\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},
as well as the Durand-Kerner implement, suffers from overflow problems. This
situation occurs, for instance, in the case where a polynomial
having positive coefficients and a large degree is computed at a
-point $\xi$ where $|\xi| > 1$, where $|x|$ stands for the modolus of a complex $x$. Indeed, the limited number in the
-mantissa of floating points representations makes the computation of p(z) wrong when z
+point $\xi$ where $|\xi| > 1$, where $|z|$ stands for the modolus of a complex $z$. Indeed, the limited number in the
+mantissa of floating points representations makes the computation of $p(z)$ wrong when z
is large. For example $(10^{50}) +1+ (- 10^{50})$ will give the wrong result
of $0$ instead of $1$. Consequently, we can not compute the roots
for large degrees. This problem was early discussed in
%%$$ \exp \bigl( \ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'}))- \ln(1- \exp(\ln(p(z)_{k})-ln(\ln(p(z)_{k}^{'})+\ln\sum_{i\neq j}^{n}\frac{1}{z_{k}-z_{j}})$$
\begin{equation}
\label{Log_H2}
-EA.EL: z^{k+1}=z_{i}^{k}-\exp \left(\ln \left(
+EA.EL: z^{k+1}_{i}=z_{i}^{k}-\exp \left(\ln \left(
p(z_{i}^{k})\right)-\ln\left(p'(z^{k}_{i})\right)- \ln
\left(1-Q(z^{k}_{i})\right)\right),
\end{equation}
\begin{equation}
\label{Log_H1}
Q(z^{k}_{i})=\exp\left( \ln (p(z^{k}_{i}))-\ln(p'(z^{k}_{i}))+\ln \left(
-\sum_{k\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right).
+\sum_{i\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right)i=1,...,n,
\end{equation}
This solution is applied when the root except the circle unit, represented by the radius $R$ evaluated in C language as:
