Find the absolute value

**wopashui** · April 6th 2010, 11:41 AM

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

**Drexel28** · April 6th 2010, 03:20 PM

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.

**FancyMouse** · April 6th 2010, 03:20 PM

$1-i\sqrt{3}=2e^{-\frac{\pi i}{3}}$

**HallsofIvy** · April 7th 2010, 03:50 AM

And then $(1- i\sqrt{3})^9= 2^9 e^{9\left(\frac{\pi i}{3}\right)}= 2^9 e^{3\pi i}$

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