\section{Introduction}

Let ${\mathcal M}$ be a finite von Neumann algebra equipped with the faithful normal tracial
state $\phi$. We denote by $\widetilde{\mathcal M}$ the set of closed, densely defined
operators affiliated with ${\mathcal M}$. 
Given $T\in \widetilde{\mathcal M}$, 
the \emph{Fuglede-Kadison determinant} of $T$, $\Delta(T)$, is defined by
\[
\Delta(T)= \exp\Big(\int_0^\infty \log t\,{\rm d}\mu_{|T|}(t)\Big).
\]
Let $\CM^\Delta$ be the set of operators $T\in\widetilde{\mathcal M}$ satisfying the condition
\[
\phi(\log^+|T|)= \int_0^\infty \log^+(t)\,{\rm d}\mu_{|T|}(t)<\infty.
\]
It is known that the set $\CM^\Delta$ is a subspace of $\widetilde{\mathcal M}$. In particular, for $T\in \CM^\Delta$ and $\lambda\in\mathbb{C}$, $T-\lambda {\bf 1}\in \CM^\Delta$. 
For $T\in\CM^\Delta$, the function $\lambda\mapsto f(\lambda):= \log \Delta(T-\lambda {\bf 1})$ taking values in $ [-\infty, \infty)$ is subharmonic on $\mathbb{C}$. The \emph{Riesz measure} associated with this subharmonic function is called the \emph{Brown measure} \cite{Brown1986, HaagerupSchultz2007} of $T$ and we denote it by $\mu_T$.  More precisely, 
\[
   {\rm d}\mu_T= \frac{1}{2\pi}\nabla^2 f \,{\rm d}\lambda
\]
taken in the distribution sense. 
Moreover, the Brown measure $\mu_T$ satisfies 
\[
        \quad \log \Delta(T-\lambda {\bf 1})= \int_{\mathbb C} \log|\lambda-z|\,{\rm d} \mu_T(z), \qquad
         \forall \lambda\in{\mathbb C}.
\]


 In a breakthrough paper \cite{Voiculescu1991}, Voiculescu discovered that the mixed moments of a family of independent Gaussian random matrices of large dimension are approximately equal to the mixed moments of a family of free random variables. This spectacular result suggests that the Brown measure of an operator in ${\mathcal M}$ arising from free probability is a natural candidate of empirical spectral distribution of some suitable random matrix model. Let $U, V$ be $N\times N$ independent Haar random unitary and let $\Sigma=\Sigma_N$ be a sequence of diagonal matrices of size $N$ that converges in moments to some operator in ${\mathcal M}$. It is known as the single ring theorem \cite{Guionnet2011singlering, Rudelson2014jams} that the empirical spectral distribution of random matrix model $U\Sigma V$ converges to the Brown measure of an $R$-diagonal operator \cite{NicaSpeicher-Rdiag}. 
 The Brown measure of an $R$-diagonal operator $T$ is supported in a single ring where the inner and outer radii, and the density formula can be expressed by the spectral measure of $|T|$, which were calculated by Haagerup--Larsen--Schultz \cite{HaagerupLarsen2000, HaagerupSchultz2007}. 
 
  Let $x_0\in\CM^\Delta$ be a random  variable with an arbitrary distribution, $*$-free from the $R$-diagonal operator $T$. The purpose of this article is to calculate the Brown measure of $T+x_0$. This question was left open in Biane--Lehner's work \cite{BianeLehner2001}, where they obtained Brown measure formula for such operators in some special cases. 
 In a work by Ho and the second author, the Brown measure of $T+x_0$ is shown to be the limit of the empirical spectral distribution of deformed single ring random matrix model \cite{HoZhong2022}. See Figure \ref{figure:one-in-intro} for a simulation of eigenvalue distribution of random matrix model. 
  


\begin{theorem}[See Theorem \ref{thm:support-Brown-x0-T} and Theorem \ref{thm:main-density-formula}] 
	\label{thm:support-Brown-x0-T-introduction}
	Let $T\in\CM^\Delta$ be $R$--diagonal and $x_0\in\CM^\Delta$ be be a operator that is $*$-free from $T$. Then the support of the Brown measure $\mu_{T+x_0}$ is contained in the closure of following set 
\begin{equation*}
		\Omega(T, x_0)=\bigg\{  \lambda\in\mathbb{C}: ||(x_0-\lambda)^{-1}||_2 ||T||_2> 1\,\,\text{and}\,\,  ||x_0-\lambda||_2 ||T^{-1}||_2>1\bigg\}.
\end{equation*}	
The Brown measure $\mu_{T+x_0}$ is absolutely continuous with respect to Lebesgue measure in $\Omega(T, x_0)$ except possibly at finitely many points. Moreover, the density formula can be expressed explicitly in terms of some subordination functions parametrized by $\lambda$. 
\end{theorem}

\begin{figure}[h]
	\begin{center}
		\begin{center}
		\begin{subfigure}[h]{0.45\linewidth}
	
					\includegraphics[scale=0.62]{HaarUplus4}
			\caption{Brown($x_0+u$)}
		
		\end{subfigure}	
    		\begin{subfigure}[h]{0.45\linewidth}
    	
    		\includegraphics[scale=0.62]{Circularplus4}
    			\caption{Brown($x_0+c$)}
    		
    		\end{subfigure}	
		\end{center}
		\caption{Brown measure simulation for sum of Haar unitary/circular operator and $x_0$ distributed as $\mu_{x_0}=0.25\delta_{-1}+0.5\delta_0+0.25\delta_{1}$}
		\label{figure:one-in-intro}
	\end{center}
\end{figure}


Fix an operator $x_0$, then the boundary of $\Omega(T,x_0)$ is completely determined by the inner and outer radii of the single ring associated with the $R$-diagonal operator $T$. See Figure \ref{figure:one-in-intro} for the comparison of the addition with a Haar unitary and a circular operator. 
The complement of $\Omega(T, x_0)$ in $\mathbb{C}$ naturally splits to the union of two disjoint closed sets. If $x_0$ is selfadjoint, then the set $\Omega(T, x_0)$ is symmetric with respect to the $x$-axis. Moreover, the complement $\mathbb{C}\backslash\Omega(T, x_0)$ has at most one bounded component, and the bounded component is a disk if $x_0$ is selfadjoint, as shown in Figure \ref{figure:one-in-intro}. 
By choosing $x_0=0$ in Theorem \ref{thm:support-Brown-x0-T-introduction}, we see immediately that the Brown measure of an $R$-diagonal operator $T$ is supported in a single ring
\[
    {\text{\rm supp}}(\mu_T)=\{\lambda\in\mathbb{C}:  1/||T^{-1}||_2  < |\lambda|<||T||_2 \}. 
\] 
Our method provides a new proof of the Haagerup-Larsen-Schultz's formula for Brown measure of $R$-diagonal operators. This was done by the second author \cite{Zhong2022Rdiag}, where the key ingredient in that work  was to study the free convolution of a symmetric probability measure with a Bernoulli distribution. In this article, the main tools are subordination functions in free convolution of two symmetric probability measures with arbitrarily distributions. 

We show that the density of the absolutely continuous part of $\mu_{T+x_0}$ is strictly positive in $\Omega(T, x_0)$ for the following cases: (1) $T$ is a Haar unitary operator, (2) $T$ is a circular operator, (3) $T$ is a circular Cauchy operator. The case when $T$ is a circular operator generalizes some results in \cite{Zhong2021Brown}, where the second author calculated the Brown measure of the sum of a circular operator and a bounded operator in ${\mathcal M}$ (see also \cite{HoZhong2020Brown} for earlier results). The circular Cauchy operator \cite{HaagerupSchultz2007} can be expressed as $xy^{-1}$, where $\{x, y\}$ is a circular family. The Brown measure of $xy^{-1}+x_0$ plays a crucial role in the work of Haagerup--Schultz's work \cite{HaagerupSchultz2009} on invariant subspace problem in type II$_1$ factors. We provide an alternative proof for the Brown measure of $xy^{-1}+x_0$ and our proof works for unbounded operator $x_0$. The extension to unbounded operator is useful for generalization of  Haagerup--Schultz's main results to unbounded operators \cite{DykemaSZ2017unbounded}. 


 Let us outline the strategy of the proof. Let $U$ be a Haar unitary operator that is $*$-free from $\{T, x_0\}$. We observe that the following identification of distributions
 \[
 |x_0+T-\lambda{\bf 1}| \underset{\ast{\mathcal D}}{\sim} |(x_0-\lambda{\bf 1})+U^*T|
 \underset{\ast{\mathcal D}}{\sim} |U(x_0-\lambda{\bf 1})+T|.
 \]
 The operator $U(x_0-\lambda{\bf 1})$ is $R$-diagonal that is $*$-free from $T$. It is well-known \cite{HaagerupSchultz2007, Nica-Speicher-1998duke} (see Proposition \ref{Rdiag-additive}) that
 the addition $U(x_0-\lambda{\bf 1})+T$ of two $R$-diagonal operators is also an $R$-diagonal operator, and 
 \[
 \tilde{\mu}_{|U(x_0-\lambda{\bf 1})+T|}=\tilde{\mu}_{|x_0-\lambda{\bf 1}|}\boxplus \tilde{\mu}_{|T|},
 \]
 where $\tilde\mu$ denote the
 {\it symmetrization} of $\mu$. That is, $\tilde\mu$ is defined by
 by
 \[
 \tilde\mu(B) = \textstyle{\frac12}(\mu(B)+\mu(-B)),
 \]
 where $B$ is any Borel measurable set of real numbers. 
 By studying the free convolution of symmetric probability measures, we obtain a formula for the Fuglede-Kadison determinant $\Delta(x_0+T-\lambda{\bf 1})$ using subordination functions parameterized by $\lambda$. As subordination functions satisfy fixed point equation, it allows us to take implicit differentiation. The calculation of Brown measure is then reduced to questions about subordination functions. 



The paper is organized as follows.  In Section 2, we review some relevant results in free convolutions and $R$-diagonal operators. In Section 3, we study the free convolution of two symmetric probability measures. In Section 4, we calculate the Fuglede-Kadison determinant formula and Brown measure formula for the sum of an $R$-diagonal operator and a free random variable. In Section 5, we study explicit formulas for three important cases.

\section{Preliminaries: free additive convolution and $R$-diagonal operators}

\subsection{Free additive convolution}
Given a selfadjoint operator $x\in \mathcal{M}$, let $\mu_x$ 
be the probability distribution of $x$ in the sense that $\mu_x(B)=\phi(E_x(B))$,
for any Boreal measurable set $B$, where $E_x$ denotes the spectral measure of $x$.
For a probability measure $\mu$ on $\mathbb{R}$, its Cauchy transform is defined as
\[
G_\mu(z)=\int_{\mathbb R} \frac{1}{z-x}d\mu(x), \qquad z\in\mathbb{C}^+;
\]
and the $F$-transform is its reciprocal, $F_\mu(z)=1/G_\mu(z)$. The $R$-transform of $\mu$ is defined as $R_\mu(z)=G_\mu^{\langle-1\rangle}(z)-1/z$ in some domain of the form $\Gamma_{\alpha,\beta}=\{ z=x+iy: y>\beta, |x/y|<\alpha \}$ for some $\alpha, \beta>0$ (see \cite{BercoviciVoiculescu1993}). 
Given two selfadjoint random variables $x,y$, freely independent in $(\mathcal{M},\phi)$, denote $\mu_1=\mu_x$ and $\mu_2=\mu_y$. The free additive convolution $\mu_1\boxplus\mu_2$ of $\mu_1$ and $\mu_2$ is defined as $\mu_1\boxplus\mu_2=\mu_{x+y}$. 
The free additive convolution is a binary convolution on probability measures on the real line $\mathbb{R}$, which is the free analogue of usual convolution in classic probability.
The $R$-transform linearizes the free additive convolution in the sense that 
$R_{\mu_1\boxplus \mu_2}(z)=R_{\mu_1}(z)+R_{\mu_2}(z)$ in some domain where all these functions are defined.   
It is known \cite{Biane1998, Voiculescu1991, DVV-general} that there are two analytic functions $\omega_1, \omega_2: \mathbb{C}^+\rightarrow\mathbb{C}^+$ such that 
\begin{equation}
  \label{eqn:2.199}
	 F_{\mu_1\boxplus\mu_2}(z)=F_{\mu_1}(\omega_1(z))=F_{\mu_2}(\omega_2(z)), \qquad z\in\mathbb{C}^+.
\end{equation}
One can show that 
\begin{equation}
 \label{eqn:2.200}
  \omega_1(z)+\omega_2(z)=F_{\mu_1\boxplus\mu_2}(z)+z, \qquad z \in\mathbb{C}^+.
\end{equation}
It turns out that \eqref{eqn:2.199} and \eqref{eqn:2.200} determine $\omega_1, \omega_2$ uniquely.
The subordination functions are defined everywhere in $\mathbb{C}^+$, while $R$-transforms are defined only in some subsets.  Hence, it is convenient to use subordination functions solely to calculate free additive convolution. To be more precise, the subordination functions are unique solutions of certain fixed point equation \cite{BB2007new}. Let $H_1(z)=F_{{\mu}_1}(z)-z$ and $H_2(z)=F_{{\mu}_2}(z)-z$.
Then $\omega_1, \omega_2$ satisfy the following fixed point equations respectively
\[
   \omega_1(z)=z+H_2(z+H_1(\omega_1(z))), \qquad 
      \omega_2(z)=z+H_1(z+H_2(\omega_2(z))).
\]
The above fixed point equations play an important role in our study. 

The subordination functions and free convolution obey some nice regularity properties. 
It is shown \cite{Belinschi2008, BelinschiBH2022} that the subordination functions $\omega_1, \omega_2$ extends as a continuous function $\omega_1, \omega_2: \mathbb{C}^+\cup\mathbb{R}\rightarrow \mathbb{C}^+\cup\mathbb{R}\cup\{\infty\}$. By using \eqref{eqn:2.200}, we see that $F_{\mu_1\boxplus\mu_2}$ has a continuous extension to $\mathbb{C}^+ \cup \mathbb{R}$. This extension satisfies $\omega_i(\infty)=\infty$ for $i=1, 2$, viewed as nontangential limits. 
We summarize some relevant results as follows. 
\begin{theorem}\cite{Belinschi2008, BercoviciVoiculescu1998}
\label{thm:regularity-free-convolution}
Let $\mu_1, \mu_2$ be two probability measures on $\mathbb{R}$ and assume that neither of them is a single delta mass. Then $\mu_1\boxplus \mu_2$ has no singular continuous part. Its absolutely continuous part of $\mu_1\boxplus\mu_2$ is always nonzero and its density is analytic whenever positive and finite. 

Moreover, the point $\gamma\in\mathbb{R}$ is an atom of the measure $\mu_1\boxplus\mu_2$ if and only if there exists $\alpha, \beta\in\mathbb{R}$ such that $\gamma=\alpha+\beta$ and 
 $\mu_1(\{\alpha\})+\mu_2(\{\beta\})>1$. In addition, $\alpha=\omega_1(\gamma), \beta=\omega_2(\gamma)$ and $(\mu_1\boxplus \mu_2)(\{\gamma\})=\mu_1(\{\alpha\})+\mu_2(\{\beta\})-1$.
\end{theorem}

\medskip


\begin{proposition}
\label{prop:regularity-subordination}
Let $\mu_1, \mu_2$ be two probability measures on $\mathbb{R}$ and assume that neither of them is a single delta mass. For $\gamma\in\mathbb{R}$, 
\begin{enumerate}[(1)]
\item $\omega_1(\gamma) \in \mathbb{C}^+$ if and only if $\omega_2(\gamma)\in\mathbb{C}^+$. 
\item   if $F_{\mu_1\boxplus\mu_2}(\gamma)=0$, then
\begin{equation}
 \label{eqn:2.30-in-proof}
     \mu_1(\{ \alpha\})+ \mu_2(\{\beta\})\geq 1.
\end{equation}
where $\alpha=\omega_1(\gamma), \beta=\omega_2(\gamma)$. 
\end{enumerate}
\end{proposition}
\begin{proof}
Given $z\in\mathbb{C}^+$, we note that $\Im F_{\mu_i}(z)>\Im z$ for $i=1,2$. If $\omega_1(\gamma)\in\mathbb{C}^+$, by using \eqref{eqn:2.200}, we have 
\[
  \Im(\omega_2(\gamma))=\Im (F_{\mu_1}(\omega_1(\gamma)) -\omega_1(\gamma) ) >0.
\]
Similarly, if $\omega_2(\gamma)\in\mathbb{C}^+$, then $\omega_1(\gamma)\in\mathbb{C}^+$. This established item (1). 

If $F_{\mu_1\boxplus\mu_2}(\gamma)=0$,
the condition \eqref{eqn:2.30-in-proof} was proved in the proof of Part 3 of \cite[Theorem 4.1]{Belinschi2008} (see Pages 147-149).
\end{proof}



\subsection{Unbounded $R$-diagonal operators}
The $R$-diagonal operators, introduced by Nica-Speicher \cite{NicaSpeicher-Rdiag}, satisfy a number of remarkable symmetric properties.  
In \cite{HaagerupSchultz2007}, the class of $R$-diagonal operators was extended to unbounded case.
Recall that \cite{NicaSpeicher-Rdiag, HaagerupSchultz2007} $T\in\widetilde{\mathcal M}$ is $R$-diagonal if there exist a von Neumann algebra ${\mathcal N}$, with a faithful normal tracial state, and $*$-free elements $U, H\in {\mathcal N}$ with $U$ Haar unitary, $H\geq{0}$, and such that $T$ has the same $*$-distribution as $UH$. If $T\in\widetilde{\mathcal M}$ is $R$-diagonal with $ker(T)=0$, let the polar decomposition of $T$ be $V|T|$, where $|T|=(T^*T)^{1/2}$,  then $V$ is a Haar unitary which is $*$-free from $|T|$. 



\begin{lemma}\label{lemma:lambda-mu-1-2} 
	\cite{HaagerupSchultz2007}
Let $\mu$ be a probability measure on $[0,\infty)$ or a symmetric probability measure on $\mathbb{R}$, and
  put
  \[
  \lambda_1(\mu) = \Bigg(\int_\mathbb{R}
  \frac{1}{u^2}\,{\rm d}\mu(u)\Bigg)^{-\frac12} \quad {\rm and} \quad \lambda_2(\mu) = \Bigg(\int_\mathbb{R}
  u^2\,{\rm d}\mu(u)\Bigg)^{\frac12},
  \]
  with the convention that $\infty^{-\frac12}=0$. 
Then $0\leq \lambda_1(\mu)\leq\lambda_2(\mu)\leq \infty$. 
If its symmetrization $\tilde{\mu}$ is supported at least three points, then 
$\lambda_1(\mu)<\lambda_2(\mu)$. 
\end{lemma}

For a probability measure $\mu$ on $[0,\infty)$ that is not a delta mass at zero, define $\psi_\mu:\mathbb{C}\backslash (0,\infty)\rightarrow\mathbb{C}$ by
\[
\psi_\mu(z)=\int_{[0,\infty)} \frac{zt}{1-zt}\,{\rm d}\mu(t), \qquad z\in
{\mathbb C}\setminus (0,\infty).
\]
One can check that: (i) $\psi_\mu'(x)>0, x\in (-\infty, 0)$; (ii) $\psi_\mu(x)\rightarrow \delta -1$ as $x\rightarrow -\infty$ where $\delta=\mu(\{ 0\})$; (iii) $\psi_\mu(x) \rightarrow 0$ as $x\rightarrow 0$ (see \cite{BercoviciVoiculescu1993}). Hence, in a neighborhood ${\mathcal V}_\mu$ of $(\delta-1,0)$, the inverse $\chi_\mu$ of $\psi_\mu$ exists: $\chi_\mu(z)=\psi_\mu^{\langle-1\rangle}(z), z\in {\mathcal V}_\mu$. Then Voiculescu's $S$-transform is defined by
\[
S_\mu(z)=\frac{z+1}{z}\chi_\mu(z), \quad z\in {\mathcal V}_\mu.
\] 
Given two probability measures $\mu,\nu$ on $[0,\infty)$, 
the $S$-transform satisfies $S_{\mu\boxtimes\nu}(z)=S_{\mu}(z)S_\nu(z)$ in a region where all these functions are defined. 


The Brown measure of bounded $R$-diagonal operators was calculated by Haagerup-Larsen \cite{HaagerupLarsen2000} and the result was further extend by Haaagerup-Schultz \cite{HaagerupSchultz2007} for all $R$-diagonal operators in $\CM^\Delta$. See \cite{BSS2018} for a new proof based on operator-valued free probability and \cite{Zhong2022Rdiag} for another proof based on free convolution of symmetric measures. 

\begin{theorem}
\label{thm-main-Rdiag}
  Let $T$ be an $R$--diagonal element in $\CM^\Delta$ with Brown measure
  $\mu_T$, and suppose $\mu_{T}$ is not a Dirac measure. The Brown measure $\mu_{|T|}$ is the unique rotationally invariant probability measure such that
  \begin{align*}
    \mu_{|T|}\{ \lambda\in\mathbb{C}: |\lambda|\leq r\}=
      \begin{cases}
         0, \quad & \text{for} \quad  r<\lambda_1(\mu_{|T|});\\
         1+S^{\langle -1 \rangle}_{\mu_{\mu_{T^*T}}}(r^{-2}),\quad &\text{for} \quad \lambda_1(\mu_{|T|})<r<\lambda_2(\mu_{|T|});\\
         1,\quad &\text{for}\quad r\geq \lambda_2(\mu_{|T|}).
      \end{cases}
  \end{align*}
When $\mu_{|T|}(\{0\})=\delta>0$, the Brown measure $\mu_T$ has a mass at the origin with the mass $\mu_T(\{0\})=\delta$. 
\end{theorem}

\vspace{.2cm}


The following result is proved in \cite[Proposition 3.5]{HaagerupSchultz2007} (see also \cite{Nica-Speicher-1998duke} for the original proof for bounded case). It is a powerful tool to calculate the distribution of the sum of $R$-diagonal operators. 




\begin{proposition}\label{Rdiag-additive} 
	Let $S,T\in\widetilde{\mathcal M}$ be $\ast$--free $R$--diagonal
	elements. Then $S+T\in\widetilde{\mathcal M}$ is $R$-diagonal and 
	\begin{equation}
	\tilde\mu_{|S+T|}= \tilde\mu_{|S|}\boxplus \tilde\mu_{|T|}.
	\end{equation}
\end{proposition}



\section{The free convolution of symmetric probability measures}
\label{section:free-convolution-symmetric}

In this section, we will study the subordination functions appearing in the free convolution of symmetric probability measures on $\mathbb{R}$. Those results will be useful in later sections. 
Throughout this section, symmetric probability measures are not the single delta mass at zero.
For such a symmetric probability measure $\mu$ on $\mathbb{R}$, we set 
  \begin{equation}
  \label{defn:h-mu-s}
  h_\mu(s)=\int_\mathbb{R}\frac{s}{s^2+u^2}\,{\rm d}\mu(u), \qquad s>0
  \end{equation}
and 
\begin{equation}
\label{defn:f-mu-s}
  f_\mu(s)=\frac{1}{h_\mu(s)}-s, \qquad s>0.
\end{equation}
By symmetry of ${\mu}$, for $s>0$, we have 
\begin{equation}
\label{eqn:G_tilde_h_mu}
   G_{{\mu}}(is)=\int_\mathbb{R}\frac{1}{is-x}d{\mu}(x)=\int_\mathbb{R}\frac{-is}{s^2+x^2}d{\mu}(x)=-ih_\mu(s),
\end{equation}
and consequently, 
\begin{equation}
  \label{eqn:f-and-F}
	  f_\mu(s)=\Im (F_\mu(is)-is), \qquad s>0. 
\end{equation}


Let $\mu_1, \mu_2$ be symmetric probability measures on $\mathbb{R}$, neither of them is a single delta mass at zero. Let $\omega_1, \omega_2$ be subrodination functions such that 
\begin{equation}
\label{eqn:F-tilde-subordination}
  F_{{\mu}_1\boxplus {\mu}_2}(z)=F_{{\mu}_1}(\omega_1(z))=F_{\mu_2}(\omega_2(z)), \qquad z\in \mathbb{C}^+.
\end{equation}

\begin{proposition}
\label{prop:h-mu-omega-s}
For any $t>0$, the functions $\omega_1(it), \omega_2(it)$ are pure imaginary. Set $\omega_1(it)=is_1(t)$ and $\omega_2(it)=is_2(t)$. We have $s_1(t)>t, s_2(t)>t$, and 
\[
    h_{{\mu}_1\boxplus {\mu}_2}(t)=h_{\mu_1}(s_1(t))=h_{\mu_2}(s_2(t)).
\]
Moreover, 
\begin{equation}
\label{eqn:sum-s-1-2}
 s_1(t)+s_2(t)=t+\frac{1}{h_{{\mu}_1\boxplus {\mu}_2}(t)}. 
\end{equation}
\end{proposition}
\begin{proof}
In some domain of the form $\Gamma_{\alpha,\beta}=\{ z=x+iy: y>\beta, |x/y|<\alpha \}$, we have 
\[
    \omega_1(z)=F^{\langle -1\rangle}_{{\mu}_{1}}\left (F_{{\mu}_{1}\boxplus{\mu}_2}(z) \right), 
    \qquad z\in\Gamma_{\alpha,\beta}.
\]
Hence, by \eqref{eqn:G_tilde_h_mu} and \eqref{eqn:F-tilde-subordination}, $\omega_1(it)$ is purely imaginary for $t>\beta$ and therefore $\omega_1(it)$ is purely imaginary for all $t>0$ by analytic extension. 
Let $s_1(t)=\Im \omega_1(it)$, we have $s_1(t)=\Im \omega_1(it)>\Im (it)=t$ (the strict inequality holds because ${\mu}_1$ is a non-trivial probability measure). Finally, 
\[
h_{{\mu}_1\boxplus {\mu}_2}(t)=-\Im G_{{\mu}_1\boxplus {\mu}_2}(it)
=-\Im G_{{\mu}_1}(\omega_1(it))=h_{{\mu}_1}(s_1(t)).
\]
Same argument works for $\omega_2$. 

From free convolution of ${\mu}_1$ and ${\mu}_2$, we have 
\begin{equation}
 \label{eqn:3.70-in-proof}
     \omega_1(z)+\omega_2(z)=z+F_{{\mu}_1\boxplus{\mu}_2}(z), \quad z\in\mathbb{C}^+.
\end{equation}
It follows that \eqref{eqn:sum-s-1-2} holds by letting $z=it$.
\end{proof}



We have 
\begin{equation}
  \label{eqn:s-f-mu}
    sf_\mu(s)=\frac{s}{h_\mu(s)}-s^2=\frac{n_\mu(s)}{d_\mu(s)}
\end{equation}
where 
\[
  n_\mu(s)=\int_\mathbb{R} \frac{u^2}{s^2+u^2}d\mu(u) \qquad \text{and}\qquad 
     d_\mu(s)=\int_\mathbb{R} \frac{1}{s^2+u^2}d\mu(u).
\]
The following result is taken from the proof of \cite[Lemma 4.8]{HaagerupSchultz2007} (see also \cite{Zhong2022Rdiag}). We provide a new proof for reader's convenience.
\begin{lemma}
\label{lemma:monotonicity-s-f-mu}
Let $\mu$ be a symmetric probability measure on $\mathbb{R}$, and $\lambda_1(\mu), \lambda_2(\mu)$ be as in Lemma \ref{lemma:lambda-mu-1-2}, and let $f_\mu$ be as in \eqref{defn:f-mu-s}. Then the function $s\mapsto sf_\mu(s)$ is a strictly increasing bijection of $(0,\infty)$ onto $(\lambda_1(\mu)^2, \lambda_2(\mu)^2)$. 
\end{lemma}
\begin{proof}
Sine $\mu$ is a symmetric probability measure, then $F_\mu$ has the following Nevanlinna representation
\[
   F_\mu(z)=z+\int_\mathbb{R}\frac{1+uz}{u-z}d\rho(u),
\]
where $\rho$ is a finite measure on $\mathbb{R}$. It is easy to see that $\rho$ is also a symmetric measure. Hence, by \eqref{eqn:f-and-F}, we have
\[
   sf_\mu(s)=s\cdot\left( \Im \int_\mathbb{R}\frac{1+ius}{u-is}d\rho(u) \right)
   =\int_\mathbb{R} \frac{(1+u^2)s^2}{u^2+s^2}d\rho(u).
\]
This implies that the function $s\mapsto sf_\mu(s)$ is strictly increasing function on $(0,\infty)$. By \eqref{eqn:s-f-mu}, we can verify that
\[
   \lim_{s\rightarrow 0} sf_\mu(s)=\frac{1}{\int_\mathbb{R}\frac{1}{u^2}d\mu(u)}=\lambda_1^2(\mu),
\]
and 
\[
  	\lim_{s\rightarrow 0} sf_\mu(s)=\int_\mathbb{R}u^2 d\mu(u)=\lambda_2^2(\mu). 
\]
This completes the proof. 
\end{proof}

Recall that $\omega_1, \omega_2$ satisfy the following fixed point equations respectively
\[
   \omega_1(z)=z+H_2(z+H_1(\omega_1(z))), \qquad 
      \omega_2(z)=z+H_1(z+H_2(\omega_2(z))),
\]
where $H_1(z)=F_{{\mu}_1}(z)-z$ and $H_2(z)=F_{{\mu}_2}(z)-z$.
Then,
for $z=it$, we have $H_1(it)=\frac{1}{G_{{\mu}_1}(it)}-it=i\left(\frac{1}{h_{\mu_1}(t)}-t\right)=if_{\mu_1}(t)$
and $H_2(it)=i\left(\frac{1}{h_{\mu_2}(t)}-t\right)=if_{\mu_2}(t)$. Using Proposition \ref{prop:h-mu-omega-s}, the above fixed point equations for subordination functions are then rewritten as the following equations for $s_1(t)$ and $s_2(t)$, 
\begin{equation}
\label{eqn:fixed-point-s-1-2}
s_1(t)=t+f_{\mu_2}(t+f_{\mu_1}(s_1(t))), \qquad
 s_2(t)=t+f_{\mu_1}(t+f_{\mu_2}(s_2(t))).
\end{equation}
Thanks to \cite[Theorem 4.1]{BB2007new}, the above equation \eqref{eqn:fixed-point-s-1-2} determines unique solutions $s_1(t), s_2(t)$ for any $t>0$. 


Recall that the subordination functions $\omega_1, \omega_2$ extends as continuous functions on $\mathbb{C}^+\cup\mathbb{R}$ taking values in $\mathbb{C}^+\cup\mathbb{R}\cup\{\infty\}$. Hence, $\Im \omega_i(x)\in [0, \infty]$ and  the function $s_1(t), s_2(t)$ extends as continuous function on $[0,\infty)$ taking values in $[0,\infty]$. We set 
\[
s_1(0)=\lim_{t\downarrow 0}s_1(t), \qquad s_2(0)=\lim_{t\downarrow 0}s_2(t).
\]





\begin{lemma}
\label{lemma:s-0-all-cases}
Assume that $\mu_1, \mu_2$ are symmetric probability measures and neither of them is a single delta mass at zero. Then, at least one of $s_1(0), s_2(0)$ is finite. Let $f_{\mu_1\boxplus\mu_2}$ be the density function of the absolutely continuous part of $\mu_1\boxplus \mu_2$. Then,
\begin{enumerate}[(1)]
\item $s_1(0)\in (0,\infty)$ if and only if $s_2(0)\in (0,\infty)$, and in this case
\[
   f_{\mu_1\boxplus\mu_2}(0)\in (0,\infty);
\]
\item if $s_1(0) =0, s_2(0)=\infty$, then $\lambda_1(\mu_1)\geq \lambda_2(\mu_2)$ and $   f_{\mu_1\boxplus\mu_2}(0)=0$. 
Moreover, 
\begin{equation}
\lim_{t\downarrow{0}} \frac{t}{s_1(t)}= \frac{\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2)}{\lambda_1^2(\mu_1)},
\end{equation}
and 
\begin{equation}
 \lim_{t\downarrow {0}} t\cdot s_2(t)=\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2). 
\end{equation}

\item if $s_1(0)=\infty, s_2(0)=0$, then $\lambda_1(\mu_2)\geq \lambda_2(\mu_1)$ and $ f_{\mu_1\boxplus\mu_2}(0)=0$.  Moreover, 
\begin{equation}
 \lim_{t\downarrow{0}} \frac{t}{s_2(t)}=\frac{\lambda_1^2(\mu_2)-\lambda_2^2(\mu_1)}{\lambda_1^2(\mu_2)},
\end{equation}
and 
\begin{equation}
\lim_{t\downarrow{0}} t\cdot s_1(t)=\lambda_1^2(\mu_2)-\lambda_2^2(\mu_1). 
\end{equation}

\item if $s_1(0)=s_2(0)=0$, then 
\[
   \lim_{\varepsilon\rightarrow 0} \Im G_{\mu_1\boxplus \mu_2}(i\varepsilon) =\infty.
\]
Moreover, $\lambda_1(\mu_1)=\lambda_1(\mu_2)=0$ and 
\begin{equation}
  \mu_1(\{0\})+\mu_2(\{0\})\geq 1.
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof}
Recall that $s_i(t)=\omega_i(t)$ for $t\in [0,\infty)$. We have  
\[
  s_2(t)=\frac{1}{h_{\mu_1}(s_1(t))}-s_1(t)+t
     =\Im (F_{\mu_1}(is_1(t)))-is_1(t)) +t.
\]
Hence, if $s_1(0)=\infty$, then $s_2(0)=0$ by the asymptotics of $F$-transform. Similarly, 
if $s_2(0)=\infty$, then $s_1(0)=0$. 

By item (1) in Proposition \ref{prop:regularity-subordination}, we see that $s_1(0)\in (0,\infty)$ if and only if $s_2(0)\in (0,\infty)$. In this case,
\[
   F_{\mu_1\boxplus\mu_2}(0)=i( s_1(0)+s_2(0)),
\]
and by Stieltjes inversion formula, we obtain
\[
f_{\mu_1\boxplus\mu_2}(0)=-\frac{1}{\pi}\Im G_{\mu_1\boxplus\mu_2}(0)=\frac{1}{\pi}\frac{1}{s_1(0)+s_2(0)}>0.
\]
This proves part (1). 

We multiply the factor $(t+f_{\mu_1}(s_1(t)))$ to the first equation in \eqref{eqn:fixed-point-s-1-2} and then rewrite it as follows
\begin{align}
\label{eqn:3.9000}
  (t+f_{\mu_1}(s_1(t))(s_1(t)-t)=(t+f_{\mu_1}(s_1(t))) f_{\mu_2}(t+f_{\mu_1}(s_1(t))).
\end{align}
We recall that for a symmetric measure $\mu$, we have 
\begin{equation}
\label{eqn:limit-s-fmu-in-proof}
   \lim_{s\downarrow 0} s f_{\mu}(s)=\lambda_1^2(\mu), \qquad \lim_{s\uparrow \infty} s f_{\mu}(s)=\lambda_2^2(\mu).
\end{equation}
Hence, if $s_1(0)=0$, then
\begin{equation} 
\label{eqn:3.109}
\begin{aligned}
\lim_{t\downarrow 0} (t+f_{\mu_1}(s_1(t))(s_1(t)-t)&=\lim_{t\downarrow{0}}\left(s_1(t)f_{\mu_1}(s_1(t))-tf_{\mu_1}(s_1(t)) \right)\\
&\leq \lim_{s\downarrow 0}sf_{\mu_1}(s)= \lambda_1^2(\mu_1). 
\end{aligned}
\end{equation}
And if $s_1(0)=\infty$, then 
\begin{equation}
 \label{eqn:3.100}
 \lim_{t\downarrow {0}}f_{\mu_1}(s_1(t))=\lim_{s\uparrow\infty}f_{\mu_1}(s)=0,
\end{equation}
and hence
\begin{align}
  \label{eqn:3.110}
 \lim_{t\downarrow 0} (t+f_{\mu_1}(s_1(t))(s_1(t)-t)&= \lim_{t\downarrow 0} ts_1(t)
     +\lim_{t\downarrow 0}s_1(t)f_{\mu_1}(s_1(t))\nonumber\\
     &\geq \lim_{t\downarrow 0}s_1(t)f_{\mu_1}(s_1(t)) =\lambda_2^2(\mu_1).
\end{align}

We now proceed the proof case by case. 

(2) Assume that $s_1(0)=0, s_2(0)=\infty$, then by \eqref{eqn:sum-s-1-2}, 
we have 
\[
    f_{\mu_1}(s_1(t))=\frac{1}{h_{\mu_1}(s_1(t)}-s_1(t)=s_2(t)-t.
\]
and hence $\lim_{t\downarrow 0}(t+f_{\mu_1}(s_1(t)))=\infty$. We then have 
\begin{equation} \label{eqn:1.200}
	\lim_{t\downarrow 0}(t+f_{\mu_1}(s_1(t))) f_{\mu_2}(t+f_{\mu_1}(s_1(t)))=
	\lim_{s\uparrow \infty} sf_{\mu_2}(s)=\lambda_2^2(\mu_2).
\end{equation}
By combing \eqref{eqn:3.100}, we deduce that $\lambda_1(\mu_1)\geq \lambda_2(\mu_2)$. 

It follows that $\lambda_1(\mu_1)>0$. Hence, 
\[
 \begin{aligned}
 \lim_{t\downarrow 0} (t+f_{\mu_1}(s_1(t))(s_1(t)-t)&=\lim_{t\downarrow{0}}\left(s_1(t)f_{\mu_1}(s_1(t))-tf_{\mu_1}(s_1(t)) \right)\\
   &=\lim_{t\downarrow{0}} s_1(t)f_{\mu_1}(s_1(t))\left[ 1-\frac{t}{s_1(t)}  \right]\\
   &=\left(\lim_{s\downarrow {0}} s f_{\mu_1}(s)\right)\cdot\left(\lim_{t\downarrow{0}}\left[ 1-\frac{t}{s_1(t)}  \right]\right)\\
   &=\lambda_1^2(\mu_1) \cdot\left(\lim_{t\downarrow{0}}\left[ 1-\frac{t}{s_1(t)}  \right]\right), 
 \end{aligned}
\]
which implies that $\lim_{t\downarrow 0} t/s_1(t)$ exists. By comparing \eqref{eqn:1.200}, we obtain
\[
 \lim_{t\downarrow{0}} \frac{t}{s_1(t)}= \frac{\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2)}{\lambda_1^2(\mu_1)}. 
\]

We multiply the factor $(t+f_{\mu_2}(s_2(t)))$ to the second equation in \eqref{eqn:fixed-point-s-1-2} and then rewrite it as follows
\begin{align}
\label{eqn:3.2001}
(t+f_{\mu_2}(s_2(t))(s_2(t)-t)=(t+f_{\mu_2}(s_2(t))) f_{\mu_1}(t+f_{\mu_2}(s_2(t))).
\end{align}
Then, 
\[
 \lim_{t\downarrow {0}}f_{\mu_2}(s_2(t))=\lim_{s\uparrow\infty}f_{\mu_2}(s)=0,
\]
which yields 
\begin{equation}
  \label{eqn:3.2002}
 \lim_{t\downarrow {0}} (t+f_{\mu_2}(s_2(t))) f_{\mu_1}(t+f_{\mu_2}(s_2(t)))
 =\lim_{s\downarrow {0}} sf_{\mu_1}(s)=\lambda_1^2(\mu_1). 
\end{equation}
Taking the limit for the left hand side of \eqref{eqn:3.2001} similar to \eqref{eqn:3.110} (by switching roles of $\mu_1$ and $\mu_2$), we deduce that
\begin{equation}
   \label{eqn:3.2003}
	 \begin{aligned}
		\lim_{t\downarrow {0}} (t+f_{\mu_2}(s_2(t))(s_2(t)-t)
		&=\lim_{t\downarrow {0}} t\cdot s_2(t) +\lim_{s\uparrow \infty} s f_{\mu_2}(s)\\
		&=\lim_{t\downarrow {0}} t\cdot s_2(t)+\lambda_2^2(\mu_2). 
	\end{aligned}
\end{equation}
By comparing \eqref{eqn:3.2002} with \eqref{eqn:3.2003}, we deduce that
\[
  \lim_{t\downarrow {0}} t\cdot s_2(t)=\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2). 
\]

(3) Assume that $s_1(0)=\infty, s_2(0)=0$, then we may switch roles of $\mu_1$ and $\mu_2$ and results follow from Case (2).  

 (4) Assume that $s_1(0)=s_2(0)=0$, then by \eqref{eqn:sum-s-1-2}, 
 \begin{equation}
  \label{eqn:3.1300}
       \lim_{t\downarrow 0} h_{\mu_1\boxplus\mu_2}(t)=\lim_{t\downarrow 0}\frac{1}{s_1(t)+s_2(t)+t}=\infty. 
 \end{equation}
 This means $-\lim_{\varepsilon\rightarrow 0} \Im G_{\mu_1\boxplus \mu_2}(i\varepsilon) =\infty$. Hence, 
 \begin{equation}
  \label{eqn:3.120}
       \lim_{t\rightarrow 0}\frac{1}{h_{\mu_1\boxplus \mu_2}(t)}=0,
 \end{equation}
 and equivalently, $\lim_{t\rightarrow 0}F_{\mu_1\boxplus \mu_2}(i t)=0$. 
 By \eqref{eqn:sum-s-1-2}, we see that
 \[
     \lim_{t\rightarrow 0}(s_1(t)+s_2(t)) = 0.
 \]
 Hence,  $\lim_{t\rightarrow 0}s_j (t)=0$ for $j=1,2$. Recall that $\omega_j(i\varepsilon)$ is a pure imaginary number for $\varepsilon>0$ and $j=1,2$. It folllows that $\omega_j(0)=\lim_{t\rightarrow 0}\omega(i t)=0$. By item (2) of Proposition \ref{prop:regularity-subordination}, we have 
 \[
  \mu_1(\{0\})+\mu_2(\{0\})\geq 1.
 \]
 
 Finally, \eqref{eqn:3.120} and $s_1(0)=0$ also imply that 
 \[
   f_{\mu_1}(s_1(t))=\frac{1}{h_{\mu_1}(s_1(t)}-s_1(t) \longrightarrow 0
 \]
 as $t\rightarrow 0$. Hence, 
 \[
    \lim_{t\downarrow 0}(t+f_{\mu_1}(s_1(t))) f_{\mu_2}(t+f_{\mu_1}(s_1(t)))=
      \lim_{s\uparrow 0} sf_{\mu_2}(s)=\lambda_1^2(\mu_2).
\]
On the other hand, we have 
\begin{align*}
  \lim_{t\downarrow 0} (t+f_{\mu_1}(s_1(t))(s_1(t)-t)&=\lim_{t\downarrow{0}}\left(s_1(t)f_{\mu_1}(s_1(t))-tf_{\mu_1}(s_1(t)) \right)\\
   &= \lim_{s\downarrow 0}sf_{\mu_1}(s)+\lim_{t\downarrow 0} t f_{\mu_1}(s_1(t))\\
   &= \lambda_1^2(\mu_1).
\end{align*}
It follows from \eqref{eqn:3.9000} that $\lambda_1(\mu_1)=\lambda_1(\mu_2)$. 

We next show that $\lambda_1(\mu_1)= 0$ by contradiction. If $\lambda_1(\mu_1)=\lambda_1(\mu_2)=r\neq 0$, then the assumption $s_1(0)=s_2(0)=0$ implies that 
 \begin{align*}
    \lim_{t\downarrow 0}h_{\mu_1}(s_1(t))&=\lim_{s\downarrow 0} h_{\mu_1}(s)\\
        &=\lim_{s\downarrow 0} \int_\mathbb{R} \frac{
        s}{s^2+u^2}d\mu_1(u)\\
       &\leq \lim_{s\downarrow 0} \int_\mathbb{R} \frac{
       	s}{u^2}d\mu_1(u)=0.
 \end{align*}
 This contradicts to \eqref{eqn:3.1300}. Hence, $\lambda_1(\mu_1)=\lambda_1(\mu_2)=0$ when $s_1(0)=s_2(0)=0$. 
\end{proof}




\begin{lemma}
\label{lemma:s-0-finite}
Let $\mu_1, \mu_2$ be two symmetric probability measures on $\mathbb{R}$, neither of them is a single delta mass. 
If $\lim_{t\downarrow 0}s_1(t)\in (0,\infty)$, then the two open intervals $(\lambda_1(\mu_1),\lambda_2(\mu_1))$ and $ (\lambda_1(\mu_2),\lambda_2(\mu_2))$ must have nonempty intersection:
\[
   (\lambda_1(\mu_1),\lambda_2(\mu_1))\cap  (\lambda_1(\mu_2),\lambda_2(\mu_2))\neq \emptyset.
\]
\end{lemma}
\begin{proof}
We set $s=\lim_{t\downarrow 0}s_1(t)\in (0,\infty)$. Notice now $f_{\mu_1}(s)=-i \left( \frac{1}{G_{{\mu}_1}(is)}-is\right)=\Im \left( F_{{\mu}_1}(is)-is\right)>0$ and also $f_{\mu_1}(s)<\infty$. Similarly, $0<f_{\mu_2}(s)<\infty$. By taking the limit as $t$ goes to zero for both side of the fixed equation \eqref{eqn:fixed-point-s-1-2}, we obtain
\[
  s=f_{\mu_2}(f_{\mu_1}(s))=\frac{1}{f_{\mu_1}(s)}{f_{\mu_1}(s) f_{\mu_2}(f_{\mu_1}(s))}=
    \frac{1}{f_{\mu_1}(s)} \frac{n_{\mu_2}(f_{\mu_1}(s))}{d_{\mu_2}(f_{\mu_1}(s))}.
\]
Hence, we must have 
\begin{equation}
 \label{eqn:balance-n-d}
   sf_{\mu_1(s)}=\frac{n_{\mu_1}(s)}{d_{\mu_1}(s)}=
 \frac{n_{\mu_2}(f_{\mu_1}(s))}{d_{\mu_2}(f_{\mu_1}(s))}.
\end{equation}
If $ (\lambda_1(\mu_1),\lambda_2(\mu_1))\cap  (\lambda_1(\mu_2),\lambda_2(\mu_2))=\emptyset$, we have either $\lambda_1(\mu_1)\geq\lambda_2(\mu_2)$ or $\lambda_1(\mu_2)\geq\lambda_2(\mu_1)$. If $\lambda_1(\mu_1)\geq\lambda_2(\mu_2)$, by Lemma \ref{lemma:monotonicity-s-f-mu}, we have 
\[
  \frac{n_{\mu_1}(s)}{d_{\mu_2}(s)}>\lim_{t\downarrow 0} \frac{n_{\mu_1}(t)}{d_{\mu_1}(t)}=\lambda_1^2(\mu_1)
     \geq \lambda_2^2(\mu_2)= \lim_{t\uparrow \infty} \frac{n_{\mu_2}(t)}{d_{\mu_2}(t)}> \frac{n_{\mu_2}(f_{\mu_1}(s))}{d_{\mu_2}(f_{\mu_1}(s))}
\]
which contradicts to \eqref{eqn:balance-n-d}. Similarly, if $\lambda_1(\mu_2)\geq \lambda_2(\mu_1)$, we have 
\[
    \frac{n_{\mu_1}(s)}{d_{\mu_2}(s)}<\lim_{t\uparrow\infty} \frac{n_{\mu_1}(t)}{d_{\mu_1}(t)}
     \leq\lim_{t\downarrow 0} \frac{n_{\mu_2}(t)}{d_{\mu_2}(t)}<\frac{n_{\mu_2}(f_{\mu_1}(s))}{d_{\mu_2}(f_{\mu_1}(s))}
\]
which again contradicts to \eqref{eqn:balance-n-d}. Hence if $\lim_{t\downarrow 0}s_1(t)\in (0,\infty)$, the two intervals must have non-empty intersection.
\end{proof}

\begin{corollary}
\label{cor:s-1-2-zero}
Let $\mu_1, \mu_2$ be two symmetric probability measures on $\mathbb{R}$, neither of them is a single delta mass at zero.
If $\mu_1(\{0\})+\mu_2(\{0\})=1$, then $s_1(0)=s_2(0)=0$. 
\end{corollary}
\begin{proof}
For a probability measur $\nu$ on $\mathbb{R}$. 
It is known (see \cite{BercoviciVoiculescu1998} or \cite[Lemm 2.17]{Belinschi2008}) that $\nu(\{x\})=\lim_{t\downarrow 0}(it)G_\nu(x+it)$. If $\mu_1(\{0\})+\mu_2(\{0\})=1$, then $\mu_j(\{0\})>0$ for $j=1,2$. Hence, 
$\lim_{t\downarrow 0}G_{\mu_j}(it)=\infty$ and $\lim_{t\downarrow 0}F_{\mu_j}(it)=0$. 
Then, by \cite[Lemma 2.13]{Belinschi2008}, the Julia-Carath\'{e}odory derivative at zero is given by
\[
    \liminf_{z\rightarrow 0}\frac{\Im F_{\mu_j}(z)}{\Im z}=\lim_{t\downarrow 0}\frac{F_{\mu_j}(it)}{it}=\frac{1}{\mu_j(\{0\})}.
\]
Therefore the Julia-Carath\'{e}odory of $H_j$ at zero are given by
\begin{equation*}
      H_j'(0)=\frac{1-\mu_j(\{0\})}{\mu_{(0)}}, \qquad j=1, 2,
\end{equation*}
where $H_j(z)=F_{\mu_j}(z)-z$, and $H_j(it)$ tends to zero nontangentially as $t\downarrow 0$. Consequently, the function $H_2\circ H_1$ has  Julia-Carath\'{e}odory derivative 
\begin{equation}
 \label{eqn:3.100-in-proof}
    (H_2\circ H_1)'(0)=\frac{1-\mu_1(\{0\})}{\mu_1(\{0\})}\times \frac{1-\mu_2(\{0\})}{\mu_2(\{0\})}=1.
\end{equation}

Let $g(z, w)=z+H_2(z+H_1(2))$. For any $z\in\mathbb{C}^+\cup \mathbb{R}$, it is known \cite{BB2007new, BelinschiBH2022} that $\omega_1(z)$ is the Denjoy-Wolf point of the function $w\rightarrow g(z,w)$. In particular, $\omega_1(0)$ is the Denjoy-Wolff point of $H_2\circ H_1$. Because $\lim_{t\downarrow 0} H_2(H_1(it))=0$ and \eqref{eqn:3.100-in-proof}, we deduce that the Denjoy-Wolf point of $H_2\circ H_1$ at zero is zero. Hence, $\omega_1(0)=0$. By using $H_1\circ H_2$, the same argument shows $\omega_2(0)=0$. We conclude that $s_1(0)=s_2(0)=0$. 
\end{proof}

\begin{theorem}
\label{thm:lambda-1-2-cases}
Let $\mu_1, \mu_2$ be two symmetric probability measures on $\mathbb{R}$, neither of them is a single delta mass, and let $s_1, s_2$ be subordination function defined in Proposition \ref{prop:h-mu-omega-s}. Then we have
\begin{enumerate}[(1)]
 \item If $\lambda_1(\mu_1)\geq\lambda_2(\mu_2)$, then $\lim_{t\downarrow 0}s_1(t)=0$
 and $\lim_{t\downarrow 0} s_2(t)=\infty$.
 \item If $\lambda_1(\mu_2)\geq\lambda_2(\mu_1)$, then $\lim_{t\downarrow 0}s_1(t)=\infty$
 and $\lim_{t\downarrow 0} s_2(t)=0$.
 \item If $(\lambda_1(\mu_1),\lambda_2(\mu_1))\cap  (\lambda_1(\mu_2),\lambda_2(\mu_2))\neq \emptyset$, and either $\lambda_1(\mu_1)>0$ or $\lambda_1(\mu_2)>0$, then $s_1(0)=\lim_{t\downarrow 0}s_1(t)\in (0,\infty)$ and $s_2(0)=\lim_{t\downarrow 0}s_2(t)\in (0,\infty)$. Moreover, 
 \begin{equation}
      s_1(0)+s_2(0)=\frac{1}{h_{\mu_1}(s_1(0))}=\frac{1}{h_{\mu_2}(s_2(0))}.
 \end{equation}
 \item If $\lambda_1(\mu_1)=\lambda_1(\mu_2)=0$, then $s_1(0), s_2(0)\in [0, \infty)$. Moreover, $s_1(0)=s_2(0)=0$ if and only if $\mu_1(\{0\})+\mu_2(\{0\})\geq 1$. 
\end{enumerate}
\end{theorem}
\begin{proof}
If $\lambda_1(\mu_1)\geq \lambda_2(\mu_2)$ then the limit $\lim_{t\downarrow 0}s_1(t)$ can not be finite due to Lemma \ref{lemma:s-0-finite}. If $s_1(0)=s_2(0)=0$, then by item (4) of Lemma \ref{lemma:s-0-all-cases}, we have 
\[
    \lambda_1(\mu_2)= \lambda_1(\mu_1) \geq \lambda_2(\mu_2) >\lambda_1(\mu_2)
\]
which contradicts to the assumption that $\mu_2$ is not a single delta mass. By comparing (2) and (3) in Lemma \ref{lemma:s-0-all-cases}, we see that $\lim_{t\downarrow 0}s_1(t)=0$
 and $\lim_{t\downarrow 0} s_2(t)=\infty$ which proves (1). Then (2) follows in the same way.

Assume now $(\lambda_1(\mu_1),\lambda_2(\mu_1))\cap  (\lambda_1(\mu_2),\lambda_2(\mu_2))\neq \emptyset$. By (1) in Lemma \ref{lemma:s-0-all-cases}, at least one of $s_1(0), s_2(0)$ must be finite. If $s_1(0)=0, s_2(0)=\infty$ or $s_1(0)=\infty, s_2(0)=0$, then we will have $\lambda_1(\mu_1)\geq \lambda_2(\mu_2)$ or $\lambda_1(\mu_2)\geq \lambda_2(\mu_1)$, which contradicts to our assumption. By (4) in Lemma \ref{lemma:s-0-all-cases}, $s_1(0), s_2(0)$ can not be both equal to zero. We then conclude that $0<s_1(0), s_2(0)<\infty$. This proves (3). 

Assume that $\lambda_1(\mu_1)=\lambda_1(\mu_2)=0$. If $s_1(0)=\infty$, 
we deduce that $\lambda_1(\mu_1)=\lambda_2(\mu_1)$, which contradicts to the assumption that $\mu_1$ is not a single delta mass. Hence, $0\leq s_1(0) <\infty$. Similarly, $0\leq s_2(0)<\infty$. By (4) in Lemma \ref{lemma:s-0-all-cases}, $s_1(0)=s_2(0)=0$ implies that $\mu_1(\{0\})+\mu_2(\{0\})\geq 1$. 

For the necessary part of (4), if $\mu_1(\{0\})+\mu_2(\{0\})> 1$, then, by Theorem \ref{thm:regularity-free-convolution}, we have 
\[
  (\mu_1\boxplus \mu_2) (\{0\})=\mu_1(\{0\})+\mu_2(\{0\})-1>0.
\]
Hence, $F_{\mu_1\boxplus\mu_2}(0)=0$. By using $\frac{1}{h_{\mu_1\boxplus\mu_2}(0)}=\Im F_{\mu_1\boxplus\mu_2}(0)=0$ and
\[
    s_1(t)+s_2(t)=\frac{1}{h_{\mu_1\boxplus\mu_2}(t)}-t,
\]
we deduce that $s_1(0)=s_2(0)=0$. Finally, if $\mu_1(\{0\})+\mu_2(\{0\})= 1$, then  $s_1(0)=s_2(0)=0$ thanks to Corollary \ref{cor:s-1-2-zero}.
\end{proof}

We end this section with an application to the inner and outer radii of the support of the sum of two $R$-diagonal operators. 

\begin{proposition}
	\label{lemma:lambada-1-T-1-2}
	Let $T_1, T_2 \in\CM^\Delta$ be $R$--diagonal that are freely independent. Denote by $r_1=\lambda_1(\mu_{|T_1+T_2|})$ the inner radii of the support of the Brown measure of $T_1+T_2$.
	\begin{enumerate}[(1)]
	\item If\, $0< \lambda_2(\mu_{|T_2|})\leq \lambda_1(\mu_{|T_1|})$, then
	$r_1={\lambda_1(\mu_{|T_1|})^2-\lambda_2(\mu_{|T_2|})^2}$.
	\item If\, $0< \lambda_2(\mu_{|T_1|})\leq \lambda_1(\mu_{|T_2|})$, then
  $r_1={\lambda_1(\mu_{|T_2|})^2-\lambda_2(\mu_{|T_1|})^2}$.
	\item If\, $(\lambda_1(\mu_{|T_1|}), \lambda_2(\mu_{|T_1|}))\cap (\lambda_1(\mu_{|T_2|}), \lambda_2(\mu_{|T_2|}))\neq \emptyset$, then 
	$r_1=0$.
	\end{enumerate}
\end{proposition}
\begin{proof}
	With $\mu_1=\tilde{\mu}_{|T_1|}$ and $\mu_2=\tilde{\mu}_{|T_2|}$, let the function $h_{\mu_1}, h_{\mu_2}$ and the subordination function $s_1(t), s_2(t)$ be as in Section \ref{section:free-convolution-symmetric}. When $0< \lambda_2(\mu_{|T_2|})\leq \lambda_1(\mu_{|T_1|})$, by Theorem \ref{thm:lambda-1-2-cases}, $\lim_{t\rightarrow 0
		^+}s_1(t)=0$ and $\lim_{t\rightarrow 0
		^+}s_1(t)=\infty$. 
By item (2) in Lemma \ref{lemma:s-0-all-cases}, we have
\[
    \lim_{t\downarrow 0} \frac{t}{s_1(t)} =\frac{\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2)}{\lambda_1^2(\mu_1)}.
\]
Therefore, 
	\begin{align*}
	\lambda_1(\mu_{|T_1+T_2|})&=
	\frac{1}{\phi\large( [(T_1+T_2)^*(T_1+T_2)]^{-1} \large)}\\
	&=\lim_{t\rightarrow 0^+}\frac{1}{ \phi\large( [(T_1+T_2)^*(T_1+T_2)+t^2]^{-1} \large)}\\
	&=\lim_{t\rightarrow 0^+} \frac{t}{h_{{\mu}_1\boxplus {\mu}_2}(t)}\\
	&=\lim_{t\rightarrow 0^+} \frac{s_1(t)}{h_{\mu_1}(s_1(t))}\frac{t}{s_1(t)}\\
	&=\lim_{t\rightarrow 0^+} \frac{s_1(t)}{h_{\mu_1}(s_1(t))}\lim_{t\rightarrow 0^+}\frac{t}{s_1(t)}\\
	&=\lambda_1^2(\mu_1) \frac{\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2)}{\lambda_1^2(\mu_1)}\\
	&={\lambda_1^2(\mu_1)-\lambda_2^2(\mu_2}).
	\end{align*}
	The case when $0< \lambda_2(\mu_{|T_1|})\leq \lambda_1(\mu_{|T_2|})$ follows by symmetric role of $T_1$ and $T_2$.
	
	When $ (\lambda_1(\mu_{|T_1|}), \lambda_2(\mu_{|T_1|}))\cap (\lambda_1(\mu_{|T_2|}), \lambda_2(\mu_{|T_2|}))\neq \emptyset$, by part (3) and (4) in Theorem \ref{thm:lambda-1-2-cases}, we have $s_1(0), s_2(0)\in [0,\infty)$.
	\begin{itemize}
	 \item If $s_1(0), s_2(0)\in (0,\infty)$, then
	\begin{align*}
	\frac{1}{\phi\large( [(T_1+T_2)^*(T_1+T_2)]^{-1} \large)}=\lim_{t\rightarrow 0^+}\frac{s_1(t)}{h_{\mu_1}(s_1(t))}\frac{t}{s_1(t)}=0.
	\end{align*}
	\item If $s_1(0)=s_2(0)=0$, then $\lambda_1(\mu_{|T_1|})=\lambda_1(\mu_{|T_2|})=0$. We note that 
	\[
	   \lim_{s\downarrow 0} \frac{s}{h_{\mu_1}(s)}=\lim_{s\downarrow 0} \frac{1}{\int_\mathbb{R}\frac{1}{s^2+u^2}d\mu_1(u)}=\lambda_1^2(\mu_1)=0.
	\]
	Since $t\leq s_1(t)$, we have 
	 \begin{align*}
	\frac{1}{\phi\large( [(T_1+T_2)^*(T_1+T_2)]^{-1} \large)}=\lim_{t\rightarrow 0^+}\frac{s_1(t)}{h_{\mu_1}(s_1(t))}\frac{t}{s_1(t)}=0.
	\end{align*}
	\end{itemize}
This finishes the proof. 
\end{proof}

\begin{proposition}
	Let $T_1, T_2 \in{\mathcal M}$ be bounded $R$--diagonal that are freely independent. We have 
	\[
	\lambda_2^2(T_1+T_2)=\lambda_2^2(T_1)+\lambda_2^2(T_2).
	\]
\end{proposition}


\section{The sum of an $R$-diagonal operator and a free random variable}
\label{section:sum-Rdiag-free-rv}
In this section, we outline an approach to the calculation of the Brown measure of the sum of an $R$-diagonal operator and a free random variable with an arbitrary distribution using subordination functions. This answers some questions asked by Biane--Lehner \cite{BianeLehner2001}.


Let $T\in\CM^\Delta$ be $R$--diagonal and $x_0\in\CM^\Delta$ be be a operator that is freely independent from $T$.  Let $U$ be a Haar unitary that is freely independent from $\{x_0, T\}$. Then, for any $\lambda\in\mathbb{C}$, we have
\begin{equation}
\label{replacement-x0-Rdiag}
|x_0+T-\lambda{\bf 1}| \underset{\ast{\mathcal D}}{\sim} |x_0-\lambda{\bf 1}+U^*T|
\underset{\ast{\mathcal D}}{\sim} |U(x_0-\lambda{\bf 1})+T|.
\end{equation}
The operator $U(x_0-\lambda{\bf 1})$ is $R$-diagonal. Hence, for any $\lambda$ fixed, the distribution of $|x_0+T-\lambda{\bf 1}|$ can be calculated by studying the sum of two $R$-diagonal operators. This observation appearing in \cite{BianeLehner2001} is the key for us to study the Brown measure of $x_0+T$. We denote $\mu_1=\widetilde{\mu}_{|x_0-\lambda{\bf 1}|}$ and $\mu_2=\widetilde{\mu}_{|T|}$, and $\mu=\widetilde{\mu}_{|x_0+T-\lambda{\bf 1}|}$. 
By Proposition \ref{Rdiag-additive}, we have $\mu=\mu_1\boxplus\mu_2$. That is,
\[
   \widetilde{\mu}_{|x_0+T-\lambda{\bf 1}|}=\widetilde{\mu}_{|x_0-\lambda{\bf 1}|}\boxplus \widetilde{\mu}_{|T|}.
\] 
Let $x\in \CM^\Delta$ and $\varepsilon>0$. The following observation provides a link between Fuglede-Kadison determinant and the Cauchy transform, 
\[
   \frac{1}{2}\frac{\partial \log\Delta(|x|^2+\varepsilon^2)}{\partial \varepsilon}=\int_0^\infty \frac{\varepsilon}{u^2+\varepsilon^2}d\tilde{\mu}_{|x|}(u)=h_{\widetilde{\mu}_{|x|}}(\varepsilon).
\] 
Recall we also have $G_{\widetilde{\mu}_{|x|}}(i\varepsilon)=-ih_{\widetilde{\mu}_{|x|}}(\varepsilon)$. 




We restate the main theorem \ref{thm:support-Brown-x0-T-introduction} as follows.

\begin{theorem}
	\label{thm:support-Brown-x0-T}
	Let $T\in\CM^\Delta$ be $R$--diagonal and $x_0\in\CM^\Delta$ be be a operator that is freely independent from $T$. 
	Then the support of the Brown measure $\mu_{T+x_0}$ is contained in the closure of following open set 
\begin{equation}
		\Omega(T, x_0)=\big\{  \lambda\in\mathbb{C}: ||(x_0-\lambda)^{-1}||_2 ||T||_2> 1\,\,\text{and}\,\,  ||x_0-\lambda||_2 ||T^{-1}||_2>1\big\}.
\end{equation}	
Moreover, there is no atom in the complement of $\Omega(T, x_0)$. 
\end{theorem}


We can write the complement of $\Omega(T, x_0)$ as 
\begin{equation*}
\begin{aligned}
\mathbb{C}\backslash \Omega(T, x_0)
&=\left\{  \lambda\in\mathbb{C}: \lambda_1(\mu_{|x_0-\lambda|})\geq \lambda_2(\mu_{|T|}) \,\,\text{or}\,\,
\lambda_2(\mu_{|x_0-\lambda|})\leq \lambda_1(\mu_{|T|}) \right\}\\
&=\left\{  \lambda\in\mathbb{C}: \lambda_1(\mu_{|x_0-\lambda|})\geq \lambda_2(\mu_{|T|})  \right\}
\cup \left\{  \lambda\in\mathbb{C}:
\lambda_2(\mu_{|x_0-\lambda|})\leq \lambda_1(\mu_{|T|}) \right\}.
\end{aligned}
\end{equation*}
By Lemma \ref{lemma:lambda-mu-1-2}, the two sets on the right hand side in above identity are disjoint. 
The function $\lambda\mapsto \lambda_1(\mu_{|x_0-\lambda|})$ is lower semicontinuous (see the Proof for Theorem \ref{proof-for-thm:main-Brown-x0-T}) and the function $\lambda\mapsto \lambda_2(\mu_{|x_0-\lambda|})$ is continuous. Hence, both sets are closed.

\begin{example}	\label{remark:support-selfadjoint}
	If $x_0$ is selfadjoint, then a direct calculation shows that the complement can be decomposed as the union of two close sets
	$\mathbb{C}\backslash \Omega(T,x_0)=A\cup B$, where
	\[
	A=\bigg\{ \lambda=a+bi: \int_\mathbb{R}\frac{1}{(u-a)^2+b^2}d\mu_{x_0}(u)\leq  \frac{1}{\phi(|T|^{2})}\bigg\}.
	\]
	and 
	\[
	B=\left\{ \lambda=a+bi:(a- \phi(x_0) )^2+b^2\leq \frac{1}{\phi(|T^{-1}|^2)}-\text{var}({x_0})  \right\},
	\]
	where $\text{var}({x_0})=\phi(x_0^2)-(\phi(x_0))^2$ is the variance of $x_0$. 
	Then $B$ is the empty set if and only if $\lambda_1(\mu_{|T|})^2-\text{var}(x_0)<0$.
	This explains that the bounded component in Figure \ref{figure:one-in-intro} is a disk, in which case $B$ is nonempty. 
\end{example}


\begin{remark}
\label{remark:def-Omega}
Let $\mu_1=\widetilde{\mu}_{|x_0-\lambda{\bf 1}|}$ and $\mu_2=\widetilde{\mu}_{|T|}$, and let $s_1, s_2$ be the subordination functions as in Proposition \ref{prop:h-mu-omega-s}. Then, by Theorem \ref{thm:lambda-1-2-cases}, the following statements are equivalent:
\begin{enumerate}[(1)]
\item $\lambda\in \Omega(T, x_0)$;
\item $\lambda_1(\mu_{|x_0-\lambda|})<\lambda_2(\mu_{|T|})$ and $\lambda_1(\mu_{|T|})< \lambda_2(\mu_{|x_0-\lambda|})$;
\item $s_1(0), s_2(0)\in [0,\infty)$.
\end{enumerate}
\end{remark}

\begin{lemma}
	Let $T_1, T_2 \in\CM^\Delta$ be $R$--diagonal that are freely independent. With $\mu_1=\tilde{\mu}_{|T_1|}$ and $\mu_2=\tilde{\mu}_{|T_2|}$, the function $h_{\mu_1}, h_{\mu_2}$ and the subordination function $s_1(t), s_2(t)$ as in Section \ref{section:free-convolution-symmetric}, we then have 
	\begin{align}
	\label{eqn:Delta-T-1-2-general}
	\Delta\large( (T_1+T_2)^*(T_1+T_2)+t^2{\bf 1} \large)=\frac{\Delta(T_1^*T_1+s_1(t)^2)\Delta(T_2^*T_2+s_2(t)^2)}{(s_1(t)+s_2(t)-t)^2}.
	\end{align}
\end{lemma}
\begin{proof}
	We set 
	\[
	H_{T_1+T_2}(t)=\frac{1}{2}\log\Delta\large( (T_1+T_2)^*(T_1+T_2)+t^2{\bf 1}\large)
	\]
	and for $i=1,2$, 
	\[
	H_{T_i}(t)=\frac{1}{2}\log\Delta\large( T_i^*T_i+t^2{\bf 1} \large).
	\]
	With $\mu_1=\tilde\mu_{|T_1|}$ and $\mu_2=\tilde\mu_{|T_2|}$, set $\mu=\tilde\mu_{|T_1+T_2|}=\mu_1\boxplus\mu_2$. We have 
	\[
	H_{T_1+T_2}(t)=\frac{1}{2}\int_0^\infty \log(u^2+t^2)d\mu(u),
	\]
	and
	\[
	H_{T_i}(t)=\frac{1}{2}\int_0^\infty\log(u^2+t^2)d\mu_i(u).
	\]
	We have 
	\begin{equation}
	\label{limit-H-T-1-2-log-t}
	\lim_{t\rightarrow\infty}(H_{T_1+T_2}(t)-\log t)=\frac{1}{2}\lim_{t\rightarrow \infty}\log\left(1+\frac{u^2}{t^2}\right)d\mu(u)=0
	\end{equation}
	and similarly
	\begin{equation}
	\label{limit-H-i-log-t}
	\lim_{t\rightarrow\infty} (H_{T_i}(t)-\log t)=0. 
	\end{equation}
	
	Fix $t_0>0$ and recall that
	\[
	s_1(t)+s_2(t)=t+\frac{1}{h_\mu(t)}.
	\]
	By definition, we have
	\[
	H_{T_1+T_2}'(t)=h_\mu(t), \quad H_{T_i}'(t)=h_{\mu_i}(t).
	\]
	It follows that 
	\begin{align*}
	\int_{t_0}^th_\mu &(v)dv\\
	&=\int_{t_0}^t h_\mu(v) \bigg(s_1(v)+s_2(v)-\frac{1}{h_\mu(v)} \bigg)'dv+\\
	&=\int_{t_0}^t h_{\mu_1}(s_1(v))s_1'(v)dv+\int_{t_0}^t h_{\mu_2}(s_2(v))s_2'(v)dv+
	\int_{t_0}^t  \frac{h_\mu'(v)}{h_\mu(v)}dv\\
	&=H_{T_1}(s_1(t))-H_{T_1}(s_1(t_0))+H_{T_2}(s_2(t))\\
	 &\qquad\qquad\qquad-H_{T_2}(s_2(t_0))+\log h_\mu(t)-\log h_\mu(t_0).
	\end{align*}
	Hence, there exists a constant $C$ such that
	\begin{equation}
	H_{T_1+T_2}(t)=H_{T_1}(s_1(t))+H_{T_2}(s_2(t))+\log h_\mu(t)+C.
	\end{equation}
	The asymptotics \eqref{limit-H-T-1-2-log-t} and \eqref{limit-H-i-log-t} as $t\rightarrow \infty$ implies that $C$ must be zero. Finally, note
	\[
	h_\mu(t)=\frac{1}{s_1(t)+s_2(t)-t},
	\]
	we obtain
	\[
	\exp(2H_{T_1+T_2}(t))=\exp(2H_{T_1}(t))\exp(H_{T_2}(t))\frac{1}{(s_1(t)+s_2(t)-t)^2}
	\]
	which is \eqref{eqn:Delta-T-1-2-general}.
\end{proof}


\begin{theorem}
	\label{thm:Delta-T-1-2-general}
	Let $T_1, T_2 \in\CM^\Delta$ be $R$--diagonal that are freely independent. 
	\begin{enumerate}
		\item If $\lambda_1(\mu_{|T_1})\geq\lambda_2(\mu_{|T_2|})$, then
		$\Delta(T_1+T_2)=\Delta(T_1)$.
		\item If $\lambda_1(\mu_{|T_2})\geq\lambda_2(\mu_{|T_1|})$, then
		$\Delta(T_1+T_2)=\Delta(T_2)$.
		\item If $(\lambda_1(\mu_{|T_1|}), \lambda_2(\mu_{|T_1|}))\cap (\lambda_1(\mu_{|T_2|}), \lambda_2(\mu_{|T_2|}))\neq \emptyset$ and $s_1(0), s_2(0)\in (0,\infty)$, then  
		\begin{equation}
		\label{eqn:Delta-T-1-2-general-t0}
		\Delta(T_1+T_2)=\frac{\big( \Delta(T_1^*T_1+s_1(0)^2)\Delta(T_2^*T_2+s_2(0)^2) \big)^{\frac{1}{2}}}{s_1(0)+s_2(0)}.
		\end{equation}
	\end{enumerate}
\end{theorem}

\begin{proof}
	By Theorem \ref{thm:lambda-1-2-cases}, when $\lambda_1(\mu_{|T_1})\geq\lambda_2(\mu_{|T_2|})$, we have $\lim_{t\rightarrow 0^+}s_1(t)=0$ and $\lim_{t\rightarrow 0^+}s_2(t)=\infty$. We rewrite the right hand side of \eqref{eqn:Delta-T-1-2-general} as
	\begin{align*}
	&\Delta(T_1^*T_1+s_1(t)^2)\frac{\Delta(T_2^*T_2+s_2(t)^2)}{s_2(t)^2}\frac{s_2(t)^2}{(s_1(t)+s_2(t)-t)^2},
	\end{align*}
	where the first factor converges to $\Delta(T_1^*T_1)$,and both the second and third factor converge to $1$ as $t\rightarrow 0^+$. Now \eqref{eqn:Delta-T-1-2-general-t0} follows from this as 
	\[
	\Delta(T_1+T_2)^2=\lim_{t\rightarrow 0^+} \Delta\large( (T_1+T_2)^*(T_1+T_2)+t^2{\bf 1} \large).
	\]
	The case when $\lambda_1(\mu_{|T_2})\geq\lambda_2(\mu_{|T_1|})$ is similar, in which case $\lim_{t\rightarrow 0^+}s_1(t)=\infty$ and $\lim_{t\rightarrow 0^+}s_2(t)=0$. 
	
	Finally, if $(\lambda_1(\mu_{|T_1|}), \lambda_2(\mu_{|T_1|}))\cap (\lambda_1(\mu_{|T_2|}), \lambda_2(\mu_{|T_2|}))\neq \emptyset$ and $s_1(0), s_2(0)\in (0,\infty)$, then \eqref{eqn:Delta-T-1-2-general-t0} follows from \eqref{eqn:Delta-T-1-2-general} by letting $t\rightarrow 0^+$. 
\end{proof}

\begin{remark}
\label{remark:4.50}
If $(\lambda_1(\mu_{|T_1|}), \lambda_2(\mu_{|T_1|}))\cap (\lambda_1(\mu_{|T_2|}), \lambda_2(\mu_{|T_2|}))\neq \emptyset$ and $s_1(0)=s_2(0)=0$, then by Lemma \ref{lemma:s-0-all-cases} and Theorem \ref{thm:lambda-1-2-cases}, we must have 
\[
     \lambda_1(\mu_{|T_1|})=\lambda_1(\mu_{|T_2|})=0,
\]
and $\mu_{|T_1|}(\{0\})+\mu_{|T_2|}(\{0\})\geq 1$.
\end{remark}

\begin{proposition}[\cite{HaagerupSchultz2007}]
	\label{prop:Brown-inverse-pushforward}
 If $T\in\CM^\Delta$ with
  \begin{equation}\label{integration-logt-Brown}
    \int_0^1 \log t\,{\rm d}\mu_{|T|}(t) >-\infty,
  \end{equation}
then $\mu_T(\{0\})=\mu_{|T|}(\{0\})=0$, and $T$ has an inverse
$T^{-1}\in\CM^\Delta$. Moreover, $\mu_{T^{-1}}$ is the push-forward 
measure of $\mu_T$ via the map $z\mapsto z^{-1}$.
\end{proposition}

The following result was proved in \cite[Section 4.2]{Zhong2021Brown}. We provide a proof for completeness. 
\begin{lemma}
	\label{lemma:support-Brown-general}
	Let $\lambda\in \mathbb{C}$ and $x_0\in \CM^\Delta$. If $\phi(|x_0-\lambda{\bf 1}|^{-2})<\infty$, then 
	\[
	\int_\mathbb{C}\frac{1}{|z-\lambda|^2}d\mu_{x_0}(z)\leq \phi(|x_0-\lambda{\bf 1}|^{-2}). 
	\]
	If  $\phi(|x_0-\lambda{\bf 1}|^{-2})<L$ in some neighborhood $U$ of $\lambda_0$ for some $L>0$, then $\mu_{x_0}(U)=0$. In particular, $\lambda_0\notin {\text{\rm supp}}(\mu_{x_0})$.
\end{lemma}
\begin{proof}
Observe that for any $t\in (0,1)$, we have $2\log t>-\frac{1}{t^2}$. Hence, 
\[
\begin{aligned}
2 \int_0^1 \log t d\mu_{|x_0-\lambda{\bf 1}|}(t)
&>-\int_0^1 \frac{1}{t^2}d\mu_{|x_0-\lambda{\bf 1}|}(t)\\
&>-\int_0^\infty \frac{1}{t^2}d\mu_{|x_0-\lambda{\bf 1}|}(t)\\
&=-\phi(|x_0-\lambda{\bf 1}|^{-2})>-\infty. 
\end{aligned}
\]
By Proposition \ref{prop:Brown-inverse-pushforward}, the operator $x_0-\lambda{\bf 1}$ has an inverse $(x_0-\lambda{\bf 1})^{-1}\in \CM^\Delta$ and the Brown measure $\mu_{(x_0-\lambda{\bf 1})^{-1}}$ of $(x_0-\lambda{\bf 1})^{-1}$ can be defined. Moreover, $\mu_{(x_0-\lambda{\bf 1})^{-1}}$ is the push-forward measure of $\mu_{x_0}$ via the map $z\mapsto (z-\lambda)^{-1}$. Hence, by \cite[Theorem 2.19]{HaagerupSchultz2007}, we obtain
\begin{equation}
\label{eqn:inequality-L2-inverse}
\begin{aligned}
\int_\mathbb{C}\frac{1}{|z-\lambda|^2}d\mu_{x_0}(z) & =\int_\mathbb{C} |z|^2 d\mu_{(x_0-\lambda{\bf 1})^{-1}}(z)\\
&\leq ||(x_0-\lambda{\bf 1})^{-1}||_2^2=\phi(|x_0-\lambda{\bf 1}|^{-2}).   
\end{aligned}
\end{equation}
If  $\phi(|x_0-\lambda{\bf 1}|^{-2})<L$ in some neighborhood $U$ of $\lambda_0$ for some $L>0$.
It then follows that
\[
  \int_\mathbb{C}\frac{1}{|z-\lambda|^2}d\mu_{x_0}(z)<L
\]
for $\lambda\in U$, which yields that $\mu_{x_0}(U)=0$ (see \cite[Lemma 4.8]{Zhong2021Brown} for example). Hence, $\lambda_0 \notin {\text{\rm supp}}(\mu_{x_0})$. 
\end{proof}


We are now ready to prove the main result of this section
\begin{proof}[Proof of Theorem \ref{thm:support-Brown-x0-T}]
	\label{proof-for-thm:main-Brown-x0-T}
	Observe that $||(x_0-\lambda)^{-1}||_2=\phi( |x_0-\lambda|^{-2} )
	  =\lim_{\varepsilon\downarrow 0}\phi( (|x_0-\lambda|^{2}+\varepsilon)^{-1} )$.
	  The function $\{ \phi( (|x_0-\lambda|^{2}+1/n)^{-1} )\}_{n}$
	  is a increasing family of continuous functions of $\lambda$. Hence, 
	  the function $\lambda\mapsto ||(x_0-\lambda)^{-1}||_2$ is lower semicontinuous. 
	  The other function $\lambda\mapsto ||x_0-\lambda||_2$ is a continuous function. Hence, the set $\Omega(T, x_0)$ is open. 
	  
	Let $U$ be a Haar unitary operator that is free from $\{T, x_0\}$. For any $\lambda\in\mathbb{C}$, as we observed earlier that
	\[
	|x_0+T-\lambda{\bf 1}| \underset{\ast{\mathcal D}}{\sim}   |U^*T+x_0-\lambda{\bf 1}| \underset{\ast{\mathcal D}}{\sim} |T+U(x_0-\lambda{\bf 1}|,
	\]
	which yields
	\[
	\Delta(x_0+T-\lambda{\bf 1})=\Delta(T+U(x_0-\lambda{\bf 1}).
	\]
	We now apply Theorem \ref{thm:Delta-T-1-2-general} to our setting (for $T_1=U(x_0-\lambda{\bf 1})$ and $T_2=T$). Then, for any $\lambda\in \mathbb{C}\backslash \Omega(T,x_0)$, we have either 
	\[
	  ||(x_0-\lambda)^{-1}||_2 ||T||_2\leq1
	\]
	or 
	\[
	  ||x_0-\lambda||_2 ||T^{-1}||_2\leq 1.
	\]
	We note
	\[
	||(x_0-\lambda)^{-1}||_2 ||T||_2\leq1\Longleftrightarrow \lambda_2(\mu_{|T|})\leq\lambda_1(\mu_{|x_0-\lambda{\bf 1}|}))
	\]
	and
	\[
	||x_0-\lambda||_2 ||T^{-1}||_2\leq 1\Longleftrightarrow \lambda_2(\mu_{|x_0-\lambda{\bf 1}|}))\leq\lambda_1(\mu_{|T|})).
	\]
	

	Hence, if $\lambda_0$ is in the interior of $\mathbb{C}\backslash \Omega(T, x_0)$ and
	 $||(x_0-\lambda{\bf 1})^{-1}||_2 ||T||_2\leq1$ in some neighborhood $U$ of $\lambda_0$. That is,
	 \begin{equation}
	  \label{eqn:4.11-in-proof}
	   	   \phi(|x_0-\lambda{\bf 1}|^{-2})\leq \frac{1}{\phi(|T|^2)}
	 \end{equation}
	 for $\lambda\in U$. By Lemma \ref{lemma:support-Brown-general}, we have $\mu_{x_0}(U)=0$. 	By Theorem \ref{thm:Delta-T-1-2-general}, in this case, we also have 
	 \begin{equation}
	 \label{eqn:det-x0-T-in-proof}
	\Delta(T+U(x_0-\lambda{\bf 1}))=\Delta(U(x_0-\lambda))=\Delta(x_0-\lambda{\bf 1}),
	 \end{equation}
	 which yields that $\mu_{x_0+T}$ and $\mu_{x_0}$ coincides in $U$ by taking the Laplacian. Hence, $\mu_{x_0+T}(U)=0$ and $\lambda_0$ is not in the support of the $\mu_{x_0+T}$. 	
	 
 Moreover, if $||(x_0-\lambda{\bf 1})^{-1}||_2 ||T||_2\leq1$, then \eqref{eqn:4.11-in-proof} implies that
 $\phi(|x_0-\lambda{\bf 1}|^{-2})<\infty$. Consequently, by observing that $2\log t>-\frac{1}{t^2}$ for $0<t<1$, we have 
 \begin{align*}
    2\int_0^1\log t \,d\mu_{|x_0-\lambda|}(t) & >-\int_0^1\frac{1}{t^2}\, d\mu_{|x_0-\lambda|}(t)\\
          &>-\int_0^\infty\frac{1}{t^2}\, d\mu_{|x_0-\lambda|}(t)
           =-\phi(|x_0-\lambda|^{-2})>-\infty.
 \end{align*}
 Then, this together \eqref{eqn:det-x0-T-in-proof} imply that 
 \[
    \int_0^1 \log t \, d\mu_{|x_0+T-\lambda{\bf 1}|}>-\infty.
 \]
 By \cite[Proposition 2.16]{HaagerupSchultz2007}, it follows that $\mu_{x_0+T-\lambda}(\{0\})=0$. Therefore, we deduce $\mu_{x_0+T}(\{\lambda\})=0$
	   
	Similarly, if $\lambda_0$ is in the interior of $\mathbb{C}\backslash \Omega(T, x_0)$ and $||x_0-\lambda||_2 ||T^{-1}||_2< 1$, we have
	\begin{equation}
	  \label{eqn:4.13-in-proof}
	  	\Delta(x_0+T-\lambda{\bf 1})=\Delta(T).
	\end{equation}
	Hence, within some neighborhood $U$ of $\lambda_0$, we have 
	\[
	\nabla^2( \Delta(x_0+T-\lambda{\bf 1}))=\nabla^2(\Delta(T))=0,
	\]
	which means that $\mu_{x_0+T}(U)=0$ and $\lambda_0$ is not in the support of $\mu_{x_0+T}$. 
If $||x_0-\lambda||_2 ||T^{-1}||_2< 1$, then 
\[
   \lambda_1(\mu_{|T|}) =\frac{1}{||T^{-1}||_2}>||x_0-\lambda||_2>0,
\]
because $x_0$ is not a scalar of identity. Hence, $\mu_{T}(\{0\})=0$ and $\Delta(T)>0$. Therefore, by \eqref{eqn:4.13-in-proof}, $\Delta(x_0+T-\lambda{\bf 1})>0$. By \cite[Proposition 2.16]{HaagerupSchultz2007}, it follows that $\mu_{x_0+T-\lambda}(\{0\})=0$. Therefore, we deduce $\mu_{x_0+T}(\{\lambda\})=0$.
	
We then conclude that the support of the Brown measure of $x_0+T$ is contained in the closure of $\Omega(T, x_0)$, and there is no atom at $\lambda\in \mathbb{C}\backslash \Omega(T, x_0)$. 
\end{proof}


\begin{definition}
Let $T\in\CM^\Delta$ be $R$--diagonal and $x_0\in\CM^\Delta$ be be a operator that is freely independent from $T$. Denote 
\[
  S(T, x_0)=\{\lambda\in\mathbb{C}:\mu_{|T|}(\{0\})  + \mu_{|x_0-\lambda|}(\{0\})\geq 1 \}.
\]
\end{definition}

\begin{remark}
We point out that the set $S(T,x_0)$ consists of finitely many complex numbers. If $\lambda\in S(T, x_0)$, then
$\mu_{|x_0-\lambda|}(\{0\})\geq 1-\mu_{|T|}(\{0\})>0$. Hence, $\lambda$ is an eigenvalue of $x_0$ and its eigenspace $p_\lambda$ has trace $\phi(p_\lambda)=\mu_{|x_0-\lambda|}(\{0\})$ with positive value. It follows that
$S(T, x_0)$ is a finite set. The points in $S(T,x_0)$ are candidates for the atoms of the Brown measure $\mu_{T+x_0}$. 
\end{remark}



\begin{proposition}
	\label{prop:boundary-values-s1-s2-analytic}
	Let $\mu_1=\widetilde{\mu}_{|x_0-\lambda{\bf 1}|}$ and $\mu_2=\widetilde{\mu}_{|T|}$, and let $s_1, s_2$ be the subordination functions as in Proposition \ref{prop:h-mu-omega-s}. Note that $s_1, s_2$ are functions of $\lambda$. The following statements are equivalent:
	\begin{enumerate}[(1)]
	 \item $\lambda\in \Omega(T, x_0)\backslash S(T, x_0)$;
	 \item $s_1(0), s_2(0)\in (0,\infty)$;
	 \item the density function $f_{\mu_1\boxplus \mu_2}$ of the absolutely continuous part of $\mu_1\boxplus\mu_2$ is finite and strictly positive $0<f_{\mu_1\boxplus \mu_2}(0)<\infty$.
	\end{enumerate}
\end{proposition}
\begin{proof}
The equivalence of $(1), (2)$ and $(3)$ follows from Remark \ref{remark:def-Omega}, Lemma \ref{lemma:s-0-all-cases} and Theorem \ref{thm:lambda-1-2-cases}.
\end{proof}



\begin{proposition}
\label{prop:partial-lambda-general}
Let $\mu_1=\widetilde{\mu}_{|x_0-\lambda{\bf 1}}$ and $\mu_2=\widetilde{\mu}_{|T|}$, and let $s_1, s_2$ be the subordination functions as in Proposition \ref{prop:h-mu-omega-s}. Then for any $\lambda\in \Omega(T, x_0)$, we have 
\begin{equation}
\label{eqn:partial-lambda-general}
  \begin{aligned}
  \phi((\lambda{\bf 1}-x_0-T)^*&(|\lambda{\bf 1}-x_0-T|^2)^{-1})\\
      &=\phi((\lambda{\bf 1}-x_0)^*(|\lambda{\bf 1}-x_0|^2+s_1(0)^2)^{-1})
  \end{aligned}
\end{equation}
\end{proposition}
\begin{proof}
Under these conditions, we have \eqref{eqn:Delta-T-1-2-general-t0}, which is rewritten as
		\begin{equation}
\label{eqn:Delta-T-1-2-t0-in-proof}
\Delta(x_0+T-\lambda{\bf 1})^2=\frac{ \Delta(|x_0-\lambda{\bf 1}|^2+s_1(0)^2)\Delta(|T|^2+s_2(0)^2) }{(s_1(0)+s_2(0))^2}.
\end{equation}
Taking the logarithmic derivative, we obtain
\begin{align*}
 &\frac{\partial}{\partial\lambda}\log\Delta(|x_0-\lambda{\bf 1}|^2+s_1(0)^2)\\
   &=\frac{\partial}{\partial\lambda}\phi( \log(|x_0-\lambda{\bf 1}|^2+s_1(0)^2 ) )\\
   &=\phi((\lambda{\bf 1}-x_0)^*(|x_0-\lambda{\bf 1}|^2+s_1(0)^2)^{-1})\\
     &\qquad\qquad+\phi ( (|x_0-\lambda{\bf 1}|^2+s_1(0)^2)^{-1})\cdot 2s_1(0) \frac{\partial s_1(0)}{\partial\lambda},
\end{align*}
and
\begin{align*}
  &\frac{\partial}{\partial\lambda}\log\Delta(|T|^2+s_2(0)^2)\\
   &=\phi ((|T|^2+s_2(0)^2)^{-1})\cdot 2s_2(0) \frac{\partial s_2(0)}{\partial\lambda},
\end{align*}
and
\begin{align*}
 \frac{\partial}{\partial\lambda}\log(s_1(0)+s_2(0))^2=\frac{2}{s_1(0)+s_2(0)}\left( \frac{\partial s_1(0)}{\partial\lambda}+\frac{\partial s_2(0)}{\partial\lambda} \right).
\end{align*}
In addition, by subordination relations, we have 
\begin{align*}
  \frac{1}{s_1(0)+s_2(0)}&=h_{\mu_1}(s_1(0))=h_{\mu_2}(s_2(0)\\
     &=s_1(0)\phi ( (|x_0-\lambda{\bf 1}|^2+s_1(0)^2)^{-1})\\
     &=s_2(0)\phi( (|T|^2+s_2(0)^2)^{-1}).
\end{align*}
Putting these identities together, we obtain
\begin{align*}
  \frac{\partial}{\partial\lambda}\log\Delta(x_0+T-\lambda{\bf 1})^2
     &=\frac{\partial}{\partial\lambda}\log(\text{Right hand side of}\, \eqref{eqn:Delta-T-1-2-t0-in-proof})\\
   &=\phi((\lambda{\bf 1}-x_0)^*(|x_0-\lambda{\bf 1}|^2+s_1(0)^2)^{-1}),
\end{align*}
which is the identity claimed. 
\end{proof}




\begin{theorem}
\label{thm:main-density-formula}
The Brown measure of $x_0+T$ is absolutely continuous in the set $\Omega(T, x_0)\backslash S(T,x_0)$. The density is real analytic whenever it is strictly positive. Moreover, the density can be expressed as
\begin{equation}
   \frac{1}{\pi} s_1^2\phi(h^{-1}k^{-1})+ \frac{1}{\pi} \frac{2s_1 \big(s_1-p'_{\mu_2}(s_2) ) \big)
     \left| \frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2} \right|^2  }{\left[ \frac{1}{s_1} p_{\mu_1}(s_1) p'_{\mu_2}(s_2)  +(s_1-p'_{\mu_2}(s_2) )p'_{\mu_1}(s_1)    \right]}
\end{equation}
where $h=h(\lambda,s_1(0))$ and $k=k(\lambda,s_1(0))$ for 
\[
h(\lambda,w)=(\lambda{\bf 1}-x_0)^*(\lambda{\bf 1}-x_0)+w^2
\]
and
\[
k(\lambda,w)=(\lambda{\bf 1}-x_0)(\lambda{\bf 1}-x_0)^*+w^2,
\]
and
\[
   p_\mu(s)=\frac{n_\mu(s)}{d_\mu(s)}.
\]
\end{theorem}
\begin{proof}
We put 
\[
g(\lambda)= \log \Delta \big( (x_0+c_t-\lambda{\bf 1})^*(x_0+c_t-\lambda{\bf 1}) \big)
\]
The Brown measure can be calculated as
\[
d\mu_{x_0+c_t}(\lambda)=\frac{1}{\pi}\frac{\partial^2}{\partial\overline{\lambda
	}\partial\lambda} g(\lambda).
\]
We rewrite \eqref{eqn:partial-lambda-general} as
\[
   \frac{\partial}{\partial\lambda} g(\lambda)=\phi\left( h^{-1} \frac{\partial h}{\partial\lambda} \right)
\]
where $\frac{\partial h}{\partial\lambda}=(\lambda{\bf 1}-x_0)^*$. By applying \cite[Lemma 3.2]{HaagerupT2005Annals}, we have
\begin{align}
\frac{\partial^2}{\partial\overline{\lambda
	}\partial\lambda}g(\lambda)
&=\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}\frac{\partial h}{\partial\lambda}\right)\nonumber\\
&=\phi\left( -h^{-1} \left( \frac{\partial h}{\partial\overline{\lambda}}+2s_1 \frac{\partial {s_1}}{\partial\overline{\lambda}}  \right) h^{-1} \frac{\partial h}{\partial\lambda}   + h^{-1} \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}  \right) \label{eqn:4.16-in-proof}\\
&=\phi\left( -h^{-1}  \frac{\partial h}{\partial\overline{\lambda}} h^{-1} \frac{\partial h}{\partial\lambda}   + h^{-1} \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}  \right)
-2s_1 \frac{\partial {s_1}}{\partial\overline{\lambda}} 
\phi\left( \frac{\partial h}{\partial {\lambda}} (h^{-1})^2  \right)\label{eqn:density-413}
\end{align}
and
\[
\frac{\partial h}{\partial\overline{\lambda}}= \frac{\partial h(\lambda,s_1)}{\partial\overline{\lambda}} =\lambda{\bf 1} -x_0; 
\]
and 
\[
\frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda} 
=\frac{\partial}{\partial\overline{\lambda}} (\lambda{\bf 1}-x_0)^*=1.
\]

We now apply the identity $x(x^*x+\varepsilon{\bf 1})^{-1}=(xx^*+\varepsilon{\bf 1})^{-1}x$
to $x=\lambda{\bf 1}-x_0$ and $\varepsilon=s_1^2$, we find that
\begin{align*}
-\frac{\partial h}{\partial\overline{\lambda}} h^{-1} \frac{\partial h}{\partial\lambda}   +  \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}
&=1-x(x^*x+s_1^2{\bf 1})^{-1}x^*\\
&=1-(xx^*+s_1^2{\bf 1})^{-1}xx^*\\
&=s_1^2 (xx^*+s_1^2{\bf 1})^{-1}\\
&=s_1^2 k^{-1}.
\end{align*}
Hence the firs term in the right hand side of \eqref{eqn:density-413} is 
\begin{equation} \label{eqn:4.2000}
  \phi\left( -h^{-1}  \frac{\partial h}{\partial\overline{\lambda}} h^{-1} \frac{\partial h}{\partial\lambda}   + h^{-1} \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}  \right)=s_1(0)^2\phi(h^{-1}k^{-1}).
\end{equation}

We now find a formula for $\frac{\partial s_1}{\partial\overline{\lambda}}$ by implicit differentiation. By adopting notations in Section \ref{section:free-convolution-symmetric} (in particular Lemma \ref{lemma:s-0-all-cases}) and \eqref{eqn:balance-n-d}, the function $s_1(0)$ is the unique solution of 
\begin{equation}
\frac{n_{\mu_1}(s)}{d_{\mu_1}(s)}=
\frac{n_{\mu_2}(f_{\mu_1}(s))}{d_{\mu_2}(f_{\mu_1}(s))}
\end{equation}
for $\mu_1=\widetilde{\mu}_{|x_0-\lambda{\bf 1}}$ and $\mu_2=\widetilde{\mu}_{|T|}$. Set $p_\mu(s)= \frac{n_{\mu}(s)}{d_{\mu}(s)}$ and recall that $f_\mu(s)=\frac{1}{h_\mu(s)}-s$. Then, 
\begin{equation}
\label{eqn:p-two-sub}
	    p_{\mu_1}(s_1(0))=p_{\mu_2}(f_{\mu_1}(s_1(0))),
\end{equation}
where $f_{\mu_1}(s_1(0))$ can be expressed as
\[
   f_{\mu_1}(s_1(0))=\frac{1}{s_1(0) \phi(h^{-1})}-s_1(0). 
\]
Note that
\[
    f_{\mu_1}(s_1(0))=\frac{1}{s_1(0) \phi(h^{-1})}-s_1(0)=s_2(0).
\]
Taking implicit differentiation of \eqref{eqn:p-two-sub}, we have 
\begin{equation}
 \label{eqn:417-implicit-diff-eqn}
   p_{\mu_1}'(s_1)\frac{\partial s_1}{\partial\overline{\lambda}}
    +\frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2}
     =p_{\mu_2}'(f_{\mu_1}(s_1)) \frac{\partial}{\partial\overline{\lambda}} f_{\mu_1}(s_1).
\end{equation}
We calculate
\begin{align*}
	  \frac{\partial}{\partial\overline{\lambda}} f_{\mu_1}(s_1)
	  &=   -\frac{1}{s_1^2 \phi(h^{-1})}\frac{\partial s_1}{\partial\overline{\lambda}}
	    -\frac{1}{s_1}\frac{\phi\left(-h^{-1} \left( \frac{\partial h}{\partial\overline{\lambda}}+2s_1 \frac{\partial {s_1}}{\partial\overline{\lambda}}  \right) h^{-1}\right) }{(\phi(h^{-1}))^2}
	    -\frac{\partial s_1}{\partial\overline{\lambda}}\\
	   &=-\frac{\partial s_1}{\partial\overline{\lambda}}
	   \left[1+\frac{1}{s_1^2 \phi(h^{-1})} -2\frac{\phi((h^{-1})^2)}{(\phi(h^{-1}))^2}  \right]
	     +\frac{1}{s_1} \frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2}.
\end{align*}
Alternatively, we have 
\[
   f_{\mu_1}(s_1)=\frac{1}{s_1} p_{\mu_1}(s_1).
\]
Hence,
\begin{align*}
  \frac{\partial}{\partial\overline{\lambda}} f_{\mu_1}(s_1) 
    &=-\frac{1}{s_1^2}p_{\mu_1}(s_1) \frac{\partial s_1}{\partial\overline{\lambda}}
      +\frac{1}{s_1}p'_{\mu_1}(s_1)\frac{\partial s_1}{\partial\overline{\lambda}}
       +\frac{1}{s_1} \frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2}.
\end{align*}
Plugging it to \eqref{eqn:417-implicit-diff-eqn}, we have 
\begin{align*}
 -\frac{\partial s_1}{\partial\overline{\lambda}}\times &\left[ \frac{1}{s_1} p_{\mu_1}(s_1) p'_{\mu_2}(s_2)  +(s_1-p'_{\mu_2}(s_2) )p'_{\mu_1}(s_1)    \right]\\
     &=\big(s_1-p'_{\mu_2}(s_2) ) \big)
     \frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2}.
\end{align*}
It follows that
\begin{equation} \label{eqn:4.20002}
	 -\frac{\partial s_1}{\partial\overline{\lambda}}=\frac{\big(s_1-p'_{\mu_2}(s_2) ) \big)
	}{\left[ \frac{1}{s_1} p_{\mu_1}(s_1) p'_{\mu_2}(s_2)  +(s_1-p'_{\mu_2}(s_2) )p'_{\mu_1}(s_1)    \right]}\times 
	\frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2},
\end{equation}
We note that
\[
  \phi\left( \frac{\partial h}{\partial {\lambda}} (h^{-1})^2  \right)=\overline{ \phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}. 
\]
Plugging \eqref{eqn:4.2000} and \eqref{eqn:4.20002} to \eqref{eqn:density-413}, then we have 
\[
 \frac{\partial^2}{\partial\overline{\lambda
 	}\partial\lambda}g(\lambda)=  s_1^2\phi(h^{-1}k^{-1})+\frac{2s_1 \big(s_1-p'_{\mu_2}(s_2) ) \big)
     \left| \frac{\phi\left ( (h^{-1})^2 \frac{\partial h}{\partial\overline{\lambda}} \right)}{(\phi(h^{-1}))^2} \right|^2  }{\left[ \frac{1}{s_1} p_{\mu_1}(s_1) p'_{\mu_2}(s_2)  +(s_1-p'_{\mu_2}(s_2) )p'_{\mu_1}(s_1)    \right]}
\]
This finishes the proof. 
\end{proof}



\begin{example}
Let $c$ be a standard circular operator and let $x_0$ be a selfadjoint operator that is $*$-free from $c$. Note that $\lambda_1(\mu_{|c^n|})=0$ and $\lambda_2(\mu_{|c^n|})=1$ for any $n\in\mathbb{N}$.  Then for any $n\in\mathbb{N}$, the set $\Omega(c^n, x_0)=\Omega(c, x_0)$ and is expressed as
\[
  \Omega(c, x_0)=\bigg\{ \lambda=a+bi: \int_\mathbb{R}\frac{1}{(u-a)^2+b^2}d\mu_{x_0}(u)> 1\bigg\}.
\]
See Figure \ref{figure1-circular} for simulations. 
\end{example}

\begin{figure}[t]
	\begin{center}
	\begin{subfigure}[h]{0.45\linewidth}
		\includegraphics[width=\linewidth]{circular_1power}
		\caption{$x_0+c$}
		\end{subfigure}
		\begin{subfigure}[h]{0.45\linewidth}
		\includegraphics[width=\linewidth]{circular_2square}
				\caption{$x_0+c^2$}
		\end{subfigure}	
		\begin{subfigure}[h]{0.45\linewidth}
		\includegraphics[width=\linewidth]{circular_3cubic}
				\caption{$x_0+c^3$}
		\end{subfigure}
				\begin{subfigure}[h]{0.45\linewidth}
		\includegraphics[width=\linewidth]{circular_4power}
				\caption{$x_0+c^4$}
		\end{subfigure}
	\end{center}
\begin{center}
	\caption{$2\,000\times 2\,000$ Matrix simulations for the Brown measure of $x_0+c^n$  for $x_0$ distributed as $\frac{1}{4}\delta_{-0.8}+\frac{3}{4}\delta_{0.8}$. The boundary of $\Omega(c, x_0)$ is red dashed on top and blue dashed on bottom.}\label{figure1-circular}

\end{center}
\end{figure}



\begin{corollary}
If $S(T, x_0)=\emptyset$, then the Brown measure of $T+x_0$ is absolutely continuous in $\Omega(T, x_0)$. 
In particular, if $\lambda_1(\mu_{|T|})>0$, then $\mu_{|T|}(\{0\})=0$. Consequently, $S(T, x_0)=\emptyset$
and the Brown measure of $T+x_0$ is absolutely continuous in $\Omega(T, x_0)$. 
\end{corollary}




\section{Addition with $R$-diagonal operators: examples}


\subsection{The sum of a Haar unitary operator and a free random variable}
Let $u$ be a Haar unitary operator and $x_0\in\CM^\Delta$ be a random variable that is $*$-free from $u$. In this section, we study the Brown measure of $x_0+u$. For any $\lambda\in\mathbb{C}$, we denote $\mu_1=\widetilde\mu_{|x_0-\lambda{\bf 1}|}$, $\mu_2=\frac{1}{2}(\delta_\gamma+\delta_{-\gamma})$ for some $\gamma>0$, and $\mu=\mu_1\boxplus\mu_2$. Because of $\mu_2$ is so simple, we can obtain more explicit formula of subordination functions $s_1$ and $s_2$. 

We adopt notations in Section \ref{section:free-convolution-symmetric} and set two $R$-diagonal operators $T_1=V(x_0-\lambda{\bf 1})$ and $T_2=\gamma u$, where $V$ is a Haar unitary operator that is $*$-free from $\{x_0, u\}$. 
We shall review the subordination function appearing in the free additive convolution of an arbitrary symmetric distribution $\mu_1$ and a Bernoulli distribution $\mu_2$. We provide a proof for completeness. 
\begin{lemma}
Using notations in Section \ref{section:free-convolution-symmetric}, the subordination function $s_1(t)$ is the unique solution $s\in (t,\infty)$ to the following equation
\begin{equation}
  \label{eqn:fixed-eqn-s1-Bernoulli}
   (s-t)^2-\frac{s-t}{h_{\mu_1}(s)}+\gamma^2=0.
\end{equation}
\end{lemma}
\begin{proof}
 We calculate that 
$G_{\mu_2}(z)=\frac{1}{2}\left( \frac{1}{z-\gamma}+\frac{1}{z+\gamma} \right)$. Hence
$H_2(z)=F_{\mu_2}(z)-z=\frac{-\gamma^2}{z}$ and $f_{\mu_2}(t)=\Im H_2(it)=\frac{\gamma^2}{t}$. 
Recall Equation \eqref{eqn:fixed-point-s-1-2} satisfied by $s_1(z)$
\[
   s_1(t)=t+f_{\mu_2}(t+f_{\mu_1}(s_1(t)))
\]
 which is now expressed as
\[
  s_1(t)=t+\frac{\gamma^2}{t-s_1+1/h_{\mu_1}(s_1(t))}.
\]
The above equation can be further reduced to 
\[
(s_1(t)-t)^2-\frac{s_1(t)-t}{h_{\mu_1}(s)}+\gamma^2=0.
\]
The uniqueness of the solution for \eqref{eqn:fixed-point-s-1-2}  implies that the function $s_1(t)$ must be  the unique solution of the above equation.
\end{proof}
 
 

\begin{theorem}
\label{thm:det-0-Bernoulli}
Let $u$ be a Haar unitary operator and $x_0\in\CM^\Delta$ be a random variable that is $*$-free from $u$. Given $\lambda\in \mathbb{C}$ and $\gamma>0$, we have 
 \begin{enumerate}[(1)]
    \item If $\lambda_1(\mu_{|x_0-\lambda{\bf 1}|})<\gamma<\lambda_2(\mu_{|x_0-\lambda
    {\bf 1}|})$, then
      \begin{equation}
        \label{eqn:det-t0-Bernoulli}
        \Delta(x_0+\gamma u-\lambda{\bf 1})=
      \Bigg(\frac{\gamma^2}{\gamma^2+s_1(0)^2}\,
      \Delta(|x_0-\lambda{\bf 1}|^2 +s_1(0)^2{\bf 1})\Bigg)^\frac12,
      \end{equation}
      where $s_1(0)$ is a real analytic function of $\lambda$ which satisfies
      \begin{equation}
      \label{eqn:fixed-eqn-s1-0-Bernoulli-in-proof}
         s^2-\frac{s}{h_{\mu_1}(s)}+\gamma^2=0.
      \end{equation}
    \item If $\gamma\leq \lambda_1(\mu_{|x_0-\lambda{\bf 1}|})$, then
    $\Delta(x_0+\gamma u-\lambda{\bf 1})= \Delta(x_0-\lambda{\bf 1})$.
    \item  If $\gamma\geq \lambda_2(\mu_{|x_0-\lambda{\bf 1}|})$, then
    $\Delta(T-\lambda{\bf 1})=\gamma$.
  \end{enumerate}
\end{theorem}
\begin{proof}
Following Theorem \ref{thm:Delta-T-1-2-general}, we only need to verify part (i).
If $\lambda_1(\mu_{|x_0-\lambda{\bf 1}|})<\gamma<\lambda_2(\mu_{|x_0-\lambda
    {\bf 1}|})$, we know that $s_1(0)$ is a real analytic function of $\lambda$ which satisfies \eqref{eqn:fixed-eqn-s1-0-Bernoulli-in-proof}. We hence have
    \begin{equation}
     \label{eqn:05.4}
     s_2(0)=\frac{1}{h_{\mu_1}(s_1(0)}-s_1(0)=\frac{\gamma^2}{s_1(0)}.
    \end{equation}
    Divide \eqref{eqn:fixed-eqn-s1-0-Bernoulli-in-proof} by $s_1^2$, we have 
    \[
       1+\frac{\gamma^2}{s_1^2}=\frac{1}{s_1h_{\mu_1}(s_1)}.
    \]
    Hence, and by \eqref{eqn:05.4}, 
     \begin{equation}
      \label{eqn:05.5}
            s_1+s_2=\frac{\gamma^2}{s_1}+s_1=\frac{1}{h_{\mu_1}(s_1(0)}=\frac{1}{h_{\mu_1}(s_2(0)}. 
     \end{equation}
    Note that, by applying \eqref{eqn:05.4}, we have 
    \begin{align*}
      \Delta(|T_2|^2+s_2(0)^2)&=\gamma^2+s_2(0)^2
        =\gamma^2\frac{s_1(0)^2+\gamma^2}{s_1(0)^2}.
    \end{align*}
   We then deduce
    \begin{align*}
     \frac{\Delta(|T_2|^2+s_2(0)^2}{(s_1(0)+s_2(0))^2}= 
     \gamma^2\frac{s_1(0)^2+\gamma^2}{s_1(0)^2} \left( \frac{s_1^2}{\gamma^2+s_1^2} \right)^2
        =\frac{\gamma^2}{\gamma^2+s_1^2}. 
    \end{align*}
Hence \eqref{eqn:det-t0-Bernoulli} follows from \eqref{eqn:Delta-T-1-2-general-t0}.
\end{proof}

\begin{remark}
 Let $T\in\CM^\Delta$ be a $R$-diagonal operator and $\gamma\in \mathbb{C}$. Then 
\[
  |T-\gamma| \underset{\ast{\mathcal D}}{\sim}  |T-\gamma u|  \underset{\ast{\mathcal D}}{\sim} |T+|\gamma| u|.
\]
Replacing $|x_0-\lambda{\bf 1}|$ by $T$, we see that Theorem \ref{thm:det-0-Bernoulli} is essentially equivalent to \cite[Theorem 4.15]{HaagerupSchultz2007}.
\end{remark}

In \cite[Section 4]{BianeLehner2001}, a Brown measure formula for $u+x_0$ was obtained for bounded, normal operator $x_0$ (see \cite[formula (4.8)]{BianeLehner2001}). The following result generalizes their formula for $x_0\in\CM^\Delta$ with an arbitrary distribution. 
\begin{theorem}
The Brown measure of $x_0+\gamma u$ is absolutely continuous on the set
\[
  \Omega(\gamma u, x_0)=\{\lambda: 1/\phi(|x_0-\lambda{\bf 1}|^{-2}) <\gamma^2<\phi(|x_0-\lambda{\bf 1}|^2) \}
\]
and the density is strictly positive. Moreover, the density is expressed as
\[
 \frac{1}{\pi}\left( s_1(0)^2 \phi(h^{-1}k^{-1})+ \frac{ |\phi \left( \frac{\partial h}{\partial\overline{\lambda} } (h^{-1})^2 |^2 \right) }{ \phi((h^{-1})^2) -1/(s_1(0)+\gamma^2)^2  } \right)
\]
where 
\[
h=h(\lambda,s_1)=(\lambda{\bf 1}-x_0)^*(\lambda{\bf 1}-x_0)+s_1(0)^2
\]
and
\[
k=k(\lambda,s_1)=(\lambda{\bf 1}-x_0)(\lambda{\bf 1}-x_0)^*+s_1(0)^2.
\]
\end{theorem}
\begin{remark}
By the second identity in \eqref{eqn:05.5}, we have 
\[
    \frac{1}{s_1^2+\gamma^2}=\phi(h^{-1}).
\]
Hence, by Cauchy-Schwartz inequality, we have 
\[
  \phi((h^{-1})^2) -1/(s_1(0)+\gamma^2)^2 =\phi((h^{-1})^2) - (\phi(h^{-1}))^2>0. 
\]
\end{remark}
\begin{proof}
By Proposition \ref{prop:boundary-values-s1-s2-analytic}, for $\lambda\in \Omega(u,x_0)$, the limit values of subordination functions $s_1(0), s_2(0)$ are strictly positive and are real analytic function of $\lambda$. We put
\[
  g(\lambda)=\log\Delta\big( (x_0+u-\lambda{\bf 1})^*(x_0+u-\lambda{\bf 1}) \big).
\]
By Proposition \ref{prop:partial-lambda-general}, we have 
 \[
 \frac{\partial}{\partial\lambda} g(\lambda)=\phi\left(h^{-1}\frac{\partial h}{\partial\lambda}\right)
 \]
 where $\frac{\partial h}{\partial\lambda}=(\lambda{\bf 1}-x_0)^*$. 
 We adapt the calculation the proof of Theorem \ref{thm:main-density-formula} (see also \cite[Theorem 4.6]{Zhong2021Brown} and \cite[Lemma 2.8]{HaagerupSchultz2007}) to get the density formula. By \cite[Lemma 3.2]{HaagerupT2005Annals}, since $g(\lambda)$ is a real analytic function of $\lambda$, we have
\begin{equation}
 	\begin{aligned}
\label{eqn:Laplacian-g-lambda}
&\frac{\partial^2}{\partial\overline{\lambda
	}\partial\lambda}g(\lambda)
=\frac{\partial}{\partial\overline{\lambda}}\phi\left(h^{-1}\frac{\partial h}{\partial\lambda}\right)\\
&=\phi\left( -h^{-1} \left( \frac{\partial h}{\partial\overline{\lambda}}+2s_1(0) \frac{\partial {s_1(0)}}{\partial\overline{\lambda}}  \right) h^{-1} \frac{\partial h}{\partial\lambda}   + h^{-1} \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}  \right) \\
&=\phi\left( -h^{-1}  \frac{\partial h}{\partial\overline{\lambda}} h^{-1} \frac{\partial h}{\partial\lambda}   + h^{-1} \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}  \right)
-2s_1(0) \frac{\partial {s_1(0)}}{\partial\overline{\lambda}} 
\phi\left( \frac{\partial h}{\partial {\lambda}} (h^{-1})^2  \right)
	\end{aligned}
\end{equation}
 where
 \[
 \frac{\partial h}{\partial\overline{\lambda}}= \frac{\partial h(\lambda,s_1)}{\partial\overline{\lambda}} =\lambda{\bf 1} -x_0; 
 \]
 and 
 \[
 \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda} 
 =\frac{\partial}{\partial\overline{\lambda}} (\lambda{\bf 1}-x_0)^*=1.
 \]
 
 We now apply the identity $x(x^*x+\varepsilon{\bf 1})^{-1}=(xx^*+\varepsilon{\bf 1})^{-1}x$
 to $x=\lambda{\bf 1}-x_0$ and $\varepsilon=s_1(0)^2$, we find that
  \begin{equation}
   \label{eqn:identity-in-proof-5.6}
  \begin{aligned}
   -\frac{\partial h}{\partial\overline{\lambda}} h^{-1} \frac{\partial h}{\partial\lambda}   +  \frac{\partial^2 h}{\partial\overline{\lambda} \partial\lambda}
   &=1-x(x^*x+s_1(0)^2{\bf 1})^{-1}x^*\\
   &=1-(xx^*+s_1(0)^2{\bf 1})^{-1}xx^*\\
   &=s_1(0)^2 (xx^*+s_1(0)^2{\bf 1})^{-1}\\
   &=s_1(0)^2 k^{-1}.
  \end{aligned}
  \end{equation}
 For $\lambda\in\Omega(u,x_0)$, the function $s_1(0)$ is determined by  (taking limit for \eqref{eqn:fixed-eqn-s1-Bernoulli} as $t$ goes to zero)
 \[
  s_1(0)^2-\frac{s_1(0)}{h_{\mu_1}(s_1(0))}+\gamma^2=0. 
 \]
 which can be rewritten as
 \[
   \frac{1}{s_1(0)^2+\gamma^2}=\frac{h_{\mu_1}(s_1(0))}{s_1(0)}=\phi (h^{-1}). 
 \]
 Taking implicit differentiation $\frac{\partial}{\partial\overline{\lambda}}$ and applying again \cite[Lemma 3.2]{HaagerupT2005Annals}, we then obtain
 \[
 -\phi\left(h^{-1} \left( \frac{\partial h}{\partial\overline{\lambda}}+ 2s_1(0) \frac{\partial {s_1(0)}}{\partial \overline{\lambda}} \right)  h^{-1}   \right)=-\frac{1}{(s_1(0)^2+\gamma^2)^2}2s_1(0)\frac{\partial s_1(0)}{\partial \overline{\lambda}},
 \]
 where $\frac{\partial h}{\partial\overline\lambda}=\lambda{\bf 1}-x_0$.
 This implies 
 \begin{equation}
 	\label{eqn:partial-s1-in-proof}
 	 - 2s_1(0)\frac{\partial {s_1(0)}}{\partial \overline{\lambda}}=\frac{ \phi \left( \frac{\partial h}{\partial\overline{\lambda} } (h^{-1})^2  \right) }{ \phi((h^{-1})^2) -1/(s_1(0)+\gamma^2)^2  }.
 \end{equation}
 By the tracial property, we observe that 
 \[
 \phi \left( \frac{\partial h}{\partial\overline{\lambda} } (h^{-1})^2  \right)
 =\overline{ \phi \left( \frac{\partial h}{\partial{\lambda} } (h^{-1})^2  \right)}.
 \]
 Hence, by plugging identities \eqref{eqn:identity-in-proof-5.6} and \eqref{eqn:partial-s1-in-proof} into \eqref{eqn:Laplacian-g-lambda}, we obtain
 \begin{align*}
  d\mu_{x_0+u}&=\frac{1}{\pi}\frac{\partial^2}{\partial\overline{\lambda
  	}\partial\lambda}g(\lambda)\\
   &=\frac{1}{\pi}\left( s_1(0)^2 \phi(h^{-1}k^{-1})+ \frac{ |\phi \left( \frac{\partial h}{\partial\overline{\lambda} } (h^{-1})^2 |^2 \right) }{ \phi((h^{-1})^2) -1/(s_1(0)+\gamma^2)^2  } \right) \\
    &=\frac{1}{\pi}\left( s_1(0)^2 \phi(h^{-1}k^{-1})+ \frac{ |\phi \left( \frac{\partial h}{\partial\overline{\lambda} } (h^{-1})^2 |^2 \right) }{ \phi((h^{-1})^2) -(\phi(h^{-1}))^2  } \right). 
 \end{align*}
 which is strictly positive, where we used  
 \[
 \frac{1}{s_1^2+\gamma^2}=\phi(h^{-1})
 \]
 due to the second identity in \eqref{eqn:05.5}.
\end{proof}

\subsection{The sum of a circular operator and a free random variable}
Denote by $c$ a standard free circular operator in ${\mathcal M}$, and by $c_\varepsilon$ a free circular operator with variance $\phi(c_\varepsilon^* c_\varepsilon)=\varepsilon$. Let $x_0$ be a random variable with an arbitrary distribution that is freely independent with $c$ and $c_\varepsilon$. We give a new proof for a main result in \cite{HoZhong2020Brown} and \cite{Zhong2021Brown}, where the case when $x_0$ is a bounded operator was studied. Our approach allows us to describe the Brown measure of $x_0+c_\varepsilon$ with $x_0\in\CM^\Delta$ having an arbitrary distribution.

Let $x_0\in \CM^\Delta$ be a random variable with an arbitrary distribution that is freely independent with $\{c_\varepsilon\}_{\varepsilon\geq 0}$. Fix $\varepsilon>0$ and for any $\lambda\in\mathbb{C}$, we denote $T_1=V(x_0-\lambda{\bf 1})$ and $T_2=c_\varepsilon$,
where $V$ is a Haar unitary $*$-free from $x_0$ and $c_\varepsilon$. Denote $\mu_1=\tilde\mu_{|x_0-\lambda{\bf 1}|}$ and $\mu_2=\tilde\mu_{|c_\varepsilon|} $ and $\mu=\mu_1\boxplus\mu_2=\tilde\mu_{|T_1+T_2|}$. 


\begin{lemma}
\label{lemma:fixed-pt-ct-simplified}
The subordination function $s_1$ satisfies 
\[
  \varepsilon h_\mu(t)=s_1(t)-t, \qquad \forall t>0.
\]
\end{lemma}
\begin{proof}
Note that $\mu_2=\mu_{|c_\varepsilon|}$ has semicircle distribution with variance $\varepsilon$. Hence its $R$-transform is given by $R_{\mu_2}(z)=\varepsilon z$. 
The additive of $R$-transform yields
\[
   R_{\mu_1}(z)+R_{\mu_2}(z)=R_\mu(z)
\]
which yields
\[
  G_{\mu_1}^{-1}(z)+R_{\mu_2}(z)=G_{\mu_1}^{-1}(z)+\varepsilon z=G_\mu^{-1}(z).
\]
By replacing $z$ by $G_\mu(z)$, we obtain
$\omega_1(z)=z-\varepsilon G_\mu(z)$, which holds for any $z\in\mathbb{C}^+$ by analytic continuation. In particular, when $z=it$, we have 
\[
  \omega_1(it)-it=-\varepsilon G_\mu(it).
\]
Recall that $G_{\mu}(it)=-ih_{\mu}(t)$ and $\omega_1(it)=i s_1(t)$. The result follows. 
\end{proof}

\begin{lemma}
	\label{lemma:det-s2-circular}
For any $t>0$, we have 
\begin{equation}
	\label{eqn:det-s2-circular}
	  \Delta(|c_\varepsilon|^2+s_2(t)^2)(h_\mu(t))^2=\exp\left(-\varepsilon h_\mu(t)^2 \right)
\end{equation}
\end{lemma}
\begin{proof}
The logarithmic derivative of \eqref{eqn:det-s2-circular} with respect to $t$ is given by
\begin{equation}
 	\label{eqn:det-s2-circular-derivative}
  \begin{aligned}
 \phi((|c_\varepsilon|^2+s_2(t)^2)^{-1}) s_2(t) s_2'(t)+\frac{h_\mu'(t)}{h_\mu(t)}=-\varepsilon h_\mu(t)h_\mu'(t).
  \end{aligned}
\end{equation}
We will first verify that \eqref{eqn:det-s2-circular-derivative} holds. 
Using notations in Section \ref{section:free-convolution-symmetric}, we have 
\[
   \phi((|c_\varepsilon|^2+s_2(t)^2)^{-1}) s_2(t)=h_{\mu_2}(s_2(t))=h_\mu(t),
\]
which allows us to rewrite \eqref{eqn:det-s2-circular-derivative} as
\begin{equation}
 \label{eqn:derivative-5.10-in-proof}
	  h_\mu(t) s_2'(t)+\frac{h_\mu'(t)}{h_\mu(t)}=-\varepsilon h_\mu(t)h_\mu'(t).
\end{equation}
We also have $s_1(t)+s_2(t)=t+1/h_\mu(t)$. Hence, 
\[
  s_2'(t)=(1/h_\mu(t)+t-s_1(t))'=-\frac{h_\mu'(t)}{h_\mu(t)^2}+1-s_1'(t).
\]
Plugging this into \eqref{eqn:derivative-5.10-in-proof}, we see that to prove \eqref{eqn:det-s2-circular-derivative}, it suffices to show that 
\begin{equation}
	\label{eqn:derivative-5.11-in-proof}
	 h_\mu(t)-h_\mu(t)s_1'(t)=-\varepsilon h_\mu(t)h_\mu'(t).
\end{equation}
Lemma \ref{lemma:fixed-pt-ct-simplified} tells that $s_1$ satisfies 
\begin{equation}
 	\label{eqn:derivative-5.12-in-proof}
 \varepsilon h_\mu(t)=s_1(t)-t.
\end{equation}
We note that \eqref{eqn:derivative-5.12-in-proof} actually implies \eqref{eqn:derivative-5.11-in-proof} by plugging the derivative of \eqref{eqn:derivative-5.12-in-proof} to \eqref{eqn:derivative-5.11-in-proof}. Hence,  
\eqref{eqn:det-s2-circular-derivative} holds for all $t>0$. 

We next observe that $ \lim_{t\rightarrow \infty}h_\mu(t)=0$, and 
  \[
     \lim_{t\rightarrow \infty}\log\Delta(|c_\varepsilon|^2+s_2(t)^2)-2\log t=0,
  \]
and 
\[
 \lim_{t\rightarrow \infty}\log(h_\mu(t))+2\log t=0.
\]
Hence \eqref{eqn:det-s2-circular} holds for all $t>0$. 
\end{proof}

\begin{lemma}
	\label{lemma:det-circular-t}
We have, for any $t>0$, 
\begin{equation}
  \Delta(|x_0+c_\varepsilon-\lambda{\bf 1}|^2+t^2)=\Delta(|x_0-\lambda{\bf 1}|^2+s_1(t)^2)\exp\left( -\varepsilon h_\mu(t)^2\right)
\end{equation}
where $s_1(t)$ is the subordination as in Section \ref{section:free-convolution-symmetric}
and 
\[h_\mu(t)=t\phi ((|x_0+c_\varepsilon-\lambda{\bf 1}|^2+t^2)^{-1}).\]
\end{lemma}
\begin{proof}
Note that $s_1(t)+s_2(t)-t=1/h_\mu(t)$. Then result then follows from Theorem \ref{thm:Delta-T-1-2-general} and Lemma \ref{lemma:det-s2-circular}. 
\end{proof}

The following result extends \cite[Theorem 3.12]{Zhong2021Brown} and \cite[Theorem 3.10]{HoZhong2020Brown} to unbounded operators in $\CM^\Delta$.
\begin{theorem}
	\label{thm:det-0-circular-general}
	Let $x_0\in \CM^\Delta$ be a random variable freely independent from the circular operator $c_\varepsilon$ with variance $\varepsilon>0$. 
	\begin{enumerate}[(1)]
		\item   If $||(x_0-\lambda{\bf 1})^{-1}||_2^2> \frac{1}{\varepsilon}$, then
		\begin{align}
		\label{eqn:Delta-x0-c-0-case1}
		\Delta(x_0+c_\varepsilon-\lambda{\bf 1})=\bigg( \Delta\big((x_0-\lambda{\bf 1})^*(x_0-\lambda{\bf 1})+s_1(0)^2{\bf 1}\big) \bigg)^{\frac{1}{2}}\exp\left(-\frac{s_1(0)^2}{2\varepsilon}\right)
		\end{align}
		where $s_1(0)$ is determined by 
		\begin{equation}
		\label{eqn:s-1-0-x0-vt}
		\phi\bigg[ \big((x_0-\lambda{\bf 1})^*(x_0-\lambda{\bf 1})+s_1(0)^2{\bf 1} \big)^{-1}\bigg]=\frac{1}{\varepsilon}.
		\end{equation}
		\item If $||(x_0-\lambda{\bf 1})^{-1}||_2^2\leq \frac{1}{\varepsilon}$, then
		\begin{align}
		\label{eqn:Delta-x0-c-0-case2}
		\Delta(x_0+c_\varepsilon-\lambda{\bf 1})=\Delta(x_0-\lambda{\bf 1}).
		\end{align}
	\end{enumerate}
\end{theorem}
\begin{proof}
We have $\lambda_2(\mu_2)=\lambda_2(\mu_{|c_\varepsilon|})=\varepsilon$ and $\lambda_1(\mu_1)=\frac{1}{||(x_0-\lambda{\bf 1})^{-1}||_2^2}$.  Hence
\begin{equation}
\label{eqn:Omega-circular-1}
		||(x_0-\lambda{\bf 1})^{-1}||_2^2> \frac{1}{\varepsilon} \Longleftrightarrow \lambda_1(\mu_1)< \lambda_2(\mu_2).
\end{equation}
	Since $\lambda_1(\mu_2)=0$,
	when $||(x_0-\lambda{\bf 1})^{-1}||_2^2> \frac{1}{\varepsilon}$,  Case 3 in Theorem \ref{thm:lambda-1-2-cases} applies, in which case $0<s_1(0)<\infty$. Hence, \eqref{eqn:Delta-x0-c-0-case1} follows from Lemma \ref{lemma:det-circular-t} by taking the limit $t\rightarrow 0$. Now, as in Lemma \ref{lemma:fixed-pt-ct-simplified}, $s_1(t)$ satisfies 
	\begin{equation*}

	h_{\mu_1}(s_1(t))=\frac{s_1(t)-t}{\varepsilon}.
	\end{equation*}
	Hence, $s_1(0)\in (0,\infty)$ satisfies
	\[
	\phi\bigg[ \big((x_0-\lambda{\bf 1})^*(x_0-\lambda{\bf 1})+s_1(0)^2{\bf 1} \big)^{-1}\bigg]= \frac{h_{\mu_1}(s_1(0))}{s_1(0)}=\frac{1}{\varepsilon}.
	\]
	
	Finally, rewrite the condition $||(x_0-\lambda{\bf 1})^{-1}||_2^2\leq \frac{1}{\varepsilon}$ as  $\lambda_2(\mu_2)\leq \lambda_1(\mu_1)$. We apply Case 1 and Case 4 in Theorem \ref{thm:lambda-1-2-cases} and have $s_1(t)\rightarrow 0$ as $t\rightarrow 0$. Then \eqref{eqn:Delta-x0-c-0-case2} from Lemma \ref{lemma:det-circular-t}.
\end{proof}

The following result extends \cite[Theorem 4.6]{Zhong2021Brown} to all $x_0\in\CM^\Delta$.  
\begin{theorem}
	\label{thm:density-circular-general}
	Let $x_0\in \CM^\Delta$ be a random variable freely independent from the circular operator $c_\varepsilon$ with variance $\varepsilon>0$. The Brown measure of $x_0+c_\varepsilon$ is supported in the closure of the set 
	\[
	\Xi_\varepsilon=\left \{\lambda\in\mathbb{C}: 	\phi\bigg[ \big((x_0-\lambda{\bf 1})^*(x_0-\lambda{\bf 1})+s_1(0)^2{\bf 1} \big)^{-1}\bigg]> \frac{1}{\varepsilon}\right\}.
	\]
	The Brown measure in this set is given by 
	\begin{equation}
	\frac{1}{\pi}\left( \frac{|\phi((\lambda{\bf 1}-x_0)(h^{-1})^2)|^2}{\phi((h^{-1})^2)}
	+s_1(0)^2\phi(h^{-1}k^{-1})  \right)
	\end{equation}
	where $h=h(\lambda,s_1)$ and $k=k(\lambda,s_1)$ for 
	\[
	h(\lambda,s_1)=(\lambda{\bf 1}-x_0)^*(\lambda{\bf 1}-x_0)+s_1(0)^2
	\]
	and
	\[
	k(\lambda,s_1)=(\lambda{\bf 1}-x_0)(\lambda{\bf 1}-x_0)^*+s_1(0)^2.
	\]
	In particular, the density of the Brown measure of $x_0+c_\varepsilon$ is strictly positive within the domain. 
\end{theorem}
\begin{proof}
The set $\Xi_\varepsilon=\Omega(c_\varepsilon, x_0)$ as we observed in the proof for Theorem \ref{thm:det-0-circular-general}, in particular \eqref{eqn:Omega-circular-1}. Hence, the Brown measure of $x_0+c_\varepsilon$ is supported in the closure of $\Xi_\varepsilon$ by Theorem \ref{thm:support-Brown-x0-T}. 
See  \cite[Theorem 4.6]{Zhong2021Brown} for a proof for the density formula. 
\end{proof}



The above result allows us to express the Brown measure of $x_0+c_\varepsilon$ and recover results in \cite[Section 3]{HoZhong2020Brown}. 

\begin{remark}
	\label{remark:x0-c-5.25}
	For $x_0$ self-adjoint, let $\mu_x$ be the spectral measure of $\mu_x$ and $\lambda=a+bi$, then the set $\Xi_\varepsilon$ in Theorem \ref{thm:density-circular-general} is expressed as
	\begin{equation}
	\label{eqn:5.34-in-proof}
	\int_{\mathbb{R}}\frac{1}{(u-a)^2+b^2}d\mu_x(u)> \frac{1}{\varepsilon}.
	\end{equation}
	Note the map $b\mapsto \int_{\mathbb{R}}\frac{1}{(u-a)^2+b^2}d\mu_x(u)$ is monotonically decreasing. If $\lambda=a+bi$ satisfies \eqref{eqn:5.34-in-proof}, we let $v_\varepsilon(a)$ be the unique value so that
	\begin{equation}
	  \label{eqn:identity-v-t-5.20}
	\int_{\mathbb{R}}\frac{1}{(u-a)^2+v_\varepsilon(a)^2}d\mu_x(u)= \frac{1}{\varepsilon}.
	\end{equation}
	Then, by comparing with \eqref{eqn:s-1-0-x0-vt}, we have 
	\[
	v_\varepsilon(a)^2=b^2+s_1(0)^2.
	\]
	This formulation appeared in Biane work \cite{Biane1997}. See \cite[Section 2.3 and Theorem 3.10]{HoZhong2020Brown}.
\end{remark}

\begin{theorem}
	\label{thm:x0-c-main-density}
	Let $x_0\in \CM^\Delta$ be a self-adjoint random variable freely independent from the circular operator $c_\varepsilon$ with variance $\varepsilon>0$. Let $\mu_{x_0}$ be the spectral measure of $x_0$. The Brown measure of $x_0+c_\varepsilon$ is supported in the closure of the set 
\begin{equation}
\label{eqn:support-circular-x0}
 \Lambda_\varepsilon=	\left \{\lambda=a+bi\in\mathbb{C}: \int_{\mathbb{R}}\frac{1}{(u-a)^2+b^2}d\mu_{x_0}(u)> \frac{1}{\varepsilon}\right\}.
\end{equation}
	The Brown measure in this set is given by 
	\begin{equation}
	d\mu_{x_0+c_\varepsilon}(\lambda)=\frac{1}{\pi\varepsilon}\left(1-\frac{\varepsilon}{2}\frac{d}{da}\int_\mathbb{R}\frac{u}{(u-a)^2+v_\varepsilon(a)^2}d\mu_{x_0}(u) \right).
	\end{equation}
	Moreover, the density is strictly positive and analytic within the interior of the support. 
\end{theorem}

\begin{proof}
The positivity of the Brown measure density follows from Theorem \ref{thm:density-circular-general}. 
By Proposition \ref{prop:partial-lambda-general}, we have 
\[
  \begin{aligned}
    &\frac{\partial}{\partial\lambda}\log\Delta(|x_0+c_\varepsilon-\lambda{\bf 1}|^2)  \\ 
       &=\phi( (x_0-\lambda{\bf 1})^* (|x_0-\lambda{\bf 1}|^2+s_1(0)^2)^{-1} )\\
       &=\int_\mathbb{R}\frac{u-a+ib}{(u-a)^2+b^2+s_1(0)^2}d\mu_{x_0}(u)\\
       &=\int_\mathbb{R}\frac{u-a}{(u-a)^2+v_\varepsilon(a)^2}d\mu_{x_0}(u)+\frac{ib}{\varepsilon}
  \end{aligned}
\]
where we used \eqref{eqn:identity-v-t-5.20}. Note that the real part of the above formula depends only on $a$. Hence, 
\[
 d\mu_{x_0+c_\varepsilon}(\lambda)=\frac{1}{\pi}\frac{\partial}{\partial\overline{\lambda}}
   \left(\int_\mathbb{R}\frac{u-a}{(u-a)^2+v_\varepsilon(a)^2}d\mu_{x_0}(u)+\frac{ib}{\varepsilon}\right)
\]
which yields the density formula. Using the same proof as \cite[Lemma 3.11]{HoZhong2020Brown}, we also see that $\mu_{x_0+c_\varepsilon}(\Lambda_\varepsilon)=1$. 
\end{proof}

The following generalizes \cite[Theorem 6.10]{Zhong2021Brown} where the support and density of the Brown measure of $g_t+u$ was studied. 
\begin{corollary}
	Let $T\in\CM^\Delta$ be an $R$-diagonal operator and $g_t$ be a semicircular operator. The support of the Brown measure of $T+g_t$ is contained in a ring where the inner boundary is a circle with radius $\sqrt{(\lambda_1(T)^2-t)_+}$ and the outer boundary is an ellipse with semi-axes $\big(\frac{2t+\lambda_2(T)^2}{t+\lambda_2(T)^2}, \frac{\lambda_2(T)^2}{t+\lambda_2(T)^2} \big)$. 
\end{corollary}
\begin{proof}
	We apply Theorem \ref{thm:support-Brown-x0-T} for $x_0=g_t$. We have 
	\[
	||g_t-\lambda{\bf 1}||_2=\phi\big( (g_t-\lambda{\bf 1})^*(g_t-\lambda{\bf 1}) \big)=t+|\lambda|^2.
	\]
	Hence, condition $||g_t-\lambda{\bf 1}||_2||T^{-1}||_2>1$ is written as
	\begin{equation}
	|\lambda|^2>1/||T^{-1}||_2-t=\lambda_1(T)^2-t.
	\end{equation}
	For other condition $||(g_t-\lambda{\bf 1})^{-1}||_2 ||T||_2>1$. We replace $T$ by a circular operator $c_s$ with $s=||T||_2^2=\lambda_2(T)^2$. This condition can then be rewritten as
	\begin{equation}
	\label{eqn:semicircular-solution}
	\int_\mathbb{R} \frac{1}{(u-a)^2+b^2}d\sigma(u)>\frac{1}{\lambda_2(T)^2}=\frac{1}{s}
	\end{equation}
	where $\sigma$ is the semicircular distribution with variance $t$. We recognize this is the same equation (see \eqref{eqn:support-circular-x0}) when we study the Brown measure of the elliptic operator $g_t+c_s$ whose support is known to be the ellipse with semi-axes $(\frac{2t+s}{\sqrt{t+s}}, \frac{2s}{\sqrt{s+t}})$. That means the solution of \eqref{eqn:semicircular-solution} is the region enclosed by the ellipse with semi-axes 
	$(\frac{2t+s}{\sqrt{t+s}}, \frac{2s}{\sqrt{s+t}})$ where $s=\lambda_2(T)^2$. 
\end{proof}


\subsection{The sum of a circular Cauchy operator and a free random variable}
In this section, we study $R$-diagonal operator $z$ with $\tilde{\mu}_z$ as the Cauchy distribution. We give a short proof of various results of Haaagerup-Schultz \cite{HaagerupSchultz2009} on this operator. Those results played a key role in their seminal paper \cite{HaagerupSchultz2009} on the invariant subspace problem in type II$_1$ factors.  

\begin{definition}
	The operator $z\in \CM^\Delta$ is called free circular Cauchy operator if $z$ has the same distribution as $uh$ where $u$ and $h$ are $*$-free,  $u$ is a Haar unitary and $h=h^*$ is Cauchy distributed,
	\[
	d\mu_h=\frac{1}{\pi}\frac{1}{x^2+1}dx.
	\]
\end{definition}
In particular, $z$ is $R$-diagonal. A simple computation shows that its Cauchy transform is 
\[
G_{\mu_h}(z)=\frac{1}{z+i}
\]
and its $R$-transform is 
\[
\mathcal{R}_{\mu_h}(z)=-i.
\]
Let $\{x, y\}$ be a circular system and $\{s_1, s_2, s_3, s_4\}$ be a semicircular system and $U$ be a  Haar unitary operator free from $\{s_1, s_2, s_3, s_4\}$ in the finite von Neumann algebra ${\mathcal M}$ equipped with faithful normal trace state $\phi$ . Then $x\underset{\ast{\mathcal D}}{\sim} \frac{s_1+is_2}{\sqrt{2}}\underset{\ast{\mathcal D}}{\sim}  Us_1$. The probability measure $\tilde{\mu}_{|x|}$ of the symmetrization of $\mu_{|x|}$ has the semicircle distribution
\[
d\tilde{\mu}_{|x|}=\frac{1}{2\pi}\sqrt{4-t^2}1_{[-2,2]}(t)dt.
\]
The following result was mentioned inexplicitly in \cite[Section 5]{HaagerupSchultz2007}. 
\begin{proposition}
	The free circular Cauchy operator $z$ has the same distribution as $xy^{-1}$ and $s_1s_2^{-1}$. 
	The $S$-transform of $|z|^2$ is given by 
	\[
	S_{|z|^2}(t)=-\frac{t}{1+t}, \quad t\in (-1,0),
	\]
	and the Cauchy transform 
	\begin{equation}
	\label{eqn:Cauchy-transform-Cauchy}
	G_{\mu_{|z|^2}}(\lambda)=\frac{1}{\lambda-\sqrt{-\lambda}}, \quad \lambda<0.
	\end{equation}
\end{proposition}
\begin{proof}
	Let $T=xy^{-1}$. The operators $x$ and $y^{-1}$ are $*$-free and hence $T$ is $R$-diagonal. The distribution of $|T|$ is given by \cite[Theorem 5.2 (b)]{HaagerupSchultz2007} as 
	\[
	d\mu_{|T|}(t)=\frac{2}{\pi}\frac{1}{1+t^2}1_{(0,\infty)}(t)dt. 
	\]
	Hence the symmetric measure $\tilde{\mu}_{|T|}$ is Cauchy distributed. Hence, 
	\[
	z\underset{\ast{\mathcal D}}{\sim} U|T| \underset{\ast{\mathcal D}}{\sim} xy^{-1}.
	\]
	The semicircular operators $s_1$ and $s_2$ are even operators, hence the product $s_1s_2^{-1}$ is $R$-diagonal. Observe that $|s_1s_2^{-1}|$ has the same distribution as $|xy^{-1}|$. Hence, 
	$z$, $xy^{-1}$ and $s_1s_2^{-1}$ have the same distribution. 
	The $S$-transform and Cauchy transform of $|z|^2$ was calculated in the proof of \cite[Theorem 5.2]{HaagerupSchultz2007}.
\end{proof}
From the distribution of $|z|^2$, one can check that (see \cite[Theorem 5.4]{HaagerupSchultz2007}),
for any $n\in\mathbb{N}$, 
\begin{equation}
\label{eqn:h-mu-Cauchy-power}
h_{\mu_{|z|^n}}(s)=\int_0^\infty \frac{s}{s^2+u^2}d\mu_{|z|^n}(u)=\left(s+s^{\frac{n-1}{n+1}}\right)^{-1}.
\end{equation}
In particular, 
\begin{equation}
\label{eqn:h-mu-Cauchy}
h_{\mu_{|z|}}(s)=\frac{1}{s+1}.
\end{equation}
We then study the Brown measure of $x_0+z$ where $x_0\in \CM^\Delta$ is $*$-free from $z$ with an arbitrary distribution. We now set
\[
\mu_1=\tilde\mu_{|x_0-\lambda{\bf 1}|}, \quad\text{and}\quad \mu_2=\tilde\mu_{|z|}.
\]
and $\mu=\mu_1\boxplus \mu_2$. We hence have
\[
  \tilde\mu_{|x_0+z-\lambda{\bf 1}|}=\mu_1\boxplus\mu_2.
\]
For $\lambda\in\mathbb{C}$ fixed, following notations in Section \ref{section:free-convolution-symmetric}, we set in this subsection
\[
 T_1=V(x_0-\lambda{\bf 1}), \qquad T_2=z.
\]
where $V$ is a Haar unitary that is $*$-free from $\{x_0, z\}$. 
We recall that for any probability measure $\mu$ on $\mathbb{R}^+$, we have $h_\mu(s)=h_{\tilde{\mu}}(s)$, where
\[
h_\mu(s)=\int_\mathbb{R}\frac{s}{s^2+u^2}d\mu(u), \quad s>0.
\]


\begin{lemma}
	\label{lemma:det-s2-Cauchy-t}
We have $s_1(t)=t+1$ and 
\begin{equation}
 \label{eqn:det-s2-Cauchy-t}
	 \Delta(|T_2|^2+s_2(t)^2)h_\mu(t)^2=1.
\end{equation}
\end{lemma}
\begin{proof}
The function $s_1(t)$ satisfies the following fixed point equation from \eqref{eqn:fixed-point-s-1-2}
\[
s_1(t)=t+f_{\mu_2}(t+f_{\mu_1}(s_1(t))), \quad t> 0
\]
where $f_{\mu_i} (i=1,2)$ was defined in \eqref{defn:f-mu-s}. In particular, due to \eqref{eqn:h-mu-Cauchy}, we have 
\[
f_{\mu_2}(s)=1.
\]
Consequently, 
\begin{equation}
\label{eqn:subordination-Cauchy}
s_1(t)=t+1
\end{equation}
and
\[
  s_2(t)=t+\frac{1}{h_\mu(t)}-s_1(t)=\frac{1}{h_\mu(t)}-1. 
\]
Taking the derivative with respect to $t$, we obtain
\begin{equation*}
  s_2'(t)=-\frac{h_\mu'(t)}{h_\mu(t)^2}
\end{equation*}
and therefore (by using $h_\mu(t)=h_{\mu_2}(s_2(t))$)
\begin{equation}
 \label{eqn:derivative-det-s2-Cauchy}
	h_{\mu_2}(s_2(t))s_2'(t)+\frac{h_\mu'(t)}{h_\mu(t)}=0.
\end{equation}
The above equation is exactly the one half of the logarithmic derivative of \eqref{eqn:det-s2-Cauchy-t}. Finally, observe that 
\[
  \lim_{t\rightarrow \infty}\Delta(|T_2|^2+s_2(t)^2)h_\mu(t)^2=1.
\]
The result then follows. 
\end{proof}

\begin{lemma}
	\label{lemma:Delta-T-z-T-t}
	Let $T\in \CM^\Delta$ be $*$-free from the circular Cauchy operator $z\in\CM^\Delta$. We then have 
	\begin{equation}
	\label{eqn:Delta-T-z-T-t}
	\Delta\large( (T+z)^*(T+z)+t^2{\bf 1}\large)=\Delta\large( T^*T+(t+1){\bf 1} \large)
	\end{equation}
\end{lemma}
\begin{proof}
It follows from Theorem \ref{thm:Delta-T-1-2-general} and Lemma \ref{lemma:det-s2-Cauchy-t}. 
\end{proof}


We are now be able to extend \cite[Proposition 4.5]{HaagerupSchultz2009} to any unbounded free random variable in $\CM^\Delta$. 
\begin{theorem}
	\label{thm:Delta-T-z-0}
	Let $T\in \CM^\Delta$ and $z=xy^{-1}$, where $\{x,y\}$ is a circular system. Then for any $a>0$,  we have 
	\begin{equation}
	\label{Delta-T-z-0}
	\Delta(T+az)=\Delta(T^*+a^2{\bf 1})^\frac{1}{2}, \quad a>0.
	\end{equation}
\end{theorem}
\begin{proof}
	By rescaling with a factor of $1/a$, it suffices to prove \eqref{Delta-T-z-0} when $a=1$. We have 
	\[
	\lim_{t\rightarrow 0} \Delta\large( (T+z)^*(T+z)+t^2{\bf 1}\large)=\large(\Delta(T+z)\large)^2
	\]
	and
	\[
	\lim_{t\rightarrow 0}\Delta(T^*T+(t+1){\bf 1})=\Delta(T^*T+{\bf 1}).
	\]
	Then \eqref{Delta-T-z-0} follows from Lemma \ref{lemma:Delta-T-z-T-t}.
\end{proof}

The following result generalizes \cite[Corollary 4.6]{HaagerupSchultz2009} to unbounded operators.
\begin{corollary}
	Let $T\in \CM^\Delta$ be $*$-free from the circular Cauchy operator $z\in\CM^\Delta$. The Brown measure of $T+az$ is given by
	\begin{equation}
	\label{eqn:Brown-T-z}
	d\mu_{T+az}(\lambda)=\frac{a^2}{\pi}\phi\large( (T(\lambda)^*T(\lambda)+a^2{\bf 1})^{-1} (T(\lambda)T(\lambda)^*+a^2{\bf 1})^{-1}\large)d\lambda_1d\lambda_2,
	\end{equation}
	where $\lambda_1=\Re \lambda, \lambda_2=\Im \lambda$, and $T(\lambda)=T-\lambda{\bf 1}$. 
	In particular, the Brown measure of $z$ is given by 
	\[
	d\mu_z(\lambda)=\frac{1}{\pi(1+|\lambda|^2)^2}d\lambda_1d\lambda_2.
	\]
\end{corollary}
\begin{proof}
	The proof is identical to the proof of \cite[Corollary 4.6]{HaagerupSchultz2009}, thanks to Theorem \ref{thm:Delta-T-z-0}.
\end{proof}





\bigskip





\bibliographystyle{acm}
