I'm trying to solve the problem: If $\phi \colon S_1(0)\to \Bbb C$ is a continuous function on the unit circle, its Cauchy transform $C\phi(z) =\frac 1{2\pi i}\int_{S_1(0)}\frac{\phi(w)}{w-z}dw$ is analytic on $\Bbb C \setminus S_1(0)$.
If $\phi$ is holomorphic on closure of $B_1(0)$ then it follows from Cauchy's formula that $C\phi(z) = \begin{cases}\phi(z)& \mbox{if }z \in B_1(0);\\0&\mbox{if }z \in C\setminus B_1(0).\end{cases}$ What happens if $\phi(z) =\bar z$?
Hint: On the unit circle $\bar z = 1/z$. I tried to substitute in the integral by $\phi(w)=1/w$ and solve the integral by writing it as $(-1/zw + 1/z(w-z))$. I got the integral is $\log$, but I'm not sure at all of my solution. Can anyone please help me? Also, I'm trying to read more in Cauchy transform but I couldn't find any resource with the same definition. The books that I found have other definition, can anyone please give me more information about this transformation to read with the same definition that I have here?
Thank you in advance.