Suppose $g(z)$ is an analytic function on the upper plane, in fact is constructed by Hilbert transform, \begin{equation} g(z) = g_R(x,y) + i g_I(x,y) \quad g_I = \mathcal{H}( g_R) + \text{real constant } \end{equation} the reason to have the real constant is that $\int_{-\infty}^{\infty} \mathcal{H}( g_R) = 0$, so we don't necessary have $g_I(\pm \infty) =0$; by adding this constant, we can demand the asymptotic behavoir for $g(z)$ on the real axis, \begin{equation} \lim_{x\rightarrow \pm \infty} g(x) = 1 \end{equation} (this and Hilbert transform relation shows that $g_I(z)\rightarrow 0$ like $1/|z|$ or faster in the upper plane.)
It satisfies the following equation, where $g(x)$ is the value of the function on the real axis, \begin{equation} g(x) \bar{g}(x) = 1 \end{equation} obviously the following set of solutions solve the equation, \begin{equation} g(z) = \frac{(z- \bar{z}_1) (z- \bar{z}_2 ) \cdots ( z- \bar{z}_N) }{(z- z_1) (z- z_2 ) \cdots ( z- z_N) }, \quad \text{Im}(z_i) < 0 \quad z_i \ne z_j \end{equation} my question is whether this set exhaust all the solutions.
Thanks.
EDIT The periodic solution seems to be a new type, \begin{equation} g(z) = \frac{\sin\pi(z-\bar{z}_1')\sin\pi(z-\bar{z}_2')\cdots \sin\pi(z-\bar{z}_N')}{\sin\pi(z-z_1')\sin\pi(z-z_2')\cdots\sin\pi(z-z_N')} \end{equation}