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Winding number questions

  1. Mar 24, 2008 #1
    I have some questions about this proof. I have numbered the equations (1), (2), ... so I can ask about them.

    If the piecewise differentiable closed curve [tex]\gamma[/tex] does not pass through the point a then:

    [tex](1) \displaystyle\oint_{\gamma}\frac{dz}{z-a}[/tex]

    is s multiple of [tex]2 \pi i[/tex]

    [tex]\gamma[/tex] is given by z(t) [tex]\alpha \leq t \leq \beta[/tex] consider:

    [tex](2) h(t)=\displaystyle\int^{t}_{\aplha}\frac{z'(t)}{z(t)-a}dt[/tex]

    h(t) is defined and continuos on the closed interval [tex](\alpha, \beta)[/tex]

    [tex](3) h'(t)=\frac{z'(t)}{z(t)-a}[/tex]

    where z'(t) is continuos.

    [tex](4) k=e^{-h(t)}(z(t)-a)[/tex]

    (5) Hence k' = 0.

    [tex](6) e^{h(t)}=\frac{(z(t)-a)}{k}[/tex]

    [tex](7) e^{h(t)}=\frac{(z(t)-a)}{z(\alpha)-a}[/tex]

    Since [tex]\gamma[/tex] is a closed curve [tex]z(\beta)=z(\alpha)[/tex]

    [tex](8) e^{h(\beta)}=1[/tex]

    [tex]h(\beta)[/tex] must be a multiple of [tex]2 \pi i[/tex].



    1. (3) why do we need to know about h'(t)?

    2. (5) why is k' = 0?

    3. How did we replace k with [tex]z(\alpha)-a[/tex] in step (7) ?

    Thanks for any help you can give. I need to understand this proof for a test and I'd rather not memorize these bits, but instead know what I'm doing! Thanks!
  2. jcsd
  3. Mar 24, 2008 #2
    You need to know [itex] h^\prime (t) [/itex] in order to answer question 2. For question 2, calculate [itex] \frac{dk}{dt}[/itex]. You should get

    [tex] \frac{dk}{dt} = -h^\prime (t) e^{-h(t) } (z(t) -a ) + e^{-h (t)} z^\prime (t) [/tex]

    And substitute what you're given for [itex] h^\prime (t) [/itex].
  4. Mar 24, 2008 #3
    Thanks! I think I get it now!
  5. Mar 24, 2008 #4
    I'm not 100% about part 3, but I think it's because [itex] \frac{dk}{dt}=0 [/itex]. Thus we know that k must be a constant. Since k is a constant, if we know it's value at one place, we know it at any other. Thus [itex] k \equiv k(\alpha) = e^{-h(\alpha)} ( z(\alpha) - a) [/itex] but then there's an extra [itex] e^{-h(\alpha)} [/itex] and I'm not sure where it disappears.
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