Simplification of $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$

Question

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....

Answer 1

Oh, I figured it out:

$$\begin{align} (\sqrt{b+c}+\sqrt{b-c})^2 &= (b+c)+2\sqrt{b^2-c^2}+(b-c) \\ &= 2b+2\sqrt{b^2-c^2} \end{align}$$

and in this case $b^2 - c^2 = 4\zeta^2-4\zeta+1 - 4\zeta^4 +4\zeta^2 = 1$

so

$$\begin{align} (\sqrt{b+c}+\sqrt{b-c})^2 &= (b+c)+2\sqrt{b^2-c^2}+(b-c) \\ &= 2b+2 \\ &= 4\zeta^2 \end{align}$$

Answer 2

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.

Answer 3

Note

\begin{align} & \sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}\\ = & \sqrt{\left(\zeta+\sqrt{\zeta^2-1}\right)^2} + \sqrt{\left(\zeta-\sqrt{\zeta^2-1}\right)^2}\\ = & \left|\zeta+\sqrt{\zeta^2-1}\right| +\left|\zeta-\sqrt{\zeta^2-1}\right| =2|\zeta|\end{align}