I am studying chapter 14 in Rudin "real and complex analysis" and I am trying to solve number 4 page 293, in which he asks to find the form of rational functions that are positive on the unit circle.
I started from the fact that these functions have same number of zeros as poles,as he mentions, but i couldn't elaborate. Rudin mentions considering functions of the form
$$f_{a,b}(z) = \frac{(z-a)(1-\bar a z)}{(z-b)(1-\bar b z)},$$
where $|a|,|b|<1.$
Then I tried to approach this problem to the form of rational functions which map unit circles to unit circles , and thought that maybe be they will have almost the same form with some rotation or something ,but i couldn't solve it . So any help is appreciated .
Hint: First show each $f_{a,b}$ is positive on the unit circle. (And the hint for that is to multiply top and bottom by $\bar z.$)
Added later: Let $\mathbb {D}$ be the open unit disc. Let $r(z)$ be a rational function with no poles on $\partial D$ such that $r>0$ on $\partial D.$ Then as mentioned, the number of poles of $r$ in $\mathbb {D}$ equals the number of zeros of $r$ in $\mathbb {D},$ counted by muliplicity. Let $a_1, \dots, a_n$ be the poles, $b_1, \dots, b_n$ be the zeros. Then
$$g(z) = r(z)\prod_{k=1}^n f_{a_k,b_k}(z)$$
defines a holomorphic function on some $D(0,R),R>1.$ Since each $f_{a_k,b_k}$ is positive on $\partial D,$ and $r$ is assumed to have the same property, $g$ is positive on $\partial D.$ That implies $g \equiv c$ on $D(0,R),$ where $c$ is a positive constant. It follows that
$$r(z) = c\prod_{k=1}^n\frac{1}{f_{a_k,b_k}(z)} , z \in D(0,R)\setminus \{a_1,\dots ,a_n\}.$$
Now a simple argument using the identity principle shows that the above holds on all of $\mathbb {C}$ minus the set of poles of this product, i.e., it holds on $\mathbb {C}\setminus \{a_1,\dots , a_n, 1/\overline {a_1},\dots ,1/\overline {a_n}\}.$