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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

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2 Answers 2

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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 $$d\bar z= d\left({1\over z}\right)=-{1\over z^2}\>dz\ .$$

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

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  • $\begingroup$ 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 $\endgroup$ Nov 25, 2018 at 22:56
  • $\begingroup$ @ProfessorPyg, the exact value of $\varphi$ is irrelevant. Edited, can you continue or do the same for the RHS? $\endgroup$ Nov 26, 2018 at 9:22

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