I’m submitting this question because my previous one was perhaps too verbose and did not get the point very quickly.
A well known $\sqrt{2}$ nested radical has the following solutions as found in the literature here and/or here:
$$R_n(x)= \sqrt{2+\sqrt{2+\sqrt{2\ldots+\sqrt{2+x}}}}=\left\{ \begin{array} \\2\cos(\cos^{-1}(x/2)\cdot2^{-n}) & \mbox{for} & |x|\le2\\ \\ b(x)^{2^{-n}}+b(x)^{-2^{-n}} & \mbox{for} & |x|\ge2\\ \end{array}\right. $$
where
$$b(x)=\frac{x+\sqrt{x^2-4}}{2}$$
However, we have found that for all $z\in\mathbb{C}$
$$R_n(z)= \sqrt{2+\sqrt{2+\sqrt{2\ldots+\sqrt{2+z}}}}=\left\{ \begin{array} \\2\cos(\cos^{-1}(z/2)\cdot2^{-n}) \\ \mbox{or eqivalently,}\\ b(z)^{2^{-n}}+b(z)^{-2^{-n}} \end{array}\right. $$
How can this be proven? This was derived from a recurrence relation, thence analytically continued. Is that sufficient? Otherwise, how would you go about proving it?