Let $ w \in \Bbb C $, and let $ \gamma, \delta : [0,1] \rightarrow \Bbb C $ be closed curves such that for all $ t \in [0,1], |\gamma(t) - \delta(t)| < |\gamma(t) - w| $. By computing the winding number $n_\sigma(0)$ about the origin for the closed curve $\sigma(t) = (\delta(t) - w)/(\gamma(t) - w) $, show that $n_\gamma(w) = n_\delta(w)$.

This seems intuitively clear to me. (Informally, the inequality tells us that $\delta$ and $\gamma$ can never be on "opposite sides" of $w$, and hence their winding numbers must be equal. Making this rigorous isn't enough, however, since I need to use the winding number of $\sigma$ about 0.

I can also show that $n_\sigma(0) = n_\gamma(w) - n_\delta(w)$ (fairly elementarily), so it remains to show that $n_\sigma(0) = 0$. This is the bit that I'm stuck on.

Can someone give me a (small) hint (not a major hint)? If I still can't get anywhere, then I'll probably ask for a larger hint!

There is also a second part to this question - dependent on whether I'm able to do the first part with a hint, I may well update this question to include the second part.

(This is an example sheet question - completely non-examinable.)

Thanks! :)


Look at

$$\lvert \sigma(t) - 1\rvert$$

in order to see $n_\sigma(0) = 0$. Since

$$\lvert \sigma(t) - 1\rvert = \biggl\lvert \frac{\delta(t) - w}{\gamma(t) - w} - 1 \biggr\rvert = \biggl\lvert \frac{\delta(t) - \gamma(t)}{\gamma(t) - w}\biggr\rvert < 1$$

we have $\operatorname{Re} \sigma(t) > 0$ for all $t \in [0,1]$. On the right half-plane there is a holomorphic branch of the logarithm, so

$$\int_{\sigma} \frac{dz}{z} = \int_0^1 \frac{\sigma'(t)}{\sigma(t)}\,dt = \log \sigma(1) - \log \sigma(0) = 0$$

since $\sigma$ is a closed curve.

  • $\begingroup$ I was trying to think of something similar, but I was thinking about just $| \sigma(t) |$, but then that doesn't show anything because it can still rotate around. If I can show that $| \sigma(t) - 1 | < 1$ for all $t$, then I'm done right? $\endgroup$ – Sam T Mar 8 '14 at 12:26
  • 1
    $\begingroup$ Right. If it stays in the right half-plane, it can't wind around $0$. $\endgroup$ – Daniel Fischer Mar 8 '14 at 12:28
  • $\begingroup$ Thanks. :) I should have got that... ¬_¬> I very nearly did that at one point, but then didn't for some reason... $\endgroup$ – Sam T Mar 8 '14 at 12:29
  • $\begingroup$ Anyway, thank you! :) $\endgroup$ – Sam T Mar 8 '14 at 12:29

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