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.