Expressing product in polar and retangular form:

**Luckegrl_16** · Dec 19th 2007, 08:06 PM

I have to convert this to polar and rectangular form:
(8 cis pi/3)(1/2 cis (-2pi/3)) = 4 cis 5pi/3, 2-2i (sq. root 3)

I just don't know how they got that answear...

I know you use the formula: rs cis (a + b) so i understand how they got 4 cis, but idk how they got the 5pi/3. And then from there i know you use r (cos theta + (sin theta)i) but i jst don't get this prob...

thanks in advance for the help

**Plato** · Dec 20th 2007, 03:19 AM

$\begin{array}{l}<br /> \frac{\pi }{3} + \frac{{ - 2\pi }}{3} = \frac{{ - \pi }}{3} \approx \frac{{5\pi }}{3} \\ <br /> \cos \left( {\frac{{ - \pi }}{3}} \right) = \cos \left( {\frac{{5\pi }}{3}} \right) = \frac{1}{2}\;\& \;\sin \left( {\frac{{5\pi }}{3}} \right) = \frac{{ - \sqrt 3 }}{2} \\ <br /> \end{array}$

**Luckegrl_16** · Dec 20th 2007, 10:14 AM

oh WOW!! Now I just feel stupid, lol, that was way easier than I made it. THANK YOU for helping me

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