Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Complex Functions

  1. Jun 12, 2008 #1
    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
     
  2. jcsd
  3. Jun 12, 2008 #2
    Last Qiustion::biggrin:
    http://www.a7bk-a-up.com/pic/Nww95795.bmp
     
  4. Jun 15, 2008 #3
    please wait:zzz:
     
  5. Jun 16, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    But what? Isn't the limit still 1?
     
  7. Jun 17, 2008 #6
    By imposing
    I said let w=i===>w^2=-1
     
  8. Jun 17, 2008 #7
    Yes,
    Find this limit
     
  9. Jun 17, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  13. Jun 17, 2008 #12
    Like this
    http://www.a7bk-a-up.com/pic/ydi48834.bmp
     
  14. Jun 17, 2008 #13
    Infact:
    http://www.a7bk-a-up.com/pic/LpB47687.jpg
    delta=???
     
  15. Jun 17, 2008 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure. |(sin(z*)/z*|=|sin(z)/z|. Because (sin(z)/z)*=(sin(z*)/z*) and |z|=|z*|. So you don't need analyticity, correct?
     
  16. Jun 17, 2008 #15
    Dear: Dick
    correct 100%.


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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure. Sorry, I'm not good at the script. afwan.
     
  18. Jun 17, 2008 #17
    O. My Dod
    afwan:surprised
    very very Excellent
    Rather than to learn English
    You have learned Arabic
     
  19. Jun 17, 2008 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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
  20. Sep 24, 2008 #19
    show that :
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook