radicals - complex modulus and square root

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?

Answer

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
$$
|\sqrt{a^2}|=\lvert-a\rvert=a,
\qquad
\sqrt{|a^2|}=\sqrt{a^2}=-a
$$