- #1
Incand
- 334
- 47
Homework Statement
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}.##
Homework 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}##.
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)##