# Conjugate of contour integrals

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

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 $$d\bar z= d\left({1\over z}\right)=-{1\over z^2}\>dz\ .$$
Hint: take the obvious parametrization $$z = e^{it}, t\in [0,2\pi]$$ and do $$\overline{\int_{\gamma}\varphi(z)\,dz} = \overline{\int_0^{2\pi}\varphi(e^{it})ie^{it}\,dt} = \cdots$$
• I know how I would do this if i knew $\varphi (z)$ but as I don't know the actual function what would go in that second integral, thanks Nov 25, 2018 at 22:56
• @ProfessorPyg, the exact value of $\varphi$ is irrelevant. Edited, can you continue or do the same for the RHS? Nov 26, 2018 at 9:22