Evaluating given expression using roots of given equation

Question

This is a question from an entrance exam for a research institute that was asked in 2019. I am not sure how to begin to be honest with you so I'd appreciate a step by step tutorial. This is to be solved using highshool (grade 12) maths only. I am from India so the syllabus may vary to some extent from other countries. I would assume you'd have to use some sort of property of complex roots to solve this but I really don't know how to begin. So if possible please no undergrad mathematics. Thank you.

Let 1, $\zeta_2$, $\zeta_3$, . . ., $\zeta_n$ be the roots of the equation $x^{n} = 1$, for $n\geq3$. Then the value of the expression $$\frac{1}{2-\zeta_2} + \frac{1}{2-\zeta_3} + . . . + \frac{1}{2-\zeta_n}$$ is -

A) $\frac{1+(n-2)2^{n}}{2^{n}-1}$

B) $\frac{1+n2^{n-1}-2^{n}}{2^{n}-1}$

C) $\frac{1+n2^{n-1}-2^{n}}{2^{n-1}+1}$

D) $\frac{1+(n-1)2^{n}-2^{n-1}}{2^{n-1}+1}$

I did do this, but the only way I was able to do it was kind of lame. I just put n = 4 and then checked which of the options gave the correct result of the expression. This is fair and something I can do in the actual exam because it is purely multiple choice, but I want to learn the proper way of doing it as well.

Also, I am aiming to give this exam this June, if anyone knows any books that can help in learning how to solve such questions of similar difficulty and format, that would be a great help, thank you and please let me know of any.

Answer 1

Let $\dfrac1{2-x}=y \implies x = 2-\dfrac1y$.
Thus the terms whose sum we want are among the roots of $\displaystyle \left( 2-\frac1y\right)^n=1 \iff (2y-1)^n=y^n$ $\iff (2^n-1)y^n-n\cdot (2y)^{n-1}+\cdots = 0\implies$ the sum of roots is $\dfrac{n\cdot2^{n-1}}{2^n-1}$ by Vieta.
As $\zeta_1=1$ is excluded from the sum, removing $\dfrac1{2-1}=1$ from this answers the MCQ.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\iverson}[1]{\left[\left[\,{#1}\,\right]\right]} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Note that $\ds{\left.\rule{0pt}{5mm}x^{n} - 1\right\vert_{n\ \in\ \mathbb{N}_{\,\geq\ 1}}\,\,\, = \prod_{k = 1}^{n}\pars{x - \zeta_{k}}}$ where $\ds{\zeta_{k}}$ is a root of

$\ds{x^{n} - 1}$, i.e. $\ds{\zeta_{k}^{n} - 1 = 0}$ with $\ds{k = 1,2,3,\ldots,n}$. \begin{align*} \mbox{Therefore,}\ \ln\left(x^{n} - 1\right) & = \sum_{k = 1}^{n}\ln\left(x - \zeta_{k}\right) \\[2mm] \substack{\mbox{Derivating both sides}\\\mbox{respect of}\ \ds{x}}\ {nx^{n - 1} \over x^{n} - 1} & = \sum_{k = 1}^{n}{1 \over x - \zeta_{k}} \\[2mm] \mbox{Set x = 2,}\quad {n2^{n - 1} \over 2^{n} - 1} & = 1 + \sum_{k = 2}^{n}{1 \over 2 - \zeta_{k}} \\[2mm] \implies \bbx{\color{#44f}{\sum_{k = 2}^{n}{1 \over 2 - \zeta_{k}}} = \color{#44f}{n2^{n - 1} - 2^{n} + 1 \over 2^{n} - 1}} & \implies \left.\mbox{B}\ option\right) \end{align*}

Answer 3

I'll present here a somewhat brute force solution. I'll be using a Taylor series. If that is too advanced for your purposes, I can delete this answer.

For convenience, I define the quantity $\omega_m = e^{2\pi i m/n}$ as the $m^\text{th}$ root of the equation $x^n = 1$, where $m$ ranges from $0$ to $n-1$. So $\omega_m = \zeta_{m+1}$. Note that $\omega_0 = 1$. The sum in the original statement of the problem is equivalent to \begin{align} S &= \sum_{m=0}^{n-1}\frac{1}{2-\omega_m} \,-\, 1\\ &= \frac{1}{2}\sum_{m=0}^{n-1}\frac{1}{1 - \frac{1}{2}\omega_m} \,-\,1 \end{align} Now we use the Taylor series expansion \begin{equation} \frac{1}{1-x} = \sum_{k=0}^{\infty} x^k\, , \end{equation} which yields: \begin{align} S &= \frac{1}{2}\sum_{m=0}^{n-1}\sum_{k=0}^{\infty}\frac{1}{2^k}\omega_m^k \,-\,1\\ &= \frac{1}{2}\sum_{m=0}^{n-1}\sum_{k=0}^{\infty}\frac{1}{2^k}e^{2\pi i m k/n} \,-\,1\\ &= \frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{2^k}\sum_{m=0}^{n-1}e^{2\pi i m k/n}\,-\,1 \end{align} When $k$ is a multiple of $n$, the sum $\sum_{m=0}^{n-1}e^{2\pi i m k/n} = n$. When $k$ is not a multiple of $n$, the sum is $0$. (Perhaps you can convince yourself of this latter fact by picturing the numbers $e^{2\pi i m k/n}$ in the complex plane. They are uniformly distributed around the unit circle, and so their sum is $0$.) Thus we are left only with the terms where $k=\ell n$: \begin{align} S &= \frac{1}{2}\sum_{\ell=0}^{\infty}\frac{1}{2^{\ell n}}n \,-\,1\\ &= \frac{1}{2}\frac{n}{1 - \frac{1}{2^n}}\,-\,1\\ &= \frac{1+n2^{n-1} - 2^n}{2^n-1} \end{align} This corresponds to answer B).

Answer 4

General strategy. Let $r_k=\frac{1}{2-z_k}$ where the $z_k$ are the roots of $z^n-1=0$. We wish to evaluate $\sum_k r_k$. We seek to treat the $r_k$ as the roots of some polynomial $R(s) =\Pi_k (1-s r_k)$ whose expansion in powers of $s$ is $R(s) = 1- (\sum_k r_k)s + \ldots (s^2)+\ldots $. (The format of this polynomial uses the normalization condition that the constant term is $R(0)= 1$.)

Note that the sum of the $r_k$ can be found by determining the coefficient of $s$ in the expansion of this polynomial $R(s)$. So we want to find an explicit description of $R(s)$. This can be done in two steps. First we look at $w_k=2-z_k$, then we look at $r_k=1/w_k$.

Step 1. First note that if $P(z)=z^N-1=0$ then there is an obvious polynomial satisfied by $w_k= 2-z_k$: namely after setting $z=2-w$ we see $=P(2-w)-1=0$.

Expand this out by the binomial theorem to obtain the explicit coefficients of the (monic) polynomial $Q(w)=\Pi_k (w-w_k)$ (Monic means normalized by the condition that the power of $w^n$ is one. This last detail will require some easy arithmetic.)

Step 2. Observe that the roots of $R(s)=s^n Q(1/s)$ are precisely these $r_k=1/w_k$ and $R(s)$ satisfies the desired normalization condition $R(0)=1$. In effect, you just need to read off the coefficients of $Q$ in reverse order.