Exer 5.7 Prove $\overline{z}^2$ has no antiderivative on a non-empty region.
In the textbook, the discussion after Cor 5.9 (*) suggests the way to prove Exer 5.7 is to note that $\int_{\gamma} \overline{z}^2 \, dz$ is path dependent with the idea that existence of antiderivative implies path independent.
Now, I will attempt to do prove Exer 5.7 both along the suggestion (Way (1)) and alternatively (Way (2)). Please verify.
Proof of Exer 5.7: (Both ways start the same)
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 noting path dependence of $\int_{\gamma} \overline{z}^2 \, dz$
Observe $G$ is open $\because G$ is a region.
Therefore, by a corollary (Cor 4.13) to the complex analogue of the Fundamental Theorem of Calculus Part II (Thm 4.11), $\overline{z}^2$ has no antiderivative on $G$.
QED via Way (1)
Way (2): Noting path dependence of $\int_{\gamma} \overline{z}^2 \, dz$
As in this 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. $$0 \ne \int_\gamma \overline{z}^2 \, dz = \int_{\gamma_1} \overline{z}^2 \, dz - \int_{\gamma_2} \overline{z}^2 \, dz \implies \int_{\gamma_1} \overline{z}^2 \, dz \ne \int_{\gamma_2} \overline{z}^2 \, dz$$
Recall that while the existence of an antiderivative of a holomorphic function on a simply-connected region implies path independence of integrals over said function over piecewise smooth paths in the region, we also have that
the existence of an antiderivative of a continuous function on an open subset implies path independence of integrals over said function over piecewise smooth paths in the subset. $\tag{**}$
Therefore, by $(**)$, $\overline{z}^2$ has no antiderivative on $G$.
QED via Way(2)
(*) (Cor 5.9) If $f$ is a holomorphic function on a simply-connected region $G \subseteq \mathbb C$, then $\forall \gamma \subset G$ piecewise smooth, $\int_{\gamma} f$ is path independent.
Way (1): Right, but if you want to avoid noting path dependence, let's not use Morera's Thm (Thm 5.6). Instead, use Cor 5.5 in the textbook, as Mark McClure points out.
By Cor 5.5, if a function $f$ has an antiderivative $F$, then both the $F$ and $f$ are holomorphic. However, an example in the textbook, Eg 2.8, proves nowhere holomorphicity.
So actually, if a function is nowhere holomorphic, it has an antiderivative only in $\emptyset$.
Way (2): Right by Eric Wofsey's confirmation of weaker assumption.