# Simplification Square Root, Complex Numbers

I am trying to simplify the following expression, where $\beta$ is complex-valued a function taking to real parameters.

If I do the math per hand it is clear that the whole expression equals unity, however using PowerExpand and FullSimplify do not simplify this expression properly down to 1. What functions would I need to use? I am hesitant using ComplexExpand as this will assume that $\beta$ is real, (or won't help either when I specify that $\beta$ is complex).

$$\frac{\left(\sqrt{\frac{1+\frac{1}{\sqrt{\left| \beta (\text{qx},\text{qy})\right| ^2+1}}}{\beta (\text{qx},\text{qy})}}\right)^* \sqrt{\left(\left| \beta (\text{qx},\text{qy})\right| ^2-\sqrt{\left| \beta (\text{qx},\text{qy})\right| ^2+1}+1\right) \beta (\text{qx},\text{qy})^*}}{\left| \beta (\text{qx},\text{qy})\right| } \overset{!}{=}1$$

(Conjugate[Sqrt[(
1 + 1/Sqrt[1 + Abs[\[Beta][qx, qy]]^2])/\[Beta][qx,
qy]]] Sqrt[(1 + Abs[\[Beta][qx, qy]]^2 - Sqrt[
1 + Abs[\[Beta][qx, qy]]^2]) Conjugate[\[Beta][qx,
qy]]])/Abs[\[Beta][qx, qy]]


First

tmp=(Conjugate[Sqrt[(
1 + 1/Sqrt[1 + Abs[\[Beta][qx, qy]]^2])/\[Beta][qx,
qy]]] Sqrt[(1 + Abs[\[Beta][qx, qy]]^2 - Sqrt[
1 + Abs[\[Beta][qx, qy]]^2]) Conjugate[\[Beta][qx,
qy]]])/Abs[\[Beta][qx, qy]]


I got it like this:

FullSimplify[(tmp /. \[Beta][qx, qy] -> a + I b // ComplexExpand //
FullSimplify) //. (a^2 + b^2) -> c - 1, c > 1] // FullSimplify


1

As explanation how i got this result consider the following:

I replace the complex $\beta$ with (a+I b) and then use ComplexExpand, as,

(tmp /. \[Beta][qx, qy] -> a + I b // ComplexExpand // FullSimplify)


$\frac{\sqrt{\frac{1}{\sqrt{a^2+b^2+1}}+1} \sqrt{\sqrt{a^2+b^2} \sqrt{\left(-\sqrt{a^2+b^2+1}+a^2+b^2+1\right)^2}}}{\left(a^2+b^2\right)^{3/4}}$

However, you still need to help mathmatica by eliminating one variable, which is possible since the output has only terms with $a^2+b^2$ in them. So i replace $a^2+b^2+1 =c$ or $a^2+b^2=c-1$ with $c>1$. To get the result.