Question

Please help! :)

What are the three roots of 27(cos (9π/5) i sin (9π/5))?
Express the answers in trigonometric form.

Answer 1

im guessing you mean the 3 solutions of the cube root of given complex number
\[\sqrt[3]{r(\cos \theta +i \sin \theta)} = \sqrt[3]{r}(\cos \frac{2\pi k +\theta}{3} +i \sin \frac{2\pi k+\theta}{3})\]
where k = 0,1,2

Answer 2

so for k=1
\[\frac{2\pi+ \frac{9\pi}{5}}{3} = \frac{\frac{10\pi}{5}+ \frac{9\pi}{5}}{3} = \frac{19\pi}{15}\]
\[\rightarrow 3(\cos \frac{19\pi}{15}+ i \sin \frac{19\pi}{15})\]
that is 1 root

Answer 3

How do you substitute the next value of k in?

Answer 4

well k=0 is easy ... the "2pi" part goes away

for k=2, just replace k with 2 ..... 2*2pi = 4pi

Answer 5

Wait... I got 3π/5 and 29π/15 as the other 2 roots

Answer 6

correct

Answer 7

Cool... thanks. I think I understand it now

Answer 8

no problem