# A question about a closed continuous curve in the complex plane

Let $\gamma:[a,b]\longrightarrow\mathbb{C}$ be a closed continuous curve in the complex plane satisfies:

$(1)\ \gamma(t)\neq 0,\ \forall t\in [a,b]$;

$(2)\ \{\frac{\gamma(t)}{|\gamma(t)|}:t\in [a,b]\}=\{z\in \mathbb{C}:|z|=1\}$.

For any given positive integer $n$, is that right that we can find $a\leq t_1<t_2\leq b$ such that $\gamma(t_1)^n=\gamma(t_2)^n?$

• Are you thinking $n \ge 2$ here? It seems to be false for $[a,b] = [0, 2\pi]$, $\gamma(t) = e^{it}$, $n=1$. (I assume you also want to rule out $t_1 = a, t_2 = b$.) – Nate Eldredge Sep 7 '17 at 16:43
• Intuitively, shouldn't this follow from the fact that $\gamma^n$ has winding number divisible by $n$, but a curve without self-intersections has winding number $\pm 1$? – Nate Eldredge Sep 7 '17 at 16:48
• Such a curve does not exist. Condition (1) implies that $\arg\gamma(t)$ is continuous on $[a,b]$ and we know that the image of a compact set under a continuous function is compact. And you have $[0,2\pi)$. – Alexandre Eremenko Sep 7 '17 at 17:18
• Alexandre Eremenko: I have revised my question. – user173856 Sep 7 '17 at 17:31
• @NateEldredge Why isn't your comment an answer? – Igor Rivin Sep 7 '17 at 18:35

If $\gamma$ has non-zero winding number, then Nate Eldridge's comment shows that the answer is positive. However, note that for (for example) $n = 2,$ the logarithmic spiral (which is not a closed curve) is a counterexample To make it closed, take the boundary of a thin tubular neighborhood of a long piece of said spiral.