\subsection{Vector $Z^{(0)}$ Initialization}
\label{sec:vec_initialization}
-As for any iterative method, we need to choose $n$ initial guess points $z^{(0)}_{i}, i = 1, . . . , n.$
+As for any iterative method, we need to choose $n$ initial guess points $z^{0}_{i}, i = 1, . . . , n.$
The initial guess is very important since the number of steps needed by the iterative method to reach
a given approximation strongly depends on it.
In~\cite{Aberth73} the Ehrlich-Aberth iteration is started by selecting $n$
\begin{equation}
\label{Eq:Hi}
-EA2: z^{k+1}=z_{i}^{k}-\frac{\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}}
+EA2: z^{k+1}_{i}=z_{i}^{k}-\frac{\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}}
{1-\frac{p(z_{i}^{k})}{p'(z_{i}^{k})}\sum_{j=1,j\neq i}^{j=n}{\frac{1}{(z_{i}^{k}-z_{j}^{k})}}}, i=0,. . . .,n
\end{equation}
It can be noticed that this equation is equivalent to Eq.~\ref{Eq:EA},
