Roots and its relations

Let $y_n=\frac1{1-x_n}$ , so $x_n=1-\frac1{y_n}$ . Therefore, since all of $x_1,x_2,\dots,x_{10}$ satisfy the equation $1+x+x^2\dots+x^{10}=0$ , all of $y_1,y_2,\dots,y_{10}$ must satisfy the following equation:

$1+\left(1-\frac1y\right)+\left(1-\frac1y\right)^2+\left(1-\frac1y\right)^3+\dots+\left(1-\frac1y\right)^{10}=0$

$\begin{aligned}\therefore\quad&1 \\+\,&1-\frac1y\\ +\,&1-{2\choose1}\frac1y+{2\choose2}\frac1{y^2}\\ +\,&1-{3\choose1}\frac1y+{3\choose2}\frac1{y^2}-{3\choose3}\frac1{y^3}\\ &\vdots&=0\\ +\,&1-{10\choose1}\frac1y+{10\choose2}\frac1{y^2}-{10\choose3}\frac1{y^3}+\dots+{10\choose10}\frac1{y^{10}}\end{aligned}$

$\begin{aligned}\therefore\quad&y^{10} \\+\,&y^{10}-y^9\\ +\,&y^{10}-{2\choose1}y^9+{2\choose2}y^8\\ +\,&y^{10}-{3\choose1}y^9+{3\choose2}y^8-{3\choose3}y^7\\ &\vdots&=0\\ +\,&y^{10}-{10\choose1}y^9+{10\choose2}y^8-{10\choose3}y^7+\dots+{10\choose10}\end{aligned}$

Now, by Vieta's formulas, the sum of the roots of this equation must be $-\frac ba$ , where $a$ is the leading coefficient and $b$ is the second coefficient. $\begin{aligned}\therefore\sum_{n=0}^{10}y_n&=-\frac{-1-{2\choose1}-{3\choose1}-{4\choose1}-\dots-{10\choose1}}{\underbrace{1+1+1+\dots+1}_{\text{Eleven }1\text{'s}}}\\ \therefore\sum_{n=0}^{10}\frac1{1-x_n}&=\frac{1+2+3+4+\dots+10}{11}\\ &=\frac{10(10+1)}{2\times11}\\ &=\boxed{5}\end{aligned}$

If $x$ is a (non-real) root, then so is its complex conjugate $\bar{x}$ . Also, $|x|=1$ since $x$ is an 11th root of unity. We can conclude that $\frac{1}{1-x}+\frac{1}{1-\bar{x}}=\frac{2-x-\bar{x}}{1-x-\bar{x}+x\bar{x}}=1$ . Since there are five such conjugate pairs, the answer is $\boxed{5}$ .

$\begin{aligned} f(x) & = \prod_{n=1}^{10}{\left ( x-x_n \right )}\\ \ln\left ( f(x) \right )&=\ln\left ( \prod_{n=1}^{10}{\left ( x-x_n \right )} \right )=\sum_{n=1}^{10}{\ln\left ( x-x_n \right )} \\ \frac{d}{dx} \ln\left ( f(x) \right )&=\frac{d}{dx}\sum_{n=1}^{10}{\ln\left ( x-x_n \right )} \\ \frac{f^{'}(x)}{f(x)} & =\sum_{n=1}^{10}{\frac{1}{x-x_n}} \\ & \color{#D61F06} \text{put } x=1 \\ \sum_{n=1}^{10}{\frac{1}{1-x_n}} &= \frac{f^{'}(1)}{f(1)} = \frac{1+2+\cdots+10}{\underbrace{1+1+\cdots+1}_{\text{Eleven } 1}}=5 \end{aligned}$

$\begin{aligned} S & = \sum_{n=1}^{10} \frac 1{1-x_n} \\ & = \frac {\sum_{k=1}^{10} \prod_{n=1, n \ne k}^{10}(1-x_n)}{\color{#3D99F6}\prod_{n=1}^{10}(1-x_n)} & \small \color{#3D99F6} \text{If }f(x) = x^{10}+x^9+\cdots + x + 1 = 0 \implies f(x) = \prod_{n=1}^{10}(x-x_n) \\ & = \frac {\displaystyle 10\sum_{n=0}^9 (-1)^n \frac {\binom 9n}{\binom {10}n} \color{#D61F06}S_n}{\color{#3D99F6}f(1)} & \small \color{#D61F06} \text{where }S_n \text{ symmetric sums with }S_0 = 1 \text{ (See note.)} \\ & = \frac {\sum_{n=0}^9 (-1)^n(10-n)\color{#D61F06}S_n}{\color{#3D99F6}11} & \small \color{#D61F06} \text{By Vieta;s formula }S_n = (-1)^n \text{ (See reference)} \\ & = \frac {\sum_{n=0}^9 (10-n)}{11} \\ & = \frac {\sum_{n=1}^{10} n}{11} \\ & = \frac {\frac {10(10+1)}2}{11} \\ & = \boxed 5 \end{aligned}$

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• Note: Symmetric sums: $\displaystyle S_1 = \sum_{k=1}^n x_k,\ S_2 = \sum_{j \ne k}^n x_j x_k,\ S_3 = \sum_{i \ne j \ne k}^n x_i x_j x_k, \cdots S_n = \prod_{k=1}^n x_k$ .
• Reference: Vieta's formula

Note that $f(x) = x^{10} + x^9 + \cdots + x + 1$ is such that $(x-1)f(x) = x^{11} - 1$ , so that $-xf(1-x) = (1-x)^{11} - 1$ , and hence $\begin{aligned} xf(1-x) &= \; (x-1)^{11} + 1 \; = \; \sum_{j=1}^{11}\binom{11}{j}(-1)^{11-j}x^j \\ g(x) \; =\; f(1-x) & = \; \sum_{j=0}^{10} (-1)^{10-j}\binom{11}{j+1}x^j \; = \; x^{10} - \binom{11}{10}x^9+ \cdots - \binom{11}{2}x + \binom{11}{1} \end{aligned}$ and $g(x)$ has roots $1 - x_j$ for $1 \le j \le 10$ . Thus we deduce that $\sum_{j=1}^{10}\frac{1}{1-x_j} \; = \; \frac{\binom{11}{2}}{\binom{11}{1}} \; = \; \boxed{5}$