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anonymous

  • 5 years ago

Find all solutions of sin(z)=0?

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  1. amistre64
    • 5 years ago
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    all of them? and is z the complex variable or just some random variable that is real?

  2. anonymous
    • 5 years ago
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    z= complex

  3. amistre64
    • 5 years ago
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    every odd iteration of pi makes sin(x) = 0 ..... but Z eh

  4. amistre64
    • 5 years ago
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    thought I erased that odd part lol

  5. angela210793
    • 5 years ago
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    z=2*k*pi

  6. anonymous
    • 5 years ago
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    z=2kpi? Where K is a real no.? :D

  7. angela210793
    • 5 years ago
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    \[k \in Z\]

  8. amistre64
    • 5 years ago
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    I dont know of any sin(z) =0 that exists in the complex part

  9. angela210793
    • 5 years ago
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    Idk how u may call them...Sorry...

  10. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    so at 1+0i and -1+0i times whatever scalar you wanna used for the unit circle

  12. anonymous
    • 5 years ago
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    But a+0i isn't a complex number, So you're saying every integer is a complex number, aren't you?

  13. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    the complex plane includes the real number line; so yes... all real numbers are a subset of the complex plane

  15. anonymous
    • 5 years ago
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    OK, if so, then the answer is just one, which is 0.

  16. angela210793
    • 5 years ago
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    \[R \subset complex no.s\]

  17. amistre64
    • 5 years ago
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    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
    • 5 years ago
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    when Z = any complex number of the form: a+0i; we get sin(z) = 0

  19. anonymous
    • 5 years ago
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    But only integer which is of the form npi is 0

  20. anonymous
    • 5 years ago
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    So the answer should be just one

  21. anonymous
    • 5 years ago
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    But now what confuses me is, x-1=0 has a complex solution?

  22. amistre64
    • 5 years ago
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  23. amistre64
    • 5 years ago
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    Z is not retricted to "off of the real number line" so just sating '1' is useless

  24. angela210793
    • 5 years ago
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    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
    • 5 years ago
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    the coordinate system is like the reals; but you define Z as: z = (x + yi)

  26. amistre64
    • 5 years ago
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    as opposed to: point = (x,y)

  27. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    exactly :)

  29. amistre64
    • 5 years ago
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    but to say npi means to say only those z values that are a real multiple of pi is wrong

  30. amistre64
    • 5 years ago
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    every point along the real number line is valid

  31. amistre64
    • 5 years ago
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    pi is a number of the real number line; its a measure of the diatmer of a circle to its circumference

  32. anonymous
    • 5 years ago
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    Is the polar form correct, bcuz you are saying sin(1) = 0?

  33. amistre64
    • 5 years ago
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    sin(1 +0i) = 0 at least :)

  34. anonymous
    • 5 years ago
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    sin(1+0i) = 0?

  35. amistre64
    • 5 years ago
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    sin(z) = 0 means at least: sin(a + bi) = 0

  36. angela210793
    • 5 years ago
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    sin(1)=0....? how come?

  37. amistre64
    • 5 years ago
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    sin(a + 0i) = 0 for any given a

  38. anonymous
    • 5 years ago
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    But polar form just gives you a way to express

  39. anonymous
    • 5 years ago
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    Not a complex plane

  40. amistre64
    • 5 years ago
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    polar form can be used to interpret the complex plane into a real plane;

  41. anonymous
    • 5 years ago
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    Hmm., check this out http://www.wolframalpha.com/input/?i=sin%281%2B0i%29

  42. amistre64
    • 5 years ago
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    it wold resemble the polar form yes...

  43. angela210793
    • 5 years ago
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    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
    • 5 years ago
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    than*

  45. anonymous
    • 5 years ago
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    Don't you think, we are leaving out someone in this discussion? The asker!

  46. amistre64
    • 5 years ago
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    i havent blocked them lol

  47. amistre64
    • 5 years ago
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    wolfram is good but still has issues with interpreting data

  48. angela210793
    • 5 years ago
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    haha lol :D.....maybe he has got it...:)

  49. anonymous
    • 5 years ago
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    hey..i'm still here,get confuse @_@

  50. angela210793
    • 5 years ago
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    Oops....@Amistre do u agree with me that the answer is Z=2kpi?

  51. amistre64
    • 5 years ago
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    im open to me being wrong :) but I would have to see how sin(z) is sposed to be interpreted :)

  52. angela210793
    • 5 years ago
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    K belongs to Z

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

  54. amistre64
    • 5 years ago
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    angela, I would have to research it to tell :)

  55. angela210793
    • 5 years ago
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    Ok.....I have always thought math is same everywhere...but...o.O

  56. amistre64
    • 5 years ago
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    maybe sin(z) = 0 where e^(ix) = e^(-ix) :)

  57. angela210793
    • 5 years ago
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    maybe....ur the smart guy here anyway :P

  58. angela210793
    • 5 years ago
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    u have Z={..........-3,-2,-1,0,1,2,3.........} right?

  59. amistre64
    • 5 years ago
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    no; z is beig used in this context to indicate the complex variable and not the set of integers

  60. amistre64
    • 5 years ago
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    its like using x for a real variable; they use z for the complex variable

  61. angela210793
    • 5 years ago
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    Really??????? well in my answer i meant for this kind of Z.....and wht do u call ''my Z''?

  62. amistre64
    • 5 years ago
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    but all signs point to the answer being npi where n is any integer

  63. anonymous
    • 5 years ago
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    Integers!

  64. amistre64
    • 5 years ago
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    your Z would indicate "any integer" ;

  65. angela210793
    • 5 years ago
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    ahaa...that's wht i meant....:S:S

  66. angela210793
    • 5 years ago
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    Thanks :):):) ^_^

  67. anonymous
    • 5 years ago
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    Using 0 = (e^(iz) - e^(-iz)) / 2i I get stucked when 2iz= log 1 what's next...

  68. anonymous
    • 5 years ago
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    @angela210793: You never seemed to be not knowing that though! lol

  69. amistre64
    • 5 years ago
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    http://jwbales.us/precal/part6/part6.3.html this is more my thinking

  70. amistre64
    • 5 years ago
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    sin(z) = bi/|z|

  71. amistre64
    • 5 years ago
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    when bi=0; sin(z) = 0

  72. angela210793
    • 5 years ago
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    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
    • 5 years ago
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    I saw one textbook have it as J and I was like... whats that? lol

  74. anonymous
    • 5 years ago
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    Why can't they keep it to just one universal symbol~!

  75. angela210793
    • 5 years ago
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    37 mins ago I've written:''Idk how u may call them...Sorry...'' So it's not my fault..:(:(:(

  76. anonymous
    • 5 years ago
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    ho ho ho

  77. amistre64
    • 5 years ago
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    i blame the chinese... cant trust anyone theat squints all the time..shifty beady little eyes lol

  78. angela210793
    • 5 years ago
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    hahahaha :D :D Lool :D :)

  79. anonymous
    • 5 years ago
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    \[z=k\pi, k\in Z\]

  80. anonymous
    • 5 years ago
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    what is with the latex???

  81. angela210793
    • 5 years ago
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    O.o

  82. amistre64
    • 5 years ago
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    its new since they upgraded today; nothing works lol,

  83. anonymous
    • 5 years ago
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    @satellite73, are u confident with that answer? which method u're using?

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