Exercise 8.26

Since

we have

Hence $P_1(t)$ and $P_2(t)$, $0\le g\le 2\pi$, define continuous closed differentiable curves in the complex plane which are nowhere 0, so we can define $(P_j)$. We have

so by Exercise 25 we have $(P_1)=(P_2)$.

Before showing Exercise 24 for the continuous case, I want to show a general theorem which is pretty standard and may have been done in a previous exercise, or even the text, although I couldn’t find it. Let $X$ be a metric space with disjoint subsets $F$ and $K$ such that $F$ is closed and $K$ is compact. Then there is a positive minimum distance between them, that is, there is a $\delta >0$ such that $d(x,y)>\delta$ for all $x\in F$ and $y\in K$. If not, then there are subsequences $\{x_n\}\subset F$ and $\{y_n\}\subset K$ such that $\lim <!--\; FUNCTION\; APPLICATION\; -->d(x_n,y_n)=0$. Since $K$ is compact, $\{y_n\}$ has a subsequence $\{y_{n_m}\}$ converging to $y\in K$. Then

which, since $F$ is closed, implies that $y\in F$, contradicting the fact that $F$ and $K$ are disjoint. (If you prefer a direct proof, for each $y\in K$ let $N_y$ be an open neighborhood of $y$ disjoint from $F$. Then $\{N_y\}$ is a cover of $K$, so you can pass to a finite subcover whose sets have a positive minimum radius and go from there.)

Suppose $\gamma$ is a closed, continuous curve in the complex plane with domain $[0,2\pi ]$, whose range does not intersect the negative real axis. Since the negative real axis is a closed set and the range of $\gamma$ is compact, by the preceding paragraph there is a $\delta >0$ such that if $P$ is a trigonometric polynomial satisfying $|P(t)-\gamma (t)|<\delta$ for all $t\in [0,2\pi ]$, so that the range of $P$ also does not intersect the negative real axis. Since $P$ is differentiable, $(P)=0$ by Exercise 25, hence $(\gamma )=0$.

Now suppose ${\gamma }_1$ and ${\gamma }_2$ are closed, continuous curves in the complex plane with domain $[0,2\pi ]$ such that $|{\gamma }_1(t)-{\gamma }_2(t)|<|{\gamma }_1(t)|$ for all $t\in [0,2\pi ]$. Then by Theorem 4.16 there is a $\delta >0$ such that

on the compact set $[0,2\pi ]$. Let $P_1$, $P_2$ be trigonometric polynomials such that $|P_j(t)-{\gamma }_j(t)|<\delta ∕4$ for all $t\in [0,2\pi ]$, $j=1,2$. Then, for all $t\in [0,2\pi ]$,

Hence $|P_1(t)|-|P_1(t)-P_2(t)|>\delta ∕4$ for all $t\in [0,2\pi ]$. Hence, by Exercise 25,