Let f(z) be an analytic function within $\lvert z \rvert\leq R$. Show that
$\iint_{\lvert z \rvert\leq R}f(z)dxdy=\pi R^2f(0)$.
I solved the problem using $z=re^{i\theta}$ and Cauchy's integral formula, But someone said that there are at least two ways to solve this problem. But I cannot find the other way around. Can someone help me?
Best Answer
Another way without using Cauchy's integral formula:
First we prove
$$\iint_{|z|\le \rho } z^n dxdy=0\quad(n\ge 1)$$
for every $\rho >0$. Using $z=re^{i\theta }$ we have
$$\iint_{|z|\le \rho } z^n dxdy=\int_0^\rho \left(\int_0^{2\pi}e^{in\theta }d\theta\right)r^{n+1}dr =0,$$
since $\int_0^{2\pi}e^{in\theta }d\theta=0$ for $n\ge 1$.
Let
$$f(z)=\sum_{n=0}^\infty a_nz^n$$
be its Taylor expansion in $|z|<R.$ Since the series $\sum_0^\infty a_nz^n$ converges uniformly on the disk $|z|<r<R$, we can change the order of integral and summation in the following:
$$\iint_{|z|\le r} \left(\sum_{n=0}^\infty a_nz^n \right) dxdy=\sum_{n=0}^\infty \iint_{|z|\le r} a_nz^ndxdy.$$
So we have
$$\iint_{|z|\le r} \left(\sum_{n=0}^\infty a_nz^n \right) dxdy=\iint_{|z|\le r} a_0\,dxdy=\pi r^2a_0.$$
In other words
$$\iint_{|z|\le r} f(z)dxdy=\pi r^2f(0).$$
Letting $r\to R$ we have
$$\iint_{|z|\le R} f(z)dxdy=\pi R^2f(0).$$