Let's consider the following $n \times n$ cyclic stochastic matrix

$$ M=   \begin{pmatrix} 
0 & a_2 &   & & &b_n \\\
b_1 & 0& a_3& &&& \\\
& b_2 & 0& \ddots &  &  \\\
&   &\ddots&\ddots &a_{n-1} & \\\
&   && &0 &a_n \\\
a_1 &  &   & &b_{n-1} &0 
\end{pmatrix}
 $$


such that $\forall i$, $a_i,\,b_i$ are positive real number, $a_i+b_i = 1$ and all other component of the matrix are zeros. This is a cyclic matrix in the sense that the associated graph is cyclic.

From the Perron-Frobenius theorem, the eigenvalues $\lambda$ of such matrix all belong to the unit circle.
$$(\Re \lambda )^2 + (\Im \lambda )^2 \leq 1 $$

From numerical explorations, I believe that all eigenvalues  of $M$ belong to the ellipse 
$$(\Re \lambda )^2 + \frac{(\Im \lambda )^2}{(\tanh p)^2} \leq 1 $$

where $p$ denote $p = \frac{1}{2}\ln \frac{\sqrt[n]{\prod_i a_i}}{\sqrt[n]{\prod_i b_i}}$, assumed to be positive, otherwise inverse $a_i$ and $b_i$.

One of the extremal case is the symmetric case $a_i=b_i$ where $p=0$ and all eigenvalues are real. The equality is reached in the uniform case of all $a_i$ to being equal to some value and all $b_i$ being equal to another value, the matrix being then a circulant matrix.

I can already prove that the imaginary part of the eigenvalue is bounded by $\tanh p$ (see below), but I am unable to extend the prove to include the real part. 
I also try to play with the Brauer theorem about oval of Cassini exposed into [Horn & Johnson, Matrix Analysis], but it did not get me anywhere

Do you have any hints or suggestions to prove the inclusion of the eigenvalue into the ellipse?

-----------------
Proof for the imaginary part:

Denote $z$ the left eigenvector associated with eigenvalue $\lambda$, we have from the eigenvalue equation $\lambda z = z M $, 
$$\forall i,\quad \lambda  = a_i \frac{z_{i-1}}{z_i} + b_i\frac{z_{i+1}}{z_i} = \frac{a_i}{a_i+b_i} \frac{z_{i-1}}{z_i} + \frac{b_i}{a_i+b_i} \frac{z_{i+1}}{z_i} $$,
where $i+1$ and $i-1$ ar evaluated modulo $n$, ad the second equality follow from $a_i+b_i=1$.

By taking the product of the imaginary part of all previous equation and denoting $p_i= \ln \sqrt{\frac{a_i}{bi}}$ , we get
$$ \Im \lambda = \sqrt{\prod_i \,a_i b_i \Im \frac{z_{i-1}}{z_i} \Im \frac{z_{i+1}}{z_i} }\prod_i \frac{\sinh (p_i+\frac{1}{2}\ln \Im\frac{ z_{i+1} }{z_{i}} \Im\frac{z_{i} }{z_{i-1}} )}{\cosh p_i} \leq \prod_i \frac{\sinh (p_i+\frac{1}{2}\ln \Im\frac{ z_{i+1} }{z_{i}} \Im\frac{z_{i} }{z_{i-1}} )}{\cosh p_i}$$ 
The inequality use that $ \prod_i \Im \frac{z_{i-1}}{z_i}\leq 1$. The concavity of $\ln \sinh$ and the convexity of $\ln \cosh$, give the result 
$$ \Im \lambda \leq \tanh p.$$