Express a given continuous path $\alpha(t):[a,b] \rightarrow \mathbb{C} \setminus a$ in the form: $\alpha(t)=a+r(t)e^{i\theta (t)}$

Question

Let $a \in \mathbb{C}$. Given a continuous path $\alpha(t):[b,c] \rightarrow \mathbb{C} \setminus a$, how can we express the path in the form of $$\alpha(t)=a+r(t)e^{i\theta (t)}$$.

First of all, a theorem says for this given continuous path, there exist unique continuous functions $r(t),\theta (t):[b,c] \rightarrow \mathbb{R}$ with $r(t)>0$ for all $t \in [b,c]$ and $\theta(b) \in [0,2\pi)$ such that $$\alpha(t)=a+r(t)e^{i\theta(t)}$$

My question is, how do we find the expressions for $r(t)$ and $\theta(t)$ if given such $\alpha(t)$?

Answer 1

Writing $\alpha (t)$ as $a+r(t)e^{i\theta t}$ is just the polar form of $\alpha (t) -a$, but it takes some work to get continuity of $\theta (t)$. [ $r(t)=|\alpha (t)-a|$ is certainly continuous]. A well known result in complex analysis says that in any open ball in $\mathbb C$ not containing $0$ there is a continuous logarithm. Using this we get the following: for any given $s \in [a,b]$ there is an open interval around $s$ in which we can choose a continuous $\theta (t)$ such that $\alpha (t)=a+r(t)e^{i\theta t}$. By compactness we can find a finite number of these intervals that cover $[a,b]$. Starting from the left we can alter the choice of $\theta (t)$ so that the function agrees on adjacent intervals: If we have $\theta_1 (t)$ in one interval and $\theta_2 (t)$ in an adjacent interval then $e^{\theta_1 (t)}=e^{\theta_2 (t)}$ in the intersection of these intervals, so $\theta_1 (t)-\theta_2 (t)$ is of the form $2n\pi $ for some integer $n$. By continuity this $n$ does not depend on $t$ and we can replace $\theta_2 (t)$ by $\theta_2 (t)+2n\pi$ to ensure that $\theta_2 (t)=\theta_1 (t)$ on the intersection. In a finite number of steps we can get one continuous function $\theta (t)$.

Answer 2

With the restrictions in your question it is in general impossible: You cannot always achieve $\theta([b,c]) \subset [0,2\pi)$. Note that I used $[b,c]$ instead of $[a,b]$ because $a \in \mathbb{C}$.

If you start with any two continuous maps $r, \theta : [b,c] \to \mathbb{R}$ such that $r([b,c]) \subset (0,\infty)$, you get a path $\alpha_{r,\theta}(t) = a + r(t)e^{i\theta(t)}$. Now take for example $[b,c] = [0,2\pi], r(t) = 1, \theta(a) = t$. Then it is impossible to find a continuous $\theta'$ with $\theta'([0,2\pi]) \subset [0,2\pi)$ such that $\alpha_{r,\theta'} = \alpha_{r,\theta}$, simply because not all points of $[0,2\pi)$ can be in the image of $\theta'$.

Conversely, let us start with $\alpha$. As Kavi Rama Murthy explained in his answer, you necessarily have $r(t) = \lvert \alpha(t) - a \rvert$ which is continuous. Consider the path $\beta : [b,c] \to \mathbb{C}, \beta(t) = \frac{\alpha(t) - a}{r(t)}$. This is a path in $S^1$ = unit circle around $0$. The map $e : \mathbb{R} \to S^1, e(t) = e^{it}$, is a covering map. This implies that there exists a unique continuous map $\theta : [b,c] \to \mathbb{R}$ such that $e \circ \theta = \beta$ and $\theta(b) \in [0,2 \pi)$. The last condition is only "scaling". Then you get $\alpha = \alpha_{r,\theta}$.

Answer 3

If

$\alpha(t) = a + r(t) e^{i \theta(t)}, \tag 1$

then

$\alpha(t) - a = r(t) e^{i \theta(t)}, \tag 2$

$\bar \alpha - \bar a = r(t) e^{-i \theta(t)}; \tag 3$

thus,

$\vert \alpha(t) - a \vert^2 = (\alpha(t) - a)(\bar \alpha(t) - \bar a) = (r e^{i \theta(t)}) (r(t) e^{-i \theta(t)} = (r(t))^2, \tag 4$

or

$r(t) = \vert \alpha(t) - a \vert, \tag 5$

which expresses $r(t)$ in terms of $\alpha(t)$ and $a$.

So far, so good; obtaining a formula for $\theta(t)$, however, is a little more difficult. Knowing $r(t)$ as we do, from (2) we may write

$e^{i \theta(t)} = \dfrac{\alpha(t) - a}{r(t)}; \tag 6$

it will make things a little easier to write if we encapsulate the right-hand side of (6) with a symbol of its own:

$\beta(t) = \dfrac{\alpha(t) - a}{r(t)}; \tag 7$

thus,

$e^{i \theta(t)} = \beta(t); \tag 8$

we set

$\beta(t) = \sigma(t) + i \omega(t), \tag 9$

and recall that

$e^{i \theta(t)} = \cos \theta(t) + i \sin(\theta(t)), \tag{10}$

then conbining (8)-(10) we find

$\cos \theta(t) + i \sin(\theta(t)) = \sigma(t) + i \omega(t), \tag{11}$

so that

$\cos(\theta(t)) = \sigma(t), \; \sin \theta(t)) = \omega(t); \tag{12}$

it follows that, as long as $\sigma(t) \ne 0$, we may write

$\tan \theta(t) = \dfrac{\sin \theta(t)}{\cos \theta(t)} = \dfrac{\omega(t)}{\sigma(t)}, \tag{13}$

and/or with $\omega \theta(t) \ne 0$,

$\cot \theta(t) = \dfrac{\cos \theta(t)}{\sin \theta(t)} = \dfrac{\sigma(t)}{\omega(t)}; \tag{14}$

since $\tan$ and $\cot$ are invertible on their respective domains of definition, which as the reader will recall overlap but are offset one from the other by $\pi / 2$, (13) and (14) may be used to define a continuous $\theta(t)$ uniquely, provided the initial value $\theta(b)$ is known; but this value is available to us via

$\alpha(b) = a + r(a) e^{i \theta(a)} \tag{15}$

by following the steps presented above in (1)-(14). Of course, (13)-(14) only determine $\theta(t)$ within an additive term of $2\pi n$, $n \in \Bbb Z$.

The preceding discussion demonstrates how $r(t)$ and $\theta(t)$ may be found given $\alpha(t) \in \Bbb C \setminus \{ a \}$; indeed, we have effectively given a high-level description of an algorithm for computing $r(t)$, $\theta(t)$ from $\alpha(t)$, which "works" for continuous $\alpha$ and returns continuous $r(t)$ and $\theta(t)$. In this sense, we have fulfilled the quest of the question; however, when $\alpha(t)$ is differentiable in $t$, we may present a different method for determining $\theta(t)$ which in some sense may be easier and more effective than the above; to wit, we write, from (9),

$i \dot \theta(t) e^{i \theta(t)} = \dot \beta(t), \tag{16}$

whence

$\dot \theta(t) = -i e^{-i \theta(t)} \dot \beta(t) = -i \dfrac{\dot \beta(t)}{\beta(t)}; \tag{17}$

from this equation we may, in principle at least, have $\theta(t)$ by direct integration:

$\theta(t) - \theta(b) = \displaystyle -i \int_b^t \dfrac{\dot \beta(s)}{\beta(s)} \; ds, \tag{18}$

$\theta(t) = \theta(a) - i \displaystyle \int_b^t \dfrac{\dot \beta(s)}{\beta(s)} \; ds. \tag{19}$

Of course whichever of these two strategems we choose for finding $\theta(t)$, we have to admit the possibility that the restriction $\theta(t) \in [0, 2 \pi)$ must be relaxed if we are in fact to obtain a continuous solution.

Finally, it is I reckon worth pointing out that the integrand in (19) may in fact be expressed as $\dot (\ln\beta(s))$ provided we are prepared to allow the complex logarithm to enter in to our discussion.