Question

Find the fifth roots of the complex number.
cos(12pi/7)+isin(12pi/7)

The fifth root of the smallest argument is __?
form a+bi

Answer 1

demorgans

Answer 2

\[r^5\sqrt{cos(5t)+sin(5t)}\]
if i remember it right

Answer 3

From Euler's formula
\[re ^{i \theta}=\cos(\theta)+isin(\theta)\]
\[r=1, \theta=12\pi/7\]
therefore\[e ^{i(12\pi/7)}= \cos(12\pi/7) +isin(12\pi/7)\]

Answer 4

5th root of the smallest arguement?

Answer 5

yes, thats what it's asking for.

Answer 6

demovers is what i was thinking; and 5 would be 1/5

Answer 7

using de Moivre's theorem
\[z ^{n}=(re ^{i \theta})^{n}\]
\[n=5, \theta=12\pi/7\]
\[\therefore z ^{5}=1^{5}e ^{i(5(12\pi/7))}= 1^{5}(\cos(5(12\pi/7))+isin(5(12\pi/7))\]

Answer 8

Please refer my ans on your earlier post

Answer 9

however, polar form can be used where\[r ^{1/n}<(\theta+360k)/n\]

Answer 10

http://openstudy.com/study#/updates/4f37eb78e4b0fc0c1a0d6c9b

Answer 11

12pi/35 * 180/pi

12*180/35 =abt 60

cos(60) is smaller than sin(60) but i got no idea if thats a good answer :/

Answer 12

\[\theta=12\pi/7 = 308.6^{o}, r= 1, n=5, k=0\]
\[@ k =0 \therefore z ^{5}= 1^{1/5}(308.6/5)=1<61.72^{o}\]
\[@ k =1\therefore z ^{5}= 1<(308.6+360)/5= 1<133.72^{0}\]
\[...@k = 4, z=1<(308.6+(360\times4))/5)=1<349.72^{?}\]
convert back to rectangular form