# Area integral in complex plane

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?

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 $$|z| Since the series $$\sum_0^\infty a_nz^n$$ converges uniformly on the disk $$|z|, 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

$$\iint_{|z|\le R} f(z)dxdy=\pi R^2f(0).$$

• Thank you. This way is slightly longer, but more illuminating and intuitive! Commented Oct 21, 2017 at 22:22
• @Septacle, you should definitely upvote and accept this answer. Commented May 9, 2018 at 8:46
• @garej Thanks. I didn't realize that. Commented May 10, 2018 at 9:47
• Fixed a typo. You had $r\to\infty$ rather than $r\to R$. Commented Jul 23, 2019 at 13:54
• @Cameron Williams Thank you. Commented Jul 24, 2019 at 21:07