# Cauchy's theorem and formula

1. Feb 14, 2016

### Incand

1. The problem statement, all variables and given/known data
Verify that
a) $\frac{1}{2\pi} \int_0^{2\pi} f(e^{it})\frac{e^{it}\bar z}{1-\bar z e^{it}}dt = 0$, if $f(w)$ is analytic for $|w|<1+\epsilon$, and that
b) $\frac{1}{2\pi} \int_0^{2\pi} f(e^{it})\frac{e^{it}}{e^{it}-z}dt = f(z).$
for $z = re^{i\theta}$ with $r < 1$.

Use the above to verify $f(x) = \frac{1}{2\pi} \int_0^{2\pi} f(e^{it}) P_r(\theta) dt, \; \; z = re^{i\theta}.$

2. Relevant equations
Cauchy's theorem:
Suppose that $f$ is analytic on a domain $D$. Let $\gamma$ be a piecewise smooth simple closed curve in $D$ whose inside $\Omega$ also lies in $D$. Then
$\int_\gamma f(z)dz = 0.$

Cauchy's formula:
Suppose that $f$ is analytic on a domain $D$ and that $\gamma$ is a piecewise smooth, positively oriented simple closed curve in $D$ whose inside $\Omega$ also lies in $D$. Then
$f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z}d\xi, \; \; \forall z \in \Omega.$

Poisson kernel:
$P_r(\theta) = \frac{1-r^2}{1-2r\cos (\theta-t)+r^2} = \frac{e^{it}}{e^{it}-z}+\frac{\bar z e^{it}}{1-e^{it}\bar z}$.
3. The attempt at a solution
For a) it seems to follow directly from Cauchy theorem since both $f$ and the other part is analytic and the product of two analytic functions is analytic. However in this case the argument, $|w| = |e^{it}| = 1$, isn't less than one? Perhaps the $\epsilon$ fixes this? But I don't understand the reason for the $\epsilon$, is that a typo?

I don't really see the difference from a) here. For the same region if a) is zero then so is b). However If I compute it with Cauchy's formula I indeed get the desired result:

Substituting $s = e^{it}$ we have $dt = \frac{ds}{ie^{it}}$ and hence we get
$\frac{1}{2\pi} \int_0^{2\pi} f(e^{it})\frac{e^{it}}{e^{it}-z}dt = \frac{1}{2\pi i} \int \frac{f(s)}{s-z} = f(z)$

2. Feb 15, 2016

### Samy_A

To understand the difference between a) and b), consider the following functions defined for $s \in \mathbb C, |s| \lt 1+\epsilon$, where $z \in \mathbb C, |z| \lt 1$ is fixed:
$f_a(s)=\frac{f(s)\bar z}{1-\bar zs}$
$f_b(s)=\frac{f(s)}{s-z}$

Which of these function is holomorphic in an open region that includes $\{s \in \mathbb C | \ |s|\leq 1 \}$, and which is not? That will hopefully explain why in one case you apply the Cauchy integral theorem, and in the other the Cauchy integral formula.

3. Feb 15, 2016

### Incand

Thanks I believe I understand now! Since the $r < 1$ $f_a$ doesn't have a singularity in $\Omega$ while the second one does since $|s| = r$ is allowed.

Do you have any idea about the $\epsilon$? Is that pretty much just another way of writing $|s| \le 1$ avoiding the less equal sign?

4. Feb 15, 2016

### Samy_A

Exactly.
No.

You need that $1+\epsilon$ in order to apply the Cauchy integral theorem.

From Wikipedia:
Let $U$ be an open subset of $\mathbb C$ which is simply connected, let $f : U \to \mathbb C$ be a holomorphic function, and let $\gamma$ be a rectifiable path in $U$ whose start point is equal to its end point. Then
$$\oint_{\gamma} f(z) dz =0$$

In your case, it means that the unit circle must lie inside the open subset in which the function is holomorphic. So they take an open subset just a little bit larger, with radius $1+\epsilon$. That way you can safely apply the Cauchy integral theorem with the unit circle as $\gamma$.
Just using $|s| \le 1$ doesn't define an open subset.

5. Feb 15, 2016

### Incand

Thanks, nicely explained!

6. Feb 15, 2016

### Samy_A

You are welcome. Complex analysis is pure beauty!
/nerd mode