## anonymous 5 years ago Find all solutions of sin(z)=0?

1. amistre64

all of them? and is z the complex variable or just some random variable that is real?

2. anonymous

z= complex

3. amistre64

every odd iteration of pi makes sin(x) = 0 ..... but Z eh

4. amistre64

thought I erased that odd part lol

5. angela210793

z=2*k*pi

6. anonymous

z=2kpi? Where K is a real no.? :D

7. angela210793

$k \in Z$

8. amistre64

I dont know of any sin(z) =0 that exists in the complex part

9. angela210793

Idk how u may call them...Sorry...

10. amistre64

if we start out as the [0,inf) real line as our initial ray, then any tine a + 0i occurs we get sin=0

11. amistre64

so at 1+0i and -1+0i times whatever scalar you wanna used for the unit circle

12. anonymous

But a+0i isn't a complex number, So you're saying every integer is a complex number, aren't you?

13. amistre64

every integer is a complex number in the same sense that every integer is a rational number; because they are inherently a part of the set..

14. amistre64

the complex plane includes the real number line; so yes... all real numbers are a subset of the complex plane

15. anonymous

OK, if so, then the answer is just one, which is 0.

16. angela210793

$R \subset complex no.s$

17. amistre64

its only '1' if we make a unit circle in the complex plane and choose the +Reals as our initial ray; then rotate around and terminate everytime we hit the real line again...

18. amistre64

when Z = any complex number of the form: a+0i; we get sin(z) = 0

19. anonymous

But only integer which is of the form npi is 0

20. anonymous

So the answer should be just one

21. anonymous

But now what confuses me is, x-1=0 has a complex solution?

22. amistre64

23. amistre64

Z is not retricted to "off of the real number line" so just sating '1' is useless

24. angela210793

Not 1....4 example...k=-1---> Z=2(-1)*pi=-2pi=angle 0 k=0--->Z=angle 0 k=1--->=2*1*pi=angle 0

25. amistre64

the coordinate system is like the reals; but you define Z as: z = (x + yi)

26. amistre64

as opposed to: point = (x,y)

27. anonymous

Oh, so a+bi is a complex number, so is a + 0i so is R+0i Thus the solution to this would be npi+0i, wouldn't it?

28. amistre64

exactly :)

29. amistre64

but to say npi means to say only those z values that are a real multiple of pi is wrong

30. amistre64

every point along the real number line is valid

31. amistre64

pi is a number of the real number line; its a measure of the diatmer of a circle to its circumference

32. anonymous

Is the polar form correct, bcuz you are saying sin(1) = 0?

33. amistre64

sin(1 +0i) = 0 at least :)

34. anonymous

sin(1+0i) = 0?

35. amistre64

sin(z) = 0 means at least: sin(a + bi) = 0

36. angela210793

sin(1)=0....? how come?

37. amistre64

sin(a + 0i) = 0 for any given a

38. anonymous

But polar form just gives you a way to express

39. anonymous

Not a complex plane

40. amistre64

polar form can be used to interpret the complex plane into a real plane;

41. anonymous

Hmm., check this out http://www.wolframalpha.com/input/?i=sin%281%2B0i%29

42. amistre64

it wold resemble the polar form yes...

43. angela210793

Don't take me wrong PLEASE....but....why do u explain a very easy problem in this way?....O.o....ur making it harder then it really is....O.o

44. angela210793

than*

45. anonymous

Don't you think, we are leaving out someone in this discussion? The asker!

46. amistre64

i havent blocked them lol

47. amistre64

wolfram is good but still has issues with interpreting data

48. angela210793

haha lol :D.....maybe he has got it...:)

49. anonymous

hey..i'm still here,get confuse @_@

50. angela210793

Oops....@Amistre do u agree with me that the answer is Z=2kpi?

51. amistre64

im open to me being wrong :) but I would have to see how sin(z) is sposed to be interpreted :)

52. angela210793

K belongs to Z

53. anonymous

i'm reading this post, try to digest, what do u guys think: http://answers.yahoo.com/question/index?qid=20080529224140AAGdnOg

54. amistre64

angela, I would have to research it to tell :)

55. angela210793

Ok.....I have always thought math is same everywhere...but...o.O

56. amistre64

maybe sin(z) = 0 where e^(ix) = e^(-ix) :)

57. angela210793

maybe....ur the smart guy here anyway :P

58. angela210793

u have Z={..........-3,-2,-1,0,1,2,3.........} right?

59. amistre64

no; z is beig used in this context to indicate the complex variable and not the set of integers

60. amistre64

its like using x for a real variable; they use z for the complex variable

61. angela210793

Really??????? well in my answer i meant for this kind of Z.....and wht do u call ''my Z''?

62. amistre64

but all signs point to the answer being npi where n is any integer

63. anonymous

Integers!

64. amistre64

your Z would indicate "any integer" ;

65. angela210793

ahaa...that's wht i meant....:S:S

66. angela210793

Thanks :):):) ^_^

67. anonymous

Using 0 = (e^(iz) - e^(-iz)) / 2i I get stucked when 2iz= log 1 what's next...

68. anonymous

@angela210793: You never seemed to be not knowing that though! lol

69. amistre64

http://jwbales.us/precal/part6/part6.3.html this is more my thinking

70. amistre64

sin(z) = bi/|z|

71. amistre64

when bi=0; sin(z) = 0

72. angela210793

Hey...X( X( I know what that is X( is just that i thought integers in english was Z as in my place is...

73. amistre64

I saw one textbook have it as J and I was like... whats that? lol

74. anonymous

Why can't they keep it to just one universal symbol~!

75. angela210793

37 mins ago I've written:''Idk how u may call them...Sorry...'' So it's not my fault..:(:(:(

76. anonymous

ho ho ho

77. amistre64

i blame the chinese... cant trust anyone theat squints all the time..shifty beady little eyes lol

78. angela210793

hahahaha :D :D Lool :D :)

79. anonymous

$z=k\pi, k\in Z$

80. anonymous

what is with the latex???

81. angela210793

O.o

82. amistre64

its new since they upgraded today; nothing works lol,

83. anonymous

@satellite73, are u confident with that answer? which method u're using?