Conjugate of contour integrals

Question

Let $\varphi : \{ z \in C | |z| = 1\} \rightarrow C$ be a continuous function and let $\gamma$ be the circular contour centered at 0 of radius 1 positively oriented. Prove that

$$\overline{\int_{\gamma}\! \ \varphi (z) \ \mathrm{d}z} = - \int_\gamma\! \overline{\varphi (z)} \frac{\mathrm{d}z}{z^2}$$

I think you might be able to use Cauchy's residue theorem but I'm not too sure, any help is greatly appreciated

Answer 1

You don't have to bother about the $\phi$. (Note that $\phi$ is only continuous.) The magic is with the $dz$. On $\partial D_1$ one has $\bar z={1\over z}$. It follows that $$d\bar z= d\left({1\over z}\right)=-{1\over z^2}\>dz\ .$$

Answer 2

Hint: take the obvious parametrization $z = e^{it}, t\in [0,2\pi]$ and do $$ \overline{\int_{\gamma}\varphi(z)\,dz} = \overline{\int_0^{2\pi}\varphi(e^{it})ie^{it}\,dt} = \cdots $$