APPENDIX H - Complex Numbers - H Exercises - Page A 64: 37

$w_0=1$ $w_1=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i$ $w_2=i$ $w_3=-\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i$ $w_4=-1$ $w_5=-\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}i$ $w_6=-i$ $w_7=\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}~i$

As we know that, $1=1+0i$ $=1(cos~0+isin~0)$ By using the equ $(3)$, with $ r=1, n=8$ and $\theta =0$, we get $w_k=1^{\frac{1}{8}}[cos(\frac{0+2k\pi}{8})+isin(\frac{0+2k\pi}{8})]$ $w_k=cos(\frac{k\pi}{4})+isin(\frac{k\pi}{4})$ where, $k=0,1,2,3,4,...........,7.$ Now, $w_0=1(cos~0+isin~0)$ $w_0=1$ $w_1=1(cos(\frac{\pi}{4})+isin(\frac{\pi}{4}))$ $=cos(\frac{(1)\pi}{4})+isin((1)\frac{\pi}{4})$ $w_1=\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i$ $w_2=1(cos(\frac{(2)\pi}{4})+isin(\frac{(2)\pi}{4}))$ $=cos(\frac{\pi}{2})+isin(\frac{\pi}{2})$ $=(0)+i(1)$ $w_2=i$ $w_3=1(cos(\frac{(3)\pi}{4})+isin(\frac{(3)\pi}{4}))$ $=cos(\frac{3\pi}{2})+isin(\frac{3\pi}{2})$ $w_3=-\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}i$ $w_4=1(cos(\frac{(4)\pi}{4})+isin(\frac{(4)\pi}{4}))$ $=cos(\pi)+isin(\pi)$ $=(-1)+(0)$ $w_4=-1$ $w_5 =1(cos(\frac{(5)\pi}{4})+isin(\frac{(5)\pi}{4}))$ $=cos(\frac{5\pi}{2})+isin(\frac{5\pi}{2})$ $w_5=-\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}i$ $w_6 =1(cos(\frac{(6)\pi}{4})+isin(\frac{(6)\pi}{4}))$ $=cos(\frac{6\pi}{2})+isin(\frac{6\pi}{2})$ $=cos\frac{3\pi}{\sqrt 2}+isin\frac{3\pi}{\sqrt 2}$ $=(0)+(-1)i$ $w_6=-i$ $w_7=1(cos(\frac{(7)\pi}{4})+isin(\frac{(7)\pi}{4}))$ $=cos(\frac{7\pi}{4})+isin(\frac{7\pi}{4})$ $w_7=\frac{1}{\sqrt 2}-\frac{1}{\sqrt 2}~i$