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The image of $[0,1]$ is compact and so must contain the purported accumulation point. It makes no loss to assume that $\gamma(t)=t$, $f(0)=0$ is the accumulation point, and the line in question is the …

Why don't you start with a function with zeros at the integers, for instance $\sin\pi z$, and then somehow eliminate the zero at $r$?

Rather obviously not: if $f(z)=\sqrt{z}$ on $U$, the plane slit along the negative real axis, then $|f(z)|^2=|z|$ is real analytic on $V$ the plane with the origin removed but $f$ does not analyticall …

This is a bit basic for MO, but I'll present my solution. I don't understand what the OP's notation is so I'll use my favourite model for hyperbolic space, the Poincaré upper half plane ('cos I like m …

I don't know if there are any general results about these, but when $f$ is a polynomial, these must be in essence results about symmetric functions. If $f(z)=z^n+a_{n-1} z^{n-1}+\cdots+a_1z+a_0$ then …

These probabilities equal $\zeta(2)^{-1}$ and $\zeta_{\mathbb{Q}(i)}(2)^{-1}$ where $\zeta$ is the Riemann zeta-function and $\zeta_K$ is the Dedekind zeta function for the number field $K$. The proof …