# DeMoivre's theorem

• Mar 13th 2010, 07:42 PM
ConfusedMath
DeMoivre's theorem
"use DeMoivre's theorem to find the indicated power of the complex number. Write the answer in rectangular form."

(squareroot{3(cos 5pie/6 + isin 5pie/6))^4

Am I doing this correctly?

(sqrt3)^4(cos(4*5pie/6)+isin(4*5pie/6)
9(cos 20pie/6)+isin(20pie/6)
9(-.5)+(-.866i)
=-4.5-7.8i

I'm worried I might have done this wrong due to the step in reducing 20pie/6 I couldn't find anything to reduce it by so just left it in that form.
• Mar 13th 2010, 07:50 PM
Prove It
Quote:

Originally Posted by ConfusedMath
"use DeMoivre's theorem to find the indicated power of the complex number. Write the answer in rectangular form."

(squareroot{3(cos 5pie/6 + isin 5pie/6))^4

Am I doing this correctly?

(sqrt3)^4(cos(4*5pie/6)+isin(4*5pie/6)
9(cos 20pie/6)+isin(20pie/6)
9(-.5)+(-.866i)
=-4.5-7.8i

I'm worried I might have done this wrong due to the step in reducing 20pie/6 I couldn't find anything to reduce it by so just left it in that form.

$\left[\sqrt{3}\left(\cos{\frac{5\pi}{6}} + i\sin{\frac{5\pi}{6}}\right)\right]^4 = (\sqrt{3})^4 \left(\cos{\frac{4\cdot 5\pi}{6}} + i\sin{\frac{4\cdot 5\pi}{6}}\right)$

$= 9\left(\cos{\frac{10\pi}{3}} + i\sin{\frac{10\pi}{3}}\right)$

$= 9\left(\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}}\right)$

$= 9\left(-\cos{\frac{\pi}{3}} - i\sin{\frac{\pi}{3}}\right)$

$= 9\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)$

$= -\frac{9}{2} - \frac{9\sqrt{3}}{2}i$.