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Complex Functions

  1. Jun 12, 2008 #1

    m_s_a

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    1. The problem statement, all variables and given/known data
    Find the limit :

    http://www.a7bk-a-up.com/pic/uDs93333.bmp
    http://www.a7bk-a-up.com/pic/zAP94166.bmp


    3. The attempt at a solution
    http://www.a7bk-a-up.com/pic/g0Q93947.bmp
    1. The problem statement, all variables and given/known data






    result


    3. The attempt at a solution
     
    m_s_a, Jun 12, 2008
  2. jcsd
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  3. Jun 12, 2008 #2

    m_s_a

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    Last Qiustion::biggrin:
    http://www.a7bk-a-up.com/pic/Nww95795.bmp
     
    m_s_a, Jun 12, 2008
  4. Jun 15, 2008 #3

    m_s_a

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    please wait:zzz:
     
    m_s_a, Jun 15, 2008
  5. Jun 16, 2008 #4

    Dick

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    cos(pi/2)+i*sin(pi/2)=i, not i^(1/2). ???
     
    Dick, Jun 16, 2008
  6. Jun 16, 2008 #5

    Dick

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    But what? Isn't the limit still 1?
     
    Dick, Jun 16, 2008
  7. Jun 17, 2008 #6

    m_s_a

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    By imposing
    I said let w=i===>w^2=-1
     
    m_s_a, Jun 17, 2008
  8. Jun 17, 2008 #7

    m_s_a

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    Yes,
    Find this limit
     
    m_s_a, Jun 17, 2008
  9. Jun 17, 2008 #8

    Dick

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    I don't understand that at all.
     
    Dick, Jun 17, 2008
  10. Jun 17, 2008 #9

    Dick

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    Substitute zbar for z in the power series. What's wrong with that?
     
    Dick, Jun 17, 2008
  11. Jun 17, 2008 #10

    m_s_a

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    m_s_a, Jun 17, 2008
  12. Jun 17, 2008 #11

    Dick

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    It doesn't have to be analytical. You've shown using the series (or l'Hopital) that if z_n is a series of complex numbers approaching 0, then sin(z_n)/z_n->1. z_n* is also a series of complex numbers approaching 0. The series expansion holds for ANY z.
     
    Dick, Jun 17, 2008
  13. Jun 17, 2008 #12

    m_s_a

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    Like this
    http://www.a7bk-a-up.com/pic/ydi48834.bmp
     
    m_s_a, Jun 17, 2008
  14. Jun 17, 2008 #13

    m_s_a

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    Infact:
    http://www.a7bk-a-up.com/pic/LpB47687.jpg
    delta=???
     
    m_s_a, Jun 17, 2008
  15. Jun 17, 2008 #14

    Dick

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    Sure. |(sin(z*)/z*|=|sin(z)/z|. Because (sin(z)/z)*=(sin(z*)/z*) and |z|=|z*|. So you don't need analyticity, correct?
     
    Dick, Jun 17, 2008
  16. Jun 17, 2008 #15

    m_s_a

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    Dear: Dick
    correct 100%.


    Thank you for answering me
    Thank you very much:blushing:
    In Arabic:
    :biggrin:شكرًا جزيلاً
     
    m_s_a, Jun 17, 2008
  17. Jun 17, 2008 #16

    Dick

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    Sure. Sorry, I'm not good at the script. afwan.
     
    Dick, Jun 17, 2008
  18. Jun 17, 2008 #17

    m_s_a

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    O. My Dod
    afwan:surprised
    very very Excellent
    Rather than to learn English
    You have learned Arabic
     
    m_s_a, Jun 17, 2008
  19. Jun 17, 2008 #18

    Dick

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    shukran, afwan, is about as far as I go. Oh, and salam alekum. That's it. I don't even remember how to count, even though this is a math site. So you might want to keep learning english. :)
     
    Last edited: Jun 17, 2008
    Dick, Jun 17, 2008
  20. Sep 24, 2008 #19

    salemghazou

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    show that :
     
    salemghazou, Sep 24, 2008
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