## anonymous 4 years ago u=[root2(1-i)/2]30 Find the number u.

1. anonymous

$\sqrt{\frac{1-i}{2}}$?

2. anonymous

where does the "30" go?

3. anonymous

wait,I will use equation

4. myininaya

u=$\sqrt{\frac{(1-i)}{2}}*30$

5. anonymous

$[(\sqrt{2}(1-i)/2 ]^{30}$

6. anonymous

aaah much easier!

7. anonymous

$(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^{30}$

8. myininaya

$2^\frac{30}{2}*\frac{(1-i)^{30}}{2^{30}}=2^{15}*\frac{(1-i)^{30}}{2^{30}}=\frac{(1-i)^{30}}{2^{15}}$

9. anonymous

makes the idea even easier than what was written

10. anonymous

oh dear no my myininaya

11. anonymous

i bet it is what i wrote, because that is an easy one. rewrite in trig (polar, whatever) form. then multiply the angle by 30. the modulus (absolute value) is 1 so nothing to worry about since you are on the unit circle.

12. myininaya

leave me alone i don't know math

13. myininaya

its all been guesses

14. anonymous

wrong answer,I need to found 1-i. :S

15. anonymous

ok do you know how to write this in polar form?

16. anonymous

because once you do it is really easy

17. anonymous

the number is $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$

18. anonymous

which sits right on the unit circle

19. anonymous

do you prefer degrees or radians for these?

20. anonymous

The answer is 1-i. But I don't know how to find it

21. anonymous

actually that is not the answer, but we will find it. it is not hard

22. anonymous

but before we begin, do you know how to write this in trigonometric form?

23. anonymous

Do you mean A(root2/2,-root2/2) ?

24. anonymous

because the idea is just to multiply the angle by 30 and see what you get. the absolute value of this number is 1, and 1 to the power of 30 is 1

25. anonymous

no i mean as $r(\cos(\theta)+i\sin(\theta))$

26. anonymous

in this case $r=1$ because we are right on the unit circle

27. anonymous

$|\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i|=1$

28. anonymous

I haven't learned anything like this yet.

29. anonymous

so all we need it $\theta$ which should be easy because we see these numbers all the time

30. anonymous

well if you have not, you have no way to raise this to the power of 30!

31. anonymous

i mean you are not expected to multiply this number by itself 30 times !

32. anonymous

Wait,see what is extcally writes :

33. anonymous

ok. but if you can write in trig from raising to the power of 30 is easy. maybe this is something else

34. anonymous

u=($(\sqrt{2/2}\times w)^{30}$

35. anonymous

and we know that w=1-i

36. anonymous

yeah well that is what i wrote

37. anonymous

38. anonymous

$(\frac{\sqrt{2}}{2}(1-i))^{30}$ like this yes?

39. anonymous

yes

40. anonymous

well we can do it easily, but if you have not seen trig from i don't know how to explain it

41. anonymous

write it

42. anonymous

i mean you certainly are not going to multiply this number by itself 30 times

43. anonymous

ok this number is $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ $=\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})$

44. anonymous

I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}

45. anonymous

so when you raise to the power of 30 you just multiply the angle by 30

46. anonymous

I have an idea. We can do this: (\frac{\sqrt{2}}{2}(1-i))^{29}*(\frac{\sqrt{2}}{2}(1-i)

47. anonymous

omg!

48. anonymous

$(\frac{\sqrt{2}}{2}(1-i))^{29}*\frac{\sqrt{2}}{2}(1-i)$

49. anonymous

ok that is what you wrote yes?

50. anonymous

YE!

51. anonymous

then what?

52. anonymous

*-(-$\sqrt{2}/2+\sqrt{2}/2$

53. anonymous

ok you have to write $\frac{7\pi}{4}\times 30=\frac{105\pi}{2}$ and realize that this is the same place on the unit circle as $\frac{\pi}{2}$ and so the answer is $i$

54. anonymous

without polar form or trig form i really don't know how you are expected to do this problem.

55. anonymous

sorry that was not much help.

56. anonymous

thanks

57. anonymous

yw

58. myininaya

omg you guys made this totally harder than is should be $(1-i)*(1-i)=1-2i-i^2=1-2i-(-1)=1-2i+1=-2i$ $(1-i)^2=-2i$ $(1-i)^{2*15}=(-2i)^{15}$ $(1-i)^{30}=(-2i)^{15}$ $(1-i)^{30}=(-2)^{15}i^{15}=-2^{15}i^{4*3+2}=-2^{15}i^2=-2^{15}(-1)=2^{15}$ $(1-i)^{30}=2^{15}$ $(\frac{\sqrt{2}}{2}(1-i))^{30}=\frac{2^{15}}{2^{30}}(1-i)^{30}=\frac{1}{2^{15}}*2^{15}=1$

59. myininaya

i made a typo in first line $(1-i)(1-i)=1-2i+i^2=1-2i+(-1)=-2i$

60. myininaya

:)

61. anonymous

actually i am not kicking myself yet

62. myininaya

lol

63. anonymous

since this is wrong

64. myininaya

what?

65. anonymous

oh wait. i was looking at the first line.

66. anonymous

but you are right it must be a "typo"

67. anonymous

you do get $(1-i)(1-i)=-2i$

68. myininaya

yep

69. anonymous

70. anonymous

71. anonymous

because i happen to know that it is i, not 1

72. myininaya

i will look at it again

73. myininaya

but the approach looks good

74. anonymous

taht is ok you just forgot the $i^{15}$ in your answer

75. anonymous

no it is right, you just forgot one part i think at the end. much much easier in polar form really

76. myininaya

$i^{4*3+3}$ omg i put a 2 where that 3 was

77. anonymous

to hell with IE9!

78. myininaya

yes we get i :)

79. myininaya

did you see my mistake above $(-2)^{15}*i^{15}=-2^{15}*i^{4*3+3}=-2^{15}*i^3=-2^{15}*(-i)=2^{15}*i$ $\frac{1}{2^{15}}*(1-i)^{30}=\frac{1}{2^{15}}*(2^{15}*i)=i$