Exercise 15.2.1

Proof. By section 15.2, we know that $\phi$ is $2\varpi$ periodic,

Moreover, for $-1\le r\le 1$, and $\frac{-\varpi }{2}\le s\le \frac{\varpi }{2}$,

Write $r^{\prime }=\phi (-s)\in [-1,1]$. Then for every $s\in [-\frac{\varpi }{2},\frac{\varpi }{2}]$,

This proves that

Write $M(s)$ the point on the lemniscate with signed arc length $s$. Consider $M^{\prime }=M(s^{\prime })$ the symmetric point of $M(s)$ about the origin. Since the lemniscate is symmetric about the origin, Consider first the case where $0\le s\le \varpi$, then the signed arc length is the positive arc length. Let $M(s)$ the point on the lemniscate with arc length $s$. Then the symmetric point $M(s^{\prime })$ about the $x$-axis is such that $r^{\prime }=\mathit{OM}(s^{\prime })=\mathit{OM}(s)=r$, thus, by definition of $\phi$, $\phi (s)=\phi (s^{\prime })$. The total arc length from $O=M(0)$ to $O=M(\varpi )$ in the first loop is $\varpi$, and the symmetry of the lemniscate about the $x$-axis implies that the arc length $\varpi -s$ between $M(s)$ and $O=M(\varpi )$ is equal to the arc length $s^{\prime }$ between $O=M(0)$ and $M(s^{\prime })$, thus $s^{\prime }=\varpi -s$. This proves

Now, if $\frac{\varpi }{2}\le s\le \varpi$, then $0\le \varpi -s\le \frac{\varpi }{2}$, thus, using (1), (2), (3)

Therefore $\phi (-s)=-\phi (s)$ if $\frac{\varpi }{2}\le s\le \varpi$. Now, if we suppose $-\varpi \le s\le -\frac{\varpi }{2}$, then $\frac{\varpi }{2}\le -s\le \varpi$, so we can apply the last equality to $-s$: $\phi (s)=\phi (-(-s))=-\phi (-s)$. This proves

Using the periodicity, if $s\in ℝ$, the is some $n\in \mathbb{Z}$ and $s^{\prime }\in [-\varpi ,\varpi [$ such that $s=2\mathit{n\varpi }+s^{\prime }$. Then

We have proved

We can now complete (2) to $-\varpi \le s\le 0$. Then $0\le -s\le \varpi$, and by (2) applied to $-s$, $\phi (s+\varpi )=\phi (-s)=-\phi (s)$, thus

.

We have proved, for all $s\in ℝ$,