## anonymous one year ago Which expression is a cube root of -1 + i sqrt(3)? A. cubert(2) (cos(120degrees) + i sin(120 degrees)) B. cubert(2) (cos(40 degrees) + i sin(40 degrees)) C. cubert(2) (cos(280 degrees) + i sin(280 degrees)) D. cubert(2) (cos(320 degrees) + i sin(320 degrees)) ***My Answer: D*** Please Help!!!

1. anonymous

$-1+\sqrt{3}=r \left( \cos \theta+\iota \sin \theta \right)$ $r \cos \theta =-1,r \sin \theta =\sqrt{3}$ square and add and find r divide find theta

2. anonymous

So I would do r = -1-cos(theta), r = 3sin^2(theta)?

3. anonymous

$r^2\cos ^2\theta =1,r^2\sin ^2\theta=3$ $r^2\left( \cos ^2\theta+\sin ^2\theta \right)=1+3=4$ r=2

4. anonymous

Oh ok

5. anonymous

r is always positive ,so $\cos~ \theta~ is~ negative~and~\sin \theta~is~positive.$ so angle lies in second quadrant. now divide to find theta

6. anonymous

Would $\theta = 120$ be the angle?

7. anonymous

no guess ,show me your calculations.

8. anonymous

Ok $r ^{2}(\cos ^{2}\theta + \sin ^{2}\theta)$$2 (\cos ^{2}\theta + \sin ^{2}\theta)$ $-1 + i \sqrt{3}$ $\theta = \frac{ \sqrt{3} }{ -1}$$\theta = -\frac{ \pi }{ 3}$

9. anonymous

That is all I know of. Not exactly sure if it is correct. Sorry

10. anonymous

$\frac{ r \sin \theta }{ r \cos \theta }=\frac{ \sqrt{3} }{ -1 }=-\sqrt{3}=-\tan 60=\tan \left( 180-60\right) =\tan 120$

11. anonymous

$\tan \theta=\tan 120,\theta=120$

12. anonymous

Oh wow I was far off but at the same time I kinda sorta knew what I was doing. Haha. I answer would now be A I'm assuming. Thank you for taking the time to explain this to me. I desperately needed the help.

13. anonymous

$-1+i \sqrt{3}=2\left( \cos 120+i \sin 120 \right)$

14. anonymous

yw

15. anonymous

Oh wait the answer would be B

16. anonymous

oh sorry you wanted cube root ,i have not noticed

17. anonymous

It's ok :) I already hit the submit button haha

18. anonymous

Wouldn't you just divide 120/3 to get 40?

19. anonymous

$1+i \sqrt{3}=2\left( \cos( 120+360n)+i \sin \left( (120+360n \right) \right)$ $or Z ^{\frac{ 1 }{ 3 }}=2^1/3e ^{\frac{ 360n+120 }{ 3 }}$ put n=0,1,2 ,you get all the three cube roots

20. anonymous

$\left( 1+i \sqrt{3} \right)^{\frac{ 1 }{ 3 }}=2^{\frac{ 1 }{ 3 }}\left\{ \cos \left( \frac{ 360n+120 }{ 3 }+i \sin \left( \frac{ 360n+120 }{ 3 } \right)\right)\right\}$ put n=0,1,2 you get all the values.

21. anonymous

Thank you so much! I very much appreciate it!

22. anonymous

yw