# Find the absolute value

• Apr 6th 2010, 11:41 AM
wopashui
Find the absolute value
Find the absolute value and the argument $\theta$ in the range $0 <= \theta < 2\pi$ for $(1 - i\sqrt{3})^9$

I dun remember how to do this type of question, I wonder if someone can show me the process
• Apr 6th 2010, 03:20 PM
Drexel28
Quote:

Originally Posted by wopashui
Find the absolute value and the argument $\theta$ in the range $0 <= \theta < 2\pi$ for $(1 - i\sqrt{3})^9$

I dun remember how to do this type of question, I wonder if someone can show me the process

Use polar coordinates.
• Apr 6th 2010, 03:20 PM
FancyMouse
$1-i\sqrt{3}=2e^{-\frac{\pi i}{3}}$
• Apr 7th 2010, 03:50 AM
HallsofIvy
And then $(1- i\sqrt{3})^9= 2^9 e^{9\left(\frac{\pi i}{3}\right)}= 2^9 e^{3\pi i}$