Question

HELP PLEASE!

use Demoivre's theorem to indicate the power of the complex numbers

a. (5+√ 3i) ^6
b. (-5+5i) ^4

Answer 1

= r^6 cis 6θ
Find r = √( a² + b²)
θ = arctan (b/a)

Answer 2

@eatrainbows Can you continue?

Answer 3

nooo, im not really sure

Answer 4

you need \(r=\sqrt{a^2+b^2}\) and \(\theta\) where \(\tan(\theta)=\frac{b}{a}\)

Answer 5

for the second one \(\theta =\frac{3\pi}{4}\) and \(r=\sqrt{5^2+5^2}=5\sqrt{2}\) and therefore
\[-5+5i=5\sqrt{2}\left(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4})\right)\]

Answer 6

to raise to the power of 4, take \((5\sqrt{2})^4\) and multiply \(\frac{3\pi}{4}\) by 4

Answer 7

you get
\[(5\sqrt{2})^4\left(\cos(3\pi)+i\sin(3\pi)\right)\] evaluate and you are done

Answer 8

btw it is not necessarily the case that \(\theta=\tan^{-1}(\frac{b}{a})\) this only works if you are in quadrant 1 or 4