$z=r(\cos (\theta) + i \sin (\theta))$

$\text{where}$ $r ≥ 0$ & $0 ≤ \theta ≤ 2 \pi$

$z=r(\cos (\theta +2k\pi) +i\sin (\theta+2k\pi))$

$k \in \Z$

$1-i=\sqrt2\biggl(\cos (\frac {7\pi} {4})+i\sin \frac{7\pi} {4}\biggl)$

$1-i=\sqrt2\biggl(\cos \biggl(\frac {-\pi} {4}\biggl)+i\sin\biggl(\frac{-\pi} {4}\biggl)\biggl)$

$\cos\frac {\pi} {4} = \frac {1} {\sqrt2}\sin\biggl(\frac {\pi} {4}\biggl) =\frac {1} {\sqrt2}$

$z=r(\cos \theta + i \sin \theta)$

$\text{where}$ $r ≥ 0$ & $0 ≤ \theta ≤ 2 \pi$

$z=r(\cos (\theta +2k\pi) +i\sin (\theta+2k\pi))$

$k \in \Z$

$\text{De moivre’s formula}$

$(\cos \theta + i \sin \theta)^n = \cos n\theta +i\ \sin n\theta$

Division

$z_1 = r_1(\cos \theta_1 +i \sin \theta_1)$

$z_2 = r_2(\cos \theta_2 +i\sin \theta_2)$

$\frac {z_1} {z_2}=\frac{r_1( \cos \theta_1+ i \sin \theta_1)} {r_2(\cos \theta_2+ i \sin \theta_2)}= \frac{r_1( cos \theta_1+ i \sin \theta_1) (\cos \theta_2- i \sin \theta_2)} {r_2(\ cos^2\theta_2+ i \sin^2 \theta_2)}$

$\frac {r_1} {r_2} = [\cos (\theta_1 -\theta_2)+ i \sin (\theta_1 -\theta_2)]$

$z_1 =1+i \quad z_2=i$

$\frac {z_1} {z_2} =\frac {\sqrt2} {1}[\cos(\frac {-\pi} {4})+i \sin(\frac {-\pi} {4})]$

$\frac {z_1} {z_2} =(1-i)$

${z_1} = {\sqrt2} \biggl(\cos \frac {\pi} {4}+i \sin \frac {\pi} {4}\biggl)$

${z_2} = 1 \biggl(\cos \frac {\pi} {2}+i \sin \frac {\pi} {2}\biggl)$

(i) $\left|\frac{z_1}{z_2}\right|=\frac{r_1}{r_2}=\frac{\left|z_1\right|}{\left|z_2\right|} \quad\left(z_2 \neq 0\right)$

(ii) $z^n=(\cos \theta+i \sin \theta)^n$

$n=-1,-2, \ldots$

$n=-1 \quad z^{-1}=\frac{1}{z}=\frac{1}{\cos \theta+ i \sin \theta}$

$= \frac{1}{1}[\cos (0-\theta)+i \sin (0-\theta)]$

$= \cos (-\theta)+i \sin (-\theta)$

$n=-2;$

$z^{-2}=\frac {1} {z_2}=\frac{1}{z}\times\frac{1}{z}=[\cos(-\theta)+i\sin(-\theta)]^2$

$z^{-2}=\cos(-2\theta)+i\sin(-2\theta)$

$z^{n}=\cos\ n\theta+i\sin\ n\theta \quad n\epsilon \Z.$

$z=(1+i\sqrt3)^n+(1-i\sqrt3)^n$

$z={2\biggl[\cos\frac{\pi}{3}+i \sin\frac{\pi}{3}\biggl]}^n+{2\biggl(\cos\biggl(\frac{-\pi}{3}\biggl)+i \sin\biggl(\frac{-\pi}{3}\biggl)\biggl)}^n$

$\tan^{-1}(\frac{\sqrt3} {1})=\frac {\pi}{3}$

$z=2^n\biggl(\cos\frac{n\pi}{3}+i \sin\frac{n\pi}{3}\biggl)+2^n[\biggl(\cos\biggl(\frac{-n\pi}{3}\biggl)+i \sin\biggl(\frac{-n\pi}{3}\biggl)\biggl)]$

$z=2^{n+1} \cos \frac {n\pi}{3}$

Find the $n^{th}$ roots of unity

$z^n =1$

observation :suppose there is a $z_0\epsilon$ C

such that $\quad z_0^n =1. \quad then |z_0|=1.$

$|z_0^n|=|1|$

$|z_0|^n=1$

$\Leftrightarrow|z_0|=1$

Fix $n=1 ; \quad z^1=1 \quad \Rightarrow z=1$

$n=2 ; \quad z^2=1$

$\Rightarrow z_1=1, \quad z_2=-1$

$n=3 ; \quad z^3=1$

$z_1=1$

$z_2=\cos 120^{\circ}+i \sin 120^{\circ}$

$z_2^3=\left(\cos 120^{\circ}+i \sin 120^{\circ}\right)^3$

$z_2^3=\cos 360^{\circ}+i \sin 360^{\circ}$ (By De Moivre formula)

$=1$

$z_3^3=\left(z_2 \times z_2\right)^3=1$

$n=4 ; \quad z^4=1$

$z_1=1, z_2=i \quad z_3=-1 \quad z_4=-i$

$n^{t h} \text { roots of unity }$

$z^n=1$

$z^n=\cos 2 k \pi+i \sin 2 k \pi$

$(\cos \theta+i \sin \theta)^n=\cos 2 k \pi+i \sin 2 k \pi \quad k \in \mathbb{Z}$

$\cos n \theta+i \sin n \theta=\cos 2 k \pi+i \sin 2 k \pi \quad \quad k = 0,1,2,3…..$

For $\theta=\frac{2 k \pi}{n}, k=0,1,2, \cdots$.

the equation (1) holds.

Define $\theta_k=\frac{2 k \pi}{n}, k=0,1, \ldots, n-1$.

& $z_k=\cos \theta_k+i \sin \theta_k, k=0,1, \ldots, n-1$

$\Rightarrow \quad \left(z_k\right)^n=1, k=0,1, \ldots, n-1$

Claim : {$z_k$ : $k \in$ $\Z$} = {$z_k: k=0,1, \ldots n-1$}

Let $r \in$ $\Z$ $\quad \theta_r=\frac{2 r \pi}{n} \rightarrow z_r$

By quotient reminder theorem .

$r=q n+k \text { where } q \in Z, \quad k \in{0,1, \ldots n-1}$

$\theta_r=\frac{2 \pi(q n+k)}{n}=2 \pi q+\frac{2 k \pi}{n}=2 \pi q+\theta_k$

$z_r=\cos \left(2 \pi q+\frac{2 k \pi}{n}\right)+i \sin \left(2 \pi q+\frac{2 k \pi}{n}\right)$

$z_r=\cos \left(\frac{2 k \pi}{n}\right)+i \sin \left(\frac{2 k \pi}{n}\right)$

$z_r=z_k$

The $n^{th}$ roots of unity $:z^n=1$

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

$n=2 \quad , \quad z^2=1$

$z_0 =1\quad z_1=\cos\pi +i\sin\pi$

$z_1=-1$

$n=3 \quad , \quad z^3=1$

$z_0 =1\quad z_1=\cos\frac{2\pi}{3} +i\sin\frac {2\pi}{3}$

$z_2=\cos\frac{4\pi}{3} +i\sin\frac {4\pi}{3}.$

$\theta_1 =\frac{2\pi} {n}$

$\theta_2 =\frac{2\pi} {n}+\frac{2\pi} {n}$

$n=8$

(1) The geometric image of the $n^{th}$ roots of unity are the vertices of a regular polygon with n sides inscribed in the unit circle with one of the vertices at 1.

(2) $z_1 =\cos\frac {2\pi}{n} + i\sin\frac{2\pi}{n}$

$z_1^2 =\cos\frac {4\pi}{n} + i\sin\frac{4\pi}{n}$

$z_1^k = \cos\frac {2k\pi}{n} + i\sin\frac{2k\pi}{n} ,\quad$ (By De Moivre’s theorem )

$z_1^k =z_k$

$\text{The}\ n^{th} \text {root of unity}$

$\{z_0,z_1,z_2 …,z_{n-1}\} =\{z_1^0,z_1^1,z_1^2,….z_1^{n-1}\}$