## anonymous one year ago Can someone explain how to express -4 in trigonometric form? Really lost. Thank you!

1. jim_thompson5910

Polar form: $$\Large (r,\theta)$$ Trig form: $$\Large r[\cos(\theta) + i*\sin(\theta)]$$ The two forms are very similar

2. jim_thompson5910

write -4 as -4+0i that is in the form a+bi

3. anonymous

Ok.

4. jim_thompson5910

a = -4 b = 0 plug those values into the formulas r = sqrt(a^2 + b^2) theta = arctan(b/a)

5. anonymous

So r = 4 and theta = 0?

6. jim_thompson5910

r = 4 is correct but theta = 0 is not

7. jim_thompson5910

|dw:1437696080574:dw|

8. anonymous

Isn't 0/-4 equal to 0, and the arctan(0) just 0?

9. jim_thompson5910

arctan(b/a) = arctan(0/(-4)) = 0 is true but we're in the wrong spot add on 180 degrees |dw:1437696139033:dw|

10. anonymous

So is r = 4 and theta = -4...?

11. jim_thompson5910

theta = 180 degrees = pi radians

12. anonymous

so theta = pi ?

13. jim_thompson5910

yeah

14. anonymous

Ok so let me try and see if I can get the answer.

15. anonymous

A. 4(cos0degrees + i sin270degrees) B. 4(cos90degrees + i sin90degrees) C. 4(cos0degrees + i sin0degrees) D. 4(cos180degrees + i sin180degrees) Would it be D?

16. jim_thompson5910

The trig form, in radians, is $$\Large 4[\cos(\pi) + i*\sin(\pi)]$$ that's equivalent to $$\Large 4[\cos(180^{\circ}) + i*\sin(180^{\circ})]$$

17. jim_thompson5910

you are correct

18. anonymous

Thank you! I'll Medal you (:

19. jim_thompson5910

you're welcome

20. anonymous

Could you help me do the same thing for -6i?

21. anonymous

@jim_thompson5910

22. jim_thompson5910

-6i = 0 + (-6i) now a = 0 and b = -6

23. jim_thompson5910

use the formulas r = sqrt(a^2 + b^2) theta = arctan(b/a)

24. jim_thompson5910

draw a graph to make sure you get the theta in the right spot

25. anonymous

So would theta = pi?

26. jim_thompson5910

what is the value of b/a this time?

27. anonymous

Undefined actually

28. jim_thompson5910

where is tan(theta) undefined?

29. anonymous

Isn't it at 90 degrees and 270 degrees?

30. jim_thompson5910

correct

31. jim_thompson5910

|dw:1437697203957:dw|

32. jim_thompson5910

|dw:1437697245770:dw|

33. anonymous

So theta = 3pi/2

34. jim_thompson5910

yep

35. anonymous

Therefore making the answer: 6(cos270degrees + i sin270degrees)

36. jim_thompson5910

correct

37. anonymous

Thank you so much, could you help me with just one more? I think I understand how to do a single number, but am not sure how to do an equation?

38. jim_thompson5910

what's the question

39. anonymous

3-3i

40. jim_thompson5910

that's not an equation. An equation needs an = sign

41. anonymous

Sorry, meant expression.

42. jim_thompson5910

compare a+bi with 3-3i to see that a = 3 and b = -3

43. anonymous

Oh ok! So then r = 3SqRt2 and theta = -1 ... but I'm still confused as to how you find the theta degree value.

44. jim_thompson5910

theta is not -1

45. jim_thompson5910

b/a = -1, that is true

46. jim_thompson5910

arctan(b/a) = arctan(-1) = ??

47. anonymous

-45...

48. anonymous

So 315 degrees?

49. jim_thompson5910

yes

50. anonymous

So would the answer be: 3SqRt2(cos7pi/4 + i sin7pi/4) ?

51. jim_thompson5910

correct

52. anonymous

Thank you so much. I finally understand this!! (: You really helped me understand this, if I could give you more medals I definitely would.

53. jim_thompson5910

that would be the radian form

54. jim_thompson5910

you're welcome

55. anonymous

Yes, that is what the answers are listed as.