finding the principal argument

Printable View

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 ( $\text{Arg}(\sqrt{3}-i)^6)=\pi$

All times are GMT -8. The time now is 09:44 AM.