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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.

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    Do you know how to put tex in this site?2012-09-25
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    I`ll try to do that but for now, no I don`t know.2012-09-25
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    So I will edit the post, and I think it will be OK.2012-09-25
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    Isn't a bar missing from your $\phi$: $\phi(z)=\bar z$, no?2012-09-25
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    yes, the bar is missing.2012-09-25
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    I wonder how different the "other definition" is. Notice that one can rewrite your definition as $$\frac{1}{2\pi i}\frac{1}{w-z}dw = \frac{1}{2\pi i}\frac{1}{1-z/w}\frac{dw}{w}=\frac{1}{2\pi}\frac{1}{1-z\bar w} |dw|$$2012-09-25
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    Also, why not use the residue theorem to calculate the integral? You can even avoid getting you hands dirty with calculations if you use the residues *outside* of the unit circle.2012-09-25
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    is there phi missing in the second equation?2014-05-09

1 Answers 1

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So far so good, $\int\displaystyle\frac1wdw=\log w$, but what is $\log$? It is the 'inverse' of $\exp$, though $\exp$ is not at all a bijection, it is a $\mathbb Z$-fold cover to $\mathbb C\setminus\{0\}$, periodic with $2\pi i$ period, as $e^{2\pi i}=1$. So, the $\log$ is unique only up to $2\pi ki$ for some $k\in\mathbb Z$.

If you calculate the integral only on the upper semicircle, from angle $0$ to $\pi$, you will get $\pi i$. Similarly, for the other semicircle, so $$\int_{S_1(o)}\frac1wdw = 2\pi i$$ However, if you calculate the integral of $\log$ on a circle that doesn't surround the pole (the denominator, now zero), then the result is $0$.

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    I feel kind of confused. I`ll try to reread it and see if that works.2012-09-26