Let $z_1$, $z_2$, $z_3$, $\cdots$, $z_{2021}$ be the roots of the polynomial $z^{2021}+z-1$. Evaluate$$\frac{z_1^3}{z_{1}+1}+\frac{z_2^3}{z_{2}+1}+\frac{z_3^3}{z_{3}+1}+\cdots+\frac{z_{2021}^3}{z_{2021}+1}.$$
In this problem, I saw that $$\frac{x^3}{x+1} = x^2-x+1 – \frac{1}{x+1}.$$ Then,
$$\sum z_i^2 = \left(\sum z_i\right)^2 – 2\sum z_iz_j = 0$$ and $\sum -z_i=1$from Vieta's formulas $\sum 1=2021$.
However, I am not able to find $\sum \frac{1}{z_i+1}$. Can you help me with that?
Best Answer
Lemma: If $z$ is a root of $p(z)=\sum_{i=0}^n a_i z^i$ where $z$ is non-zero, then $z^{-1}$ is a root of $q(z) = \sum_{i=0}^n a_{n-i}z^i$.
Proof:
Suppose $z$ is a root of $p$, $$\sum_{i=0}^n a_iz^i=0,$$ we divide by $z^n$, then we have
$$\sum_{i=0}^n a_iz^{i-n}=\sum_{i=0}^n a_i(z^{-1})^{n-i}=0.$$
Let $n-i=j$, $$\sum_{j=0}^n a_{n-j}(z^{-1})^{j}=0$$
That is $z^{-1}$ is a root of $q$.
What the lemma says is given a polynomial with $z$ as a non-zero root, then $z^{-1}$ is a root for the polynomial obtained by reversing the coefficient order. Let's apply this result on our problem.
$$f(z)=z^{2021}+z-1$$
If $z$ is a root of $f$, then $z+1$ is a root of
$$g(z)= f(z-1)=(z-1)^{2021}+(z-1)-1=\sum_{i=0}^{2021}a_iz^i$$
where $a_0=-3, a_{1}=2021+1=2022$.
Hence, $$\sum \frac1{z_i+1}=-\frac{a_1}{a_0}=\frac{2022}{3}=674$$