a) (sqrt3 - i)^6 b) (2-2i)^5
If you plot the complex number $\displaystyle \sqrt 3 - i $ on the complex plain and convert to polar coordinates you get $\displaystyle R = 2$ and $\displaystyle \theta= \frac {- \pi}6$. Stated another way: $\displaystyle \sqrt 3 - i = 2(\cos (\frac {-\pi} 6}) + i \sin (\frac {-\pi} 6}) ) = 2 cis (\frac {-\pi} 6})$. Now to raise it to the 6th power apply de Moivre's theorem: $\displaystyle (R cis \theta) ^n = r^n cis(n\theta)$