# 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....

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}

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.

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}

• true for $\zeta \ge 1$ but the use of absolute values is not true for $\zeta < 1$ – Jason S Jun 18 '20 at 1:47