Question

Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form a + bi.)
(3(sqrt)3 + 3i)^-5

Answer 1

@credmond Do you know what De'Moivre's Theorem is and what it's used for?

Answer 2

mhm!

Answer 3

do you need it?

Answer 4

is there an "i" as part of the exponent?

Answer 5

oh shoot there shouldn't be. sorry that was a typo! just raised to the -5 power!

Answer 6

start by rewriting in polar form
\[\rightarrow [6(\cos \frac{\pi}{6} +i \sin \frac{\pi}{6})]^{-5i}\]
...
oh shoot i had it all worked out even with the "i"
haha

Answer 7

so anyway ignore the "i"
then applying the theorem
\[\rightarrow 6^{-5} (\cos \frac{-5\pi}{6} +i \sin \frac{-5\pi}{6})\]

Answer 8

\[= -\frac{\sqrt{3}}{2}6^{-5} -\frac{1}{2}6^{-5} i\]

Answer 9

Thank you so much!! Sorry about the typo! haha @dumbcow

Answer 10

yw

Answer 11

(3(sqrt)3 + 3i)^-5
3^-5 ((sqrt)3 + i)^-5
3^-5 4^-5 (cos(PI/6) + i sin(PI/6))^-5
(cos(-5 PI/6) + i sin(-5 PI/6)) 12^-5

Answer 12

thank you @KenLJW !