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user173856
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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)\ \{\arg\gamma(t):t\in [a,b]\}=[0,2\pi)$$(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 $\gamma(t_1)^n=\gamma(t_2)^n?$

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

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

$(2)\ \{\arg\gamma(t):t\in [a,b]\}=[0,2\pi)$.

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

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 $\gamma(t_1)^n=\gamma(t_2)^n?$

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user173856
  • 2k
  • 11
  • 17

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:

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

$(2)\ \{\arg\gamma(t):t\in [a,b]\}=[0,2\pi)$.

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