# finding the principal argument

• January 30th 2010, 12:56 PM
LCopper2010
finding the principal argument
Find the principal argument of ( $\sqrt{3}-i)^6$.
• January 30th 2010, 01:43 PM
girdav
Use $\arg\left(z\cdot z'\right)=\arg z +\arg z' \left[2\pi\right]$.
• January 30th 2010, 03:12 PM
Plato
Quote:

Originally Posted by LCopper2010
Find the principal argument of ( $\sqrt{3}-i)^6$

( $\sqrt{3}-i)=2\exp\left(\frac{-\pi}{6}\right)$ so ( $\text{Arg}(\sqrt{3}-i)^6)=\pi$