Question

find the indicated power using DeMoivres theorem (Express your fully simplified answer in the form
a + bi.)(3+sqrt3i)^4

Answer 1

you need two numbers, \(r=\sqrt{a^2+b^2}\) and \(\theta\) so write this in either exponential or trig form

Answer 2

\(z=3+\sqrt{3}i\) and so
\(a=3,b=\sqrt{3}, \sqrt{a^2+b^2}=\sqrt{3^2+\sqrt{3}^2}=\sqrt{12}\)

Answer 3

as for \(\theta\) since you are in quadratant 1 you can use \(\tan^{-1}(\frac{b}{a})\) which in your case is
\[\tan^{-1}(\frac{\sqrt{3}}{3})=\frac{\pi}{6}\]

Answer 4

therefore you can write the number is trig form as
\[\sqrt{12}\left(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})\right)\] unless you would prefer exponential from, not sure what class you are taking

Answer 5

to raise the number to the 4th power, take the 4th power of \(\sqrt{12}\) which is \(144\) and then multiply the angle by 4
you good from there?

Answer 6

Change (3+sqrt3i) into ploar Form

Answer 7

*Polar

Answer 8

\[r(\cos x + i sinx) ^n\]

\[r^n (Cos nx + i Sin nx)\]

Answer 9

how would I express this fully simplified in the form
a + bi.)

Answer 10

that is your job, but it is a relatively easy one
when you multiply \(\frac{\pi}{6}\) by \(4\) you get \(\frac{2\pi}{3}\)

Answer 11

so
\[\left(\sqrt{12}\left(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})\right)\right)^4\]
\[=144\left(\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})\right)\]
you only need to evaluate sine and cosine to finish

Answer 12

144(-1/2+i (sqrt3)/2

Answer 13

72+72isqrt3 is that right

Answer 14

looks good except you dropped a minus sign

Answer 15

-72+72isqrt3

Answer 16

thank you so much satllite73