I have an equation of the form:
$f(s)=g(s)\exp(ih(s))f(b-s), \qquad (b \in \mathbb{R}, s \in \mathbb{C})$
where $f:\mathbb{C} \to \mathbb{C}$, $g(s)>0$, $h(s) \in (-\pi,\pi]$.
We can prove that if $h(s) \neq 0$ then $f(s)=0$, $f(b-s)=0$ (I know that if $f(s)=0$ then $f(b-s)=0$). However the case $h(s)=0$ does not imply this case directely. But we can see also that we can have $f(s)=0$, $f(b-s)=0$ if $h(s)=0$.
(a) My question is how to deal with the case when $h(s)=0$.
(b) How I can solve this functional equation: $g(1-s)g(s)=1$ with respect to $g$ for all $s$.