$\displaystyle z = \frac{\sqrt{3}}{2} + \frac{1}{2}i$.
Changing to polars gives:
$\displaystyle |z| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2}$
$\displaystyle = \sqrt{\frac{3}{4} + \frac{1}{4}}$
$\displaystyle = \sqrt{1}$
$\displaystyle = 1$
$\displaystyle \theta = \arctan{\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}}$
$\displaystyle = \arctan{\frac{1}{\sqrt{3}}}$
$\displaystyle = \frac{\pi}{6}$.
So $\displaystyle z = \frac{\sqrt{3}}{2} + \frac{1}{2}i = 1\,\textrm{cis}\,{\frac{\pi}{6}}$.
Now, using DeMoivre's Theorem...
$\displaystyle z^{69} = 1^{69}\,\textrm{cis}\,{\frac{69\pi}{6}}$
$\displaystyle = \textrm{cis}\,{\frac{3\pi}{2}}$
$\displaystyle = \cos{\frac{3\pi}{2}} + i\sin{\frac{3\pi}{2}}$
$\displaystyle =-i$.