Find the principal argument of ( $\sqrt{3}-i)^6$.
2. Use $\arg\left(z\cdot z'\right)=\arg z +\arg z' \left[2\pi\right]$.
Find the principal argument of ( $\sqrt{3}-i)^6$
( $\sqrt{3}-i)=2\exp\left(\frac{-\pi}{6}\right)$ so ( $\text{Arg}(\sqrt{3}-i)^6)=\pi$