Let $P(x)=\prod_{i=1}^n(x-z_i)=x^n+c_{n-1}x^{n-1}+\ldots+c_0$, such that $0 < z_i < 1$. Assuming that we know the root powersums $s_i = \sum_{j=1}^n z_i^j$ we can use the Newton Identities to find the coefficients $c_i$ and then solve the polynomial equation $P(x) = 0$ to find the roots $z_i$.
The following systems of Newton Identities can be solved to find the coefficients. \begin{eqnarray*} s_k + \sum_{i=1}^{k-1} c_{n-i}s_{k-i} + k c_{n-k} =0, \quad k = 1,2, \ldots,n \\ s_{n+k} + \sum_{i=1}^{n} c_{n-i}s_{n+k-i} = 0,\quad k=1,2,\ldots,m-n \end{eqnarray*} For $m=n$ the system takes the following matrix form, where we omit zero elements. \begin{equation*}\label{equation-newton-identities} \begin{pmatrix} 1 & & & & \\ s_1 & 2 & & & \\ s_2 & s_1 & 3 & \\ \vdots & \vdots &\ddots & \ddots & \\ s_{n-1} & s_{n-2} & \dots & s_1 & n \end{pmatrix} \begin{pmatrix} c_{n-1} \\ c_{n-2} \\ c_{n-3} \\ \vdots \\ c_0 \end{pmatrix} = \begin{pmatrix} - s_{1} \\ - s_{2} \\ - s_{3} \\ \vdots \\ - s_n \end{pmatrix} \Leftrightarrow Ax=b \end{equation*} For $m>n$ we get the additional $k = m-n$ equations \begin{equation*} \begin{pmatrix} s_n & s_{n-1} & \ldots & s_2 & s_1 \\ s_{n+1} & s_n & \ldots & s_3 & s_2 \\ \vdots & \vdots & & \vdots & \vdots\\ s_{n+k-1} & s_{n+k-2} &\ldots & s_{k+1} & s_{k} \\ \end{pmatrix} \begin{pmatrix} c_{n-1} \\ c_{n-2} \\ \vdots \\ c_0 \end{pmatrix} = \begin{pmatrix} - s_{n+1} \\ - s_{n+2} \\ \vdots \\ - s_{n+k} \\ \end{pmatrix} \Leftrightarrow Cx=d \end{equation*}
Question: Is there a way to go from the root powersums $s_i$ directly to the roots $z_i$ without having to solve the polynomial equation $P(x)=0$ ?
