Let $\gamma \subset \mathbb{C}$ be a piecewise differentiable curve and let $\overline{\gamma}$ be the image of $\gamma$ under the conjugation map $\sigma(z) := \overline{z}$ Show that $$\overline{\int_\gamma f(z) \, dz} = \int_{\overline{\gamma}} \overline{f(\overline{z})}\, dz$$ and show that if $\gamma$ defines the positively oriented united circle, then $$\overline{\int_{|z| = 1} f(z) \, dz} = - \int_{{|z| = 1}} \overline{f(\overline{z})}\, \frac{dz}{z^2}$$
My attempt: I have tried showing the difference is zero, where $$\int_{\overline{\gamma}} \overline{f(\overline{z})}\, dz - \overline{\int_\gamma f(z) \, dz} = \int_{\sigma(\gamma)} \sigma(f(\sigma(z))\, dz - \overline{\int_\gamma f(z)\, dz}$$ However, I am not sure how to convert the integral over $\gamma$ instead of $\sigma(\gamma)$. Any hints are appreciated.
