How does one determine (or show) that the complex function $f(z)=\lvert z \rvert^2$ does not have an antiderivative? (I'm assuming this because contour integrals along two different curves with the same end points yield different results).
Is there no antiderivative because $\lvert z \rvert^2 = zz^\star = -z^2+2z\Re \{z\}$, and $\Re\{z\}$ is non-integrable?
The simplest way is to know that every function that is once differentiable in an open set is also twice differentiable, so $f$ cannot have an antiderivatve unless it is differentiable.
But if $|z^2|$ is differentiable, then so is $\bar z = \frac{|z|^2}z$, and that quite blatantly violates the Cauchy-Riemann equations everywhere!
$\lvert z \rvert^2$ has no antiderivative for the same reasons that $\lvert \overline{z} \rvert^2$ doesn't:
Prove $\overline{z}^2$ has no antiderivative w/ and w/o using path (in)dependence
