Thread: Expressing product in polar and retangular form:

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

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

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