<s>[INST] 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....
 [/INST] 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
$$\begin{align}
(\sqrt{b+c}+\sqrt{b-c})^2 &= (b+c)+2\sqrt{b^2-c^2}+(b-c) \\
                          &= 2b+2 \\
                          &= 4\zeta^2
\end{align}$$
</s>