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{\displaystyle P(z)=a_{n}z^{n}+a_{n-1}z^{n-1}+\cdots +a_{1}z+a_{0}\quad (a_{i}\in \mathbb {C} )}
By the Fundamental Theorem of Algebra , it has
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{\displaystyle P(z)=a_{n}(z-z_{1})(z-z_{2})\cdots (z-z_{n})}
As we know, Vieta's formulae link between the coefficients and roots of a polynomial:
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{\displaystyle {\begin{cases}z_{1}+z_{2}+\cdots +z_{n}=-{\dfrac {a_{n-1}}{a_{n}}}\\(z_{1}z_{2}+z_{1}z_{3}+\cdots +z_{1}z_{n})+(z_{2}z_{3}+z_{2}z_{4}+\cdots +z_{2}z_{n})+\cdots +z_{n-1}z_{n}={\dfrac {a_{n-2}}{a_{n}}}\\(z_{1}z_{2}z_{3}+\cdots +z_{1}z_{2}z_{n})+(z_{1}z_{3}z_{4}+\cdots +z_{1}z_{3}z_{n})+\cdots +(z_{2}z_{3}z_{4}+\cdots +z_{2}z_{n-1}z_{n})+\cdots +z_{n-2}z_{n-1}z_{n}=-{\dfrac {a_{n-3}}{a_{n}}}\\\vdots \\z_{1}\cdots z_{n}=(-1)^{n}{\dfrac {a_{0}}{a_{n}}}\end{cases}}}
As we can see, these sums are symmetric polynomial, and we will call them elementary symmetric polynomials .
The elementary symmetric polynomials in variables
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{\displaystyle X_{1},\ldots ,X_{n}}
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{\displaystyle {\begin{aligned}E_{1}(X_{1},\ldots ,X_{n})&=\sum _{1\leq i\leq n}X_{i}\\[5pt]E_{2}(X_{1},\ldots ,X_{n})&=\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!X_{i_{1}}X_{i_{2}}\\&\,\,\,\vdots \\[5pt]E_{k}(X_{1},\ldots ,X_{n})&=\sum _{1\leq i_{1}<\cdots <i_{k}\leq n}\!\!\!\!\!\!\!\!X_{i_{1}}\!\cdots X_{i_{k}}\\&\,\,\,\vdots \\[5pt]E_{n}(X_{1},\ldots ,X_{n})&=X_{1}\cdots X_{n}\end{aligned}}}
![{\displaystyle {\begin{aligned}E_{1}(X_{1},\ldots ,X_{n})&=\sum _{1\leq i\leq n}X_{i}\\[5pt]E_{2}(X_{1},\ldots ,X_{n})&=\sum _{1\leq i_{1}<i_{2}\leq n}\!\!\!\!X_{i_{1}}X_{i_{2}}\\&\,\,\,\vdots \\[5pt]E_{k}(X_{1},\ldots ,X_{n})&=\sum _{1\leq i_{1}<\cdots <i_{k}\leq n}\!\!\!\!\!\!\!\!X_{i_{1}}\!\cdots X_{i_{k}}\\&\,\,\,\vdots \\[5pt]E_{n}(X_{1},\ldots ,X_{n})&=X_{1}\cdots X_{n}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26fbb9b04910c60799d3ef82c7fea30ba87c4a69)