I managed to get to the point where there is an infinite range of values $\displaystyle theta + 2n pi $ (Problem & Ans Attached). I am confused how he put it in exp form.
I think you need to be more specific as to where you are at.
You should have
$\displaystyle
z^6 = 1 = \cos (2k\pi) + i \sin (2k\pi)
$
By deMoivre's theorem:
$\displaystyle z = \cos \frac{k\pi}{3} + i \sin \frac{k\pi}{3} $
You do not have an infinite range, because you are given the conditions that $\displaystyle -\pi < \phi \leq \pi$
So you want to pick the values k for which the angle lies between $\displaystyle -\pi$ and $\displaystyle \pi $
These values of k are: $\displaystyle 0, \pm 1, \pm 2 , 3$ (Note: it's 3, not -3 because your range includes $\displaystyle \pi$, but not $\displaystyle -\pi$)
To put it into exponential form as required, you'll need euler's formula:
$\displaystyle e^{i\theta} = \cos \theta + i \sin \theta $
The other way is to remember that all the roots are evenly spaced around a circle. So if you can figure out the angle between them, you can keep adding this angle and find all the roots.
In your case, the real roots are easy ($\displaystyle \pm 1$)...
For the complex roots...
$\displaystyle z^6 = 1$
$\displaystyle z^6 = \cos{2\pi} + i\sin{2\pi}$
$\displaystyle z = \cos{\frac{2\pi}{6}} + i\sin{\frac{2\pi}{6}}$
$\displaystyle z = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}$.
So all roots differ by an angle of $\displaystyle \frac{\pi}{3}$. From here we can say the roots are:
$\displaystyle \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}} = \frac{1+i\sqrt{3}}{2}$
$\displaystyle \cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}} = \frac{-1+i\sqrt{3}}{2}$
$\displaystyle \cos{\frac{3\pi}{3}} + i\sin{\frac{3\pi}{3}} = -1$
$\displaystyle \cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}} = \frac{-1-i\sqrt{3}}{2}$
$\displaystyle \cos{\frac{5\pi}{3}} + i\sin{\frac{5\pi}{3}} = \frac{1-i\sqrt{3}}{2}$
$\displaystyle \cos{\frac{6\pi}{3}} + i\sin{\frac{6\pi}{3}} = 1$.