1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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: 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook