if and only if $0\in\partial\Omega$ and $$0<\frac{1}{F_\mu'(\alpha)}=(F_\mu^{-1})'(0),$$ where $(F_\mu^{-1})'(0)$ is the Julia-Carathéodory derivative of $F_\mu^{-1}$ at $0$. Hence $\mu(\{\alpha\})=(F_\mu^{-1})'(0)$ and (3) holds. Next, note that for any $x\in\mathbb{R}$ we have $F_\mu(\psi(x))=x+if(x)$. Since $F_\mu$ extends continuously to $\mathbb{C}^+\cup\mathbb{R}$, applying the inversion formula (\[inversion\]) gives $$\frac{d\mu^{\mathrm{ac}}}{dx}(\psi(x))=\frac{-1}{\pi}\Im G_\mu(\psi(x))=
\frac{f(x)}{\pi(x^2+f^2(x))},\;\;\;\;\;x\in V^+,$$ which, along with \[prop2.1\](4) gives (5) and (6). As noted above, $F_\mu(x)=0$ a.e. relative to the singular part of $\mu$, from which we deduce that the singular part of $\mu$ is atomic, which gives (4). That (7) follows from (5) and the fact that $\mu$ has at most one atom.

The constants appearing in \[3.10\](1) and (2) are sharp. Indeed, by considering the standard semicircular distribution we have $\Omega=\{z\in\mathbb{C}^+:|z|>1\}$, and then taking $z_1=1$ and $z_2=-1$ shows that $1$ is the best constant in (1), whence the same conclusion for (2) follows immediately.

Recall that the compound free Poisson distribution $p(\lambda,\nu)$ with the rate $\lambda>0$ and jump distribution $\nu$is defined as the weak limit as $n\to\infty$ of $\mu_n^{\boxplus n}$, where $$\mu_n=\left(1-\frac{\lambda}{n}\right)\delta_0+\frac{\lambda}{n}\nu$$ and $\nu$ is compactly supported. The next proposition generalizes the jump distribution with compact support to any measure in $\mathcal{M}$.

\[Poisson\] Given $\nu\in\mathcal{M}$, define $$d\rho(s)=\frac{s^2}{s^2+1}\;d\nu(s).$$ Then $p(\lambda,\nu)=\mathbb{B}_{1+\lambda,1/(1+\lambda)^*}(\mu_0)$, where $\mu_0$ is a measure in 