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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?

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1 Answer 1

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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

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

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  • $\begingroup$ Thank you. This way is slightly longer, but more illuminating and intuitive! $\endgroup$
    – Septacle
    Commented Oct 21, 2017 at 22:22
  • $\begingroup$ @Septacle, you should definitely upvote and accept this answer. $\endgroup$
    – garej
    Commented May 9, 2018 at 8:46
  • $\begingroup$ @garej Thanks. I didn't realize that. $\endgroup$
    – Septacle
    Commented May 10, 2018 at 9:47
  • $\begingroup$ Fixed a typo. You had $r\to\infty$ rather than $r\to R$. $\endgroup$ Commented Jul 23, 2019 at 13:54
  • $\begingroup$ @Cameron Williams Thank you. $\endgroup$
    – ts375_zk26
    Commented Jul 24, 2019 at 21:07

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