Continuation via a functional equation

Let the function $f(z)$ be regular in some domain containing the segment $[0,1]$ and satisfy the condition $f(z+1) = f(z)$. Prove that the function $f(z)$ can be regularly continued into some strip $-\delta < \mathrm{Im}\left(z\right) < \delta$, $\delta > 0$.

Let the function $f(z)$ be regular in some domain containing the segment $[0,1]$ and satisfy the condition

where $p(z)$ is a polynomial. Prove that the function $f(z)$ can be regularly continued into some strip $-\delta < \mathrm{Im}\left(z\right) < \delta , \delta > 0$.

Let the function $f(z)$ be regular in some domain containing the segment $[1,2]$ and satisfy the condition

where $p(z)$ is a polynomial. Prove that the function $f(z)$ can be regularly continued into some angle $-\delta < \arg z < \delta , \delta > 0$.

Let the function $f(z)$ be regular in the annulus $\rho < \left|z\right| < 1, \rho > 0$, and satisfy in it the functional equation

where $g(z)$ is some given function, regular in the disk $\left|z\right| < 1$. Prove that the function $f(z)$ can be regularly continued into the annulus $0 < \left|z\right| < 1$.

Euler's beta function $\mathrm{B}(z, \zeta )$ is defined for real $z > 0$ and $\zeta > 0$ by the equality

Prove that the function $\mathrm{B}(z, \zeta )$ can be regularly continued in each of the variables to the entire complex plane, with the exception of the points $z = 0,-1,-2, \ldots ($ respectively $\zeta = 0,-1,-2, \ldots )$.