# 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?

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 $$|\sqrt{a^2}|=\lvert-a\rvert=a, \qquad \sqrt{|a^2|}=\sqrt{a^2}=-a$$
To the OP : nice analysis, I upvoted. My understanding is that it is precisely because of the anomaly that you discovered that the principal Argument is restricted to $$\;(-\pi, \pi].\;$$ Part of the idea behind complex analysis is to consider real analysis as a special case and to preserve all of the pre-existing real analysis conventions.
By forcing Arg(z) into the range as I describe, zero radians is a permissible value for $$\;(1/2) \times\;$$ Arg(z), while $$\pi$$ radians is not a permissible value. This preserves the real analysis convention that the principal square root of a positive real number is positive.