1
$\begingroup$

I have been experimenting with quadrature domains. The most obvious one is a circle. Let $f(z)$ be holomorphic on a large enough region and $\Omega= \{|z| < 1 \}$ then:

$$ \int_{\Omega} f\, dA = \pi \,f(0) $$

I almost reasoned the answer should be zero. We could change the area to polar coordinates:

$$ \int_{\Omega} f\, dA = \int_0^1 \int_{|z| = r } f(x+iy) \,dx\, dy = \int_0^1 \int_{|z| = r } f( r \, e^{i\theta}) \,r \, dr \, d\theta = 0$$

since the integral around the circle should be zero. Assuming there is no pole in side the circle $\{ |z| < 1\}$:

$$ \oint_{|z| = 1} f(z) \, dz = \int_{|z| = 1} f ( r \, e^{i\theta})\, r e^{i\theta} d\theta = 0 \cdot f(0) = 0$$

$\endgroup$
1
$\begingroup$

Short Version: $dz\ne d\theta$.

In Detail:

Forget $r$ for the moment - it's irrelevant to the problem.

The notation $\int_{|z|=1}f(e^{i\theta})\,d\theta$ doesn't really make sense, because there's no $z$ in the integrand. Assumng $f$ is hoolomorphic in a neighborhood of the closed unit disk both of the following hold: $$\int_{|z|=1}f(z)\,dz=0$$(Cauchy's Theorem) $$\frac1{2\pi}\int_0^{2\pi}f(e^{i\theta})\,d\theta=f(0)$$ (mean-value property for holmorphic (hence harmonic) functions).

The two integrals are simply not the same at all. If we parametrize the unit circle via $z=e^{i\theta}$, $0\le\theta\le2\pi$ we get $$dz=ie^{i\theta}\,d\theta,$$hence $$\int_{|z|=1}f(z)\,dz=i\int_0^{2\pi}f(e^{i\theta})e^{i\theta}\,d\theta \ne\int_0^{2\pi}f(e^{i\theta})\,d\theta.$$

Now when you do the original integral in polar coordinates you get $d\theta$, not $dz$. A correct version of your polar-coordinate calculation is this: $$\int_{\Omega} f\, dA = \int_0^1 \int_0^{2\pi} f(x+iy) \,dx\, dy = \int_0^1 \int_0^{2\pi} f( r \, e^{i\theta}) \,r \, dr \, d\theta = 2\pi\int_0^1 f(0)r\,dr=\pi f(0).$$The integral $\int_{|z|=r}f(z)\,dz$ simply doesn't come up there.

(Look up the polar coordinates formula. You don't see $\int_{|z|=r}$ there.)

$\endgroup$
0
$\begingroup$

$\int_{|z|=1} f(r e^{i \theta}) e^{i \theta}\,d\theta$ is zero as you say, but $\int_{|z|=1} f(r e^{i \theta}) \,d\theta$ doesn't have to be. You could think about the constant function $f=1$, for instance.

$\endgroup$
  • $\begingroup$ $\int_{|z|=1} f(r e^{i \theta}) e^{i \theta}\,d\theta$ doesn't really make sense, or at least not the sense you intend - if the integral is wrt $\theta$ then the limits should say something about $\theta$, not $z$. Not to be pedantic, but it seems to me that being careless about this sort of thing is what's leading to the OP's confusion. $\endgroup$ – David C. Ullrich Dec 8 '17 at 15:54
0
$\begingroup$

Note that $$\frac{1}{2\pi}\int_0^{2\pi} f(re^{it}) dt = \frac{1}{2\pi i}\int_{\partial B_r(0)} \frac{f(z)}zdz$$ and use Cauchy's theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.