## Alexis1994_13 4 years ago u=[root2(1-i)/2]30 Find the number u.

1. satellite73

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

2. satellite73

where does the "30" go?

3. Alexis1994_13

wait,I will use equation

4. myininaya

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

5. Alexis1994_13

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

6. satellite73

aaah much easier!

7. satellite73

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

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. satellite73

makes the idea even easier than what was written

10. satellite73

oh dear no my myininaya

11. satellite73

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. Alexis1994_13

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

15. satellite73

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

16. satellite73

because once you do it is really easy

17. satellite73

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

18. satellite73

which sits right on the unit circle

19. satellite73

do you prefer degrees or radians for these?

20. Alexis1994_13

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

21. satellite73

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

22. satellite73

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

23. Alexis1994_13

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

24. satellite73

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. satellite73

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

26. satellite73

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

27. satellite73

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

28. Alexis1994_13

I haven't learned anything like this yet.

29. satellite73

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

30. satellite73

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

31. satellite73

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

32. Alexis1994_13

Wait,see what is extcally writes :

33. satellite73

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

34. Alexis1994_13

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

35. Alexis1994_13

and we know that w=1-i

36. satellite73

yeah well that is what i wrote

37. Alexis1994_13

38. satellite73

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

39. Alexis1994_13

yes

40. satellite73

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

41. Alexis1994_13

write it

42. satellite73

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

43. satellite73

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. Alexis1994_13

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

45. satellite73

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

46. Alexis1994_13

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

47. Alexis1994_13

omg!

48. satellite73

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

49. satellite73

ok that is what you wrote yes?

50. Alexis1994_13

YE!

51. satellite73

then what?

52. Alexis1994_13

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

53. satellite73

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. satellite73

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

55. satellite73

sorry that was not much help.

56. Alexis1994_13

thanks

57. satellite73

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. satellite73

actually i am not kicking myself yet

62. myininaya

lol

63. satellite73

since this is wrong

64. myininaya

what?

65. satellite73

oh wait. i was looking at the first line.

66. satellite73

but you are right it must be a "typo"

67. satellite73

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

68. myininaya

yep

69. Alexis1994_13

70. satellite73

71. satellite73

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. satellite73

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

75. satellite73

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. satellite73

to hell with IE9!

78. myininaya

yes we get i :)

79. myininaya

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

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