show that $f(z) = \overline{z}^2$ has no antiderivative in any nonempty region It's been a few quarters since I've taken complex analysis and I'm reviewing for a comprehensive exam.  I ran into this problem on a sample exam and it stumped me.  
I'm guessing I would have to do something with the Cauchy integral formula, but I'm honestly not sure.  Complex wasn't one of my best classes.  
 A: When you're reviewing for an entire course, you probably know that a function cannot have an antiderivative in a region unless it is differentiable there. So it suffices to show that your $f$ is not differentiable in any nonempty open set.
A straightforward approach would be to show that it doesn't satisfy the Cauchy-Riemann equations anywhere except at the origin -- that's not a lot of computation in this case.
Alternatively you could observe that $\bar z^2 = \overline{z^2}$, if you have proved a theorem that the only way for a function and its complex conjugate both to be differentiable is if it is constant, which $z^2$ isn't. (You may know this theorem in the form: If $g$ is real-valued and differentiable everywhere in a connected open set, then it is constant. This can be applied to $g(z)=f(z)+\overline{f(z)}$ and then $g_2(z)=\frac{f(z)-\overline{f(z)}}i$).
A: If a complex function $g$ defined on a region $A$ has a derivative at each point of $A$ it is holomorphic (Goursat's theorem). The derivative of a holomorphic function is holomorphic. Since $z\mapsto \bar{z}^2$ is nowhere holomorphic…
A: A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 5.7

Exer 5.7 Prove $\overline{z}^2$ has no antiderivative on a non-empty region.

Related: Prove path independence w/o using antiderivatives and then w/o using holomorphicity
The discussion after Cor 5.9 (*) suggests the/a way to prove is with path (in)dependence. However, I believe I can prove w/o using path (in)dependence.
Pf: Let $G$ be a nonempty region. Observe that $\overline{z}^2$ is continuous in $G$, so by Morera's Thm 5.6(*), $$\exists \gamma \subset G: \int_\gamma \overline{z}^2 \, dz \ne 0$$
Now, I think we can proceed in 1 of 2 ways:

  
*
  
*Way (1): Without path (in)dependence
  

Observe $G$ is open because $G$ is a region, so by Cor 4.13, $\overline{z}^2$ has no antiderivative on $G$.
QED Way (1)

  
*
  
*Way (2): With path (in)dependence
  

As in my proof for Cor 5.9, decompose $\gamma = \gamma_1 \wedge -\gamma_2$ where $\gamma_1$ and $\gamma_2$ have the same start and end points s.t. $$0 \ne \int_\gamma \overline{z}^2 \, dz = \int_{\gamma_1} \overline{z}^2 \, dz - \int_{\gamma_2} \overline{z}^2 \, dz \implies \int_{\gamma_1} \overline{z}^2 \, dz \ne \int_{\gamma_2} \overline{z}^2 \, dz$$
In (Q1), it was attempted to prove that antiderivative implies path independent on simply-connected regions for holomorphic functions. In (Q1.1), it was attempted to prove that antiderivative implies path independent on weaker conditions namely on open subsets for continuous functions.
If the statement with the assumption weakening in (Q1.1) is right, then by the contrapositive of the statement, $\overline{z}^2$ has no antiderivative on $G$.
QED Way(2)

(*)
Ch4
Cor 4.13 to the complex analogue of the Fundamental Theorem of Calculus Part II (Thm A.3(b))



Ch5
Morera's Thm 5.6



Cor 5.9, proof and definition of path independence




