Question

Find the fifth roots of 243(cos 260° + i sin 260°)

Answer 1

Have you considered DeMoivre?

Answer 2

Idk how to do that :l @tkhunny

Answer 3

De Moivre's Theorem states that (cos x + i sin x)^n=cos nx + i sin nx, if I'm not mistaken.

Answer 4

\([243(cos 260° + i sin 260°)]^{1/5} =\)

\(243^{1/5}\left[cos\left(\dfrac{x + 2k\pi}{5}\right) + i\sin\left(\dfrac{x + 2k\pi}{5}\right)\right]\) for k = 0, 1, 2, 3, 4

Really, just fill in the values for k.

Answer 5

3[cos(x+2*0pi/5) + isin(x+2*0pi/5)]

Answer 6

Sorry, \(x = 260º\). I forgot to fill that in.

Answer 7

What goes on now?

Answer 8

3[cos(260+2) + isin(260+2)]
3[cos(262) + isin(262)]

Answer 9

oh no! it will be 3[cos(260/5) + isin(260/5)

Answer 10

Please remember your Order of Operations.

x + 2pi/5 is absolutely NOT the same as (x+2pi)/5

For k = 0, you have (260+0)/5
For k = 1, you have (260+360)/5
and so on.

Normally, I'd prefer to use \(2\pi\) rather than 360º, but that 260º is a bit annoying. You could convert that to \(\dfrac{13\pi}{9}\) if you would like!

Answer 11

3[(cos52) + isin(52)]

Answer 12

@tkhunny

Answer 13

so when k = 2, it is 260 + 720?

Answer 14

Yup. That's the first one. Four more to go.

Answer 15

You are beginning to get the idea.

Answer 16

So far:
3[cos(52) + isin(52)]
3[cos(124) + isin(124)]
@tkhunny

Answer 17

Each new value should be different by 360/5 degrees.

360/5 = 72
124-52 = 72

I believe you have it!

Answer 18

I've got it!
3[cos(52) + isin(52)]
3[cos(124) + isin(124)]
3[cos(196) + isin(196)]
3[cos(268) + isin(268)]
3[cos(340) + isin(340)]
@tkhunny

Answer 19

I wish I could give you best answer 10 times over! @tkhunny

Answer 20

Excellent work!