• The $n^{th} \text {root of unity is a number $z$ satisfying the equation}$

$z^n=1, \quad (n \ge1)$

• $The\ n^{th} \text{roots of unity are given by}$

$z_k = cos\frac{2k\pi}{n} + i\ sin \frac{2k\pi}{n}, \quad k=0,1,…n-1$

$=z_1^k$

Notation

$cis\theta\ = cos\theta +i\ sin\theta$

$(cis \theta)^k = cis k \theta$

• $\alpha =z_1 =cis \frac {2\pi}{n}$

• $\alpha^n = (cis\frac {2\pi}{n})^n = cis2\pi=1$

• $\alpha^{kn} = (\alpha^n)^k=1 \quad k\ge0$

• $k<0,\quad (\alpha^k)^n = (\alpha^n)^k =1$

• $\alpha^{kn} =1 \quad$ for all $k\in \Z$

${\{1,\alpha,\alpha^2,…..\alpha^{n-1} \}}$ are $n^{th}$ roots of unity

$\rightarrow \alpha^k satisfies\quad z^n=1$

$\rightarrow \alpha^k ~ is\ root\ of\ z^n-1, k=0,1,….(n-1)$

$z^n-1 =(z-1)(z^{n-1}+z^{n-2}+… +z+1)$ , k=1,2,…,(n-1)

$z^{n-1} +z^{n-2}+..z+1=(z-\alpha)(z-\alpha^2)…(z-\alpha^{n-2})$

$=\Pi_{k=1}^{n-1}(z-\alpha^k)$

• $\text{Find the number of ordered pairs (a,b) of real numbers such that}$

$(a+ib)^{2016}=a-ib$

Given $(a,b) \in {\R}^2$ $z=a+ib \in$ C

consider $\quad z=a+ib$

$z^{2016}=\bar{z}\quad \rightarrow (1)$

$\Rightarrow\ |z|^{2016}={|\bar z|}=|z|$

$\Rightarrow\ |z|^{2016}-|z|=0$

$\Rightarrow\ |z|(|z|^{2015}-1)=0$

$\Rightarrow$ Either $|z|=0$ or $|z|^{2015}=1$

If $z \neq 0$, then $|z|=1$

[$z \neq 0 \Rightarrow |z|^{2015}=1$
If $|z|<1 \text { or }|z|>1$ then $|z|^{2015} \neq 1$ $\Rightarrow |z|=1$]

If $|z|=1$, then $|z|^2=1, \bar{z}=\frac{1}{z} \quad \Rightarrow|z|^{2015} \neq 1$

From (1), $\quad z^{2016}=\bar{z}$

$\Rightarrow z^{2017}=1 \rightarrow \text { (2) }$

$\Rightarrow$ equation(2) has 2017 distinct solutions & $z=0$ is also solution to (1)
Equation (1) has 2018 solutions

• let $z_k=\operatorname{cis}\left(\frac{2 k \pi}{n}\right), k=0,1, \ldots, n-1$

• Then for any positive integer ’ $m$ ’ the fallowing relation holds

$\sum_{k=0}^{n-1} z_k^m= \begin{cases}n & \text { if } n \text { divides } m \\ 0 & \text { otherwise }\end{cases}$

$z_k=\alpha^k$

$\alpha=\operatorname{cis}\left(\frac{2 \pi}{n}\right), \quad k=0,1, \ldots, n-1$

Consider $\sum_{k=0}^{n-1} z_k^m=\sum_{k=0}^{n-1}\left(\alpha^k\right)^m=\sum_{k=0}^{n-1}\left(\alpha^m\right)^k$

Recall

$\sum_{k=0}^{n-1} a^k=\frac{a^n-1}{a-1}, a \neq 1$

$\Rightarrow \sum_{k=0}^{n-1}z_k^m=\frac{\left(\alpha^m\right)^n-1}{\alpha^m-1}$

$\Sigma_{k=0}^{n-1} \quad z_k^m = \frac{(\alpha^m)^n-1} {\alpha^m-1} = \frac{(\alpha^n)^m-1} {\alpha^m-1}$

If n does not divide m, $(\alpha^m \ne1)$

$\Sigma_{k=0}^{n-1} \quad z_k^m =0\ (n \cancel{|} m)$

If m=qn,

$\Sigma_{k=0}^{n-1} \quad z_k^{qn}= \Sigma_{k=0}^{n-1} \quad (\alpha^n)^{kq}=n$

$(m=qn)$

$\text{In particular}, \quad \quad m=1$

$\Sigma^{n-1}_{k=0}\quad z_k=0,\quad n>1$

$\Rightarrow\Sigma^{n-1}_{k=0}\quad cos\frac {2k\pi}{n}+i\sin \frac {2k\pi}{n}=0$

$\Rightarrow \sum_{k=0}^{n-1} \cos \frac{2 k \pi}{n}=0 \text { and } \sum_{k=0}^{n-1} \sin \frac{2 k \pi}{n}=0$

$\Rightarrow \ cos \frac {2\pi}{n}+ \ cos \frac {4\pi}{n}+….+\ cos \frac {2(n-1)\pi}{n}=-1 \&$

$\ sin \frac {2\pi}{n}+ \ sin \frac {4\pi}{n}+….+\ sin \frac {2(n-1)\pi}{n}=0$

$Let \ \alpha = cis(\frac {2\pi}{n})$

and $z$ be a complex number such that

$|z-\alpha^k|\le 1| ~ \forall~$ k=0,1,….n-1

Then show that ${z=0}$

$|z-\alpha^k|\le1\quad k=0,1,…n-1$

consider

$|z-\alpha^k|^2\ =|z-\alpha^k|\ |z-\alpha^k|\ \le1$

$\Rightarrow\ (z-\alpha^k)(\overline{z-\alpha^k})\le1$

$\Rightarrow|z|^2-z \overline{\alpha^k}-\bar{z} \alpha^k+\left|\alpha^k\right|^2 \leq 1$

$\Rightarrow \quad|z|^2 \leq z\overline{\alpha^k}+\bar{z} \alpha^k, k=0, \ldots, n-1$

$\Rightarrow\ n\ |z|^2\ \le \Sigma^{n-1}_{k=0} (z \ \bar{\alpha^k} +\bar z \alpha^k)$

$\Rightarrow n|z|^2 \leqslant z \sum_{k=0}^{n-1} \bar{{\alpha}^k}+\bar{z} \sum_{k=0}^{n-1} \alpha^k$

$=0$

$\Rightarrow |z| =0 \Rightarrow z=0$

$Let\ P_0\ P_1…\ P_{n-1}\ \text{be a regular polygon inscribed in a circle of radius 1}$

$\text {Then show that }$

• $\ P_0P_1.\ P_0P_2.\ …\ P_0P_{n-1} =n$

• $\ sin\frac {\pi}{n}\ sin\frac {2\pi}{n}\ … sin \frac {(n-1)\pi}{n} =\frac {n}{2^{n-1}}$

• $\ sin\frac {\pi}{2n}\ sin\frac {3\pi}{2n}\ … sin \frac {(2n-1)\pi}{2n} =\frac {1}{2^{n-1}}$

$W.L.O.G. P_k\ =cis(\frac{2k\pi}{n})=\alpha^k,$

$k=0,…n-1$

$z^{n-1} +z^{n-2}+..+z+1=\Pi_{k=1}^{n-1}(z-\alpha^k)$

$z=1,\quad n=\Pi_{k=1}^{n-1}(1-\alpha^k)$

i.e $\quad |\pi_{k=1}^{n-1}(1-\alpha^k)|=n$

$\Rightarrow \pi_{k=1}^{n-1}|{1-\alpha^k}|=n$

$\Rightarrow\ p_0p_1.\ p_0p_2.\ …\ p_0p_{n-1} =n$

• Consider

$|1-\alpha^k|^2 = \biggl|1-cos\frac{2k\pi}{n}-i\ sin \frac{2k\pi}{n}\biggl|^2$

$=\biggl(1-cos\frac{2k\pi}{n}\biggl)^2+ \biggl(sin \frac{2k\pi}{n}\biggl)^2$

$=2-2cos\frac{2k\pi}{n}$

$=2\biggl(2sin^2\frac{k\pi}{n}\biggl), \quad k=1, …..n-1$

$|1-\alpha^k|^2 =4\biggl(sin\frac{k\pi}{n}\biggl)^2, \quad k=1, …..n-1$

$\Rightarrow|1-\alpha^k| =2\biggl|sin\frac{k\pi}{n}\biggl|,\quad 0\le\frac{\pi}{n}\le \frac{k\pi}{n}\le\frac{(n-1)\pi}{n}\le\pi$

$P_0P_k\ = 2\ sin\frac{k\pi}{n}$

$\Rightarrow 2\ sin\frac{\pi}{n}\times2\ sin\frac{2\pi}{n}…\ 2\ sin\frac{(n-1)\pi}{n}=n$

$\ sin\frac{\pi}{n}\times2\ sin\frac{2\pi}{n}…\ \ sin\frac{(n-1)\pi}{n}=\frac {n}{2^{n-1}}$

Thank you