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Here is Prob. 18, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Let $\gamma_1, \gamma_2, \gamma_3$ be curves in the complex plane, defined on $[0, 2 \pi ]$ by $$ \gamma_1 (t) = e^{\iota t}, \qquad \gamma_2(t) = e^{2 \iota t}, \qquad e^{2\pi \iota t \sin (1/t) }. $$ Show that these three curves have the same range, that $\gamma_1$ and $\gamma_2$ are rectifiable, that the length of $\gamma_1$ is $2 \pi$, that the length of $\gamma_2$ is $4 \pi$, and that $\gamma_3$ is not rectifiable.

Here is Definition 6.26 in Baby Rudin, 3rd edition:

A continuous mappaing $\gamma$ of an interval $[a, b]$ into $\mathbb{R}^k$ is called a curve in $\mathbb{R}^k$. To emphasize the parameter interval $[a, b]$, we may also say that $\gamma$ is a curve on $[a, b]$.

If $\gamma$ is one-to-one, $\gamma$ is called an arc.

If $\gamma(a) = \gamma(b)$, $\gamma$ is said to be a closed curve.

It should be noted that we define a curve to be a mapping, not a point set. Of course, with each curve $\gamma$ in $\mathbb{R}^k$ there is associated a subset of $\mathbb{R}^k$, namely the range of $\gamma$, but different curves may have the same range.

We associate to each partition $P = \left\{ \ x_0, \ldots, x_n \ \right\}$ of $[a, b]$ and to each curve $\gamma$ on $[a, b]$ the number $$ \Lambda (P, \gamma) = \sum_{i=1}^n \left\lvert \gamma \left( x_i \right) - \gamma \left( x_{i-1} \right) \right\rvert. $$ The $i$th term in this sum is the distance (in $\mathbb{R}^k$) between the points $\gamma \left( x_{i-1} \right)$ and $\gamma \left( x_i \right)$. Hence $\Lambda (P, \gamma)$ is the length of a polygonal path with vertices at $\gamma \left(x_0\right), \gamma \left( x_1 \right), \ldots, \gamma \left( x_n \right)$, in this order. As our partition becomes finer and finer, this polygon approaches the range of $\gamma$ more and more closely. This makes it seem reasonable to define the length of $\gamma$ as $$ \Lambda (\gamma) = \sup \Lambda (P, \gamma), $$ where the supremum is taken over all partitions of $[a, b]$.

If $\Lambda (\gamma) < \infty$, we say that $\gamma$ is rectifiable.

In certain cases, $\Lambda(\gamma)$ is given by a Riemann integral. We shall prove this for continuously differentiable curves, i.e., for curves $\gamma$ whose derivative $\gamma^\prime$ is continuous.

Here is Theorem 6.27 in Baby Rudin, 3rd edition:

If $\gamma^\prime$ is continuous on $[a, b]$, then $\gamma$ is rectifiable, and $$ \Lambda ( \gamma) = \int_a^b \left\lvert \gamma^\prime(t) \right\rvert \ \mathrm{d} t. $$

My first question is, how to define $\gamma_3(0)$? The function $\sin (1/t)$ is not defined at $t=0$.

My Attempt:

We note that $$ \begin{align} \mathrm{Range} \ \gamma_1 &= \left\{ \ e^{\iota t} \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \cos t + \iota \sin t \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \left( \cos t, \sin t \right) \in \mathbb{R}^2 \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ z \in \mathbb{C} \ \colon \ \lvert z \rvert = 1 \ \right\}. \end{align} $$ Similarly, $$ \begin{align} \mathrm{Range} \ \gamma_2 &= \left\{ \ e^{2 \iota t} \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \cos 2t + \iota \sin 2t \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ \left( \cos 2t, \sin 2t \right) \in \mathbb{R}^2 \ \colon \ 0 \leq t \leq 2 \pi \ \right\} \\ &= \left\{ \ z \in \mathbb{C} \ \colon \ \lvert z \rvert = 1 \ \right\}. \end{align} $$

What about the range of $\gamma_3$?

As $\gamma_1^\prime(t) = \iota e^{\iota t}$ and $\gamma_2^\prime (t) = 2 \iota e^{2 \iota t}$ on $[0, 2 \pi ]$, and as both of $\gamma_1^\prime$ and $\gamma_2^\prime$ are continuous on $[a, b]$, so $\gamma_1$ and $\gamma_2$ both are rectifiable, by Theorem 6.27 in Rudin, and by the same theorem we obtain $$ \begin{align} \Lambda \left( \gamma_1 \right) &= \int_0^{2\pi} \left\lvert \gamma_1^\prime(t) \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} \left\lvert \iota e^{\iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} \lvert \iota \rvert \left\lvert e^{\iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} 1 \ \mathrm{d} t \\ &= 2 \pi, \end{align} $$ and $$ \begin{align} \Lambda \left( \gamma_2 \right) &= \int_0^{2\pi} \left\lvert \gamma_2^\prime(t) \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} \left\lvert 2\iota e^{2 \iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} 2 \lvert \iota \rvert \left\lvert e^{2\iota t} \right\rvert \ \mathrm{d} t \\ &= \int_0^{2\pi} 2 \ \mathrm{d} t \\ &= 4 \pi. \end{align} $$

Is there any mistake in what I've done so far? If not, then how to handle $\gamma_3$?

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You need to look at the graph of $t\sin(1/t)$. It tends to zero as $t\to0$. So take $\gamma_3(0)=\exp(0)=1$. It will have a maximum $M$ and minimum $m$ on $[0,2\pi]$. If you can show $M-m\ge1$ then $\exp(2\pi i t\sin(1/t))$ will cover the whole unit circle. To prove $M-m\ge1$ all you need are values $a$ and $b$ with $f(a)-f(b)\ge1$.

To show the path is not rectifiable, all you need is $\int|\gamma_3'|=\infty$.

Everything else is fine so far.

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