# Zeroes of $f(s)=g(s)\exp(ih(s))f(b-s)$

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 $s$.

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If $g(s)>0$ and $h(s)\in(-\pi,\pi]$, shouldn't the codomain of $g$ and $h$ be $\mathbb R$? Also, is the equation supposed to be true for all $s\in\mathbb C$ and all $b\in\mathbb R$, or for all $s$ and a particular $b$, or what? –  Rahul Dec 30 '12 at 13:06
Yes, Done. The equation hlods for all $s$ and some $b$ including $b=1$. I am searching for the roots of $f(s)=0$ when $b=1$. –  ZE1 Dec 30 '12 at 13:14

Fix $b \in \mathbb{R}$ and suppose that $f(s) = g(s)\exp(ih(s))f(b-s)$ for all $s \in \mathbb{C}$. As $g(s) > 0$ and $\exp(ih(s)) \neq 0$, $f(s) = 0$ if and only if $f(b - s) = 0$. Depending on $f$, there may not be any zeroes, but if there is a zero $s_0$, then $b - s_0$ is another zero.

Now suppose $b = 1$ as in your comment, then we have

$$f(s) = g(s)\exp(ih(s))f(1-s)$$

and

$$f(1-s) = g(1-s)\exp(ih(1-s))f(s)$$

so

\begin{align*} f(s) &= g(s)\exp(ih(s))f(1-s)\\ &= g(s)\exp(ih(s))[g(1-s)\exp(ih(1-s))f(s)]\\ &= g(s)g(1-s)\exp(ih(s) + ih(1-s))f(s). \end{align*}

Provided $f$ has isolated zeroes (as in the case of a non-zero holomorphic function), we see that

$$1 = g(s)g(1-s)\exp(ih(s) + ih(1-s))$$

and therefore

$$g(s)g(1 - s) = \frac{1}{\exp(ih(s) + ih(1-s))} = \exp(-ih(s)-ih(1-s)).$$

For an arbitrary $b$, the same manipulations give

$$g(s)g(b - s) = \exp(-ih(s)-ih(b-s)).$$

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