Question

Find the cube roots of 125(cos 288° + i sin 288°).

Answer 1

Euler's formula?

Answer 2

\[\cos{\theta}+i\sin{\theta} = e^{i\theta}\]

Answer 3

No, a little more to that: De Moivre.

Answer 4

which can be derived from Euler's, I was trying to get him to see that ;)

Answer 5

There's a handy theorem called De Moivre's Theorem that will help with problems like this. It states that for a complex number
\[\alpha(\cos \theta + i \sin \theta)^n = \alpha^n(\cos n\theta + i \sin n\theta)\]
So you have to go backwards. Clearly the magnitude (a) will be 5, but what about the angle? You could just divide by 3, and get 96, but that's not the only option. You could also add 2pi and divide by 3, for 216, or subtract and divide, for -24.

Answer 6

\[\sqrt[3]{125\cdot(\cos(288) + i\sin(288)} = \sqrt[3]{125} \cdot \sqrt[3]{\cos(288) + i\sin(288)} = 5\cdot \left(\cos(288) + i\sin(288)\right)^{1/3}\]

Answer 7

How can I put that into my calculator with the "i" and pellet?

Answer 8

Thanks for all the responses by the way

Answer 9

You don't even need a calculator for this. lol

Answer 10

i've believe i is 2nd zero on a 83 and cube root is in the math menu

Answer 11

You just have to divide 288 by 3 somewhere. Is that so hard?

Answer 12

2nd period, my bad, but yeah, you shouldn't really need a calculator

Answer 13

(you should also try figuring out De Moivre's theorem from Euler's formula)

Answer 14

I have no idea what any of this means. I have that formula but I dont use it? What?

Answer 15

Just use this identity.\[(\cos\theta + i\sin\theta)^x = \cos(x\theta) + i\sin(x\theta)\]

Answer 16

You have\[5\cdot (\cos (288^{\circ}) + i\sin (288^{\circ}))^{1/3}\]

Answer 17

Why is it 5* and not 125?

Answer 18

Scroll up...

Answer 19

So I do 5* cos((1/2)288) + isin(1/2)288)? And I get -3.4573?

Answer 20

Did I get it right @ParthKohli

Answer 21

1/3*

Answer 22

You shouldn't really use a calculator for this bro...

Answer 23

A calculator might be necessary for computing the decimal component form, and convenient for getting the angles.

Answer 24

I don't get how I get the root(s) of the original equation with that formula. Wouldn't the formula always give me a single number?

Answer 25

Are my cube roots 96, 216, and -24?