## anonymous one year ago Express the complex number in trigonometric form. -3 + 3 square root of three i

1. jim_thompson5910

I'm assuming the expression given is $$\Large -3+3\sqrt{3}*i$$ If so, then it's the same as last time but now $\Large a = -3$ $\Large b = 3\sqrt{3}$

2. anonymous

yeah, that is the correct expression

3. Plasmataco

quick question... what is the square root of i?

4. Plasmataco

I know... im a little slow with the noggin but still...

5. jim_thompson5910

the i isn't part of the square root

6. Plasmataco

i know, just curios.

7. anonymous

-1

8. Plasmataco

...

9. Plasmataco

i thought that was the square... exponent 2.

10. Plasmataco

-1*-1=1 not i.

11. jim_thompson5910

$\Large i = \sqrt{-1}$ $\Large i^2 = (\sqrt{-1})^2$ $\Large i^2 = -1$

12. anonymous

$\iota ^{2} = -1$

13. Plasmataco

am i just the stupid idiot here or u genuinely dont know or connection problemos?

14. Plasmataco

15. jim_thompson5910

sorry @Plasmataco I'm not following

16. Plasmataco

i know but like -1 under a radical of 4

17. jim_thompson5910

were you able to find r and theta, @lxoser ?

18. Plasmataco

...

19. jim_thompson5910

oh you mean $\Large \sqrt{i} = \sqrt[4]{-1}$ @Plasmataco ??

20. Plasmataco

$\sqrt[4]{-1}$

21. Plasmataco

yeah

22. anonymous

im gonna need help finding r and theta @jim_thompson5910

23. Plasmataco

sry for interrupting...

24. anonymous

its okay

25. jim_thompson5910

$\Large r = \sqrt{a^2 + b^2}$ $\Large r = \sqrt{3^2 + (3\sqrt{3})^2}$ ... ... ... $\Large r = ??$

26. anonymous

r = 30?

27. Plasmataco

dont think so.

28. Plasmataco

simplify.

29. Plasmataco

it should be somthing a lot lower

30. jim_thompson5910

Hint: $\Large (3\sqrt{3})^2=(3\sqrt{3})*(3\sqrt{3})$ $\Large (3\sqrt{3})^2=(3*3)*(\sqrt{3}*\sqrt{3})$ $\Large (3\sqrt{3})^2=(3*3)*\sqrt{3*3}$ $\Large (3\sqrt{3})^2=9\sqrt{9}$ $\Large (3\sqrt{3})^2=9*3$ $\Large (3\sqrt{3})^2=27$

31. anonymous

r = 6

32. jim_thompson5910

yes

33. Plasmataco

yup.

34. Plasmataco

horay! now my question.

35. anonymous

for theta, i got -60?

36. jim_thompson5910

incorrect

37. anonymous

is the result suppose to be in radians or degrees?

38. jim_thompson5910

it depends on what the instructions say

39. jim_thompson5910

does it say which mode they want?

40. anonymous

no, but these are the answer choices

41. jim_thompson5910

ok so they want radian form

42. anonymous

so would theta be - pi/3

43. jim_thompson5910

yes, now because our point is in Q2 (see graph) we add pi radians to our angle to land in the right quadrant |dw:1442707471746:dw|

44. jim_thompson5910

- pi/3 is in Q4 |dw:1442707591362:dw|

45. jim_thompson5910

so we have to add 180 degrees or pi radians to get it into the right quadrant |dw:1442707637525:dw|

46. anonymous

2pi/3?

47. anonymous

@jim_thompson5910

48. jim_thompson5910

correct

49. anonymous

so would the answer be 6(cos 2pi/3 + i sin 2pi/3) ?

50. jim_thompson5910

yes

51. anonymous

thank you so much!! :")

52. jim_thompson5910

no problem