+
+In order to hold into account the limit of size of floats, we propose to modifying the iterative function and compute the logarithm of:
+
+\begin{equation}
+EA: 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=1,. . . .,n
+\end{equation}
+
+This method allows, indeed, to exceed the computation of the polynomials of degree 100,000 and to reach a degree upper to 1,000,000. For that purpose, it is necessary to use the logarithm and the exponential of a complex. The iterative function of Ehrlich-Aberth method with exponential and logarithm is given as following:
+
+\begin{equation}
+\label{Log_H2}
+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}
+
+where:
+
+\begin{equation}
+\label{Log_H1}
+Q(z^{k}_{i})=\exp\left( \ln (p(z^{k}_{i}))-\ln(p'(z^{k}_{i}))+\ln \left(
+\sum_{i\neq j}^{n}\frac{1}{z^{k}_{i}-z^{k}_{j}}\right)\right)i=1,...,n,
+\end{equation}
+
+
+%We propose to use the logarithm and the exponential of a complex in order to compute the power at a high exponent.
+Using the logarithm and the exponential operators, we can replace any multiplications and divisions with additions and subtractions. Consequently, computations manipulate lower absolute values and the roots for large polynomial degrees can be looked for successfully~\cite{Karimall98}.