for all $t\ge0$. Note that for $(c_1,c_2)=(c_1^*,c_2^*)$ in , the first condition is satisfied with the strict inequality sign, while in the second condition equality occurs. Therefore, see, e.g., [@Tsuji Theorem IV.23], $$\begin{aligned}
 p_0(z,t)&=\lambda\frac{1+\mu_1 z}{1-\mu_1 z}+(1-\lambda)\frac{1+\mu_2 z}{1-\mu_2 z}\\
          &=1\,+\,2(\lambda\mu_1+(1-\lambda)\mu_2)z\,+\,2(\lambda\mu_1^2+(1-\lambda)\mu_2^2)z^2\,+\,\ldots,\quad z\in{\mathbb{D}},\end{aligned}$$ for some constants $\lambda\in(0,1)$ and $\mu_1\neq\mu_2$ on the unit circle (possibly depending on $t$). Comparing the coefficients of $z$ and $z^2$ with $c_1^*$ and $c_2^*$, we conclude that ${\lambda=1/2}$, ${\mu_{1,2}=\pm i}$, and hence $p(z,t)=(1-kz^2)/(1+kz^2)$. The corresponding function $f\in{\mathcal{S}}_k^B$ is $f(z)=f_1(z):=z/(1-kz^2)$, with $a_3|_{f=f_1}=k$.

Now suppose that $a\neq0$. Denote $t_0:=\max\big\{0,\,\log|2(1+k)/a|\big\}$. Then according to , $c_1^*(t)=-e^ta/(1+k)$ whenever ${0\le t<t_0}$, and $c_1^*(t)=-2{\mathop{\mathrm{sgn}}}a$ for all $t\ge t_0$. Substituting $c_1(t):=c_1^*(t)$ into , we get $$a=2k\left(\frac{a t_0}{1+k}+2e^{-t_0}{\mathop{\mathrm{sgn}}}a\right)=\frac{2k(1+t_0)}{1+k}a.$$ It follows that $t_0=(1-k)/(2k)$ and $$\label{EQ_a2}
a_2(+\infty)=a/2=\pm\alpha(k),\quad \text{where~$~\alpha(k):=(1+k)e^{-t_0}$.}$$ Using  and , we obtain $$a_3(+\infty)~=~a_2(+\infty)^2-\displaystyle\int\limits_0^{+\infty}\!\! e^{-2t}p^*_2(t)\,{\mathrm{d}}t,$$ where $p_2^*(t)$ is the value of $p_2$ that corresponds to $\big(c_1,c_2\big)=\big(c_1^*(t),c_2^*(t)\big)$. Elementary calculations yield $p_2^*(t)=2k\big(e^{2(t-t_0)}(1+k)-1\big)$ when ${0\le t\le t_0}$, $p_2^*(t)=2k^2$ for all ${t\ge t_0}$, and hence $$a_3(+\infty)=k\big(1+e^{1-1/k}(1+k)\big)>k=a_3|_{f=f_1}.$$

This gives the maximal value of $\Re a_3$ (and hence of $|a_3|$) in ${\mathcal{S}}_k^B$. There are two extremal functions for $\Re a_3$, which we denote by $f_\pm$, corresponding to two possible choices of the sign in . Since $z\mapsto -f(-z)$ has the same coefficient $a_3$ as $f$, it 