# Thread: finding the principal argument

1. ## finding the principal argument

Find the principal argument of ( $\sqrt{3}-i)^6$.

2. Use $\arg\left(z\cdot z'\right)=\arg z +\arg z' \left[2\pi\right]$.

3. Originally Posted by LCopper2010
Find the principal argument of ( $\sqrt{3}-i)^6$
( $\sqrt{3}-i)=2\exp\left(\frac{-\pi}{6}\right)$ so ( $\text{Arg}(\sqrt{3}-i)^6)=\pi$