## anonymous one year ago Thanks for helping! I need to simplify this equation to a+bi form. (1/2(cos(72 degrees)+isin(72 degrees))^5 To solve I should do 1/2^5 and multiply 72 degrees by 5 right? Thanks for helping!

1. anonymous

@jim_thompson5910 @ganeshie8 @Nnesha @wio @abb0t @zepdrix @Whitemonsterbunny17 @mathmate @jagr2713 @iki @Mehek14 @ikram002p

2. anonymous

@ikram002p

3. anonymous

multiply $$72$$ by $$5$$

4. anonymous

so if I do that and simplify it is right?

5. anonymous

you also have to take $$\left(\frac{1}{2}\right)^5$$

6. anonymous

if by "simplify" you mean "evaluate the functions" then yes

7. anonymous

yup, that is what I meant

8. anonymous

@satellite73 Can you help me find the complex fifth roots of 5-5sqrt(3)*i?

9. anonymous

yeah first write $5-5\sqrt{3}i$ in trig form do you now how to do that?

10. anonymous

r=sqrt((5)^2+(-5sqrt(3)^2)) for r

11. anonymous

and -5sqrt(3)/5=tan(theta) for theta

12. anonymous

and then I make a rcis(theta)?

13. anonymous

@satellite73

14. anonymous

yeah $$r=\sqrt{a^2+b^2}$$

15. anonymous

from there what do I do?

16. anonymous

did you find $$\theta$$?

17. anonymous

theta=-pi/3 Sorry for being a little late. My browser froze :(

18. anonymous

r=10

19. anonymous

yeah looks like you got it

20. anonymous

From there how do I fin the complex fifth roots?

21. anonymous

find*

22. anonymous

divide the angle by 5

23. anonymous

-pi/15

24. anonymous

yeah

25. anonymous

do I get the 5th root of r?

26. anonymous

just say it $\sqrt[5]{10}$

27. anonymous

you can't really evaluate any of this, just write it in trig form

28. anonymous

Is there more to it?

29. anonymous

It asks me for the complex fifth roots

30. anonymous

there are five of them

31. anonymous

go around the circle again, then repeat

32. anonymous

so I add 2pik to the angle?

33. anonymous

right, the original angle then divide by 5 again

34. anonymous

or else you can divide the circle in to 5 equal parts, with $-\frac{\pi}{15}$as one of them

35. anonymous

each time I add +2pi, should I divide r by 5, or is that a one time thing?

36. anonymous

add $$2\pi$$ to $$-\frac{\pi}{3}$$ then divide that one by $$5$$

37. anonymous

then repeat

38. anonymous

oh, does r change at all?

39. anonymous

no it is going to be $$\sqrt[5]{10}$$ each time

40. anonymous

okay

41. anonymous

oh wait, i think your $$r$$ is wrong, check it again

42. anonymous

So would the first one be $\sqrt[5]{10}(cis(\pi/3))?$

43. anonymous

oh, okay

44. anonymous

Are you sure, I got 10 again

45. anonymous

$\sqrt{5^2+(5\sqrt{3})^2}$

46. anonymous

$\sqrt{25+75}=10$

47. anonymous

you are right sorry

48. anonymous

it's okay, you're the one helping me :) Would the first answer be $\sqrt[5]{10}(cis(\pi/3))$

49. anonymous

?

50. anonymous

@satellite73

51. anonymous

it's fine he is still viewing the chat so just give him a minute or to

52. anonymous

oh sorry was away for a minute

53. anonymous

kool

54. anonymous

it's okay

55. anonymous

ok you found $$r=10,\theta =-\frac{\pi}{3}$$

56. anonymous

Is the thing I wrote earlier correct?

57. anonymous

so $5-5\sqrt{3}i=10\left(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3})\right)$

58. anonymous

you want the five fifth roots

59. anonymous

$\sqrt[5]{10}(cis(\pi/3))?$

60. anonymous

one is $\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)$

61. anonymous

then add $$2\pi$$ to $$-\frac{\pi}{3}$$ to get $$\frac{5\pi}{3}$$

62. anonymous

oh yeah , what you said

63. anonymous

divide by 5 and the next one is $\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)$

64. anonymous

65. anonymous

not sure what you mean

66. anonymous

you divided without adding 2pik where k=1

67. anonymous

to theta

68. anonymous

we now have two fifth roots (3 more to go) one is $\sqrt[5]{10}\left(\cos(-\frac{\pi}{15})+i\sin(-\frac{\pi}{15})\right)$ and the other is $\sqrt[5]{10}\left(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\right)$

69. anonymous

So I can +4pi and divide by 5 for another one?

70. anonymous

oh no we did add $$2\pi$$ to the original angle of $$-\frac{\pi}{3}$$ to get $$\frac{5\pi}{3}$$ then we divided that one by 5 to get $$\frac{\pi}{3}$$

71. anonymous

yeah

72. anonymous

or add $$2\pi$$ to $$\frac{5\pi}{3}$$ same thing

73. anonymous

when you add you get $$\frac{11\pi}{3}$$ divide that by 5 and get $$\frac{11\pi}{15}$$ for the third root

74. anonymous

oh kk

75. anonymous

and 2pi+11pi/15

76. anonymous

lather, rinse, repeat

77. anonymous

oh no careful!

78. anonymous

don't add $$2\pi$$ to $$\frac{11\pi}{15}$$ add it to $$\frac{11\pi}{3}$$

79. anonymous

80. anonymous

hold on a second, it is not a mystery why you are doing this both sine and cosine are periodic with period $$2\pi$$ that means the angle is NOT unique

81. anonymous

so all the original numbers are the same $5-5\sqrt{3}i=10(\cos(-\frac{\pi}{3})+i\sin(-\frac{\pi}{3}))=10(\cos(\frac{5\pi}{3})+i\sin(\frac{5\pi}{3}))=...$ then are all equal

82. anonymous

you are just expressing the original number in trig form with different values of $$\theta$$ then you divide each of them by 5 to get the different fifth roots

83. anonymous

yup, so I just continue doing what we were just doing?

84. anonymous

yes

85. anonymous

So the theta of the roots would be -pi/15, pi/3, 11pi/15, 17pi/15, and 23pi/15?

86. anonymous

@satellite73