An entire function whose integral is bounded is identically zero

Solution 1:

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.