<h3 id="successcomplex-number-br-lecture-6success"><Success>Complex number <br> lecture-6</Success></h3> <div style='float: left; width:100%;'> </div>

<h3 id="successrecallsuccess"><Success>Recall</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <ul> <li>The $n^{th} \text {root of unity is a number $z$ satisfying the equation}$</li> </ul> <p>$ z^n=1, \quad (n \ge1)$</p> <ul> <li>$The\ n^{th} \text{roots of unity are given by}$</li> </ul> <p>$ z_k = cos\frac{2k\pi}{n} + i\ sin \frac{2k\pi}{n}, \quad k=0,1,…n-1$</p> <p>$ =z_1^k$</p> <p><strong>Notation</strong></p> <p>$cis\theta\ = cos\theta +i\ sin\theta$</p> <p>$ (cis \theta)^k = cis k \theta$</p> </div> ====== <h3 id="successremark1success"><Success>Remark(1)</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <ul> <li> <p>$ \alpha =z_1 =cis \frac {2\pi}{n}$</p> </li> <li> <p>$ \alpha^n = (cis\frac {2\pi}{n})^n = cis2\pi=1 $</p> </li> <li> <p>$ \alpha^{kn} = (\alpha^n)^k=1 \quad k\ge0$</p> </li> <li> <p>$ k<0,\quad (\alpha^k)^n = (\alpha^n)^k =1 $</p> </li> <li> <p>$ \alpha^{kn} =1 \quad$ for all $ k\in \Z$</p> </li> </ul> </div>

<h3 id="successremark-2success"><Success>Remark (2)</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$ {\{1,\alpha,\alpha^2,…..\alpha^{n-1} \}}$ are $n^{th}$ roots of unity</p> <p>$ \rightarrow \alpha^k satisfies\quad z^n=1$</p> <p>$ \rightarrow \alpha^k ~ is\ root\ of\ z^n-1, k=0,1,….(n-1)$</p> <p>$z^n-1 =(z-1)(z^{n-1}+z^{n-2}+… +z+1)$ , k=1,2,…,(n-1)</p> <p>$z^{n-1} +z^{n-2}+..z+1=(z-\alpha)(z-\alpha^2)…(z-\alpha^{n-2})$</p> <p>$=\Pi_{k=1}^{n-1}(z-\alpha^k)$</p> </div>

<h3 id="successproblem-1success"><Success>Problem 1</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <ul> <li>$\text{Find the number of ordered pairs (a,b) of real numbers such that}$</li> </ul> <p>$(a+ib)^{2016}=a-ib$</p> </div> <p>======</p> <h3 id="successsolutionsuccess"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Given $(a,b) \in {\R}^2$ $z=a+ib \in$ C</p> <p>consider $\quad z=a+ib$</p> <p>$z^{2016}=\bar{z}\quad \rightarrow (1)$</p> <p>$\Rightarrow\ |z|^{2016}={|\bar z|}=|z|$</p> <p>$\Rightarrow\ |z|^{2016}-|z|=0$</p> <p>$\Rightarrow\ |z|(|z|^{2015}-1)=0$</p> </div>

<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\Rightarrow$ Either $|z|=0$ or $|z|^{2015}=1$</p> <p>If $z \neq 0$, then $|z|=1$</p> <p>[$z \neq 0 \Rightarrow |z|^{2015}=1$ <br> If $|z|<1 \text { or }|z|>1$ then $|z|^{2015} \neq 1$ $\Rightarrow |z|=1$]</p> <p>If $|z|=1$, then $|z|^2=1, \bar{z}=\frac{1}{z} \quad \Rightarrow|z|^{2015} \neq 1$</p> <p>From (1), $\quad z^{2016}=\bar{z}$</p> <p>$\Rightarrow z^{2017}=1 \rightarrow \text { (2) }$</p> <p>$\Rightarrow$ equation(2) has 2017 distinct solutions & $z=0$ is also solution to (1) <br> Equation (1) has 2018 solutions</p> </div>

<h3 id="successpropositionsuccess"><Success>Proposition</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <ul> <li> <p>let $z_k=\operatorname{cis}\left(\frac{2 k \pi}{n}\right), k=0,1, \ldots, n-1$</p> </li> <li> <p>Then for any positive integer ’ $m$ ’ the fallowing relation holds</p> </li> </ul> <p>$\sum_{k=0}^{n-1} z_k^m= \begin{cases}n & \text { if } n \text { divides } m \\ 0 & \text { otherwise }\end{cases}$</p> </div> ====== <h3 id="successproofsuccess"><Success>Proof</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$z_k=\alpha^k$</p> <p>$\alpha=\operatorname{cis}\left(\frac{2 \pi}{n}\right), \quad k=0,1, \ldots, n-1 $</p> <p>Consider $\sum_{k=0}^{n-1} z_k^m=\sum_{k=0}^{n-1}\left(\alpha^k\right)^m=\sum_{k=0}^{n-1}\left(\alpha^m\right)^k$</p> <p><strong>Recall</strong></p> <p>$\sum_{k=0}^{n-1} a^k=\frac{a^n-1}{a-1}, a \neq 1$</p> <p>$\Rightarrow \sum_{k=0}^{n-1}z_k^m=\frac{\left(\alpha^m\right)^n-1}{\alpha^m-1}$</p> </div>

<h3 id="successproofsuccess-1"><Success>Proof</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$ \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}$</p> <p>If n does not divide m, $(\alpha^m \ne1) $</p> <p>$\Sigma_{k=0}^{n-1} \quad z_k^m =0\ (n \cancel{|} m) $</p> <p>If m=qn,</p> <p>$ \Sigma_{k=0}^{n-1} \quad z_k^{qn}= \Sigma_{k=0}^{n-1} \quad (\alpha^n)^{kq}=n$</p> <p>$(m=qn)$</p> </div>

<h3 id="successproofsuccess-2"><Success>Proof</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\text{In particular}, \quad \quad m=1$</p> <p>$\Sigma^{n-1}_{k=0}\quad z_k=0,\quad n>1$</p> <p>$\Rightarrow\Sigma^{n-1}_{k=0}\quad cos\frac {2k\pi}{n}+i\sin \frac {2k\pi}{n}=0$</p> <p>$\Rightarrow \sum_{k=0}^{n-1} \cos \frac{2 k \pi}{n}=0 \text { and } \sum_{k=0}^{n-1} \sin \frac{2 k \pi}{n}=0$</p> <p>$\Rightarrow \ cos \frac {2\pi}{n}+ \ cos \frac {4\pi}{n}+….+\ cos \frac {2(n-1)\pi}{n}=-1 \&$</p> <p>$\ sin \frac {2\pi}{n}+ \ sin \frac {4\pi}{n}+….+\ sin \frac {2(n-1)\pi}{n}=0 $</p> </div>

<h3 id="successproblem-2success"><Success>Problem-2</Success></h3> <div style='float: left; width:60%;font-size:30px;text-align:left'> <p>$Let \ \alpha = cis(\frac {2\pi}{n})$</p> <p>and $z$ be a complex number such that</p> <p>$|z-\alpha^k|\le 1| ~ \forall~$ k=0,1,….n-1</p> <p>Then show that ${z=0}$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d665b7b3-2767-4b3d-48f9-fd00e3929700/public"></p> <!----> </div>

<h3 id="successsolutionsuccess-2"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$|z-\alpha^k|\le1\quad k=0,1,…n-1 $</p> <p>consider</p> <p>$|z-\alpha^k|^2\ =|z-\alpha^k|\ |z-\alpha^k|\ \le1$</p> <p>$ \Rightarrow\ (z-\alpha^k)(\overline{z-\alpha^k})\le1$</p> <p>$\Rightarrow|z|^2-z \overline{\alpha^k}-\bar{z} \alpha^k+\left|\alpha^k\right|^2 \leq 1$</p> <p>$\Rightarrow \quad|z|^2 \leq z\overline{\alpha^k}+\bar{z} \alpha^k, k=0, \ldots, n-1$</p> <p>$ \Rightarrow\ n\ |z|^2\ \le \Sigma^{n-1}_{k=0} (z \ \bar{\alpha^k} +\bar z \alpha^k) $</p> </div>

<h3 id="successsolutionsuccess-3"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\Rightarrow n|z|^2 \leqslant z \sum_{k=0}^{n-1} \bar{{\alpha}^k}+\bar{z} \sum_{k=0}^{n-1} \alpha^k$</p> <p>$=0$</p> <p>$\Rightarrow |z| =0 \Rightarrow z=0$</p> </div>

<h3 id="successproblem-3success"><Success>Problem-3</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$Let\ P_0\ P_1…\ P_{n-1}\ \text{be a regular polygon inscribed in a circle of radius 1}$</p> <p>$\text {Then show that }$</p> <ul> <li> <p>$\ P_0P_1.\ P_0P_2.\ …\ P_0P_{n-1} =n $</p> </li> <li> <p>$\ sin\frac {\pi}{n}\ sin\frac {2\pi}{n}\ … sin \frac {(n-1)\pi}{n} =\frac {n}{2^{n-1}}$</p> </li> <li> <p>$\ sin\frac {\pi}{2n}\ sin\frac {3\pi}{2n}\ … sin \frac {(2n-1)\pi}{2n} =\frac {1}{2^{n-1}}$</p> </li> </ul> </div>

<h3 id="successsolutionsuccess-4"><Success>Solution</Success></h3> <div style='float: left; width:70%;font-size:30px;text-align:left'> <p>$W.L.O.G. P_k\ =cis(\frac{2k\pi}{n})=\alpha^k,$</p> <p>$k=0,…n-1$</p> <p>$ z^{n-1} +z^{n-2}+..+z+1=\Pi_{k=1}^{n-1}(z-\alpha^k)$</p> <p>$ z=1,\quad n=\Pi_{k=1}^{n-1}(1-\alpha^k)$</p> </div> <div style='float: right; width:30%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2cd7b5bf-5961-405f-5948-9c21627d8500/public"></p> <!----> </div>

<h3 id="successsolutionsuccess-5"><Success>Solution</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>i.e $\quad |\pi_{k=1}^{n-1}(1-\alpha^k)|=n$</p> <p>$\Rightarrow \pi_{k=1}^{n-1}|{1-\alpha^k}|=n$</p> <p>$\Rightarrow\ p_0p_1.\ p_0p_2.\ …\ p_0p_{n-1} =n $</p> <ul> <li>Consider</li> </ul> <p>$|1-\alpha^k|^2 = \biggl|1-cos\frac{2k\pi}{n}-i\ sin \frac{2k\pi}{n}\biggl|^2$</p> <p>$ =\biggl(1-cos\frac{2k\pi}{n}\biggl)^2+ \biggl(sin \frac{2k\pi}{n}\biggl)^2$</p> <p>$ =2-2cos\frac{2k\pi}{n}$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$ =2\biggl(2sin^2\frac{k\pi}{n}\biggl), \quad k=1, …..n-1$</p> <p>$|1-\alpha^k|^2 =4\biggl(sin\frac{k\pi}{n}\biggl)^2, \quad k=1, …..n-1$</p> <p>$\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$</p> <p>$P_0P_k\ = 2\ sin\frac{k\pi}{n}$</p> <p>$\Rightarrow 2\ sin\frac{\pi}{n}\times2\ sin\frac{2\pi}{n}…\ 2\ sin\frac{(n-1)\pi}{n}=n$</p> <p>$\ sin\frac{\pi}{n}\times2\ sin\frac{2\pi}{n}…\ \ sin\frac{(n-1)\pi}{n}=\frac {n}{2^{n-1}}$</p> </div>

<div> <h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> </div>