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How to connect the winding number's definition to geometrical intuition?

  1. Mar 20, 2010 #1
    We know the winding number as a curve winds round a point.If a point is inside a closed curve,then the winding number is 1,this is the geometrical intuition.

    The definition of winding number of closed curve gamma with respect to a is [tex]n(\gamma ,a) = \frac{1}{{2\pi i}}\int\limits_\gamma {\frac{{dz}}{{z - a}}} [/tex].

    The textbook of Lars Ahlfors said we can write [tex]\int\limits_\gamma {\frac{{dz}}{{z - a}}} = \int\limits_\gamma {d\log (z - a) = \int\limits_\gamma {d\log |(z - a)|} } + i\int\limits_\gamma {d\arg (z - a)} [/tex],when z describes a closed curve,log|z-a| returns to its initial value and arg(z-a) increase or decreases by a mutiple of 2pi.This would seem to imply the lemma,but more careful thought shows that the reasoning is of no value unless we define arg(z-a) in a unique way.

    I don't understand this scentence,if arg(z-a) is not define in a unique way,and how does it be defined in a not unique way,the number is not multiple of 2pi?
     
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  3. Mar 20, 2010 #2

    tiny-tim

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    Hi kakarotyjn!:smile:

    We don't need arg(z-a) to be uniquely defined (which, as you say, it isn't),

    we only need d(arg(z-a)) to be uniquely defined (which it is :wink:).
     
  4. Mar 20, 2010 #3
    Hi tiny-tim!Thanks for your reply!

    Then how can we make d(arg(z-a)) uniquely defined?I'm really confused of this.
    Can we define it in the complement of z>0?
     
  5. Mar 21, 2010 #4

    tiny-tim

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    Hi kakarotyjn! :smile:

    (just got up :zzz: …)
    (i don't understand what you mean by "complement of z>0" :confused:)

    arg has an ambiguity of ±2πn, but d(arg) doesn't, because we can assume that the same value of n is used …

    so long as γ does not go through the point a, a small change in z makes a small change in arg(z-a) … ie, very nearly 0, not very nearly ±2πn. :wink:
     
  6. Mar 21, 2010 #5
    HiHi,tiny-tim:tongue2:(Just got up:rofl:)


    the complement of z>0 means the complement of positive real axis.

    and when they make some functions single value,for example log(z) ,the choose complement of negative real axis as the principle value brance,I'm not clear of that:confused:

    Well,then why Lar Ahlfors said :This would seem to imply the lemma,but more careful thought shows that the reasoning is of no value unless we define arg(z-a) in a unique way.

    Or why can't we think about the definition of winding number and just prove the formula of winding number is the winding number?Just as what he said :z describes a closed curve,log|z-a| returns to its initial value and arg(z-a) increase or decreases by a mutiple of 2pi.

    Sorry for my poor English,and thanks for your help!:smile:
     
  7. Mar 22, 2010 #6

    tiny-tim

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    Hi kakarotyjn! Good afternoon from London! :smile:

    I think the answer is that, although arg(z-a) is multi-valued, this is a curve, and so is a function z(t) for some parameter t, and we can define arg(z(t)-a) to be single-valued (which is most easily done by simply requiring arg(z(t)-a) to be a continuous function of t). :smile:

    (but using the complement of the positive real axis, or of any other line, won't work, because any closed curve will intersect that line … but it doesn't matter, because our single-value requirement is not for all z, or indeed for any open set, but only for the closed image z(R) :wink:)
     
  8. Mar 22, 2010 #7
    Yar!!!

    That is to say,if we define arg(z-a) in a unique way,then the discussion is consistent.

    Really thank you for your patience:smile:

    Well,I have to go bed.

    Good night from Weihai,haha!
     
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