An entire function whose integral is bounded is identically zero

Suppose $f$ has a power series at $0$ that converges in all of $\mathbb{C}$ and $$\int_{\mathbb{C}} |f(x+iy)|dxdy$$

Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.

A hint is given: “Use polar coordinates to show $f(0)=0$”

Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc

• Why $f(0) = 0$ makes things any easier? – xyzzyz Jan 29 '18 at 0:03
• You probably need to tell us what you do know. – user99914 Jan 29 '18 at 0:07
• @JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2 – mysatellite Jan 29 '18 at 0:13
• You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle? – anomaly Jan 29 '18 at 2:57
• @anomaly Yes I know – mysatellite Jan 29 '18 at 3:00

Let $f(z)=\sum_0 ^{\infty} a_n z^{n}$ be the power series expansion. Write $z=re^{i\theta}$ and integrate with respect to $\theta$ from 0 to $2\pi$. Integrating term by term is permitted because of uniform convergence. You get $2\pi a_0= \int_0 ^{2\pi} f(re^{i\theta}) d\theta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr d\theta =dxdy$ you will see that $|(\int_0 R rdr) 2\pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.