Solve for $\alpha$, $z^* ( B + \alpha A )^{-1} A ( B + \alpha A )^{-1} z = c$, where ${\rm rank}(A)=1$,$B$ full rank matrix 
Problem (I raised a similar problem, but in this problem $A$ and $B$ are additionally Hermitian):
Let $A \in M_{n,n}(\mathbb{C})$ be a rank one matrix and Hermitian, $B \in M_{n,n} (\mathbb{C})$ be a full rank matrix and Hermitian, $z \in M_{n,1} (\mathbb{C})$ be a column vector, and $c \in \mathbb{R}$ are given, then 
Solve for $\alpha \in \mathbb{R}$ (which is non-negative, i.e., $\alpha > 0$) such that 
  \begin{align}
z^* \left( B^* + \alpha A^* \right)^{-1} A \ \left( B + \alpha A \right)^{-1} z &= c \\
z^* \left( B + \alpha A \right)^{-1} A \ \left( B + \alpha A \right)^{-1} z &= c \ \hspace{5mm} \textrm{since } A \textrm{ and } B \textrm{ are Hermitian}. 
\end{align}
Is it possible to obtain a closed-form solution for $\alpha$?

NOTE: 


*

*If $B = I$ and $A$ is Hermitian then one can solve the problem in closed form. However, I don't know yet how to solve for any Hermitian matrix $B$. 

 A: Since $A$ is rank one and Hermitian, then we can write
\begin{equation}
 A = aa^H
\end{equation}
So the equation is now
\begin{align}
z^H \left( B^H + \alpha A^H \right)^{-1} aa^H \ \left( B + \alpha A \right)^{-1} z =c
\end{align}
which is 
\begin{equation}
 \vert a^H(B + \alpha A)^{-1}z \vert^2 = c
\end{equation}
Let's solve 
\begin{equation}
 a^H(B + \alpha A)^{-1}z = \sqrt{c}e^{i \theta} = d
\end{equation}
for $\theta \in \mathbb{R}$. This means in your solution, you will have a free $\theta$ parameter.
Now, let's apply the Matrix Inversion Lemma and since $B$ is invertible, we can say:
\begin{equation}
 (B + \alpha aa^H)^{-1}
 =
 B^{-1} - \alpha \frac{B^{-1}AB^{-1}}{1 + \alpha a^H B^{-1} a}
\end{equation}
Plugging this we get
\begin{equation}
 a^H(B^{-1} - \alpha \frac{B^{-1}AB^{-1}}{1 + \alpha a^H B^{-1} a})z =  d
\end{equation}
which is 
\begin{equation}
 a^H B^{-1} z 
 -
 \alpha \frac{a^HB^{-1}AB^{-1}z}{1 + \alpha a^H B^{-1} a}
 =
  d
\end{equation}
which is 
\begin{equation}
 [a^H B^{-1} z][1 + \alpha a^H B^{-1} a] - \alpha[a^HB^{-1}AB^{-1}z]
 =
  d[1 + \alpha a^H B^{-1} a]
\end{equation}
Let 
\begin{align}
 \beta_1 &= a^H B^{-1} z \\
 \beta_2 &= a^H B^{-1} a\\
 \beta_3 &= a^HB^{-1}AB^{-1}z = \beta_2\beta_1
\end{align}
Rearranging, we get
$$\alpha=\frac{\beta_1 - d}{d\beta_2}$$
