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Limit of ln(complex)

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data

    The limit of ln([tex]\frac{1-\frac{ix}{z}}{1+\frac{ix}{z}}[/tex]) as x goes to 0


    3. The attempt at a solution
    I am using taylor where f = [tex]\frac{1-\frac{ix}{z}}{1+\frac{ix}{z}}[/tex]
    f'(0) -[tex]\frac{2i}{z}[/tex]

    Is this correct? Can you take the derivative just as equations without imaginary numbers?
     
    Last edited: Mar 4, 2010
  2. jcsd
  3. Mar 4, 2010 #2
    write the equation clearly.
     
  4. Mar 4, 2010 #3
    I am sorry, but English is not my native language. What does "write the equation clearly" mean?
     
  5. Mar 4, 2010 #4

    Mark44

    Staff: Mentor

    rsa58 means "write the limit expression clearly."

    What you have written looks like
    [tex]\lim_{x \to 0} \frac{ln(1 - \frac{ix}{z})}{1 + \frac{ix}{z}}[/tex]

    I think that your limit expression is really this:
    [tex]\lim_{x \to 0} ln \left(\frac{1 - \frac{ix}{z}}{1 + \frac{ix}{z}}\right)[/tex]

    You can see my LaTeX script by double-clicking either of the limit expressions above.
     
  6. Mar 4, 2010 #5
    Thanks, yes I ment the last one
    [tex]
    \lim_{x \to 0} ln \left(\frac{1 - \frac{ix}{z}}{1 + \frac{ix}{z}}\right)
    [/tex]
     
    Last edited: Mar 4, 2010
  7. Mar 4, 2010 #6

    Mark44

    Staff: Mentor

    Is z a constant in your expression? If not, I think you need to use partial derivatives.
     
  8. Mar 4, 2010 #7
    Yes, it is a constant.
     
  9. Mar 4, 2010 #8

    Mark44

    Staff: Mentor

    Then I also get f'(0) = -2i/z
     
  10. Mar 4, 2010 #9
    Thanks
     
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