## Alexis1994_13 Group Title u=[root2(1-i)/2]30 Find the number u. 3 years ago 3 years ago

1. satellite73 Group Title

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

2. satellite73 Group Title

where does the "30" go?

3. Alexis1994_13 Group Title

wait,I will use equation

4. myininaya Group Title

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

5. Alexis1994_13 Group Title

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

6. satellite73 Group Title

aaah much easier!

7. satellite73 Group Title

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

8. myininaya Group Title

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

9. satellite73 Group Title

makes the idea even easier than what was written

10. satellite73 Group Title

oh dear no my myininaya

11. satellite73 Group Title

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 Group Title

leave me alone i don't know math

13. myininaya Group Title

its all been guesses

14. Alexis1994_13 Group Title

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

15. satellite73 Group Title

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

16. satellite73 Group Title

because once you do it is really easy

17. satellite73 Group Title

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

18. satellite73 Group Title

which sits right on the unit circle

19. satellite73 Group Title

do you prefer degrees or radians for these?

20. Alexis1994_13 Group Title

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

21. satellite73 Group Title

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

22. satellite73 Group Title

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

23. Alexis1994_13 Group Title

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

24. satellite73 Group Title

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 Group Title

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

26. satellite73 Group Title

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

27. satellite73 Group Title

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

28. Alexis1994_13 Group Title

I haven't learned anything like this yet.

29. satellite73 Group Title

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

30. satellite73 Group Title

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

31. satellite73 Group Title

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

32. Alexis1994_13 Group Title

Wait,see what is extcally writes :

33. satellite73 Group Title

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 Group Title

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

35. Alexis1994_13 Group Title

and we know that w=1-i

36. satellite73 Group Title

yeah well that is what i wrote

37. Alexis1994_13 Group Title

38. satellite73 Group Title

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

39. Alexis1994_13 Group Title

yes

40. satellite73 Group Title

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 Group Title

write it

42. satellite73 Group Title

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

43. satellite73 Group Title

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 Group Title

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

45. satellite73 Group Title

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

46. Alexis1994_13 Group Title

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

47. Alexis1994_13 Group Title

omg!

48. satellite73 Group Title

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

49. satellite73 Group Title

ok that is what you wrote yes?

50. Alexis1994_13 Group Title

YE!

51. satellite73 Group Title

then what?

52. Alexis1994_13 Group Title

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

53. satellite73 Group Title

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 Group Title

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

55. satellite73 Group Title

sorry that was not much help.

56. Alexis1994_13 Group Title

thanks

57. satellite73 Group Title

yw

58. myininaya Group Title

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 Group Title

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

60. myininaya Group Title

:)

61. satellite73 Group Title

actually i am not kicking myself yet

62. myininaya Group Title

lol

63. satellite73 Group Title

since this is wrong

64. myininaya Group Title

what?

65. satellite73 Group Title

oh wait. i was looking at the first line.

66. satellite73 Group Title

but you are right it must be a "typo"

67. satellite73 Group Title

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

68. myininaya Group Title

yep

69. Alexis1994_13 Group Title

70. satellite73 Group Title

71. satellite73 Group Title

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

72. myininaya Group Title

i will look at it again

73. myininaya Group Title

but the approach looks good

74. satellite73 Group Title

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

75. satellite73 Group Title

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

76. myininaya Group Title

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

77. satellite73 Group Title

to hell with IE9!

78. myininaya Group Title

yes we get i :)

79. myininaya Group Title

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