-{b1}^2_{(\mu,a), (\nu,b),3}}\]

\[\ge \frac{1}{e}-1+ \sum_{n=1}^{\infty} \frac{2(b1)_{(\mu,a), (\nu,b),3}^{4}}{\psi_{(\mu,a), (\nu,b),n}^{4}-b1_{(\mu,a), (\nu,b),3}^{4}} = \frac{1}{e}-F_g(b1_{(\mu,a), (\nu,b),3}) = \frac{1}{e}>0.\]

This gives \[\sum_{n=1}^{\infty} \frac{2\psi_{(\mu,a), (\nu,b),n}^{2}r^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{4}} \le F_g(r)-\frac{1}{e}\]

\(\forall\) \(r\le b1_{(\mu,a), (\nu,b),1}\). Observe that the zero \(b1_{(\mu,a), (\nu,b),1}\) is always smaller than \(b1_{(\mu,a), (\nu,b),2}\), as if not, then \(F_g(r)\) would lie between \(0\) and \(\frac{1}{2}\left(1+\frac{1}{e}\right)\) which is not possible as then the point \(1\) would not belong to the disk ?? . Hence Lemma 1 proves that the disk ?? is contained inside \(\Omega_e\) for all \(r\le b1_{(\mu,a), (\nu,b),1}\), that is, \[R_e^*(g_{(\mu,a), (\nu,b)})\ge b1_{(\mu,a), (\nu,b),1}.\]

Since \[\left| \log \left( \frac{b1_{(\mu,a), (\nu,b),1}\cdot g'_{(\mu,a), (\nu,b)}(b1_{(\mu,a), (\nu,b),1})}{g_{(\mu,a), (\nu,b)}(b1_{(\mu,a), (\nu,b),1})} \right) \right| =1,\]

therefore \(R_e^*(g_{(\mu,a), (\nu,b)})=b1_{(\mu,a), (\nu,b),1}\).
For part c, from the normalization and the decomposition, it is clear that \[\frac{z h'_{(\mu,a), (\nu,b)}(z)}{h_{(\mu,a), (\nu,b)}(z)} =1+\frac{1}{2}\cdot \frac{\sqrt{z}\,\mathfrak{W}'_{(\mu,a), (\nu,b)}(\sqrt{z})}{\mathfrak{W}_{(\mu,a), (\nu,b)}(\sqrt{z})} =1-\sum_{n=1}^{\infty}\frac{z}{\psi_{(\mu,a), (\nu,b),n}^{2}-z}.\]

Using 1 , we have the inequality \[\label{eq3c} \left|\frac{z h'_{(\mu,a), (\nu,b)}(z)}{h_{(\mu,a), (\nu,b)}(z)}-1+\sum_{n=1}^{\infty}\frac{r^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{2}}\right| \le \sum_{n=1}^{\infty}\frac{\psi_{(\mu,a), (\nu,b),n}^{2} r}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{2}}\tag{6}\]

valid for all \(|z|\le r<\psi_{(\mu,a), (\nu,b),1}^{2}\) with equality at \(z=-r\) because of the minimum principle for the harmonic functions. Take \(J:=(0,\psi_{(\mu,a), (\nu,b),1}^{2})\), then the function \(L_h:J\to\mathbb{R}\) defined by \[L_h(r)=1-\sum_{n=1}^{\infty}\frac{r^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{2}} =\frac{1}{2}\left(\frac{r h'_{(\mu,a), (\nu,b)}(r)}{h_{(\mu,a), (\nu,b)}(r)}-\frac{r h'_{(\mu,a), (\nu,b)}(-r)}{h_{(\mu,a), (\nu,b)}(-r)}\right)\]

is continuously decreasing on the interval \(J\). Note that \(L_h(0)=1\) and the function \(C_h\) takes the limit \(-\infty\) when \(r\uparrow\psi_{(\mu,a), (\nu,b),1}^{2}\) implying that \(L_h(l_{(\mu,a), (\nu,b),3})=0\) for some \(l_{(\mu,a), (\nu,b),3}\in J\). By intermediate value theorem, we have \(L_h(l_{(\mu,a), (\nu,b),2})=\frac{1}{2}\left(1+\frac{1}{e}\right)\) for some \(l_{(\mu,a), (\nu,b),2}\in(0,l_{(\mu,a), (\nu,b),3})\). This gives \[\frac{1}{2}\left(1+\frac{1}{e}\right)\le L_h(r)\le 1\] whenever \(0\le r\le l_{(\mu,a), (\nu,b),2}\). define the function \[\xi_h(r):=\frac{1}{e}-\frac{r h'_{(\mu,a), (\nu,b)}(r)}{h_{(\mu,a), (\nu,b)}(r)},\]

it behaves similar to the functions \(\xi_f\) and \(\xi_g\), that is, it has one root in the interval \(J\), being an increasing function of \(r\) and also observe that \[\lim_{r\searrow 0}\xi_h(r)=\frac{1}{e}-1<0 \quad\text{and}\quad \lim_{r\nearrow\psi_{(\mu,a), (\nu,b),1}^{2}}\xi_h(r)=+\infty.\]

Let \(l_{(\mu,a), (\nu,b),1}\) be the unique zero of the function \(\xi_h\). It can be easily seen that \(l_{(\mu,a), (\nu,b),1}\) is always smaller than \(l_{(\mu,a), (\nu,b),3}\) because \[\xi_h(l_{(\mu,a), (\nu,b),3}) =\frac{1}{e}-1+\sum_{n=1}^{\infty}\frac{l_{(\mu,a), (\nu,b),3}}{\psi_{(\mu,a), (\nu,b),n}^{2}-l_{(\mu,a), (\nu,b),3}}\]

\[\ge \frac{1}{e}-1+\sum_{n=1}^{\infty}\frac{l_{(\mu,a), (\nu,b),3}^{2}}{\psi_{(\mu,a), (\nu,b),n}^{4}-l_{(\mu,a), (\nu,b),3}^{2}} = \frac{1}{e}-L_h(l_{(\mu,a), (\nu,b),3}) = \frac{1}{e}>0.\]

Therefore for all \(r\le l_{(\mu,a), (\nu,b),1}\), we have \[\sum_{n=1}^{\infty}\frac{\psi_{(\mu,a), (\nu,b),n}^{2} r}{\psi_{(\mu,a), (\nu,b),n}^{4}-r^{2}} \le L_h(r)-\frac{1}{e}.\]

Hence by applying Lemma 1*, the disk (3.1) lies inside the region \(\Omega_e\) for \(r\le l_{(\mu,a), (\nu,b),1}\). Following similar observation as the previous two parts the radius \(l_{(\mu,a), (\nu,b),1}\) cannot be further improved so that \[R^{*}_{e}(h_{(\mu,a), (\nu,b)})=l_{(\mu,a), (\nu,b),1}.\] ◻

Proof. To prove the theorem’s statements, we need to show these inequalities \[\left| \log \left( 1 + \frac{zf''_{(\mu, a), (\nu, b)}(z)}{f'_{(\mu, a), (\nu, b)}(z)} \right) \right| < 1, \quad \left| \log \left( 1 + \frac{zg''_{(\mu, a), (\nu, b)}(z)}{g'_{(\mu, a), (\nu, b)}(z)} \right) \right| < 1\] and \[\left| \log \left( 1 + \frac{zh''_{(\mu, a), (\nu, b)}(z)}{h'_{(\mu, a), (\nu, b)}(z)} \right) \right| < 1\] holds respectively for all \(z\) lying in disks of some particular radius and does not holds in any disk of larger radii.

We will first consider part (a), the Weierstrass decomposition of the functions \(\Psi_{(\mu, a), (\nu, b)}\) and \(\Psi'_{(\mu, a), (\nu, b)}\): \[\Psi_{(\mu, a), (\nu, b)}(z) = \frac{z^{ab}}{\Gamma(a) \Gamma(b)} \prod_{n=1}^\infty \left( 1 - \frac{z^2}{\psi_{(\mu, a), (\nu, b), n}^2} \right) \quad \text{and} \quad \Psi'_{(\mu, a), (\nu, b)}(z) = \frac{z^{ab-1}}{\Gamma(a) \Gamma(b)} \prod_{n=1}^\infty \left( 1 - \frac{z^2}{\psi_{(\mu, a), (\nu, b), n}^{\prime 2}} \right)\] using logarithmic differentiation we get that \[\label{eq4a} 1 + \frac{zf''_{(\mu, a), (\nu, b)}(z)}{f'_{(\mu, a), (\nu, b)}(z)} = 1 - \sum_{n=1}^\infty \frac{2z^2}{\psi_{(\mu, a), (\nu, b), n}^{\prime 2} - z^2} - \left( \frac{1}{ab} - 1 \right) \sum_{n=1}^\infty \frac{2z^2}{\psi_{(\mu, a), (\nu, b), n}^2 - z^2},\tag{7}\] where \(\psi_{(\mu, a), (\nu, b), n}\) and \(\psi'_{(\mu, a), (\nu, b), n}\) denote the \(n\)th positive zero of the functions \(\Psi_{(\mu, a), (\nu, b)}\) and \(\Psi'_{(\mu, a), (\nu, b)}\) respectively, it is now clear that \[\label{eq4b} \left| 1 + \frac{zf''_{(\mu, a), (\nu, b)}(z)}{f'_{(\mu, a), (\nu, b)}(z)} - C_f(r) \right| \leq \sum_{n=1}^\infty \frac{2\psi_{(\mu, a), (\nu, b), n}^{\prime 2} r^2}{\psi_{(\mu, a), (\nu, b), n}^{\prime 4} - r^4} + \left( \frac{1}{ab} - 1 \right) \sum_{n=1}^\infty \frac{2\psi_{(\mu, a), (\nu, b), n}^2 r^2}{\psi_{(\mu, a), (\nu, b), n}^4 - r^4}\tag{8}\] for all \(|z| \leq r < \psi'_{(\mu, a), (\nu, b), 1} < \psi_{(\mu, a), (\nu, b), 1}\) and the equality holds if and only if \(z = ir\), where \[\begin{align} C_f(r) &:= 1 - \sum_{n=1}^\infty \frac{2r^4}{\psi_{(\mu, a), (\nu, b), n}^{\prime 4} - r^4} - \left( \frac{1}{ab} - 1 \right) \sum_{n=1}^\infty \frac{2r^4}{\psi_{(\mu, a), (\nu, b), n}^4 - r^4} \\ &= 1 + \frac{1}{2} \left( \frac{rf''_{(\mu, a), (\nu, b)}(r)}{f'_{(\mu, a), (\nu, b)}(r)} + \frac{irf''_{(\mu, a), (\nu, b)}(ir)}{f'_{(\mu, a), (\nu, b)}(ir)} \right). \end{align}\] Since \(0 < a,b \leq 1\), \(1/ab > 1\) and therefore \(C'_f(r) < 0\) for all \(r \in (0, \psi'_{(\mu, a), (\nu, b), 1})\). This means the function \(C_f\) is continuously decreasing on the interval \((0, \psi'_{(\mu, a), (\nu, b), 1})\) and it satisfies \(C_f(0) = 1\). Also, \(\lim_{r \nearrow \psi'_{(\mu, a), (\nu, b), 1}} C_f(r) = -\infty\) thus the function \(C_f\) has a zero, say \(s'_{(\mu, a), (\nu, b), 3}\), in the interval \((0, \psi'_{(\mu, a), (\nu, b), 1})\). Here also by using the intermediate value theorem gives the existence of a unique real number \(s'_{(\mu, a), (\nu, b), 2} \in (0, s'_{(\mu, a), (\nu, b), 3})\) that satisfies \(C_f(a'_{(\mu, a), (\nu, b), 2}) = \frac{1}{2} (1 + \frac{1}{e})\). If \(r \in [0, s'_{(\mu, a), (\nu, b), 2}]\), then \(C_f \in [\frac{1}{2} (1 + \frac{1}{e}), 1]\). Next, consider a function \(\xi_f\) defined by \[\xi_f(r) := \frac{1}{e} - 1 - \frac{rf''_{(\mu, a), (\nu, b)}(r)}{f'_{(\mu, a), (\nu, b)}(r)}.\]

Then \(\lim_{r \searrow 0} \xi_f(r) = \frac{1}{e} - 1 < 0\) and \(\lim_{r \nearrow \psi'_{(\mu, a), (\nu, b), 1}} \xi_f(r) = \infty\). Since \(0 < a,b \leq 1\), we have \(\xi'_f(r) > 0\) for all \(r \in (0, \psi'_{(\mu, a), (\nu, b), 1})\) implying that the function \(\xi_f\) is increasing in the interval \((0, \psi'_{(\mu, a), (\nu, b), 1})\). Consequently, the function \(\xi_f\) has a unique zero \(s'_{(\mu, a), (\nu, b), 1}\) in the interval \((0, \psi'_{(\mu, a), (\nu, b), 1})\). Moreover, the calculation \[\begin{align} \xi_f(s'_{(\mu, a), (\nu, b), 3}) &= \frac{1}{e} - 1 + \sum_{n=1}^\infty \frac{2 s_{(\mu, a), (\nu, b), 3}^{\prime 2}}{\psi_{(\mu, a), (\nu, b), n}^{\prime 2} - s_{(\mu, a), (\nu, b), 1}^{\prime 2}} + \left( \frac{1}{\beta} - 1 \right) \sum_{n=1}^\infty \frac{2 s_{(\mu, a), (\nu, b), 3}^{\prime 2}}{\psi_{(\mu, a), (\nu, b), n}^2 - s_{(\mu, a), (\nu, b), 3}^{\prime 2}} \\ &\geq \frac{1}{e} - 1 + \sum_{n=1}^\infty \frac{2 s_{(\mu, a), (\nu, b), 3}^{\prime 4}}{\psi_{(\mu, a), (\nu, b), n}^{\prime 4} - s_{(\mu, a), (\nu, b), 3}^{\prime 4}} + \left( \frac{1}{\beta} - 1 \right) \sum_{n=1}^\infty \frac{2 s_{(\mu, a), (\nu, b), 3}^{\prime 4}}{\psi_{(\mu, a), (\nu, b), n}^4 - s_{(\mu, a), (\nu, b), 3}^{\prime 4}} \\ &= \frac{1}{e} - C_f(s'_{(\mu, a), (\nu, b), 3}) = \frac{1}{e} > 0 \end{align}\] shows that the function \(\xi_f\) will always has its zero \(s'_{(\mu, a), (\nu, b), 1}\) within the interval \((0, s'_{(\mu, a), (\nu, b), 3})\). Hence \(\xi_f(r) \leq 0\) for all \(r \in [0, s'_{(\mu, a), (\nu, b), 1}]\), or equivalently \[\sum_{n=1}^\infty \frac{2 \psi_{(\mu, a), (\nu, b), n}^{\prime 2} r^2}{\psi_{(\mu, a), (\nu, b), n}^{\prime 4} - r^4} - \left( \frac{1}{ab} - 1 \right) \sum_{n=1}^\infty \frac{2 \psi_{(\mu, a), (\nu, b), n}^2 r^2}{\psi_{(\mu, a), (\nu, b), n}^4 - r^4} \leq C_f(r) - \frac{1}{e}\] for all \(r \leq s'_{(\mu, a), (\nu, b), 1}\). Observe that if \(r \geq a'_{(\mu, a), (\nu, b), 2}\), then \(0 \leq C_f(r) < \frac{1}{2} (1 + \frac{1}{e})\) and in this case, the point 1 does not belong to the disk 8 denying the fact that it does. Thus Lemma 2 proves that the disk 8 lies inside \(\Omega_e\) for all \(r \leq s'_{(\mu, a), (\nu, b), 1} \leq s'_{(\mu, a), (\nu, b), 2}\), that is, \(\mathcal{R}_e^c(f_{(\mu, a), (\nu, b)}) \geq s'_{(\mu, a), (\nu, b), 1}\). But since \[\left| \log \left( 1 + \frac{s'_{(\mu, a), (\nu, b), 1} \cdot f''_{(\mu, a), (\nu, b)}(s'_{(\mu, a), (\nu, b), 1})}{f'_{(\mu, a), (\nu, b)}(s'_{(\mu, a), (\nu, b), 1})} \right) \right| = 1,\] which follows that \(\mathcal{R}_e^c(f_{(\mu, a), (\nu, b)}) = s'_{(\mu, a), (\nu, b), 1}\).
Other parts can be proved similar fashion. ◻

Proof. Let \(\alpha \in [0,1)\) and \(\gamma \in \left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\), \(\mu,\nu,a,b>0\) and let \(\{\psi_{(\mu,a),(\nu,b),n}\}_{n\ge1}\) denotes the positive zero sequence of \(\mathcal{W}_{(\mu,a),(\nu,b)}(-z^2)\).
We aim to establish the spiral-like inequalities \[\mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z f'_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)}(z)}\right) > \alpha \cos\gamma,\qquad \mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z g'_{(\mu,a),(\nu,b)}(z)}{g_{(\mu,a),(\nu,b)}(z)}\right) > \alpha \cos\gamma,\] and \[\mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z h'_{(\mu,a),(\nu,b)}(z)}{h_{(\mu,a),(\nu,b)}(z)}\right) > \alpha \cos\gamma,\] valid respectively in the disks \(\mathbb{D}_{R_{sp}(\gamma,\alpha;f)}\), \(\mathbb{D}_{R_{sp}(\gamma,\alpha;g)}\), and \(\mathbb{D}_{R_{sp}(\gamma,\alpha;h)}\), with sharpness in the sense that none of the three holds in any larger disk.

By using the classical inequality from lemma 1 for every \(|z|=r<\psi_{(\mu,a),(\nu,b),1}\), \[\mathrm{Re}\!\left(\frac{z^2}{\psi_{(\mu,a),(\nu,b),n}^{2}-z^{2}}\right) \le \left| \frac{z^{2}}{\psi_{(\mu,a),(\nu,b),n}^{2}-z^{2}} \right| \le \frac{|z|^{2}}{\psi_{(\mu,a),(\nu,b),n}^{2}-|z|^{2}}.\] we obtain \[\begin{align} \mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z f'_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)}(z)}\right) &= \mathrm{Re}(e^{-i\gamma}) - \frac{1}{ab}\,\mathrm{Re}\!\left( e^{-i\gamma} \sum_{n\ge1}\frac{2z^{2}}{\psi_{(\mu,a),(\nu,b),n}^{2}-z^{2}} \right) \\ &\ge \cos\gamma - \frac{1}{ab} \sum_{n\ge1}\frac{2r^{2}}{\psi_{(\mu,a),(\nu,b),n}^{2}-r^{2}}. \end{align}\] Hence \[\mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z f'_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)}(z)}\right) \ge \cos\gamma - \frac{2r^{2}}{ab} \sum_{n\ge1}\frac{1}{\psi_{(\mu,a),(\nu,b),n}^{2}-r^{2}}.\\ =\dfrac{r f'_{(\mu,a),(\nu,b)}(r)}{f_{(\mu,a),(\nu,b)}(r)}+cos \gamma -1\] The same reasoning for \(g_{(\mu,a),(\nu,b)}, h_{(\mu,a),(\nu,b)}\) with lemma 1 yields \[\mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z g'_{(\mu,a),(\nu,b)}(z)}{g_{(\mu,a),(\nu,b)}(z)}\right) \ge \cos\gamma - 2r^{2} \sum_{n\ge1}\frac{1}{\psi_{(\mu,a),(\nu,b),n}^{2}-r^{2}}.\] we similarly obtain for \(|z|=r<\psi_{(\mu,a),(\nu,b),1}\) \[\mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z h'_{(\mu,a),(\nu,b)}(z)}{h_{(\mu,a),(\nu,b)}(z)}\right) \ge \cos\gamma - \sum_{n\ge1}\frac{r}{\psi_{(\mu,a),(\nu,b),n}^{2}-r}.\]

Consider for the \(f\)-case, the equality holds when \(|z|=r\), Thus, for \(r \in\) \((0,\, \psi_{(\mu,a),(\nu,b), 1})\) it follows that \[\inf_{|z|<r}\left\{\mathrm{Re}\!\left(e^{-i\gamma}\,\frac{z f'_{(\mu,a),(\nu,b)}(z)}{f_{(\mu,a),(\nu,b)}(z)}\right)\right\}\\ =\dfrac{r f'_{(\mu,a),(\nu,b)}(r)}{f_{(\mu,a),(\nu,b)}(r)}+(1-\alpha)cos \gamma -1\] Now, consider the auxiliary function on \(\Theta_f:(0,\psi_{(\mu,a),(\nu,b),1})\rightarrow \mathbb{R}\) defined by \[\Theta_f(r) :=\dfrac{r f'_{(\mu,a),(\nu,b)}(r)}{f_{(\mu,a),(\nu,b)}(r)}+(1-\alpha)cos \gamma -1 \\ =(1-\alpha)cos \gamma -\frac{1}{ab}\sum_{n\ge1}\frac{2r^2}{\psi_{(\mu,a),(\nu,b),n}^{2}-r^2}\] is strictly decreasing as \(\Theta'_f(r)<0\). Since, \[\Theta_f'(r) = -\frac{1}{ab} \sum_{n\ge1} \frac{4r\,\psi_{(\mu,a),(\nu,b),n}^{2}}{(\psi_{(\mu,a),(\nu,b),n}^{2}-r^{2})^{2}} < 0,\] so \(\Theta_f\) is strictly decreasing on \((0,\psi_{(\mu,a),(\nu,b),1})\). Moreover, \[\lim_{r\downarrow 0}\Theta_f(r)=(1-\alpha)\cos\gamma>0,\qquad \lim_{r\uparrow \psi_{(\mu,a),(\nu,b),1}}\Theta_f(r)=-\infty.\] The minimum principle for harmonic functions implies that the required inequality for \(f_{(\mu,a),(\nu,b)}\) holds iff \(z \in\) \(\mathbb{D}_{R_{sp}(\gamma,\alpha;f)}\), where \(\mathbb{D}_{R_{sp}(\gamma,\alpha;f)}\) is the smallest positive root of \(\dfrac{r f'_{(\mu,a),(\nu,b))}}{f_{(\mu,a),(\nu,b))}}=1-(1-\alpha)cos \gamma\). Hence there exists a unique \(R_{sp}(\gamma,\alpha;f)\in(0,\psi_{(\mu,a),(\nu,b),1})\) such that \(\Theta_f(R_{sp}(\gamma,\alpha;f))=0\), and the inequality for \(f\) holds precisely for \(|z|<R_{sp}(\gamma,\alpha;f)\). The defining equation reads \[(1-\alpha)\cos\gamma=\frac{2{R_{sp}(\gamma,\alpha;f)}^{2}}{(ab)} \sum_{n\ge1}\frac{1}{\psi_{(\mu,a),(\nu,b),n}^{2}-{R_{sp}(\gamma,\alpha;f)}^{2}}.\] Reasoning along the same lines the other two parts follows. ◻