Let $a\in\mathbb C$ and consider the matrix $$A=\begin{pmatrix}1&a&\dots&\dots &a\\\bar a&1&a&\dots&\vdots\\\vdots&\bar a&1&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots& a\\\bar a &\dots&\dots&\bar a&1\end{pmatrix},$$ i.e. with $1$ on the diagonal, $a$ above the diagonal and $\bar a$ below the the diagonal.
Is there a name for such matrices? Is there an explicit formula for the inverse $A^{-1}$?
The question is trivial if $a\in\mathbb R$ since in this case $A$ is the rank-one perturbation $$A= (1-a)I+a \mathbf e \mathbf e^t$$ of the identity $I$ and therefore $$A^{-1} = \frac{1}{1-a}\biggl(I - \frac{\mathbf e \mathbf e^t}{N+1-a} \biggr),$$ where $N$ is the dimension of $A$ and $\mathbf e$ is the constant-1 vector.