Let $\gamma$ be a piecewise-$C^1$ curve, and let $\overline{\gamma}$ be its image under the mapping $z\mapsto \overline{z}$ (symmetry in the real axis).
I am trying to show that if $f(z)$ is continuous on $\gamma$, then $z\mapsto \overline{f\left(\overline{z}\right)}$ is continuous on $\overline{\gamma}$, and
$$\overline{\int_{\gamma}f(z)\,dz}=\int_{\overline{\gamma}}\overline{f\left(\overline{z}\right)}\,dz$$
I think for the first part I just need to use the usual definition of continuity, but does this apply for continuity on the curve as well?
I'm not sure how to approach the second part.
Continuity is not relevant here, this is just a formal manipulation.
\begin{eqnarray} \overline{\int_\gamma f(z)\,dz} &=& \overline{\int_0^1 f\left(\gamma(t)\right) \, \gamma'(t)\,dt} \\ &=& \int_0^1 \overline{f\left(\gamma(t)\right) \, \gamma'(t)\,dt} \\ &=& \int_0^1 \overline{f\left(\gamma(t)\right)} \ \overline{ \gamma'(t)} \ dt \\ &=& \int_0^1 \overline{f\left(\overline{\overline{\gamma}(t)}\right)} \ \overline{ \gamma}'(t) \ dt \\ &=& \int_{\overline\gamma} \overline{f\left(\overline{z}\right)}\,dz \end{eqnarray}