I am failing to understand something about complex square roots:
If we fix the argument $\theta\in(0,2\pi],$ that is we take the positive real line as branch cut, than for $z=r\mathrm{e}^{i\theta}$, $\sqrt{z}$ has argument in the interval $(0,\pi].$ In other words, a positive real number will have a negative square root and thus $$|\sqrt{z}|\neq\sqrt{|z|}.$$ Is that true?
According to the definition, $\sqrt{1}=-1$ and so $$ |\sqrt{1}|=1 $$ whereas $$ \sqrt{|1|}=\sqrt{1}=-1 $$
For any positive real it's the same. If $a>0$, then $$ |\sqrt{a^2}|=\lvert-a\rvert=a, \qquad \sqrt{|a^2|}=\sqrt{a^2}=-a $$