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