Question

Which expression is a cube root of -2?
A(3sqrts(2) (cos(90) + i sin(90))
B(3sqrts(2) (cos(60) + i sin(60))
C(3sqrts(2) (cos(260) + i sin(260))
D(3sqrts(2) (cos(120) + i sin(120))

Answer 1

None of the above?

What does i stand for? sqrt(-1)?

Answer 2

i^2 = -1

Answer 3

Oh, wait, I see. You don't mean "3sqrt()", you mean cube root. As in root(2, 3) or something.

Answer 4

Did you try evaluating each of the expressions? I'll bet it's the first one just because cos(90) = 0, sin(90) = 1

Answer 5

i tried to evalute it i screwed up ... and yeah i meant the tiny 3 over the sqrt thing lol

Answer 6

No wait, that doesn't work:

http://www.wolframalpha.com/input/?i=cube+root+of+2+multiplied+by+%28cos%2890%29+%2B+i+sin%2890%29%29

...None of the above work.

Answer 7

cube root of negative 2

Answer 8

Nope. For that to work, we would need to multiply 3root(2) by -1.

I can't see anywhere where the -1 would come from... unless... e^(i*pi) ?

Answer 9

Not that that would work anyways. We already have a sqrt(-1) and there's no way that's turning into a -1 anytime soon.

The question is wrong.

Answer 10

your grasp of math isn't fully matured my friend.. i^2 = 1 ...its the i ...or imaginary number that lets you do negative roots sqrt(-64) = 8i for example

Answer 11

http://www.wolframalpha.com/input/?i=cube+root+of+-2 look at the polar coordinate angle..thats how i figured it out and got it right
it was 60 degrees

Answer 12

Wait wtf.

http://www.wolframalpha.com/input/?i=%28cube+root+-2%29+minus+%28cube+root+of+2+multiplied+by+%28cos%2860%29+%2B+i+sin%2860%29%29%29

Answer 13

You win this round. I will be back.

Answer 14

What I don't get is how you went from graphing sines to imaginary numbers?

Answer 15

lol alright ... my bad? ..and win this round? ..i don't get it

Answer 16

I think I should have used a smiley face... I wasn't being serious.

... :) <-- Makes everything OK. ... :)

Answer 17

haha alright...well i have another direction i need to go and understand now ...its arc lengths such :/