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Hoolomorphic functions stretching angles

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data

    If z0 is a critical point of f(z), and m is the smallest integer such that f(m)(z0)=!= 0, then the mapping given by f multiples angles at z0 by a factor of m.

    Hint: Taylor's Theorem

    2. Relevant equations



    3. The attempt at a solution

    I don't even know where to start. Any hints would be appreciated.
     
  2. jcsd
  3. Dec 9, 2009 #2

    Dick

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    Start by writing down the taylor expansion. What does the given condition tell you about the coefficients of the series?
     
  4. Dec 9, 2009 #3
    It tells me that the first couple of coefficients are zero. But what's the connection between the Taylor series of a function and the angles in its image?
     
  5. Dec 9, 2009 #4

    Dick

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    Yes, it's telling you that you can write the taylor series as f(z)=a_m*(z-z0)^m*(1+...) where the '...' stuff becomes small as z->z0. How do angles of straight lines through the origin behave under the function f(z)=z^m?
     
  6. Dec 9, 2009 #5
    I am not sure.

    Suppose f(z) = z^n, and g is a line of slope m, say g(t) = t + m t i. Then

    [tex]f(g(t)) = (t + m t i)^n = \Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k} t^n \, + \, i \, \, \Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} t^n [/tex]

    Interpreting this as a parametrized curve in R^2, I would get

    [tex]x(t) = \Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k} t^n [/tex]
    [tex]y(t) = \Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} t^n[/tex]

    Their derivatives with respect to t are

    [tex] \frac{dx}{dt} = n \Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k} t^{n-1} [/tex]
    [tex]\frac{dy}{dt} = n \Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} t^{n-1}[/tex]

    Resulting in a slope of

    [tex]\frac{dx}{dy} = \frac{\Sigma \left( \begin{array}{c} n \\ 2k \end{array} \right) (-1)^k m^{2k}}{\Sigma \left( \begin{array}{c} n \\ 2k+1 \end{array} \right) (-1)^{k} m^{2k+1} }[/tex]

    Does this look right? And if it does, does this relate to m in any obvious way?
     
  7. Dec 9, 2009 #6

    Dick

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    Yeow. Nice try. But you are taking the long way around!! Write the line in polar notation r*exp(i*theta) where r is variable and theta is the constant angle.
     
  8. Dec 9, 2009 #7
    Hahaha, yes, it should have occurred to me that I am in complex analysis and not multivariable calculus :)

    Let's see. In this case g(r) = r exp(i theta), and f(g(r)) = r^n exp(i n theta). Now my angle is n*theta. Makes sense.

    By pre- and post-composing with translations, I can WLOG assume that I am interested in a pair of lines through the origin, and that f(0) = 0. But why do higher-order terms not impact the angle?
     
  9. Dec 9, 2009 #8

    Dick

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    The angles they are talking about are the tangent angles of curves through the point z0. You really just need the angle in the limit as z->z0. You can make the higher order terms as small as you want relative to the first term. The higher order terms will make lines transform into curves away from z=z0, but they won't affect the tangent angle.
     
  10. Dec 9, 2009 #9
    It makes sense intuitively. I am still struggling to come up with a rigorous argument why higher-order terms don't matter - I guess I am not sure how tangent angles are defined in complex-analysis-land to begin with. But at least now I get the idea for the proof. I can probably figure the rest out on my own.

    Thank you very much for your help!
     
  11. Dec 9, 2009 #10

    Dick

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    I wouldn't worry about making this overly rigorous. The result isn't THAT interesting or fundamental. I would feel free to do some hand waving, as long as you communicate the idea. That's about the only interesting part.
     
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