\(Solve\ z^3 =1\)

\begin{align} & A\ polynomial\ of\ degree\ three\ has\ three\ solutions: let\ these\ be\ \omega_{1},\ \omega_{2},\ \omega_{3} \\ \\ & Begin\ by\ expressing\ our\ equation\ in\ modulus-argument\ form\\ & z^3 = 1(cos(0) +isin(0)) \\ & \quad = 1(cos(0+2k\pi)+ isin(0+2k\pi)) \\ \\ & Then\ taking\ the\ cube\ root\ of\ both\ sides,\ our\ solutions\ will\ be\ given\ by \\ & \omega_{k+1}=(z^3)^{\frac{1}{3}} = \left[1(cos(0+2k\pi)+ isin(0+2k\pi))\right]^\frac{1}{3}\ where\ k=0,\ 1,\ 2 \\ \\ & Applying\ deMoivre's\ Theorem\ provides\ our\ solutions \\ & for\ k=0, \quad \omega_{1}= \left(cos\left(\frac{0+0\pi}{3}\right) +isin\left(\frac{0+0\pi}{3}\right) \right) = 1 \\ & for\ k=1,\quad \omega_{2}=\left(cos\left(\frac{0+2\pi}{3}\right) +isin\left(\frac{0+2\pi}{3}\right) \right) = -\frac{1}{2}+ i\frac{\sqrt{3}}{2} \\ & for\ k=2,\quad \omega_{3} = \left(cos\left(\frac{0+4\pi}{3}\right) +isin\left(\frac{0+4\pi}{3}\right) \right)= -\frac{1}{2}- i\frac{\sqrt{3}}{2} \\ \end{align}