Answer
$w_0=\frac{\sqrt 3}{2}+\frac{1}{2}i$
$w_1=-\frac{\sqrt 3}{2}+\frac{1}{2}i$
$w_2=-i$
Work Step by Step
As we know that,
$1=1+0i$
$=1(cos~\frac{\pi}{2}+isin~\frac{\pi}{2})$
By using the equ $(3)$, with $ r=1, n=3$ and $\theta =\frac{\pi}{2}$, we get
$w_k=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2k\pi}{3})+isin(\frac{\frac{\pi}{2}+2k\pi}{3})]$
where, $k=0,1,2.$
Now,
for $k=0$ :
By using formula;
$w_k=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2k\pi}
{3})+isin(\frac{\frac{\pi}{2}+2k\pi}{3})]$
$w_0=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2(0)\pi}
{3})+isin(\frac{\frac{\pi}{2}+2(0)\pi}{3})]$
$=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+0}
{3})+isin(\frac{\frac{\pi}{2}+0}{3})]$
$=1(\frac{cos~\frac{\pi}{2}}{3}+\frac{isin~\frac{\pi}{2}}{3})$
$=cos~\frac{\pi}{6}+isin~\frac{\pi}{6}$
$w_0=\frac{\sqrt 3}{2}+\frac{1}{2}i$
for $k=1$ :
By using formula;
$w_k=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2k\pi}
{3})+isin(\frac{\frac{\pi}{2}+2k\pi}{3})]$
$w_1=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2(1)\pi}
{3})+isin(\frac{\frac{\pi}{2}+2(1)\pi}{3})]$
$=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2\pi}
{3})+isin(\frac{\frac{\pi}{2}+2\pi}{3})]$
$=1(\frac{cos~\frac{\pi+4\pi}{2}}{3}+\frac{isin~\frac{\pi+4\pi}{2}}{3})$
$=1(\frac{cos~\frac{5\pi}{2}}{3}+\frac{isin~\frac{5\pi}{2}}{3})$
$=cos~\frac{5\pi}{6}+isin~\frac{5\pi}{6}$
$w_1=-\frac{\sqrt 3}{2}+\frac{1}{2}i$
for $k=2$ :
By using formula;
$w_k=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2k\pi}
{3})+isin(\frac{\frac{\pi}{2}+2k\pi}{3})]$
$w_2=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+2(2)\pi}
{3})+isin(\frac{\frac{\pi}{2}+2(2)\pi}{3})]$
$=1^{\frac{1}{3}}[cos(\frac{\frac{\pi}{2}+4\pi}
{3})+isin(\frac{\frac{\pi}{2}+4\pi}{3})]$
$=1(\frac{cos~\frac{\pi+8\pi}{2}}{3}+\frac{isin~\frac{\pi+8\pi}{2}}{3})$
$=1(\frac{cos~\frac{9\pi}{2}}{3}+\frac{isin~\frac{9\pi}{2}}{3})$
$=cos~\frac{9\pi}{6}+isin~\frac{9\pi}{6}$
$=(0)+i(-1)$
$w_2=-i$
