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The following proposition is given without proof in these lecture notes:

Proposition 4.5.1.: Let $\gamma$ be a path in $\mathbb C \setminus \{0\}$. Then there exists a parametrisation $\gamma: [a,b]\to \mathbb C \setminus \{0\}$ of $\gamma$ for which $t \mapsto \arg \gamma (t)$ is a continuous function.

I tried to find a proof of this in other lecture notes online but since it does not appear to have a name I could not find anything.

If $\gamma$ is a curve in $\mathbb C \setminus \{0\}$ not passing through the origin how can I prove that there exists a continuous map $\alpha$ (the argument) so that $\gamma(t) = |\gamma (t)| e^{i\alpha(t)}$?

learner
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  • You should have a look to https://en.m.wikipedia.org/wiki/Complex_logarithm – mathcounterexamples.net Sep 18 '15 at 05:31
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    This is at least misleadingly formulated. Perhaps outright wrong. Parametrisations have nothing to do with it. What is true is that there exists a continuous $\alpha \colon [a,b] \to \mathbb{R}$ such that for all $t\in [a,b]$, $\alpha(t)$ is an argument of $\gamma(t)$, i.e. $\gamma(t) = \lvert \gamma(t)\rvert\cdot e^{i\alpha(t)}$. – Daniel Fischer Sep 18 '15 at 09:11
  • @DanielFischer But in your comment $\alpha$ is not continuous? – learner Sep 18 '15 at 11:10
  • No, $\alpha$ is continuous. What we can't necessarily do (only for some paths) is choose a continuous branch of $\arg$ on a neighbourhood of $\gamma([a,b])$. If $\gamma$ is a closed path winding around $0$, we can't even choose a discontinuous $\arg$ function such that $\arg\circ \gamma$ is continuous. Thus if we read "for which $t\mapsto \arg \gamma(t)$ is a continuous function" in that way, it becomes wrong. If we read that sentence in the way I formulated it in my comment, it is true (but has nothing to do with parametrisations). – Daniel Fischer Sep 18 '15 at 11:17
  • @DanielFischer Sorry, I misread your comment. I am going to edit my question. – learner Sep 18 '15 at 13:12

1 Answers1

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Since the trace of $\gamma$ is compact, there is an $r > 0$ such that $\lvert \gamma(t)\rvert \geqslant r$ for all $t\in [a,b]$. By uniform continuity of $\gamma$, there is an $\varepsilon > 0$ such that

$$\lvert t-s\rvert \leqslant \varepsilon \implies \lvert\gamma(t) - \gamma(s)\rvert \leqslant r.$$

Thus for every interval $I\subset [a,b]$ of length $\leqslant \varepsilon$, $\gamma(I)$ is contained in a domain on which a continuous branch of $\arg$ exists (and hence infinitely many continuous branches, all differing by multiples of $2\pi$), namely a half-plane. Pick an $n\in \mathbb{N}\setminus \{0\}$ such that $\frac{b-a}{n}\leqslant \varepsilon$, and let $I_k = \bigl[ a + (k-1)\frac{b-a}{n}, a + k\frac{b-a}{n}\bigr]$ for $1 \leqslant k \leqslant n$.

Choose an arbitrary branch of $\arg$ defined on a neighbourhood of $\gamma(I_1)$ and define $\alpha(t) = \arg \gamma(t)$ on $I_1$ with that branch. Once $\alpha$ is defined on $I_1 \cup \dotsc \cup I_k$, if $k < n$, on a neighbourhood of $\gamma(I_{k+1})$ choose the branch $\arg_{k+1}$ of $\arg$ with $\arg_{k+1} \gamma\bigl(a + k\frac{b-a}{n}\bigr) = \alpha\bigl(a + k\frac{b-a}{n}\bigr)$, and define $\alpha(t) = \arg_{k+1} \gamma(t)$ for $t\in I_{k+1}$. The choice of $\arg_{k+1}$ makes $\alpha$ well-defined and continuous on $I_1 \cup \dotsc \cup I_k \cup I_{k+1}$. After $n-1$ steps, $\alpha$ is defined on all of $[a,b]$.

If there is some topological machinery available, one can avoid an explicit construction, and note that since the interval $[a,b]$ is simply connected, and $\exp \colon \mathbb{C} \to \mathbb{C}\setminus \{0\}$ is the universal covering, there exists a lift $\tilde{\gamma} \colon [a,b] \to \mathbb{C}$ of $\gamma$ - that is, $\exp \circ \tilde{\gamma} = \gamma$ - and one can define $\alpha(t) = \operatorname{Im} \tilde{\gamma}(t)$.

Daniel Fischer
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  • Thank you for your answer! In your first sentence, did you mean $|\gamma (t)|\le r$? – learner Sep 21 '15 at 03:26
  • (Meanwhile I realised that by reparametrisation the author doesn't mean a homeomorphism $\phi: [a,b]\to[c,d]$.) – learner Sep 21 '15 at 03:26
  • No, $\lvert \gamma(t)\rvert \geqslant r$ is intended. We need to bound $\gamma$ away from $0$ in order to have a lower bound on the path length needed to "get to the other side of $0$". I note however that I wrote the condition wrong, I just wrote the distance between points on the left, and the path length on the right (the $\pi r$ is the minimal path length needed to change the argument of $\gamma(t)$ by $\pi$, when $\gamma$ traverses a half of the circle with radius $r$). Fixing that. – Daniel Fischer Sep 21 '15 at 09:21
  • Thank you, I understand that now. But how do you get the existence of a continuous branch of argument in the line after the displaystyle equation? – learner Sep 24 '15 at 01:48
  • We're looking at a subinterval $I$ of $[a,b]$ with length at most $\varepsilon$, say $I = [t_0,t_1]$. Thus we have $\lvert \gamma(t) - \gamma(t_0)\rvert \leqslant r$ for all $t\in I$. Let $z_0 = \gamma(t_0)$. Then $\gamma(I) \subset D_r(z_0) \setminus D_r(0)$, and that set is contained in a half-plane with $0$ in its boundary, and on such half-planes there exist continuous branches of the logarithm, and therefore of the argument. We get the branches of the logarithm by choosing some value $w_0$ of $\log z_0$ and setting $$h(z) = w_0 +\int_{z_0}^z \frac{dz}{z},$$ where we integrate over some – Daniel Fischer Sep 24 '15 at 09:02
  • path from $z_0$ to $z$ in the half-plane, e.g. the straight line segment. Since there are no paths in the half-plane that wind around $0$, the integral is independent of the chosen path, and only depends on the end points, so this defines a holomorphic function $h$ on the half-plane. Since $z\cdot e^{- h(z)}$ is constant (its derivative is $(1 - z\cdot h'(z))e^{-h(z)}$, and since $h'(z) = \frac{1}{z}$, that vanishes identically), and $e^{h(z_0)} = e^{w_0} = z_0$ shows $z = e^{h(z)}$ on the half-plane. Then $\operatorname{Im} \circ h$ is a branch of the argument on the half-plane. – Daniel Fischer Sep 24 '15 at 09:02