Simplification of $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ If I try to evaluate $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ numerically for real $\zeta$, it looks like it is just equal to $2|\zeta|$ for $\zeta \ne 0$ and $2j$ for $\zeta=0$, but I can't figure out how to simplify to get there...
It's of the form $\sqrt{b+c} + \sqrt{b-c}$ with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$. I can write:
$$\sqrt{b+c} + \sqrt{b-c} = \frac{(b+c) - (b-c)}{\sqrt{b+c} - \sqrt{b-c}}$$
but that doesn't seem to help either....

If $|\zeta|\ge1$, with a substitution $\zeta=\mathrm{sign}(\zeta)\cosh z$, $z>0$ you can find:
$$
\sqrt{2\cosh^2z-1+2\cosh z\sqrt{\cosh^2z-1}}+\sqrt{2\cosh^2z-1-2\cosh z\sqrt{\cosh^2z-1}} = 
\\
\sqrt{\cosh^2z+\sinh^2z+2\cosh z\sinh z}+\sqrt{\cosh^2z+\sinh^2z-2\cosh z\sinh z} = \\
(\cosh z+\sinh z)+(\cosh z-\sinh z) = 2\cosh z=2|\zeta|.
$$
If $|\zeta|<1$, the answer depends on how you define a complex square root. But a substitution $\zeta=\sin x$ might help anyway.
