Let $B$ be a cyclic upper-triangular nonnegative matrix,
$$B = \begin{bmatrix} 0 & b_1 & 0& \dots &\dots &0 \\ 0 & 0 & b_2 & 0& \dots & 0\\ \vdots &\vdots&\vdots &\ddots &\ddots &\vdots \\ 0& 0 & 0 & 0 &\dots & b_{n-1}\\ b_n& 0 & 0 & 0 &\dots & 0 \end{bmatrix}$$
with $b_i \in [0,1]$. For any $\alpha > 0$, let us define the matrix
$$C(\alpha)=\alpha B +\alpha^{-1}B^T,$$
where $\alpha$ controls the degree of asymmetry, $\alpha=1$ giving a symmetric $C(\alpha)$ .
I am interested in controlling the localization of the normalized eigenvalues of $C(\alpha)$ on the complex plane. Formally, given any $C(\alpha)$ with eigenvalues $\{\lambda_1,\dots,\lambda_n\}$, let us define the normalized eigenvalues as $$\bar \lambda_i=\frac{\lambda_i}{\max_i |\lambda_i |}$$ By definition, the normalized eigenvalue fall within a circle of radius $1$ on the complex plane. Moreover, when $\alpha=1$ and $C(\alpha)$ is symmetric, the spectrum is real so the normalized eigenvalues fall on the real axis in the interval $[-1,1]$. For larger but finite $\alpha$, I have found numerically that the normalized eigenvalues fall within an ellipse of height $\tanh(\ln \alpha)$. This ellipse is achieved when all $b_i$ are equal so that $C(\alpha)$ is a circulant matrix (as pointed out in the partial answer by @Charr).
Any ideas for how to prove this for a general $B$?
Here's a plot of normalized eigenvalues from many random matrices (random $b_i$), $\alpha=1.2$, with the ellipse shown:
Edit: Let $v$ be the eigenvector associated with the largest (in magnitude) eigenvalue $\lambda^*$ of $C(\alpha)$. By Perron-Frobenius, we may assume $\lambda^*$ and $v$ are real-valued and nonnegative, so \begin{align}\lambda^* = v^T C(\alpha)v = (\alpha + 1/\alpha) v^T B v \end{align} Then, consider any other eigenvalue $\lambda_i$ with right eigenvector $u$ ($\Vert u \Vert =1$). It may be helpful to express $\bar{\lambda_i}$ as $$\begin{align} \bar{\lambda}_i = \frac{1}{\lambda^*}{u^* C(\alpha) u} &= \frac{1}{\lambda^*}\left[(\alpha + 1/\alpha) \,u^*\frac{B+B^T}{2}u + (\alpha -1/\alpha) u^*\frac{B-B^T}{2}u\right]\\ &= \mathrm{Re} x + \mathrm{i} \tanh(\ln \alpha)\mathrm{Im}\, x \end{align}$$ where for convenience I defined $x= u^* B u/v^T B v$. Thus, the problem reduces to showing that \begin{align}\vert x\vert =\left\vert \frac{u^* B u}{v^T B v}\right\vert \le 1 \end{align} where $u,v$ are two eigenvectors of $C(\alpha)$, with $v$ being the largest (Perron-Frobenius) one.
