Showing that $\overline{\int_{\gamma}f(z)\,dz}=\int_{\overline{\gamma}}\overline{f\left(\overline{z}\right)}\,dz$

Question

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.

Answer 1

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}